Page 1
Chapter 22.1-
1 Let us denote the signal in question by q(t) and its energy by Ey . For parts (a) and (b)
/ 2"1 1 f
2r
Eg — Jsin
2t dt — - J
dt - -J
cos 2t dt = it + 0 = it
(c) Eg ~ f sin2Idl = ~ f dt-\ ( cos 2t dt = tr + 0 = tr
J2» 1 Jiw 1 JlK
(d) = y(2 sin 0
2dt = 4
Jdt~\J
cos2t dt = 4[5t + 0] = 4ir
Sign change and time shift do not affect the signal energy. Doubling the signal quadruples its energy. In the
same wav we can show that the energy of kg(t) is k 1Eg.
2.1-
2 (a) /•;, - /*( 1 ?di = 2. Eg = /J (1 )~dt + /*(- l)2dt = 2
= J (2
)
2dt = 4. £*-* - ^
(2}3dt = 4
Therefore £**„ - Ez ~ Eg.
r2r,.r r7r l"*/ 2 /•» /-a»/2 /•2»
(bl Eg = j(\)
3dt+J (
— l)2dt = 2rr. £„ = y j
(-ifdt^J
(1 )dt+
^- 1 )
dt = 2tr
,»/2 rW2 y-2.
£z^,= / (2)2</t + / (0)
2t/t + /
(-l)2d( = 47T
JO Jir/2 J%n/2
Similarly, we can show that £*.„ = 4rr Therefore £**„ = £„ + Ev . We are tempted to conclude that £,. ±1) =
Ez — Eg in general. Let us see.
fw/A r* fn
(c) £x= / (1 )
2<f<+
/(-l)
2c/« = 7T Ey = I (1 )
2dt = 77
,/o */ir/4 *'0
£*+„ = J(2
)
2./f + J (0)
2dt = 7T £*- „ = J
(0
)
2df
J(~-)
2dt. = 3tr
Therefore, in general E*±j, / Ez + £«
2.1-3
. = — / C 2cos
2(wot + 0)dt. = £- f |1 + cos (2w0 ' + 29)] dt
~o J0
z-»2 ch) rTa 1 r>2
= / dl 4. / cos (2-uo* + 29) dt = —- |To + 0] =2^0 [Jo Jo
27°£!2
2.1-4 This problem is identical to Example 2.2b. except that w, i- a* In this case, the third integral in Pg (see p. 19
is not zero. This integral is given by
~ lim2C,r2 r
r-4^ T1 -T/2
= limC1C2 '
rT/2
/T—*x. T L' -T/2
• r rT/- Irn
J—- / cos(9j — 62) dt 4- / cos(2-vjt + 0\ + 6j) dt
f [J-T/2 J-r/i
=. lim — [7 ’ cos(9i — 9j)| 4 0 — £162 cos(9i — O2)
T— *x. 7
1
Page 2
Therefore
p9- £±. + Qi + C1C2 cos(0i - #2)
2.1-5
-u; {l'fdt - 64/7 (a) P ffdt = 64/7
2
(b) P2y = Ij
2
(Ot'fdt = 4(64/7) = 256/7 («) P«. = J)
2dt = 64c
2/
7
Sign change of a signal does not affect its power. Multiplication of a signal by a constant - incieases the po
bv a factor c2
.
2 .1-6
(a) P, -if* Ja
- f c-t dt=\ (1-a ’
<b » p’~t, jy v)m
-i/'(d) p. (±i)-t/t = 1
2.1-7
<«> Ps= ^L (£)rft
5
P“ = 7 S. Tn[i)g
‘{t)d1 =
J-r,*i?JkDkD rf di
^ .> are finite because the integrands are periodic signals
The integrals of the cross-product terms ( hc" *J, d The remaining terms (.* = r) yield
(made up of sinusoids). These terms, when divided by T - 00. yield
/
T/2 n
T/2 kmm k~rr.
, (., P„.«, of . sinusoid of amplitude C is C*/2 IE,. (2-6.)) «rik. of iU fr«,u«ncv (at # °> and ph«.
JXtZ, 1 r,“. to fh. sum of tit* powet, o, .h.— |E,. «.«S)|. Thefts, in
this case P =^ = 178'
,. p _ | Ui =51.
(c) (10 + 2 sin 3f) cos 10/ = lOcos 10/ + sin 13/ - sin 3/. Hence from Eq. (2.6) ^^ 5
(d) 10 cos 5/cos lOt = 5 (cos 5/ + cos lot. Hence from Eq. (2.6b) P - +4^
- 25_
<«» ^ - d/O «./«> - «/>
2.2-1 For a real a
r_3o
‘ dt = oo
rT/2
For imaginary e. let a — j* Then
oc
P„ = IlifiT
r/2 , r7'/ 2
- I (rjx
')(r~'xt
)dt = lint — /
dt = 1
}_ T/;t-~ t J-t/2
2
Page 3
*w
Fig. S2.3-2
Clearly, if o is real. r~“ is neither energy not power signal,
power 1.
However, if a is imaginary, it is a power signal with
g3 {i) = .<?(' - 1) + .9t( t ~ !)• 9s(0 = 9( < - 1 ) + -,J'f f + 1 ^ 9*{f)-a(t O.o) + ffi(f . 0.5)
The signal can he obtained by (i) delaying ,U) by 1 «~«l *- 1 »<">
by h factor 2 (replace t with t/2). (iii) then multiply with 1 o. Thus gs {t) - l-Offlj '
2.3-2 All the signals are shown in Fig S2.3-2.
2.3-3 All the signals are shown in Fig. S2.3-3
Fig. 52.3-3
2 J'4
- £>(>.« - r.. *<•
»
/>-»’* *
=£ led - Til’d/ - J »! (*)dr > £. E.I.,1 - £w«'>]’di -
;£»V)d* ~ F'J°
£ tot- - »]’ d. - i£ ,V) d.e **,-> - £«'/•))’ d. -£«’(') dl - as,
Eoe(t) - J[ag(J)ftl* - J 8~V)* = 0
" E*
2.4-
1 Using the fact that g(r)f‘(r) = g(0)h[r). we have
(.)„ (hlPM (.)).(/) W-p('-l) i.)A«- + 3| (f) ge{w) (use L‘ Hbptlals rule)
2.4-
2 In these problems remember ,ba. impute ><*> I. located a, a . 0. Thus, an impute *<—) » '»»'“* a. r I.
and so on. . .
(a) The impulse is located at r = l and g(r) at r = Ms gU). Therefore
3
Page 4
L q(r)f<(t - r)»/r = g(t)
(b) The impulse *(t) is at r = 0 and q(1 - r) at r = 0 is s(t)- Therefore
fi[r)g(t - t) dr = ff(t)LUsing similar arguments, we obtain
(c) 1 (d) 0 (e) rs
(f) 5 (g)p(-l) (h) -e
2.4-3 Letting nf = .r. we obtain (for n > 0)
f dt = - f 0)
Similarly for <i < 0, we show that this integral is -£<£(0). Therefore
j0(t)t>(at)dt. — ~|d(0) =
jjjj J<t>(t)6(t)d1
Therefore
»<*•» - w‘">
2.5-
1 Trivial Take the derivative of |e;3 with respect to r. and equate it to zero.
2.5-
2 (a) In this case Ex = JQ'
lit = 1. and
= jrj
si'M'M' =\ J0
td, ~ 0 b
(b) Thus, git, * 0.5.r«). and the error c«) - t - 0.5 over (0 < i < 1). and zero outs.de this interval Also E„
and £, tt he energy of the error) are
E9 = Jg2 (t)dt - J
t2 dt- 1/3 and Et ~ J
— 0.5)2
rff = 1/12
The error (f - 0.5) is orthogonal to r(t) because
J(t — 05)(l)</f = 0
Note that £ = r*£x + £#. To explain these result, in terms of vector concepts we observe from Fig. 2.15
ihat the error vector e is orthogonal to the component «. Because of this orthogonally, the length-square of
Xe^ o^(O) ^ equal to the sum of the square of the lengths of cx and e [sum of the energ.es of «r(f) and
f(0l-
2.5-3 In this case Eg = J0‘ g
2(l)dt = f0t2dt - 1/3. and
c =4~ f r(t)g{t)dt = 3 ftdt= ll
Ey Jo Jo
Thus. r(t) * 1.5,(/). and the error r(l) = r(t) - 1.5»(f) = 1 - 1-W over (0 < t < 1). and zero outside this
interval Also £, (the energy of the error) is £e = /0 (1- 1-51) dt _ 1/4.
.5-4 (a) In this case Ez = j^sin7 2ntdt = 0.5, and
— ~ J «(()»(')* = djjf (sin 2irt<l( — -1/-
(b ) Thus. »(,) - -(l/*).r((). and (h. .no, .(.) - < + (•/>)»» to. o«r (0 < . < D- »d -» .«•
interval. Also £aand £, (the energy of the error) are
4
Page 5
Eg = f g, (t)dt = jf’ t*Jt = 1/3 and £, = [f - (l/*)sin 2trf.]
2(ft = f
'
The error \t + (1 /tt) sin 2itl\ is orthogonal to x (/.Jbecause
Jsin 2trf[< + (l/tr)sin 2irt]dt = 0
Note tho. e. - **£a + TO -pH.L.— in ~the error vector e is orthogonal to the component ex. Because of this orthogon y, *
, ,y,^ equal to the sum of the square of the lengths of ex and e [sum of the energies of «(t) and
2.5-5 (a) If r(f) and u(t) are orthogonal, then we can show the energy of x(t) ± y(t) is Ex + Ev
rnv
P \x(t)±y(t)\1dt =p \r(t)\
3dt +p \y(t)\
2 dt±j
J\t)v(t)dt± jj^t)y{t)dt
= J|.r(f)|
5dt + J
|«(<)t*df
( 1 )
(2 )
The last result follows from the fact, that because of orthogonality the two integrals ®f 1
r(,)y‘(t) and r'(t)y(t) are zero [see Eq. (2.40)}. Thus the energy of x(t) + y(t) is equal to that of x(t) V (t) if
(b) Using similar argument, we can show that the energy of c,r(t) + n,,(t) is equal to that of ''i*(f) “ if
.r(/) and >/(/) are orthogonal. This energy is given by |m| £* +M Ey(c) If :(() = r(t) ± y(t). then it follows from Eq. (1) in the above derivation that
E, = Ex + Ev ± (Exy + Eyx)
2.5-6 g,(2. -1). ga (-1.2). g.t(0. -2). *4(1.2). gs(2.1), and «a(3,0). From Fig. S2.5-6. we sec that pairs (gs-ge).
(gl . g4 )and (gj.gs) aie orthogonal. We can verify this also analytically.
gvg«=(0x3) + (-2x0) = 0
gi g4 = (2xi)+(-lx2) = 0
g3 gs = (-l x 2) + (2 x 1 )= 0
We can show that the corresponding signal pairs are also orthogonal.
P {-rMZx,[t)\di = Q
„i (t)q*(tult =P |2r,(f) - ra(0][*i(0 + ^(t))dt = 0
g2 lt)gi (i) dt = P + 2x a (/)][2r,(f) + x3 (t)\dt = 0
J -'V-
Page 6
2 .6-1
In deriving these results, we used the fact that = f^xfrtjdt - 1 and T,(t)xs (t.) dt 0
We shall compute , „ using Eq. (2.48) for each of the 4 cases. Let us first compute the energies of all the signals.
Ex = jsin
22irt dt = 0.5
In the same way we find = Em = Ets = Egt = 0.5.
Using Eq. (2.48). the correlation coefficients for four cases are found as
2trt sin 4trt tit = 0 (2)(1) ,
1
f sin 21 '
v/to sxo s) J0
I f 0.707 sin 2nt<it = 0 (4)’ s) Jor{t) and
2.8-1 Here To = 2. so that o = 2tr/2 = tr. and
Vg(t) = <*o +^ cosnirf + bn sinnrt.
n= l
whore
1
/ (sin 2*/)(-sin 2vt)dt = —
1
(3) i n I v.iviuu mum -» v»/ /^r:v ’v/to.sHO.s) J0 V<0!
Signals r(f) and 32 (f) provide the maximum protection against noise.
J— I / 0.707 sin 2»rtdt-j 0.S) (° S)
[70 J0.5
707 sin 2irtdt = 1.414/tt
- 1 < t < I
no -i /:Usinmrf dt = 0
Therefore1 4 A' (-1)"
3 n-1
n= 1
cos turf - 1 < f < 1
Figure S2.8-1 shows q\l) — t2
for all I and the corresponding Fourier series representing q(t
/
o\er I .1)
The power of q(l) is
Moreover, from Farseval s theorem [Eq. (2.90)j
Pg — C0 + / 9 ^3 ) 2 2—
t
l rr*n 2 ) 9 it4 ^ ” 4 9 90 5
," n«l V 7 " =1
(b) If the ;V-term Fourier series is denoted by r(t), then
V— 1
1 4 ( — l)n
n=l
The power Pz is required 10 be 99%P9 = 0.198. Therefore
p. = i +iri = 0.1989+
w 4 L* n*
- 1 < 1 < 1
6
Page 7
For tt = 1. P, = 0.1111; for .V = 2. Px = 0.19323. For N = 3. Px = 0.19837, which is greater than 0.198.
Thus, .V = 3.
2.8-2 Here T0 = 2rr. so that u>0 = 2ff/2rr = 1. and
g (r) = a0 + cosnt + fe„ sinnt - tt < t < tr
where
oo = — f tdt — 0 , a„ = f2* J-r 2* J-rt cos nt dt — 0,
2 r . ,,, 2(-irhn = _ J
^tsinr>tdt = —
Therefore
— ir < t < ir.»(*) =2(-l)"
+1 ^-BinT)tn*l
Figure S2.8-2 shows g(t) = t for all t and the corresponding Fourier series to represent g(t) over (-it, tr).
1*1^>i Loil]
t-*
Fig. S2.8-2
The power of <j(t'j is
s £<•)’* -TMoreover, fioni Parseval s theorem [Eq. (2.90)]
1 tr‘
(b) If the .Y-tetm Fourier series is denoted by r(t). then
s
.T(t) = 2(-l)”+ 1 ^isinn W t
nxl
The power Px is required to be 0.90 x t = 0.37T2
. Therefore
- n < > < n
*-*£*-"*
For .V = 1. Px » 2; for A’ = 2. Pz = 2.5. for N = 5, P, = 2.927. which is less than 0.3tt2
. For N - 6. Px -
2.9825. which is greater than 0.3jt2
. Thus. N = 6
2.8-3 Recall that
flO ;
[To/2(la)
/.9(0*
J -To,'2
[To/i
1 g(t) cos njjot dt(lb)
J-T0 /
2
[To12(lc)
1 g(l) sin dt
J-To/2
/
Page 8
Recall also that cos n-»l is an even function and sin Wjs an odd^
t. then g(i) cosW is also an even function and fl (t)sinW » an odd Junction
2 fT° /2 MM (
2a )
-t.J.,mno :
4 rTon
°""jo1q(t
)cos n*iQtAt (
2b)
(2c)
hr, = 0
Similarly, if *(*) is an odd function of t. then g(t) cos nwot is an odd function of t and g(t) sin r,^ t is an even
function of t. Therefore
ao = «» = 0(3a)
bn= ±f°'\mnn.0Ut MObserve that , because of symmetry, the intention retired to compute the coefficients need be performed over
only half the period.
2.8-4 (a) To = 4. -vo = ^ = f - Because of even symmetry, all sine terms are zero.
*X,
.<?(*)= «o + Y* a " cos \Tl
nss 1
n0 = 0 (by inspection)
«- - J [/* cos(f')
d1 -
1
cos (tO '"]= ^ s,nT
Therefoie. the Fouriei series for q{t) is
4 / at 1 3trt 1 5trf 1 rnr_7*t \
q(t) = - (cosy - gCOS— + -cos-y7cos
2 j
Here K = 0. and we allow C„ to take negative values. Figure S2.8-4a shows the plot of C„.
(b) To = lOtr. ^ = 3 . Because of even symmetry, all the sine terms are zero.
qii) = 00 + £«„ cos (Jf) +ftt.sin (t')
». = 1
n 0 = - (bv inspection)5
«. - -k£- (?•)* - & (H) (?•)L A !“n ("«'
)
. f sin f -tl dt = 0 (integrand is an odd function of t)
n10* J_„ V 5 )
Here b„ = 0. and we allow C„ to take negative values. Note that Cn = a„ for n = 0, 1, 2. 3,
shows the plot of Cn-
(c) Tq = 2tt . wo — 1*
9 (t) = oo + ^OnCosnt + b„sinnt with no = 0.5 (by inspection)
narl
—
£
cosn"" =0 -',n=
-X ^ S1
, 1
sin ntdtITT)
and
q(t) = 0.5 - ^(sinf + ^
sin 2/ + -sm3f + 4sin4M-
)
= 0.5 + i [cos (' + £) + \cos (2/ + §) + jcos (3* + 0 )
+ ']
Figure S2.8-4b
8
Page 10
Figure S2.8-4d shows the plot of C„ and fi„.
(e) To = 3. uJo = 2ir/3.
fin2 2mr 3 ,
27rn 2nn:
3 J0iCOS— tJ'=
2^ [cOS— + -T
2vn.C ——
—
3^0 3 2
Therefore Co = g and
_ 3 4tr 2n 2 ”2jtti 4jtt<
"2rrn
c-- 53^[V a+— - 2c“— —
i= 6. -<o
2 /',
2nir, 3 . , .
2nn 2nn 2nn=
3 J0(sin—<d,= 27^'sm—--T cosT
(f) To -
!2+ ~t
rr/3 . 110
and fin = tan"/ 4ir 2n 2 ”2jtti 4jtt<
"
2rrn
\/2 + -—~ 2cos- 3
= 0.5 (by inspection). Even symmetry; 6n
2sh cos - sin 2^n \
+ 2iii sin^ -1 )
, ( ^fcoVcos^
= 0.
«n4 r
1
, . nrr
ij./if
2
3
6_ir2 n 2
>s~ tit+ J
(2 - t) cos Zj-t tit
nrr 2r)7r"|
lS__ C0S _r j
6 / jr 2 1 5ir ] 7t \
1 * 0 5 ~2(cos 3' *
9cos,r ' +
20cos— f t - cos— f +
J
2.8-5
Observe that even harmonics vanish The reason is that if the dc (0.5) is subtracted from <7 (f). the resulting
function has half-wave symmetry. (See Prob. 2 8-6). Figure S2.8-4f shows the plot of Cn .
An even function qr n) and an odd function 9o(0 have the property that
Se(() — 9e(~0 and Co(t) = -flo (—0 !1 '
F.vety signal </(t) can be expressed as a sum of even and odd components because
9(0 = \ i.o(0 + 2 i.9(0- 9(-0]
#v«»n
From the definitions in Eq. (1). it can be seen that the first component on the right-hand side is an even
function, while the second component is odd. This is readily seen from the fact that replacing t by —1 in the
first component yields the same function. The same maneuver in the second component yields the negative of
that component.
To find the odd and the even components of g{t) — u(t). we have
9(0 - 9«(0 + 9o(0
where {from Eq. (1)]
<iA<) = 5 M0 + «(-/)] = |and
9o(0 = 3 MO - «(-0 ! = ^sgn(f)
The even and odd components of the signal v(l) are shown in Fig. S2.8-5a.
Similarly, to find the odd and the even components of g(t) = e.~at
n(t), we have
9(0 = 9» (0 + 9o(0
where
9,(0 = i[e— «(fJ + e«u(-‘)]
and
10
Page 11
ftOFig. S2.8-JS
The even and odd components of the signal c~tttu(t) are shown in Fig. S28-5b.
For g(f) = r''. we have
ejl = <j«(f) + flo(0
where
and
««(') = 3 K + r' J<
]* cos '
Soft) « ^[e
jt - <•"'*] * jsin t
2.8-6 (a) For half wave symmetry
and
and
<7(0 = -<7 (t ± y )
Toi* 7*0/3 /*To
= -1 [° g(t)cosn^otdt = — /
g(t)cosn«>otdt + »(f)co«W«ftTo
_/0JO
./o •'To/2
Let r = / - 7o/2 in the second integral. This gives
jf
r° /2
S(0 cos two* dt + Jg (* + y )
cos nu'o (t + y )<i*j
- JL]^J
a'
g(f) cosn^ot dt 4- J— <7(t)[- cosnwor) cfaj
,i.f r\To L/o .
fl(f) cosnwof tit
In a similar way we can show that
4 fTo/2
l,r = — jg(t) sin nuiot <lt
To Jo
(b) (i) Jo = 8. -o = j. no = 0 (by inspection). Half wave symmetry. Hence
11
Page 12
2.9-
4 / ntr rift.
ntr \ ,= ^(cosT + T smT _1)
(nod)
Therefore
Similarly
4 /r)7T.
T17T ,>I Sin — 1
|
(n odd)r>3 it
3 V 2 2 /
flu “^,-^t(¥ + 0
n = 1 , 5 . 9 . 13 .-
n = 3 , 7 . 11 , 15 ,
1 f Lsm^Lt <11 =L 2 4
Hi coslil^ « -J-sinf^)fe"~
2 n 2tr2 l
Sm2 T 2 / n 2
tr2 V 2 /
and
nir „ ,•
n7r,
a„ cos— * + on sin— f
4.9(0 = Hn*l.S.5.-
(ii) To = 2ir. wo = 1 no = 0 (by inspection). Half wave symmetry Hence
0 (f) s n„ cosnf + sin til
n= l ,1
,S .
n,, = — Jr~ t,i0
cos nt ilt
_ £ [tr j^n 2 + 0.01
,- tr/tO
(-0 1 cosnf + nsinnf)
1
(n odd)
n 2 + 0.01
o
(01)n 3 + 0.01
107r{n'J + 001)
-*/10 -D =
(-0 . 1 )
0.0465
n 2 + 0.01
and
* Joe-,/l0
sin nf </f
-t/JO_ -(— 0.1 sinnf — n cosnt) (n odd)
ir|n 2 + 0.01 Jo
2n:(e
-w/10
“ (n 2 + 0.01)
1 (a): To = 4. wo = tr/2. Also Do = 0 {by inspection)
- 1 )=
1.461n
n 2 +0,01
- 5J-£ ^r-^W, dt = |n< > 1
(b) To = 10*. wo = 2tr/10fr = 1/5
g(,)= D„r J S\ where D„ =T^t /
* ^ <H =lint
(~ 2j S'" t)" "
12
Page 13
-9 1
-J I Up
Fig. S2.9-1
(c>
</(i)~ Do + ^2 D”r'" t
- Where, by inspection D0 = 0.5
r»= -ti
Dn = — r = T^-' sothat l^-l = 2^r- and zDn = {l2ir JQ 2jt 2irn 2*71 l -i
(d) To = ir. wo = 2 and £>„ = 0
Page 14
(e) 7'o = 3.^o —
.9(0
Therefore
= where On =5^
rft = 4^ C 3+ *)
l
\
lOnl - ,,.-2
_3_
4w 2 n :j
*sin 222 \
/') -L4lr3n2
-1 i( 1
cos aS1" -t \
/7T71
-2COS-3- and iDn — tan| , r 2ftn ,
2wnsin 25a - 1 /
V2 + ~9~ 3 3 ^COS -5- + 3
(f) To = 6. *>0 = *73 Do = 0.5
Dn = -o
*(t)® 0.5+ ^n*s-rtw
,1 ,1,
/-a 1 3 ( nit 2*n \
J <
1 + ay + l'-‘*•“
+ 1(-. + 2K^<"J = ^5('“T- C“~i
H • Cna -
t1 - _U0 15 8 0-
Co)
|J>*»
TV5t *
i_i_— ..-*t i-
J m. CS «Z
-8 -3
-<s
t 5 s
n4 ajL>
2.9-2
n
1
1
)3 -Ml l"* n
« > - a? 1 ^
For a compact trigonometric form, all terms must
reason, we rewrite 9(f) as
uO i .**
Fig. S2.9-2
q( 1) = 3 cost + sin ^5t “ " 2cos
(8t ~
3 )
have cosine form and amplitudes must be positive. For this
9(f) = 3cosr + cos(5t-|-|)+2cos(8t- *)
as 3 cos I 4- cos ^5t - + 2 cos (b* - —)
Figure S2.9-2a shows amplitude and phase spectra.
“ p:_ 09 q_2a we plot the exponential spectra as shown in rig.
(b) By inspection of the trigonometric spectra in Fig. S2.9 2a, we plot in p
S2.9-2b. By inspection of exponential spectra in Fig. S2.9-2a, we
,(„ .y + +iy-*> * + ['*-*’ +
. 5,,. *(I,-*) r-« + (.->») + I.- + (j''*) + (''*)”
-jSl
14
Page 15
2.9-3 (a)
g[t) = 2 + 2cos(2< - n) + cos(3t - |)
= 2-2 cos 21 + sin 3t
(b) The exponential spectra are shown in Fig. S2.9-3.
(c) By inspection of exponential spectra
o(t) = 2 + [„<*-> + + 5[e**'*’
= 2 + 2cos(2t - it) + cos ^3t -
+ r-r(«-f)]
(d) Observe that the two expressions (trigonometric and exponential Fourier series) are equivalen
Fig. S2.9-3
2.9-4
D„ - —rTo/2
.J-Ta /2f(t) cos riuiot dt
/To/3
To/2
/(f) sin nu/of dl
If n't) is even the second term on the right-hand side is zero because its integrand is an odd function of t.
H«n«. D. rl.ll iTZiJ" If ,m is odd. ,h= first tsm on the righ.-h.nd sid. is soro boco.se » -»«»*
is an odd function of f. Hence. D n is imaginary.
15
Page 16
Chapter 3
3.1-1dt
If ,(» is an even function of .'*)-»-* *• « odd function of
<7(t)cos is an even function of t. and the first integral is twice the integral
when fl(f) is even
/,nu
(1)G(w) = 2 / fl(t)coswtdt
Jo
Similar argument shows that when g(t) is odd
CM = -2j f 9 (t)*in~>tdt(2)
If g{t) is also real (in addition to being even), the integral (1) is real. Moreover from (1)
G(-~) = 2
yg(t)cos*/tdt = GM
Hence CM is real and even function of Similar arguments can be used to prove the rest of the properties.
3 . 1-2
.(.) - ££ CM-"' - s /~ *-•
„ J- |C(j)!cos[-' + #,Mi'<-' + J j|GMI»n!.-M- #»(-)[ J-’
and therefore
9 (C) = — f |G'(w)j cos[u;t + 0*(<*>)] da?
* Jo
For 9 (c) = r-‘»(t), CM = Therefore |C(u»)| = 1/V^T^ and *„M = - tan_1 <?) Hence
3.1-3
Therefore
and
n ii— cos 1
— tan
Jo v'w 2 4- o? 1
GM = r aMc-’^dt*f ~r*j
G*M = r^dtj -~*j
G'(—->) = -*X311<
O)^4
16
Page 17
3.1-4 (a)
G(u-) =£ r-atc-J*‘ di = jf e
_0w+a)‘ = j _ f,-(j-'+°)'r
jvj + a.
(b)
3.1-5 (a)
1 — e-(jus-a)T
jul — aG(^) = JT
dt = Jr.-
(}w -a)di =
4 _ 2f~ JU’ - 2r"j2w
G(-0 = i:Ar-'^dt +
J^2r
' Jwtdt =
(b)
G(u>) . J°
dt +J'VJ“‘ dt = ^ [cos^-r + wr sinwr - 1]
This result could also be derived by observing that g(t.) is an even function. Therefore from the result .n Prob.
3.1-1
3.1-6 (a)
2 f ,2 .
Gp) = - / f cos -it dt = s '
ry0Tu;
cos ujt + sin U>T - 1]
i r 2
"0 / .2,2
(wpt2 ~ 2)sin-upf + 2uiq/ cos-upt
ir1'
a.
r-n—
f
i
1
-2. -I
A_
czz 1
- 1 [o 1
2 o
Fig. S3.1-0
(b) The derivation can be simplified by observing that G(~) ran be expressed as a sum of two gate functions
GiU) and Gj(*j) as shown in Fig. S3. 1-6. Therefore
a{,) = h JjGiM +
G
^)Viut^ = i {//“" ,U+L *** * sin 2/ + sin t
trf
3.1-T (a)
s(0 = j_ r'\
2* J-*/
2
*/2
,/wl, ^ w/2{jf cos a; + sin^)..,/2
2tr ( 1 - ts )
1
5t(l - t2)
cos(?)
(b)
<?(0_ J_ f
0
G(^)r i~,tf/wU = — [ [ G{~) cos *Jtdu/ + j f G(w) Sin attdw
2 it J2ir j_y_^ d- wo
Because G(w) is even function, the second integral on the right-hand side vanishes. Also the integrand of the
first term is an even function. Therefore
17
Page 18
.1-8 (a)
17(0 =
1 f"0 yj 1 rcostiv-MwsinU]= — / COS t.u/ cL) = -9
JT yo U^O ltU>0 L 1 J
n^at 2(cos wot + wot sin wot - M
rJ‘*,(t‘ t° ) duj
“'0
0
*rju/(t-ep)
(2n)j(t - to)
U/Q
-“'0
8inu>o(t -_to) = i±sinc[u,o(t _ <o) ]
rr(t - to) *
(b)
i r° /u'°
<?(t) = — Jjr dw + J
-jcJW< dw
1 JW «
0_ 1 r
jw.*'°
= 1 - COS wot
2tr/r
-^o 2trt* o
1.2-
1 Figure S3 2-1 shows the plots of various functions. The function in part (a) is a gate function centered at the
origin and of width 2. The function in part (b) can be expressed as A (*£75)This is a triangle pulse centered
at the origin and of width 100/3. The function in part (c) is a gate function rect(|) delayed by 10. In other
words it is a gate pulse centered at t = 10 and of width 8. The function in part (d) is a sine pulse centered
at the origin and the first zero occurring at Sf = if, that is at w = 5. The function in part (e) is a sine pulse
sincf*) delayed bv 10*. For the sine pulse sine(f ), the first zero occurs at f = ft. that is at u; - 5tr. Ther t 1
the function is a sine pulse centered at w = 10tt and its zeros spaced at intervals of Sir as shown >n the fig.
S3.2-le. The function in part (f) is a product of a gate pulse (centered at the origin) of width 10* and a sine
pulse (also centered at the origin) with zeros spaced at intervals of 5ir. This results in the sine pulse truncated
beyond the interval ±5tr (|t| > 5ir) as shown in Fig. f. .......
5.2-
2 The function rect (f - 5) is centered at t = 5. has a width of unity, and its value over this interval is u»U». Hence
-jutdt = --
i
4.3
—>5w
_ _Lir --'4 - 5*' _ r -Js 1
/*’om ‘L
(f)-"= sine
i
18
Page 19
3.2-3
.9(0 = — /°
2* Jio-
10+w WwlrJ“‘ dw =
2* (,/w)
10+ir
10— *
1 rfl
j(io+w)t _ ^(to-ion
j 2ttw*•
llOl
— [2j sin —f)= sinc(irf)fi
,,0‘
j2?rw3.2-
4 Observe that
result.
3.2-
5 Observe that
1 + Sgn(t) = 2u(t). Adding pah* 7 and 12 in Table 3.1 and then dividing by 2 yields the desired
cos(^ + #) = I[^^> + r-^0,+,)]
_ ir>Vu'0 ' + Ie--,9e"JW0 ‘
2 2
Fourier transform of the above equation yields the desired result.
3.3-1 (a)
,. 1
iy(t) <=> irt>(ui) + —0(0
Clw)
Application of duality property yields
irAlt) + — *=* 2ttu(-w)
' V1•“l/ 2wj(-u>)
6(0
3h +^]
>'(-*’)
Application of Eq. (3.28) yields
£(-t) - — <=> «(•*')jtrt
But Mf ! is an even function, that is — MO- an<*
+ -4) <=> «(*')2‘ trr
(b)
cos wo( •*==* rr[A(«*; t wo) + ^(w wq)|^
5/C*) G?(u>)
Application of duality property yields
ir[A(/ + wo) + *(f - wo)] •*==> 2tr cos (-wow) = 2 tt cos (wow)
O'(t) 2»0(-w)
Setting wo = T yields
(c)
k{1 + T) + f>(t - T) <=> 2 cos Tw
sin wof <=f
G(u/)
sin
s(t)
Application of duality property yields
jn^t + Wo) - t>{t ~ wo)j <=> 2tr Sin(-wow) = -2tr sin(w0w)W ' v——
'
G(e) 2
Setting wo = T yields
19
Page 20
b{l + T) - b(t - T) <=> 2j sin Tu>
3.3-2 Fig. (b) jji(f) = fl(-f) and
GjP) = G(-w) * - 1)
Lu
Fig. (c) fl2 (0 = g(t - 1) + 9i(* - 1)- Therefore
G,(-») = [Gp) + G 1 (-»)]e"J-’ = [Gp) + G( -u.)]e-
2-(cos»j + wsinu> — 1)
Fig. (d) fl2(0 = flU - 1) + .?»(' + J )
C4 (-) = CP)*"*' + G(-w)r*-
= -L(2 - 2 cos.*,] = sin2
|= sine
2
(f
)
Fig. (e) 94(0 = fl(* “ ?) + flit' + £)• and
G4W = GO)r-'u/S + G,(w)r^2
. . . rlu,/2
- juirr'" - 1] + " V
= 35'2^ sin
f]=sinc(^)
Fig. (f) 95(f) can be obtained in three steps: (i) time-expanding 9(0 b> a factor 2 00 then deia>ing '
-
JLl mi and multiplying it by 15 [we may int.reh.ng, the tmqu.nc. to, .up. W and (n)l. lb. «.« «P(time-expansion by a factor 2) yields
/ (5)2G(2w) = - J2^ " l >
Second step of time delay of 2 secs, yields
/ (42)—^ - .2-'”' - i)e-
~
The third step of multiplying the resulting signal by 15 yields
3.3-3 (a)
(t)rect Tsinc (!y)
rect(^)— (f)p±j»Ti2
20
Page 21
and
GU) = TsincwT/ai
Of)>= 2jTsinc
,/4 . a /^T\= T Sm (“)
wT\ .utT
s,nT
(b) Prom Fig. S3.3-3b we verify that
<t(t) = sin f u{f) + sin((. - ir)«(t - tr)
Note that sin(t - ir)u(f - tr) is sin t»i(«) delayed by tr. Now. sintu(t) <=* - 1) - + 1 ))+ i--3 and
sin(t - - ir) «=» {^-[AU - 1) - *U + 1)1 +j _35^
p J
Therefore
guo = {Jim-3 - d - *(-» + D] +
Recall that g(.r)*(.r - to) = .«(.t0 )/>(t - to). Therefore *U ± 1)0 + c~ , ’r“') = 0- and
1
GU) =1 -^ 2
(1 + r‘ J *-v)
(c) From Fig. S3.3-3c we verify that
q{ 1) = cost [»/(t) - « (f - = cosfti(f) - cost t, (f - -)
But sinft - f )= - cost. Therefore
<t(0 = cost i/(0 + sin ^t -//
(t -
GU; = |(*U -!) + «*• + !))+ + - 1) " «(*• + »)) + 7T^2}
2
Also because jj(t)A(.t — .to) — j?(to)A(t to).
^•±l)r‘ J~/J = *U± Dr*'*'2 = ±j/>(*± 1)
Therefore
(d)
GU) = J*’
-j»w/22
1 - U/J
1 - W2 1 - -t2[j-*' + <
-JWU//2I
9(0 = r-“‘|«(0 - «(* - r)l= r - T )
= c-0,(t) - c-o7V 8 (, - T,
rr(f - T)
GU) = -1
-a7-j^T
jjj + o jui + a Jut + O(i-
-(o+.i^)T]
3.3-4 From time-shifting property
q(t ± T) GU)c ±J
Therefore
g(t + T) + s(t - T) <=> GUy-T + GU)'"^T = 2GU) cos utT
\\e can use this result to derive transforms of signals in Fig. P3.3-4.
(a) Heie p(t) is a gate pulse as shown in Fig S3.3-4a.
21
Page 22
9(0 = rect <= 2sinc(u>)
Also T = 3. The signal in Fig. P3.3-4a is g(t + 3) + g(t - 3), and
g(t + 3) + g{t - 3) <=*• 4 sinc(u>) cos 3ui
(b) Here g(i
)
is a triangular pulse shown in Fig. S3.3-4b. From the Table 3.1 (pair 19)
.9(0 = A(i) <=> sine2
Also r = 3. The signal in Fig. P3.3-4b is g(t + 3) + g{t - 3), and
g(t + 3) + g(t - 3) 4-a 2 sine2
(|)cos3ut
Fig. S3.3-4
3.3-5 Frequency-shifting propeny states that
Therefore
g(t)r±, 'J°‘ *=f G(u> T- -^o)
q(1) sin - y-IfllOe'"0 ' + .9(0^“^') - jj\
G^ ~ -*) + GU ~ ^o)i
Time-siiifting property states that
git ± T) <==> G(-')ed
Theiefoie
Q(f - T) - g(t - T) <=> G(o>)cj'*T - C(-)r J~T = 2jG(w)sin»T
— ! ff (/ + r)-g(f-T)) <=> GV)sinTv12./
‘
The signal in Fig. P3.3-5 is fl(f + 3) - »(/ - 3) where
g(t) = rect «=> 2sinc(v)
Therefore
4. 3) _ 9 (/ - 3) <=> 2j[2sinc(;*>)sin3w] = 4j sinc(-) sin 3->
3.3-6 Fig. (a) The signal g(t) in this case is a triangle pulse A(£) (Fig. S3.3-6) multiplied by cos lOt.
9(0 = A cos 10/
Also from Table 3.1 (pair 19) A(£)— trsinc=(^) From the modulation property (3.35), it follows that
9 (/) = a(—)cosl0f*=* ^jsme r2
+ S)nc2 JJ
The Fourier transform in this case is a real function and we need only the amplitude spectrum in this case as
shown in Fig. S3.3-6a.
Fig. (b) The »(.) here i, .he ,™. H .he.igh.lh, Fig. (.) dd*jd by 2.. Fta» >i™ ehif.ing prep.r.y-
its Fourier transform is the same as in part (a) multiplied by <* €l€ °,e
22
Page 23
The Fourier transform in this case is the same as that in part (a) multiplied by e This multiplying factor
represents a linear phase spectrum -2 trot. Thus we have an amplitude spectrum [same as in part fa)j as well as
a linear phase spectrum ZG(v) = -2*w as shown in Fig. S3.3-6b. the amplitude spectrum in this case as shown
in Fig P3.3-6b
Note: In the above solution, we first multiplied the triangle pulse A(y;) by cos 10/ and then delayed the tesult
by 27t. This means the signal in Fig. (h) is expressed as iM^j^cos 10(/ - 2*).
We could have interchanged the operation in this particular case, that is. the triangle pulse A(^) is first delayed
by 2r and then the result is multiplied by cos 10/. In this alternate procedure, the signal in Fig. (b) is expressed
as A ( “ T"— )cos 10/ ..
This interchange of operation is permissible here only because the sinusoid cos 10/ executes tntegial numbei o
cycles in the interval 2tr Because of this both the expressions are equivalent since cos 10(/ - 2tr) = cos 10/.
Fig. (c) In this case the signal is identical to that in Fig. b. except that the basic pulse is rect(^) tnstsad o.
a triangle pulse A(£). Now
rect(27)
<=> 2/r sinc(/rut)
Using the same argument as for part (b). we obtain
C(ut) = tr{sinc[tr(ut + 10)) + sinc[7r(w - 10)]}rJ ~
3.3-7 (a)
/tj — 4 \ /lu + 4 \G(-t) = rect (—2” j
+ rect ~)
Also
isinc(f) <=> rectj
Therefore
2g(t) = — sinc(/)cos4/
7T
(b)
OM . a (2ii) + A(^i)
Also
23
Page 24
Therefore
isinc2 (t) <=> A
g(t) = — sine2(I) cos 4t
IT
3.3-8 From the frequency convolution property, we obtain
g2{t) <=> — G(-v) * C(u>)
The width property of convolution states that if t.'i(x) * cj(.t) = y(x), then the width ot y(x) is equal to the
sum of the widths of ri(.r) and C2 (r). Hence, the width of G(u>) * G(*>) is twice the width of G(ui). Repeated
application of this argument shows that the bandwidth of gn(t) is nB Hz (n times the bandwidth of g{t)).
3.3-9 (a)
(b)
<?(•") - r'^
,dt ^-^- C08 ^T] = J
Z sin2 {T)
T/2\g(t) = rect (^)—(^)
and
rect (1) «=> Tsinc|(f)
rect|
'/±T/2\
,T J
•t=> Tsinc
cu )= Tsinc
|
\rJwTI'
i - c'^T/2]
/uT\ .-tT
= 2jl sine 1 — 1 sin2
,/ 4 . 2— em— 9111
sjJ V 2 )
(c)
^ =r A(t + T) - 2#(t) + A(i - T)r/f
The Fourier transform of this equation yields
ju>G(uj) = c-'wT - 2 + r~i
‘-'T = —2[1 - cos wT] = -4sin2
Therefore
C(w)=^sin 2
(^)
3.3-10
A basic demodulator is shown in Fig. S3.3-10a. The product of the modulated signal 9(f)cos~o< with 2cos^0 f
yields
9 (/ ) cos u.‘ot X 2 cos
-
2,9(«)cas2-t0 t = 9 (/)[l + cos2^0 <| = <?(0 + 9(f)cos2*o<
The product contains the desired g(t) (whose spectrum is centered at a) = 0) and the unwanted signal g(t)caa 2wo*
with spectrum i[C(^ + 2^o!-i'C(a;-2^o!. which is centered at ±2u>0 . The two spectra are nonoverlapping because
24
Page 25
Fig. S3.4-1
-&zr
• Itdir
tvTT to-*
CrJc^^dO
MlW CO .
3.4-1
If < _’o (See Fig S3.3-l0b). We can suppress the unwanted signal by passing the product through a lowpas„
filter as shown in Fig S3.3-10a
O’i(w) = sinctjgggj) and Gi(u>) - 1
Figure S3. 4-1 shows G\(~). H\(~) and Now
Yib:)»GiMHi(u)
Vj(ui) = Gj(w)//3(ui)
The spectra YiM and V2P) are also shown in Fig. S3.4-1. Because y(t) « wt(*)va(0. t»» frequency conv°lution
property yieldsY<» = *(-/)•*(*)- From the width property ofconvo^ he band id
y(j) is the sum of bandwidths of and Y*J). Because the bandwidths of Vi(tf) and YaM are 10 kHz.
kHz. respectively, the bandwidth of Y(j0 is 15 kHz.
3.5-1*(<*>)«
Using pair 22 (Table 3.1) and time-shifting property, we get
MO = 1r-0-to)a /«fc
y/Titk
This is noncausal Hence the filter is unrealizable. Also
|In \H (u-')ll
J + 1
kj1
J1 + 1
dui = oc
25
Page 26
Figure S3.5-1
H— -H. «.» isnoncausal .... .h^tor.— Sj»« ,,(<_)» •_*-» '”"'*£as shown in the adjacent figure. Choosing to — 3v2fc, n(0)
to = 3v^f is a reasonable choice to make the filter approximately realizable.
3 -5'22 x 10
8->wi0
~u,2 + lO 10
^
From pair 3. Table 3.1 and time-shifting property, we get
,,(,) = e-10i|-‘o1
The impulse response is noncausal. and the filter is unrealizable.
t.
The exponential delays to 1.8% at 4 times constants. Hence to - 4/a = 4 x 105 - 40/is is a reasonable choice
to make this filtei approximately realizable
3.5-3 From the results in Example 3 16
\HM\
Also H 10) = 1. Hence if is the frequency where the amplitude response drops to 0.95. then
10s
|tf(u>l)| = = 0.95= Wi = 328.684
y/u? + 10‘ 2
Moreover, the time delay is given by (see Example 3.16)
. r<i (Q) = i = 10' 6
w J + a2
If is the frequency where the time delay drops to 0.98% of its value at w - 0. then
= J2l°io iS
= 0,98 x 10-6 W3 = 142,857
We select the smaller ofw, and that is u, = 142,857, where both the specifications are satisfied. This yields
a frequency of 22.736.4 Hz.
3.5-4 There is a typo in this example. The time delay tolerance should be' 1 of^ sj 01 * 10». Let
Tire band of Aj = 2000 centered at « = 10s represents the frequency range from 0.99 x 10
us consider the gains and the time delays at the band edges. From Example 3.16
!«(-•)! = .nr rf ~ ZPTJ a ~ 10
At the edges of the band
26
Page 27
l03
IHI0.99 » 10®)! - ~ 10
1
« 10~1
’ “d ‘“ 'I=
The s.i„ variation ove, the band i. only 1 .»%. SWMr. »•M «" «">* “"* “ 'h* “
= 9.901 x 10'
,MM « 10®) = ">'i “(1 01 - 10‘> *
The time deley variation over the bend ii <1%. Hence the ttenetniseton rnny
^ find
= 0.01 alt - 107)
Fig. S3.6-1
3.6-1
rU) = G(w)rect (~) <
- j{wfo+*c »m wT> )
*G(„)rect (^) U-./1: sin ^TK
This follows from the fart that r1 * 1 + t when * < 1- Moreover. Gp)rert («*g) = GU) hecause GU) i*
bandlimited to B Hz. Hence
YU) = nU)r~ J*t0 -jkGU) si" +-Tr-J“'°
Moreover, we can show that (see Ptob. 3.3-5)
lr\g(t + T) - g(t - T)] «=> GU)
Hence
I/(f) = g(t - t0 ) + |l»(t - to - T) - g(t ~*o + T)}
Figure S3.6-1 shows g(t) and ?/(/)
3.C.2 Rectth that the ttanafe, function of an ideal time deiay of T seconds is Hence, the uansf.r functton o,
the equalizer in Fig. P3.6-2 is
-jwil .-;2».At , -
HvfU) = °o + «i*3 +«!'• +• +a„r
Ideally, we require the equalizer to have
[h1‘
l
W) ] desired
1
1 + c,r
1 — or'
,wAt
J-Af ,2 -jZwAt
+ Q r.1 -jSutAI
a c + ••• + ••
The equalizer in Fig. P3.6-2 approximates this expression if we select a0 1- a, a. «2 a .
(-1)"q u.
27
Page 28
3.7-1
Letting i = ^5and consequently dt =
1 » r -x1/!, ^ - -1—
E“=
27^75 J-J 2^2** 2(ty/H
Also from pair 22 (Table 3.1)
G(u>) = c
Letting <tu? = and consequently du/ = j^dx
Viir 1
3.7-2 Consider a signal
1 1 r -.VJ, V2tr L_^=2^772 L'
T ~2wi7\/2
<j(f) = sinc(A't) and G(w) = rect(2^)
= Jsine
2(ki)dt = -^ J [
r6Ct(2*-)]
d.j
3.7-3 Recall that
Therefore
flj(0_ _L G 2(~V
W< andJ
r, . (»)»<<)*- 5;/_»(<) [/^ox “
«/ —X -»
1
dt dui -*JGl(-w)C?2(^')^
Interchanging the roles of *(«) and «(#) in the above development, we can show that
P <n(t)92(t)dt = ~ JGip)G2(-w)dv
.. u • 17.1 if we identify 01U) = sinc(2irBf - rntr) and .9a(«)
3.7-4 In the generalized Parseval s theorem in
sine (2irBt - nit), then
Gi(a' )=jB
rect and C2(w) =2B
rect(475 )
r""
Therefore
Jgi(i)<l2 (t)di = ^ [
rect(4*5)]
But rect (^g) = 1 for kl < 2tr B. and is 0 otherwise. Hence
2,
r 20 rfw
28
Page 29
1r2wB f 0
In evaluating the integral, we used the fact that r±J2wk - 1 when k is an integer
3.7-5 Application of duality property [Eq. (3.24)] to pair 3 (Table 3.1) yields
n ^ r»
n — m
2a
t* + a*2trr
The signal energy is given by
Eg = - r ^ = 4* f e' 2aw^ ^" dO do
The energy contained within the band (0 to W )is
fwEw = 4ir / r-
2au<hj = — [1 -
doa
If Eu = 0.99Ea . then
—2aW >
_ 0 012.3025 . ,
0.366 uIV = — rad/s Hz
a a
3-7-6, f 9
2(f)^ 4^.) then the outpul rU.) = AU)NM. where Jf (-») is the lowpass filter transfer function (Fig.
S3. 7-6). Because this filter band A/ 0. we may express it as an impulse function of area 47rA/. Thus.
H(*>) a [4 tt A/]A(—’) and >'(.*>) « (4tt.4p)A/]ft(w) *= [4trA(0)A/;M~)
Here we used the property q(r)hlr) = g(0)*(.r) [Eq. (1.23a)]. This yields
»/(f) = 2A(0)A/
Next because q2(t) <==> A(v). we have
Ap) = J <?
2(f)c 1*>t dt so that A(0) = r ,•(«>*-
J -flw
Hence. »/(<) = 2E9 A/.
40.f
«5-»-
Fig. S3.7-6
3.8-1 Let 9 (f) = r/i(f) + <?2 (0- Then
where
*«(r)1 /
,T/2
= lim — / [ffi(i) + ,92 (t)ji.9 i(f +t) +ff2(t + r)]d(
r— 7 J_ 77 2
= 7£9) (t) 4- ^siys(r ) 4" (T )
^ry(T)1 A
772
lim - /T(t)j/(t + r)dt
T-*oti i J - 772
29
Page 30
3 .8-
lf we let g\(1) = Ci cospit + #i) and <?a(0 = C3 cospat + d2 ). then
/T/2
C,Ca cosPit + #1 ) cosPa* +^ dt
T/3 _
According to the argument used in Example 2.2b. the integral on the right-hand side is zero 9 ,n (
Using the same argument, we have i(T )= 0. Therefore
Ha (r) = = % COSiJlT + ^2 COSU,2T
This result can be extended to a sum of any number of sinusoids as long as the frequency of each
distinct , hence, if
g (l) =£ C„ «*(*-** + #«) then H a (r) =£% C0S^°T
n — 1
Moreover, for go(0 = Co, 72bo(t )= ant^
/T/a
ngo9 l(r)= Urn i
/CoCicosP,t + wir + ei)rft = 0
T— >». i J -T/2
Thus, we can generalize the result as follows. If
and
<,(0 = Co +^ C„ cos(n~of + then 7Za (r) - Cl +^ 2
nx 1
Ok_i
Syp) = 2*Cfa(~) + | 5^_ n~‘0) + A(U; + nW0)
'
cos njjQT
figure S3 8-2a show, the waveforms r„> and - r, for e < fa* U, T -NT, 0. .ho “'v/2 pulses in ,h< waveform of dura, ion 7". The .... onto the product .<(<),« O » •'« '1-
shown in Fig. S3.8-2b. Therefore
. rT/a
H x [t)= lim = /r(f)r(f - r)rft
i j_7-/a
1 JV /T* \ WI _ JL'j T < ^= "2 U Tj 2
2 \2 Tt)1T
! < y
for a < w < n. there » no ove„.phew puis...-Wl - ° SfcBut on the average, only half pulses overlap. Hence. *«(T )
ycomponents, as shown in Fig.
transoms of„W - -M-
Hence
S.(w) = ^ aincJ
(~) + st(-0
where s,W is the Fourier Worm of the periodic triangle function, shown in Fig. S3.8-2d W, find the
exponential Fourier series for this periodic signal to be
w,(r) = Cnfinu’T'Jb = Y
n* — nu
Using Eq. (2.80). we find D„ = ^sinc 2(^). Hence, according to Eq. (3.41)
S2P) = | ^ sine2
(y)AP ~ =
l7T>= - -X.
30
Page 32
3 .8-
(C) T7(t)
(b) x'(f)
-n
2 '*' _ _ / dul = —* Jo *
and y (t)
A(uJ - \)d*i = - andIt
= - f —i— dw = i
* Jo uJ +l 4
y { )” Jo S + 1 * ^0 2 2rr
» The ideal differentiator transfer function is j-u. Hence, the transfer function of the entire system is
HW = (7^)°'w) = j^T and
~ )= hj r6Ct
= rect (l)^TT du,= -/0 dTi**’** (1_
4)=0 °.06831
32
Page 33
4.2-1 (i) For ni(t) = cos 1000'
V’dsb-sc^) rri(t) cos 10,000 / = cos 1000/ cos 10, OOOt
x [cos 9000t + cos 1 1 . 000'j
LSB USB
(ii) Foi mi(/) = '2 cos 1000/ + cos 2000/
^d=b « c (0 = ">(/) cos 10. OOOt = [2 cos lOOOt + cos 2000/] cos 10. OOOt
= cos 9000/ + cos 11. 000/ + [cos 8000/ + cos 12. 000/]
= Icos 9000/ + i cos 8000/] + [cos 11. OOOt + ^cos 12. OOOt]
1 2 .*
(iii) For mU) = cos 1000/ cos 3000/
^nsBtc (f) = m(/)cos 10.000/ = i [cos 2000f + cos 4000/jcos 10. OOOt
= - [cos 8000/ + cos 12. OOOt] + i[cos6000f + cos 14,000/]
2 2
= - [cos 8000 / + cos 6000tj + j [
cos 1 2. 000 / + cos 14 , 000/)
LSB USB
This information is summarized in a table below. Figure S4.2-1 shows various spectra.
Page 34
Fig. S4.2-2
c
USB
Fig. S4.2-3
LSB frequency USB frequency|
9000 11.000|
9000
8000
8000 12,000|
6000 14.CKK)
9000 and 11,000
9000 and 11,000
8000 and 12.000
8000 and 12.000
6000 and 14,000
4.2-
2 The relevant plots are shown in Fig. S4 2-2.
4.2-
3 The relevant plots are shown in Fig. S4.2-3.
4.2-
4 (a) The signal at point b is
f/a (i) = rn(t) cos* *jc t
= 7/1 (f f— COSU/c^ "i" *7 C06 3-Vct
Page 35
The term 2,„(0 cos~-c f is the desired modulated signal, whose spectrum is centered at ±u»c/ The remaining
,erm JtsdUw * <he unwanted^term, which JETS^"» H*"“-
*
h" ,>s“” "”k! “
desired with the output ^rn(t) cosu)ct-
(b) Figure S4.2-4 shows the spectra at points b and c.
(c) The minimum usable value of-* is 2trS in order to avoid spectral folding at dc.
(d)
in (t) cos2 wc t = [1 + cos 2a;c t|
= im(t)+ im(t) cos 2wc t
The signal at point b consists of the baseband signal |f»(0 and a modulated signal
carrier frequency 2we not the desired value *e . Both the components will be suppressed by the filter, whose
center center frequency is wt . Hence, this system will not do the desired job.
(e) The reader may verify that the identity for cosn^t contains a term cosu;ct when n is odd. This
when n is even. Hence, the system works for a carrier cosn
u-c t only when n is odd.
Fig. S4.2-4
4.2-5 Wc use the ring modulator shown in Fig. 4.6 with the carrier frequency fc = 100 kHz Pc - 200 t x 10a
). and
the output bandpass filter centered at fc = 300 kHz. The output MO is found in Eq. (4.«b) as
„,(t) = i fm(t) cos S>et - |m(r)cos 3<1-J + im(t)cos 5
w
et + •
•
]
The output bandpass filter suppresses all the terms except the one centered at 300 kHz (corresponding to the
carrier 3
i
c f). Hence, the filter output is
-4,/(*) = — m(r.) cos 3
3ir
This is the desired output km{t) cos uie t. with *• = -4/3tr.
4 2-6 The resistance of each diode is , ohms while conducting, and oo when off. When the carrier A cos^J is positive
ZSXLto* (during ,h. entire positive cycle), end »'h.n the c.rrie, Sjgm,dU ri ne the entire negative half cycle). Thus, during the positive half cycle, the voltage TfpW) aPPc* s ™2S3 theSo”* During the .eg.tive half cycle, the output volte,. » ««ro
;
Therefore, .he drod.s », es
a gate in the circuit that is basically a voltage divider with a gain 2R/(R + r). The output is therefore
onr0 (l) = = w(t)w(l )
n + r
The period of «>(*) is To = 2rr/o>c . Hence, from Eq. (2.75)
35
Page 36
!!•(>) = i^
^COS Wc' - ^cos 3wC t + g
cos *]
The output r0 (t) is
<•0(0 - ^.OMO - [5 + I (“* - 5~^ ' + 5“• 5“‘' + -
)]
(a) If .e p.v, the output 0.( 1 ) through . b.ndb.nd filter (centered .< .,.). the Biter .uPPre*«£ "t„,(f)cos nor-ct for all n # 1, leaving only the modulated term ^^rn^cos^c m
(b) Th‘ seine
1
circuit c.n be ueed „ . demodul.tor It .. « . b»ep». Biter .. the output, in .hi.« the input
is <p(t) ~ Tn(l) cosset and the output is rn(t ).
4.2-7 From the results in Prob. 4.2-6. the output co(f) = hni{t) cos u/d, where k = ln the present caSe '
m(t) = sin(^r t + ft). Hence, the output is
ro(t) = ksm(wet + 6)co&ujc i = ^[sinfl + sin(2^ct + 0))
The lowpass filter suppresses the sinusoid of frequency 2u,c and transmits only the dc term * sine.
at G>>arf (E>
-<©K 5K f©K <51
s
Of©
Fig. S4.2-8
channel b.ndwidth mua, be a, Leaf 30.000 ,.d/s (Bom
'shows’lhe receiver to recover ...i(t) and m,«) from the received modulated signal.
4.2-9 (a) S4.2-9 shows the output signal spectrum >'(-’)•.
. hi h frequencies
b Observe that W) is the same as J»/(w) with the frequency^ £l«Scy -Act™™,
are shifted to lower freq.tencies and vice versa. Thus, the scrambler in F.g, P4.2-9 tnverts the frequency
36
Page 37
Y^) s«^*nes«vf
fFig. S4.2-9
M- 1
-04;
^ 30 *
- /c -
Fig. S4.3-2
To get back the original spectrum J17U-). wo need to invert the spectrum V(*>) once again. This can be done by
passing the scrambled signal >/(/) through the same scrambler.
4.2-10 We use the ring modulator shown in Fig. 4.6 except, that the input is ,,,(/)cos(2*)10b
/ insr^d of rn(t). The
carrier frequency is 200 kHz [*, - (400»)10\]. and the output bandpass filter is centered at 400 kHz T -
output r,(t) is found in Eq. (4 7b) as
,,(t) = j»n i) eos(2?r)10S
f j't'olO = ^(Ocos^lO6
*[cos (400*)10’t - |
cos 3<400»)l0»t + \cos 5(400tr)l0\ + ]
The product of the terms" (-1/3) cos 3(400-r)10’t and (4/*)m(0cos(2rr)106
t yields the desired term
cos (800rr)10Y whose spectrum is centered at 400 kHz. It alone passes throug P
(centered at 400 kHz). All the other terms arc suppressed. The desired output is
iy(t) = rn(t) cos (8Q0n) 10* t
07T
4.3-1 QaU) — \A ‘r rn(f)] cosu-vt. Hence.
gh {t) = \A + rn(f)] COS2u>e <
= + m(t)j + + w(t)]cos2u>cf
The first term is a lowpass signal because its spectrum is centered at u; = 0. The lowpass filter allows this term
"p« to Tupp^J <1* JLd term. who* ,-c.rom i. «n«r«d .. Hoc. <ho ottptt of ,h, lowp«»
filter is
j,(t) = A + rn(t)
When this signal is passed through a dc block, the dc term A is suppressed yielding the output n.(f). Tins
shows that the system can demodulate AM signal regardless of the value of A. I h.s is a synchronous or col,
demodulation.
37
Page 38
4.3-2
(a) ,i = 0.5
(b) /t = 1.0
(c) /t = 2.0
fd) /! = OC
l/ip _ 10=> A = 20T “
Atrip _ 10
=> A = 10A
rn p __10
=>.4 = 5
>4 Ar/ip 10
=> A = 0>1 A
This means that // = oo represents the DSB-SC case. Figure S4.3-2 shows various waveform
4.3-3 (a) According to Eq. (4.10a), the carrier amplitude is .4 = mp/,i = 10/0.8 = 12 8. The camei powei is
pr = ,42/2 = 78.125.
Fig. S4.3-3
(b) The sideband power is m 2 (t)/2. Because of symmetry of amplitude values e q_
-' J m(f)
I,, (, )mav be computed by averaging the signal energy over a quarter cycle only Over a quarter cv
be represented as w(t) = 40>/T0 (see Fig. S4.3-3) Hence.
»"*(0 L. [
T° 4
[12I12
dt = 33.34
To/4 J0L To J
The sideband power is
P - mli
H
= 16.67J » rt
The efficiency is
V ~P, 16.67
+_
78.125+ 16.67x 100 = 19.66%
4.3-4 From Fig S4.3-4 i, i. *« .« d.
for demodulating AM signal using envelope detector is .4 -r m(f) > 0 for al .
27Tt
Page 39
Fig. S4.3-5
4 3-5 When an input to a DSB-Sc generator is »»(*), the corresponding output is m(t)^by^dding4
A+
", <“ the corresponding output will be \A + »«)] «**. This is precse.y theAMs.gn^ Thus, by adchng
a dc of value .4 to the baseband signall we can gyrate> AM usmg a DS' ^ ,
f wg ^ tw0ti,« rnnvprse is eenerallv not true. However, we can generate UbU-at> using a- 8
identical AM s.nfrston in > balanced scheme shown in F*. S4.3-S to cancsl not the earner component.
1.3-6 When an input to a DSB-SCWM is -«> <tnsw.t. the
input is \A r«(/)]cos-*f. the corresponding output will A + m(t). By blocking
output, we can demodulate the AM signal using a DSB-SC demodulator.. , .
(,s
,
The converse unfortunate*, is not true. This is because, when an mput to an AM demodulator rn ,
)j3S c
.
coni.po„di”5 output is |U» envelop, of ,.,(.)! Hence, unlew »«) > 0 for .H < » <* ••
demodulate DSB-SC signal using an AM demodulator.
Fig. S4.3-7
4.3-7 Observe that ,n2(1) = -4
5for all t. Hence, the time average of m 2
(t.) is also A 2. Thus
= A 2 _ ™ (0 = i_
The carrier amplitude is A = ,n p/„ = m, « A. Hence Pc = A2/2. The total power is P, = P, + P. - The
power efficiency is
„ = £ x 100 = x 100 = 0.5' Pt
A2
The AM signal for /i = 1 is shown in Fig. S4.3-7.
4.3-8 The signal r.t point a is [.4 + »,(t)|cos wct. The signal at point b is
, .42 + 2.4rn(t) + , „„„ <j ,
r(t) = 1.4 + m(0l cosset = § * l + COS ’
The low-pass filter suppresses the term containing cos 2»*,«. Hence, the signal at point, c is
A 2 + 2Am(t) + m 2(l) _ A2
«‘(0 = —2 2
14^ +
Usually. m(t)/A « 1 for most of the time. Only when ™(f) is near its peak, this condition is violated
the output at point d is
Hence.
39
Page 40
»/(() sb — + Arn(l.)
do .era, V/J. yield,., the oa.pa, *10 From *» •*“' ” “*
A blocking capacitor will suppress
that the distortion component is m {t)/2.
4.4-1 In Fig. 4.14. when the carrier is cos |(Aw)t + A] or sin l(Aw)t + *]. we have
nit) = 2[ir«»(0 M.’el : + n»a(0 sinwctjcos [(wc + Aw)t + A)
= 2m, (0 cos we t cos [(we + Aw)t + »} + 2n«a(0 sinwctcos [(w« + Aw)t + «
= rn i (t>{oos [(Aw)f + *1 + cos [(2we + Aw)t + b]} + rn 2 (t){sin !(2wc + Aw)f + 1
Similarly
.Tj(f') = "'i(t){sin|(JJc + ie)/ - ») + sin |(id)< + #1) + ».<*><«• K'WX + ‘I
-“ + a*'H + 6,1
After nit) and x»(t) are passed through lowpass filter, the outputs are
= r»j(t) cos [(Aw)t + A) - «< 2(0 sin IM* + A]
m '
a (t) = mi(f)«in [(Aw)t + *1 + m 2 (f)cos |(Aw)f + b)
, . ,... ... . I However, to generate the SSB signals
4.5-1 To generate a DSB-SC signal from "<(<)• we tnultip y
™
(
.it£ 2cosw c f.
This also avoids the nuisance
of the same relative magnitude, it is convergent to multiply () We suppress the USB spectrum
of the fractions 1/2. and yields the DSB-SC spectrum A/(w--*)+J
*
lh. Spectrum. we suppress
(above w, and below -we )to obtain the LSB
S4.M a. b and c show the three
the LSB spectrum (between -we and wc )from the DSB-SC spectrum
(a)From Fig. a. we can express *u.(0 = cos900f and *«b (0 =^os ll0° t
_ 2 cos l300 ,
(b)From Fig. b. we can express „ t..d) = 2 cos 70°t + c°s^ and , B (- co
(c)From Fig. c. we can express ,lf.(0 - il«.400t + cos600t] and *VSB (0 - 3^ 140
4.5-2 d UJ <t) - Tri(t)cosWet + m„(f)sinw c t
V^vsbIO = rr)(t)cosw e t - rr/(,(f)sinwe < and ^lsb! > v
° c«X, + .7. WO, 1000/ - cos£1000 - 100)/ —900,
,00, ,06 1000/ - .in 100/ .in 1000, . co.<1000 + 100), = =» 1 100/
(b) ,„(/) . co. 100/ .2 cos 30® and m»(<) = "in 100/ + 2.in300/. H.nc,
. (co. 100, + 2=0.300,) co. 1000, + (.in 100, + 2.1.300,) .in 1000, - co.900, + 2=0.700,
,.(„ - (cos 100, + 2CO. 300,) cca 1000, - (.in 100, + 2.1.300,) .1.1000, - co. 1100, + 2co.,300,
(cZ, - co. 100, co. 300, = 0.3 cos .00, + 0.5 cos 600, and -.(,) - 0 .5 .in.00, + 0.5.i»600,. Hence.
,„.(„ . (0.3 cos 400, + 0.3 co. 600,) co.1000, + (03.in.00. + 0.3.in600„ .in ,000, - 0 3CO..00, +0.3c«600,
dlia (,) = (0.5 cos 400, + 0.5CO6600,) cos 1000, - (0.3 sin 400/ + 0.5 .in 600,) .in 1000, — 0.5 cos 1.00, + O.Scos 1600,
40
Page 42
A«u # 0 in SSB
Aw «u—-w
Fig. S4.5-5
f m and Fic S4 5-3b shows the corresponding DSB-SC spectrum
4.5-3 (a) Figure S4.V3a shows ihe spectrum of m(t) and Fig.
'2 hi 1 1 ' cos <OrO®C>Trt
obtained by suppressing the USB spectrum
(hi Figure S4.5-3C shows -he corresponding ^ by suppressing the LSB spectrum.
(c) Figure S4.5-3d shows the corresponding ^SB spec r
from Table 3.1 (pan 16) and the
We now find the inverse Fourier transforms of the LSB and U.B sp
fret|uency shifting property as
^ (/) s* 1000 sine {lOOOtri) cos 9000ttf
Vico (/)« 1000 sine (lOOOtrf) cos 11. OOOtrf
4.5.4 ® (“> "* " * B"b*rt ,ra"Sf°"”'r “
H{*i) — -j sgn (•*>)
„ „ apply a. .h. inpp. of .1. Hilb.,. W <*» °f *te ” < ') "
VP) = - l-jJl/U) sgn Mll-JSgn (-»)]= -
f f (t) is —m(f) To show that the energies of »«(0 and '" h( ai
This shows that the Hilbert transform of »u(0 »
equal, we have
£— = r™*(»*-5r£.i*,MI’*'
»/ - *Xi
42
Page 43
4.5-5 The incoming SSB signal at the receiver is given by [Eq. (4.17b)]
flsb(0 = m(<) cosu/cf + ma(<)sin u
>
ct
Let the local carrier be cos [(wc + Au>)t + A], The prodeuct of the incoming signal and the local carrier is e d (t)
given by
ra(t) = *Lsa (f) [(^e+ Aw)« + A]
= 2[t»(t)coswjc « + ma(t) sin ta»ct] cos [(we + Au>)t + A]
The lowpass filter suppresses the sum frequency component centered at the frequency (2ujc + Au/). and passes
only the difference frequency component centered at the frequency Au>. Hence, the filter output c0 ( )is given
by
c0 (f) = m(f)cos(Au>)< + A) - tnft(t)sin(Aui)t + A)
Observe that if both Au and A are zero, the output is given by
p.0 (t) = m(t)
as expected. If only A = 0. then the output is given by
ro(l )= m(f-) cos(Aut)f - rru(t)sin(Aui)t
This is an USB signal corresponding to a carrier frequency Au as shown in Fig. S4.5-5b^ This spectrum is the
same as the spectrum MU) with each frequency component shifted by a frequency Au This changes the soun
of an audio signal slightly. For voice signals, the frequency shift within ±20 Hz is considered tolerable. Mos
svstems. however.lestrict the shift to ±2 Hz.
(b) When only A-u = 0. the lowpass filter output is
r0 {f) = m(t) cos A - rris(f) sin A
We now show that this is a phase distortion, where each frequency component of MU) is shifted in phase by
amount A. The Fourier transform of this equation yields
EoU) = MU) cos A - MhU) sin A
But from Eq. (4.14b)
MhU) = - jsgn U)MU
)
-{*—jMU)MU)
w > 0
it! < 0
and
( MU)***"
{ MU)f~J *
u> > 0
u; < 0
It follows that the amplitude spectrum of co(0 is MU)- the same as that for w(0- But the phase of each
component is shifted bv A. Phase distortion generally is not a serious problem with voice signals because h
human ear is somewhat insensitive to phase distortion. Such distortion may change the quality of speech, but
the voice is still intelligible. In video signals and data transmission, however, phase distortion may be intolerable.
4 .5-6 We showed in prob. 4.5-4 that the Hilbert transform of m h (t) is -rri(t). Hence, rn h (t) [instead of m(f)j is
applied at the input in Fig. 4.20. the USB output, is
tj(i) = nih(l) cos
—
rn(i') sin
= rn(t) cos [u;c l + ^ j+ r»/i(f) sin +
^
)
43
Page 44
C-C.7p f
*• .
»""" —»
/1 . .
2-
-2.- < ± 2_ 4
f M*Fig. S4.6-1
Thus, if we apply m*(f) at the input of the Fig. 4.20. the USB output is an LSB signal corresponding to ».(#).
The carrier also acquires a phase shift ir/2. Similarly, we can show that if we apply n, K{t) at * he input of le
Fig. 4 20. the LSB output would be an USB signal corresponding to rn(t) (with a carrier phase shifted by it 12).
4.6*1 From Eq (4.20)
Wop) -1
P| < 2rBW,p + *lc) + W,p - «.
;
e )
Figure S4.6-U shows W.p - -»e )and H,p +«.)• Figure S4.6-lb shows the reciprocal, which is Hop).
4.8-
1 A station can be heard a. its allocated frequency 1500 kHz as well as at its image frequency. The frequencies
are 2hr Hz apart. In the present case. /IF = 455 kHz. Hence, the image frequency is 2 x 455 - 910 kHz apait
Therefore, the station will also be heard if the receiver is tuned to frequency 1500-910 _ 590 kHz f°
this is as follows. When the receiver is tuned to 590 kHz. the local oscillator frequency « /*> « 590 + 455 - 045
kHz. Now this frequency /lo is multiplied with the incoming signal of frequency fc - 1500 kHz. The ou p
vields the two modulated signals whose carrier frequencies are the sum and difference frequencies, w ic
«
-
1500 -c 1045 = 2545 kHz and 1500 - 1045 = 455 kHz. The sum earner is suppressed, but the difference carnei
passes through, and the station is received.
4.8-
2 The local oscillator generates frequencies in the range 1+8=9 MHz to 30+8-38 MHr "’hen
is 10MHz Jvo = 10 + 8 = 18 MHz. Now. if there is a station at 18 + 8 = 26 MHz. it will beat (nux) with
Ao I ,8 MHz to produce two signals centered at 26 + 18 = 44 MHz and at 26 - 18 = 8 MH*. The sum
component is suppressed by the IF filter, but the difference component, which is centered at 8 MHz. passes
through the IF filter.
(P
44
Page 45
Chapter 5
Fig. S5.1-1
5.1-
1 In this case fc = 10 MHz. m p = 1 and m'p = 8000.
For FM :
A/ = k,w P!2* = 2tr X ]05/2tr = 10
sHz. Also fc = 10
T. Hence. (/,)-»« = 10 + 10 = 10.1 MHz. and
(
~
107 - 10
s = 9.9 MHz. The carrier frequency
increases linearly from 9.9 MHz to 10.1 MHz over a quarter (rising) cycle of duration n seconds. For the next a
seconds when w(f) = 1. the carrier frequency remains at 10.1 MHz. Over the next quarter (the falling) c>c e o
duration «. the carrier frequency decreases linearly from 10.1 MHz to 9.9 MHz., and over the last qua.tei cycle
when ,„(f) = -1. the carrier frequency remains at 9.9 MHz. This cycles repeats periodically with the period 4a
seconds as shown in Fig. S5 1-la.
For PM : , , „ „ w„A r _ n- = 50- x 8000/2* = 2 x 10
s Hz. Also fc = 107
. Hence. (/.)m« =10 +2x10 - J0.2 MHz.
and l f )m = 107 - 2 x 10
s = 9.8 MHz. Figure S5.1-lb shows m(t). We conclude that the frequency remains at
10 2 MHz over the (rising) quarter cycle, where rri(t) = 8000. For the next a seconds, m(f) - 0. and the carnet
frequency remains at 10 MHz. Over the next a seconds, where m(t) = -8000. the carrier frequency rem ,
9.8 MHz. Over the last quarter cycle rh(t) = 0 again, and the carrier frequency remains at 10 MHz. This cycles
repeats periodically with the period 4a seconds as shown in Fig. S5.1-1.
5.1-
2 In this case fc = 1 MHz. w p = 1 and vi'p = 2000.
Af = kttnJ2it = 2O,000t/2 tr = 104Hz. Also fc = 1 MHz. Hence, (/.)m« = 10* +- 10
4 = 101 MHz. and
(/ )min = 106 - 10
4 = 0.99 MHz. The carrier frequency rises linearly from 0.99 MHz to 1.01 MHz over the cycle
(oveTthe interval < t < ^). Then instantaneously, the carrier frequency falls to 0^ MHz and ^arts
rising linearly to 10.01MHz over the next cycle. The cycle repeats periodically with penod 10 as sho»n in
Fig. S5.1-2a.
For PM : ,
Here, because m{t) has jump discontinuities, we shall use a direct approach. For convenience, we select the
origin for *.(f) as shown in Fig. S5.1-2. Over the interval to if. we can express the message s.gnal as
rn (/) = 2000f. Hence.
^(t) = cos[2tr(10)6t+|m(0]
— ccs 2Jt(l0)*t +- |2000tj
r--- cos [2tr(10)6
f +- lOOOirt] = cos [2rr (10s
+- 500) f]
At the discontinuity, the amount
the carrier frequency is constant
of jump is u, d = 2. Hence, the phase discontinuity is kpm d = Therefore,
throughout at 10® +- 500 Hz. But. at. the points of discontinuities, there is a
45
Page 46
5.1-3
5.2-1
5.2-
5.2-:
5.2-
5.2-
Fig. S5.1-2
phase discontinuity of * radians as shown in Fig. S5.1-2b. In this case, we
is a discontinuity of the amount 2. For k
,
> *, the phase discont.nu.ty
ambiguity in demodulation.
must maintain kr < ir because there
will be higher than 2* giving rise to
(a) *pm(0 = A cos [u».f + V«(0) = 10 cosilO, 000/ + kpm(t))
W, „e .iv„, rh.r »,„(.) - 10 a, (13.0000 with A, - WOO. Cb»rlj. m(.) - 3. over ,he inter*! HI < 1-
(b) r'FM(0 = -4 COS ^ + k,f rr.(a) da = 10 cos 10.000/ + kj
Jt»(a) da
j
Therefore k, j‘ rn(a)da = 1000Jm(a) do = 3000/
3/ = Jm(a) doHence rn(/.) = 3
In this case kf = lOOOtr and *> — 1. For
...(,) = 2 cos 100' + 18 cos 2000-1 end ”<(!) “ —200 sin 1001 - 36.000* sin 2000-1
Therefore . 20 end - 36.000. + 200. Abo the braebend IpAl b.od.ld.h B - 2000./2. - 1 kHz.
Fz PM ;
'“> “ 38 06366
kHz.
V,EM (,) = 10 cos(+c / + o.l sin 20007:/). Here, the baseband signal bandwidth 2? = 20007r/27t = 1000 Hz. - so,
U7,(t) = u>c + 200/T cos 2000tr/
Therefore. A- = 200 tt and A/ = 100 Hz and Bem « 2(A/ + B) « 2(100 + 1000) = 2.2 kHz.
- It
)
= 5 cos(wc / + 20 sin 1000*/ + 10 sin 2000 trt).
Here, the baseband signal bandwidth B = 2000/7/2* = 1000 Hz. Also,
u.\(/) = uic + 20, 000/r cos 1000/r/. + 20,000* cos 2000*/
Therefore. A+ = 20,000*+20.000* = 40,000* and A/ = 20 kHz and Bem = 2(A/+B) = 2(20. 000+ 1000)
kHz.
I The baseband signal bandwidth B = 3 x 1000 = 3000 Hz._ , „
For FM : A/ = = iS^7
1 = 15-951 kHz !lnd Bfw = + S) ~ 37 831 kH
For PM • A/ =kp™'r = = 31831 kHz and Bpm = 2(A/ + B) = 66 662 kHz.
i The baseband signal bandwidth B — 5 x 1000 = 5000 Hz’
2/2 + 51-14 kHzFor FM : A/ =
k)
^r ~ 200^— = 1 kHz and Bfm = 2(A/ + B) = 2(2 + 5) - 14 kHz.
46
Page 47
For PM : To find BPM. we observe from Fig. S5.1-2 that vFM (t) is essentially a sequence of sinusoidal pulses
of width T = lO'1 seconds and of frequency fr. = 1 MHz. Such a pulse and its spectrum are depicted in Figs.
3.22c and d. respectively. The bandwidth of the pulse, as seen from Fig. 3.22d, is 4ir/T rad/s or 2/T Hz Hence.
Bpm = 2 kHz.
5.2-6 (a) For FM : A/ = = 22±™£z*± = 100 kHz and the baseband signal bandwidth B -—Therefore
Bfm = 2(A/ + B) = 202 kHz
For PM : A/ =^ = = 10 kHz and BPM = 2(A/ + B) = 2(10+1) = 22 kHz.
(b) m{t) = 2 sin 2000* r. and B = 2000?r/2tr = 1 kHz. Also mp = 2 and m'p = 4000tt.
For FM : A/ =^ - *»-<fps2 = 200 kHz. and
Bfm = 2(A/ + B) = 2(200 + 1) = 402 kHz
For PM : A/ = ^# = ^ig2te = 20 kHz and BPM = 2(A/ + B) = 2(20 + 1) = 42 kHz.
(c) m(t.) = sin 4000*/, and B = 4000tr/2?r = 2 kHz. Also w p = 1 and m'p = 4000tr.
For FM : A/ = = 2°?-'^L— = 100 kHz, and
Bfm = 2(A/ + B) = 2(100 + 2) = 204 kHz
For PM A / = = iP-jW = 20 kHz and BPm = 2(A/ + B) - 2(20 + 2) = 44 kHz.
(d) Doubling the amplitude of in It) roughly doubles the bandwidth of both FM and PM. Doubling the frequency
of /n(fi [expanding the spectrum A/(ui) by a factor 2] has hardly any effect on the FM bandwidth. However, it
roughly doubles the bandwidth of PM. indicating that PM spectrum is sensitive to the shape of the baseband
spectrum. FM spectrum is relatively insensitive to the nature of the spectrum Af(~)-
5 .2-7 From pair 22(Table 3.1). we obtain c' 1 ’ «-» The spectrum A/(ui) = a
which decays rapidly. Its 3 dB bandwidth is 1.178 rad/s=0.187 Hz. This is an extremely small bandwidth
compared to A/.3
Also Mt) = -2fr-*'». The spectrum of m(f) is A/V) = J*A/(->) = jy/*»r* >4. This spectrum also decays
rapidly away from the origin, and its bandwidth can also be assumed to be negligible compared to A/.
For FM: A/ = = 222&pl « 3 kHz and Bfm ft 2A/ = 2 x 3 = 6 kHz.
For PM : To find iii'p . we set the derivative of rri(f) •= -2fc"‘ /s equal to zero. This yields
,,,(/) = -2c-,S/a
+ 4/.ac-
,S/s = 0 t =V2
and rrij, = = 0-858. and
_ tgt - sooo^.Q.sss _ 3 432 kHz and BPW ft 2(A/) = 2(3.432) = 6.864 kHz.
5.3-1 The block diagram of the design is shown in Fig. S5.3-1.
X 1%s\Bf f
£= 1-&35N ’
/\C loRfitlUX 60
-iSL50H*T
£- q*ifi
(0 10-245 M«e
Fig. S5.3-1
47
Page 48
5 .3-2 The block diagram of the design is shown in Fig. S5.3-2.
(a) ^pm(I) = ^ cos [u'ct + kPm{t)}
When this v’pnUO *s passed through an idea) FM demodulator, the output is kpfn(t) This signal. \ h " pas
lough an tdea! iterator, yields 1^(0- Hence. FM demodulator followed by an .dea
demodulator. However, if »(*) has a discontinuity. m(t) = oo at the po,nt(s) of d.scont.nutty, and the system
will fail.
(b) i£Fm(0 = cos u#'c^ *4" lc f Jm(a)do
5 .4
-
2
5 .4
-
3
When this signal „m(0 is passed through an ideal PM demodulator, the output is*,£m(o) **
signal is passed through an ideal differentiator, the output is k/m(t). Hence. PM demodula . >
ideal differentiator, acts as FM demodulator regardless of whether m(t) has jump discontinuities
Figure S5.4-2 shows the waveforms at points b. c. d. and e. The figure is self explanatory.
From Eq (5.30). the Laplace transform of the phase error MO
>
s K‘ven b>'
+ 7i0e ' (s)
For (9,(r) = kt2
. ©,(*) = and
, , ,2*
0<(s) “SVS + AKH(*)\
The steady-state phase error [Eq. (5 33)] is
2khm (9,(t) = Jim »©.(*) = + AK)
Hence, the incoming signal cannot be tracked. If
n + oH{s) = then ©«(*) =
= oo
2k
, 3 >]
and 2k Ik
t
ljm = Jim *©«(•-) = Jim + AK[s + ~) “ Ako
Hence, the incoming signal can be tracked within a constant phase 2k/Aka radians. Now, if
s* + n* + fc- - -
H(s) = then 0*( s)—
s 2 .i +—
and 2k.-
(
lim «,(/) = Jim «©«(*) - Jim ^ + AK^ + 0 , + /,)
0
1„ this case, the incoming signal can he tracked with zero phase error.
48
Page 50
Chapter 6
0.1-2
0.1-1 The bandwidths of gi(t) and g3 (t) are 100 kHz and 150 kHz, respectively. Therefore the Nyquist sampling rates
for gi(t) is 200 kHz and for g3 {>) is 300 kHz., ... , 2,,-,
Also gi3 U) <=> ±qiU) * giM, and from the width property of convolution the bandwidth of ffi. W' >*
the bandwidth ofP 9,(0 and that of »*(*) » three times the bandwidth of g3 (t) (se also Prob. 4.3-10). Similar y
the bandwidth of ffl (/)9a (t) is the sum of the bandwidth of *,(#) and n (t). Therefore the Nyquist rate for fli (')
is 400 kHz. for <j23(f) is 900 kHz. for gi(t)g3 (t) is 500 kHz.
(a)
sinclOOtrf) <=> O.Olrect (3357)
1 he bandwidth of this signal is 100 7r rad/s or 50 Hz. The Nyquist rate is 100 Hz (samples/sec).
(b)
sinc2(100rrf) e*=> O.OlA(jj^)
The bandwidth of this signal is 200 tr rad/s or 100 Hz. The Nyquist rate is 200 Hz (samples/sec).
(c)
sinc(lOOrf) -t- sine (50irt) <==> O.Olrect 0.01 (jfc) + 0.02rect
The bandwidth of the first term on the right-hand side is 50 Hz and the second term is 25 Hz Clearly the
bandwidth of the composite signal is the higher of the two, that is. 100 Hz. The . vquist rate is
(samples/sec).
(d)
sinc(lOOTt) + 3sinc2(60trt)
The bandwidth of rect(^
0.01 rect( 555; ) + ^5 A(3^)
240* <is 50 Hz and that of A(rfb) is 60 Hz. The bandwidth of the sum is the higher of
the two. that is. 60 Hz. The Nyquist sampling rate is 120 Hz.
(•)
sinc(50trr) <=* 0.02 rect( 333^)
sinc(lOOirf) <=s 0.01 rectijg^-)
The two signals have bandwidths 25 Hz and 50 Hz respectively. The spectrum of the product of two signals is
1 /2n times the convolution of their spectra. From width property of the convolution the width of the convolu
signal is the sum of the widths of the signals convolved. Therefore, the bandwidth of tinc(50»f)amc(100irf) is
25 + 50 = 75 Hz. The Nyquist rate is 150 Hz.
6 1-3 The pulse train is a periodic signal with fundamental frequency 2B Hz. Hence, ui, — 2ir('2B) — AnB. The peiiod
is To =1/2B. It is an even function of t. Hence, the Fourier series for the pulse train can be expressed as
Using Eqs (2 72). we obtain
PT. (0 = Co +^ Cn cos ru.t
no
1 1/1601
4'
0/•1/1SB
2i( n7! \
/On = Cn = 7TTTo J
/ cos mj,t dt. =-1/16B
— sinT)7T [ 4 )
fcn =0
Hence.
g(t) = hOpt.V)
= ^ (f) +E^ sin (T) fl(,)cosW
50
Page 51
-0.2
Fig. S6.1-4
For ,{*) = sinc2(5ir/) (Fig. Sfl.Ma). the spectra is GM = 0.2A(*) (Fig- S6.1-4b). The bandwidth of this
signal is 5 Hz (TO* rad/s). Consequently, the Nyquist rate is 10 Hz, that is. we must sample the s.gnal at a rate
no less than 10 samples/s. The Nyquist interval is T = 1/2B = 0T second.
Recall that the sampled signal spectrum consists of (1 /T)G(w) = ^ *(*) repeatmg penod.cally w.th a penod
equal to the sampling frequency /, Hz. We present this information in the following Table for three samphng
rates' /, = 5 Hz (undersampling). 10 Hz (Nyquist rate), and 20 Hz (oversamphng).
sampling frequency sampling interval T ~~±G(u)[
comments|
Thz 0.2 A (afe) I
Undersampling
iolH 0T 2A(^)|
Nyquist Rate" ]
20lh 0.05 4A(jfe) |
Oversampling-]
In the first case (undersampling), the sampling rate is 5 Hz (5 samples/aec) and the ^pea.s
every 5 Hz (10* rad/sec.). The successive spectra overlap, as shown in Fig. S6.1-4d_and the speciru O )
not recove, able from £(-•) that is. g(t) cannot be reconstructed from tts samples
sampled signal is passed through an ideal lowpass filter of bandwidth 5 Hz. the output spectrum
0
1
5 Hz 0.2
10 Hz 0.1
20 Hz 0.05
Page 52
end the output signal '» lOsinc (20.1). which isnot'he <«•“! £ph3he Nyquist sampling t.te of 10 Hs ("*»-<•) T^f bfiiS“om 5(w) using a. Waal lowp~arepetitions of yGM repeating every 10 Hz. ence, (.
) Finally in the last case of oversampling
filter of bandwidth 5 Hz (Fig. S6.1-4f)_The output is,10s,nc Srt) - FmaU^m^^ (repeating every
(sampling rate 20 Hz), the spectrum CM consists of nonoveHapp g Pcan £ from ®(w)
2^ SKiSiJSi^ »
Thi.sd.Ls is analyzed fully in Problem 3 <-V“'here 30
kHz, 5 kHz. and 15 kHz, respectively. Hence, the Nyquist rates lor me
kHz. respectively._ ... _ q-y Th js acts ^ the input to
0 1-6 (a) When the input to this filter is *(<). the output of the summe ( )
the integrator. And, h(t). the output of the integrator is:
6.1-5
Hi) = J|/i(r) - b(r - T)] rfr = v{t) - «(# - T) = reel (^)
The impulse response /,(t) is shown in Fig. S6.1-6a.
(b) The transfer function of this circuit is:
and
ju>TJ2
Observe that the filter is a lowpass filter of
araa-output I fstaircase approximation of the input assU in F,g. S6.1-6c.
The amplitude response of the filter is shown in Fig. S6.1-6b
bandwidth 2ir/T rad/s or 1/T Hz.
t e*#
6 . 1-7 (a) Figure S6.1-7a shows the si§n.a* "co“,ruc* io
Jl1 ardre'^n^in^nstanf . ^he^hVight^of the* triangle is equal
roTlu-Lampie val ue* 'The'resuh hig^sfgn aTconsist s' of straight line segments joining the sample tops.
(b) The transfer function of this circuit is:/u>T\
//M = ^{^(0} = ^i^(or)| =T8inC \T) l . ,
2«ssgs
ITfiSmS' rfTS'SS? S S^Si'fcfo »u.d (Sizable). Such a delay would „«« -he
area' uudet'pM -P« - «• *«<*' “
, /'[„(.) - 2„(, - T) - u(r - 2T)|tlr = (..(1) - 2(1 - T)»(l - T) + (I - 2T)u{t -2T) - 4(—
)
shown in Fig S6.1-7b.
6 1-8 Assume a signal „(f) that is simultaneously timelimited and bandlimited. Let ff(w) = 0 for M > 2*B. Therefore
w (^.jicci( 4“'
g . ) = QU) for B' > B. Therefore from the time-convolution proper > (
q[i) = g(i) * i2B'sinc(2»rB'f)l
= 2 B'g(t) * sinc(2?rB't)
52
Page 53
Figure S6.1-7
B— *(0 i. timelimiwi. .«) - 0 b>:|»| > T. B«|
,(0 k o(“ Sl'.l
is not timelimited. It is impossible to obtain a t.me-l.m.ted s.gnal from the convolution
with a non-timelimited signal.
fl.2-1 (a) Since 128 = 27
. we need 7 bits/character.
(b)For 100,000 characters/second . we need 700 kbit.s/second.
(a) 8 bits /character and 800 kbits/second.
6.2-2
6.2-3
(a) The bandwidth is 15 kHz. The Nyquist rate is 30 kHz.
(b) 65536 = 216
. so that 16 binary digits are needed to encode each sample.
(c) 30000 x 16 = 480000 bits/s.
(d) 44100 x 16 = 705600 bits/s.
(a) The Nyquist rate is 2 x 4.5 x 106 = 9 MHz. The actual sampling rate = 1.2 x 9 = 10.8 MHz.
(bl 1024 = 210
. so that 10 bits or binary pulses are needed to encode each samp' ' f . ^ run a /—
(c) 10.8 x 106
x 10 = 108 x 106or 108 Mbits/.-..
6.2-4 If rn p is the peak sample amplitude, then
,(0.2)(t« P ) _ ru
quantization error <jqq jq
Because the maximum quantization error is ^ — 4/
’r _ Til,
500
500
,it follows that
L - 500
f n , ; _ tto _ 9® This requires a 9-bit binary code per sample The
Because L should be a power of 2. we chooM L - 512 - 2 . 1 9 signal has 2400
Nyquist rate is 2 x 1000 = 2000 Hz. 20% ^ ' j™0^ ,L 9 x 2400 = 21.6 kbits/second.
sampies/second, and each sample is encoded yf * v 21 6 = 108 kBits/second data bits. Framing
Five such signals arex 0.005 = 540 bits, yielding a total of 108540
and synchronization requires additional 0.5% blts tnat
bits/second. The minimum transmission bandwidth is ™ "
6.2-5 Nyquist rate for each signal is 200 Hz.
The sampling rate /, = 2 x Nyquist rate = 400 Hz
Total number of samples for 10 signals = 400 x 10 = 4000 samples/second.
, 0 .25m, _ m.Quantization error < ,(/,
-4S0
Moreover, quantization error - 2 — ji l «oo, ,
Because L is a power of 2. we select L = 512 = 2*. that is, 9 bits/sample.
Therefore, the minimum bit rate = 9 x 4000 = 36 kbits/second.
Tlie minimum cable bandwidth is 36/2—18 kHz.
For a sinusoid, =?? = 0.5. The SNR = 47 dB =50119. From Eq. (6.16)
m.6.2-6
So _ 3L2'jfVl _ 3l 2(0.5) = 50119 =>1=182.8
No
Because L is a power of 2. we select I « 256 = 28
. The SNR for this value of L is
So _ a ) = 3(256)* (0.5) = 98304 = 49.43 dB
No
53
Page 54
6.2-7 For this periodic tn(t). each quarter cycle takes onby averaging its energy
contributes identical energy. Consequently, we can compute the power^or^^^ = ^/Tq where A is
over a quarter cycle The equation of the first quar er eyesquared value (energy averaged over
the peak amplitude and To is the period of rn(f). The power or the mean squared
a quarter cycle) is
i fTa/A
( iA \* * - -L=(ft) 3
Hence. i1 — Jl
—:r /cici 9NR of 47 dB is a ratio of 50119- is
The rest of the solution is identical to that of Prob. 6.2-6. From Eq. (
So _ 2/2 T" = 3I
2(l/3) = 50119 => L = 223-87
A’o" "i?
B«,u„ I is , po«s, of 2. «. *1«< L - 25« - a'. The SNR for this v,h,e of L is
So _ ^L ' = 3(256)2(1 /3) = 65536 = 48.16 dB
So T"p
6.2-8 Here „ = 100 and the SNR = 45 dB= 31,622.77. From Eq- (6. 18)
So 31/31,622.77 /. = 473 83
So~
(In 101 )2
, „ i r — m 9 — 9® The SNR for this value of L is
Because L is a power of 2. we select L — ol2 —21So _ 3(512)
3
_ 3gg»>2 84 = 45.67 dBSo (In 101) ?
,. i ~ io v in6 '! = 3 x 106 Hz. Moreover. L = 256
6.2-9 (a) Nyquist rate = 2 x 106Hz. The actual sampling rate is 1.5 x (2 x 10 )
and ii = 255. From Eq. (6.18)
So 3I* = 3(256)2
^ 6394 = 33 Qg dBNo " |ln(/-+ l)!
s (In 256) 2
ZTvtZZ 7'r«d each sample is encoded by 8 bits (L = 256). Hence, the transmiss.on rate ,s
8 x 3 x 106 = 24 Mbits/second
Nyquist rate), then for the same transmission rate
™-— *- *• - io24 -,he
new SNR is
So 3L1 3(1024)2
_
% =[ln(,f"+~ (In 256)=
Clearly, the SNR is increased by more than 10 dB.
.= 102300 = 50.1 dB
Clearly, the arm is increase ^
6.2-10 Equation (6.23) shows that increasing n 20%.
by 12 dB (front 30 to 42) can be accomplished by increasing n trom iu
6.4-1 (a) From Eq. (6.33)<rf. thftt _ (ffKM^OOO) 0.0785
.4 mux =— 60 that 1 “2tr x 800
54
Page 55
(b>(0.0785)
2(350Q) = n % 10
-4
(3)(64000)
(c) Here So = = 0.5. and
(d) For uniform distribution
2
So
No
0.5
1.12 x 10-= 4.46 x 10
So = =-jj
21 =^
, Soso that — =
0.333 = 2.94 x 103
No 1.12 x 10-4
(e) For on-off signaling with a bit rate 64 kHz, we need a bandwidth of 128 kHz. For a bipolar case, we need a
bandwidth of 64 kHz
Page 56
Chapter 7
7.2-1 For full width rect pulse p(t) = rect
^j
For polar signaling [see Eq. (7.12)]
S
For on-off case [see Eq. (7.18b)]
P(eo) = Tb sine
-•(?)Sy{(o)^^-‘Tb smz
i = 0. Hence,
»*•(?)[•» _v(--¥l
But sinc2
^] = 0 for forall **0, and = 1 for n = 0.
Sy(o) *^smc2 +
For bipolar case [Eq. (7.20b)]
. . \P(*f . 2 (coTb \
5>) = LX-sm^ 2 J
= Tb sine2
sin2
The PSDs of the three cases are shown in Fig. S7.2-1. From these spectra, we find the bandwidths for all
three cases to be Rb Hz.
The bandwidths for the three cases, when half-width pulses are used, are as follows:
S^3bS&ssssss==bs=“bipolar case because the term .in
2 (^)i» S,(a.) daannin.. it. b»dwi*b.
56
Page 57
From Fig. S7.2-2, it is clear that the bandwidth is— rad / s or 2Rb Hz.
7.2-3 For differential code (Fig. 7.17)
To compute K„we observe that there ate four possible 2-bit sequences 11. 00. 01 and 10. which are
equally likely. The product at,,oM for the fust two combinations is I and ts -1 for the last two
combinations. Hence,
Similarly, we can show that Rn = 0 n > 1 Hence,
%1-i (a) Fig. S7.2-4 shows the duobinary pulse train yfr) for the sequence 1110001 101001010.
(b) To compute Bo. we observe that on the average, half the pulses have o, =0and the mmaining half
have a* — 1 or —1. Hence,
*0= lim 4f(±l)2 +f(0)]4
To determine B| ,we need to compute0*0,. t
There.*,fourpossible equ^lIM>
11,10,01,00. Since bit 0 is encoded by no pulse(ui -0),the product of »»n».l - or
these sequences. This meanson dte avor*.Hcombin«ions have -Oand only - combinations
57
Page 58
htVe nonzero akaM . Because of*0 duubuuuy rule. tb. bit *<.«»« .1« only be encoded by two
consecutive pulses of the same polarity (both positive or both negative).
This means o* endo**. are i and 1 or -I end -1 respectively. b-.—Wd-'-
these— combinations have akak+ \= 1. Therefore,
possible combinations of three bits m sequence.and/or the last bit 0.
-- --—--
combination 101 are of opposite polarity yielding akak^ .= -I. Thus on the average, ak *+2
— terms,-l for— terms, and 0 for~ terms. Hence,
m a similar way we can show that^ - 0 » > 1 ,and from Eq. (7.10c), we obtain
S,(«) = + cosair6 )=—jr" cos
{(o>Tb
'
l 2 J
yrO1 I 1 0 0 o 1 1 • * 9
mnu0 10»°
_JL_JL-n—
For half-width pulse P{t) = rect(2l / 7*)-
SJ,H = ^-sinc
2
^] cos2
(^)
From Fig. S7.2-4 we observe that the bandwidth is approximately Rb I 2 Hz.
7.3-1 From Eq. (7.32)
4000 =(l+r)6000 1
r ~3
58
Page 59
7.3-
2 Quantization error S 0.0 \mp => L £ 1 002. J-t
•
j
(a) Because I is a power of 2, we select L * 128 = 2
(b) Ibis requires 7 bit code per sample. NWaisl«e=2 x2000.4lcHsfbrMchsign.l. lb. sanplm*
rate /, = 125 x 4000 = 5 kHZ.
Eight signals require 8 x 5000 * 40,000 samples/sec.
Bit rate * 40,000 x 7 = 280 kbits/s. From Eq. (7.32)
(\+r)Rb 12 x280x 103_
Bt2 ' 2
7.3-
3 (a) BT =2Rb =»Rb • IS kbits/s.
(b) Bt - => Rb = 3 kbits/s.
(c) Bt * Hence, 3000 =— Rb => Rb = 4J kbits/s.
(d) Bt = Rb => Rb = ‘S kbits/s.
(e) Bt = /?6 => - 3 kbits/s.
7.3-
4 (a) Comparison of P{co) with that in Fig. 7.12 shows that this P(eo) does satisfy the Nyquist criterion with
<ob = 2/r x 106 and r = 1. The excess bandwidth ax = n x 10
6.
(b) From Table 3. 1 ,we find
/>{f) = sinc2 (>rx!0
6/)
From part (a), we have cob = 2n x 106 and = 10
6. Hence, 7* = 10 . Observe that
p(n Tb)<*\ n * 0
«0 n*0
Hence /*(r) satisfies Eq. (7.36).
(c) the pulse transmission rate is— * Rb = 106
bits/s.
7.3-
5 In this case = 1 MHz. Hence, we can transmit data at a rate Rb - 2 MHz.
Also, Bt * 12 MHz. Hence, from Eq. (7.32)
12 x 106 - * 10&
)^ r = ®-2
7.3_6 /2 = 700 kHz. Also, -^- = 500 kHz and /, = 700 - 500 * 200 kHz.
HenCe,r-^i
.0.4 «.<l/, -500 - 20°. 30° Mb.
7.3-
7 To obtain the inverse transform of /(»),we derive the dual of Eq. (3.35) as follows:
g(r - r)o G(co)e'jTa>
and g(/ + T)o G(o>)eyT"
Hence,
g(i + T) + g(t - T)o 2G(a>) cos7©
Now, P{eo) in Eq. (7.34a) can be expressed as
*•>4 rea(^r)
+M^]“fe)59
(2)
Page 60
7.3-8
7.3-9
Hence,
Using Pair 17 (Table 3.1) and Eq.(l) above, we obtain
P(t) - *fcsinc (2*Rbt)+&.smc [***(' + 2^]
+ ^-smc |2*^/ J
« /lysine (2/ri^/)+^sinc (2»Jty + >r)+^sinc (2ff^/-^)j
f sin(2jr/Zfe?) l sm(2xRbt + x).[;
1 M***4' -*) 1
=/?A
|_
2^^/+2 2xRbt+* 2 2nRbt-n J
* Rb sin(2nRbt)
= J?4 sin(2ff^t)j
1/2 1/2
(2ff*4f + ») (2* fy-*).
k —
2Rh cosx RbtimxRb t _ Rb cosx Rb t
sinc/ «
£
_ . _ 5 i AD.
2xRbt(\-4Rb2t2)\
inxRb t _ Rb cos* ^
2/rRt,t(\-4Rb2t2
)I-***
2'3
I,--""1*
Kt U»*w l
1 ( °> 1 1 Y
p(t) * sine (^^r) + sinc
sin/r/y(
sin(/r^-ff)
fffyf xRbt-x
s\nxRbt sinxRbt sin xR/,1
xRbt~ xRbt-*~ xRbA'-Rb*)
n,e Nyquist interval is Ts = -p - Tb . The Nyquist samples are p (±nTb )for n = 0, 1, 2,
RbFrom Eq. (7.16), it follows that
p{0) » p(Tb )= 1 and p{±nr0) = 0 for all other n.
Hence, from„ „ Rb 1
Eq. (6.10) with Ts = Tb , and B = -£ =—
.
p(r)*sinc tf^t+sinc
sin
‘
r i YxRb
-
sin sin <r/y
*~ xRbt-x xRbt{l-Rbt)
60
Page 61
The Fourier transform of Eq. (1) above yields
. J_rwf_2_'l [«/*'**•
Rt, [2xRb )^
ja>/2Rt
-jullRt,
7.3-10 (.) No^rbec.u«the!raple».lu«ofa««n,.po^^s^^ eV“"mto0fimS "‘1
the sample values ofopposite polarities are separated by^ the first digit is 1. The
(b) The first sample value is 1 because there is no pulse before this digit Hence uie &(b) The first sample
detected sequence is
uoooiooiioiioioo
7.3-10 Hie lint ample vel»« is 1. indicata, tha d,. ^ “ IS’“^ ''
The duobinary rule is violated over the digits shown by underbracket.
12000 -200 -20 2 0 0 - 2 0 2 2 0 -2
Following are possible cornea ample values in place ofIhe 4 ondertnadcel values:: S 2 0 -2. or2j» -
-2j-S.
„ 0ooj” 2 0 0 0. These ample values lepmsen. Ae following 4 digit aqnence. 1 100, or 1000, or
0100, or 1010. Hence the 4 possible correct digit sequences are
1101001001X|XjXjX4lllOO
where x^xjx, is any of the four possible sequences 1100, 1000, 0100, or 1010.
7.4-1 S * 101010100000111
From example 7.2
T » (l© D3 © D5 © Db © D9 ® D10 © Z)M © Dn ©£>,3 ©D15 ©.~)s
K = (l®£>3 ©D5
)r
7- = 101110001101001
R = 101010100000111= S
7.4-2 5=101010100000111
T « (l © D2 © Z>4 © D6 ® Z>
8 © Di0 © Dn © Z),4
©...)S
R * (l© Z>2)r (see Fig. S7.4-2)
T = 100010000000110
R
=
10101C100000111 =
S
Fig. S7.4-2
7.4-3 S = 101010100000111
T = (l<BD<BD2 <BD*®D1 ®D*®D9 ®DU <BDU )s
/? = (l®Z>®03)7' (see Fig. S7.4-3)
r= lioiiiioiooioii
R = 101010100000111 = S
61
Page 62
7.5-1 From Eq. (7.45), we obtain
’*-1
Co
«1
1 0.3 - 0.07"-1
'o'-0528"
0.1 1 0.3 1 = 1.07
-0.002 0.1 1 0 -0.113
7.6-1 (a) -2- * 5
<*n
(i) For polar case Pt 2(5) = 257 x 107
(ii) For on-off case Pt » 2(5 / 2) = 0.0062
1
(iii) For bipolar case Pt — 152(5 1 2) * 0.0093 1
5
In the following discussion, we assume = .4, the pulse amplitude.
(b) Energy of each pulse isf^, - A1Tf, / 2 and there are pulses/second for polar case and f
pulses/second for on-off and bipolar case. Hence, the received powers are
Fl -i!2L^.ii-1222^.u25Kio-#“2^2 2
. - ^*5-x =— = 05625x1 O'
6
2 2 4
*b
^on-off
'Dipolar_
.^ jSL x^- =— = 05625 x 10"^
2 2 4
(c) For on-off case:
We require /*(e) * 257 x 107 * 2(^p Hence,
For bipolar case:
/<p / 2<r„ « 5 and Ap * I0cr„ = 0.003
P{ e) - 257 x 10'7 - 15fi(^ /2a*) =>^ * 5.075
Hence
and
7.6-2 For on-off case:
A = Ap = 5.075 x 2a* = 0.003045
251x10“*
Pt - 10-6 =>-^-*4.75
2tr„
62
Page 63
<t„ = 10"3 =>Ap 2 (4.75)(2 x 10"3)= 9S x 1<T
3
For on-off case, half the pulses are zero, and for half-width rectangular pulses, the transmitted power is:
/ v
Si4
(9ixl0-3
)s 2236 x 10“* watts.
There is an attenuation of30 dB, or equivalently, a ratio of 1000 during transmission. Therefore
ST * 1000S, = 22.56 x 10-3
watts
7.6-3 For polar case:
Pt - 10-6
=g^-j=>£-= 4.75 => Ap = 4.75 x 10"3
For polar case with half-width rectangular pulse:
z-h c -(4ixlO-3
)
2« 1 128 x 10
-6watts
V ‘ J2
ST * (I000)(l 128 x lO^BK) = 11.28 x 10-3
watts
For bipolar case:LAI W(UV.
Pf = lo-6 = 1-5^“
j
=>^-» 4.835 and Ap = 4.835 x 2 x 10"3 = 9.67 x 10"3
For bipolar (or duobinary), half the pulses are zero and the receive powerS, for half-width rectangular
pulses is
2
s = * 1(9.67 x 10“3f * 23.38 x 10-6
watts' 4 4 ' ’
Sr *(1000)S, =23J8xlO"3watts
7.7-2 Sampling rate = 2 x 4000 x 125 = 10,000 Hz.
mpQuantization error =y m 0.001mp L =
Because I is a power of 2, we select L = 1024 = 210
. Hence, n - 10 bits/sample.
(a) Each 4-ary pulse conveys log2 4 = 2 bits of information. Hence, we needy = 5 4-ary pulses/sample,
and a total of5 x 10,000 = 50,0004-ary pulses/second. Therefore, the minimum transmission bandwidth is
50,000
1000
(c)
. 25kHz.
T2 2
7.7-3 (a) Each 8-ary pulse carries log2 8 = 3 bits of information. Hence, the bandwidth is reduced by a factor of
(b) The amplitudes ofthe 8 pulses used in this 8-ary scheme are ±AI 2, ±3-4/2, ±SA 12, and ± 1A 12.
Consider binary case using pulses ±A 12. Let the energy of each of these pulses (of amplitude ±AI 2)
be Eb . The power of this binary case is
Wintry* SbRb
Because the pulse energy is proportional to the square of die amplitude, die energy of a pulse ± 21S
k2Eb . Hence, the average energy ofthe 8 pulses in the 8-ary case above is
63
Page 64
Eb
£<» = '
2(±i)! +
2[±|)
2
+2(±f)
+a(±!)
= 21Eb
Hence,
Therefore,
fy.xy = £flvX pulse rate = 21£j x = 7EbRb •
^tty - 7 binary
7.7-1 (a) M - 16. Each 16-ary pulse conveys the information oflog2 16 * 4 bits. Hence, we need
12 ’0(M) - 3Q00 16-ary pulses/second.
43000
Minimum transmission bandwidth = —r~ = 1 500 Hz.
(b) From Eq. (7.32), we have Rb =— Br . Hence,
3000 *— Bt => Bt = 1 800 Hz.
(.) For polar signaling, *m^^*****<**«* 11» P"* <*
amplitude— has energy
^2
The power P is given by f = EbRb *— *i>=~
8 8
(b) The energy of a pulse ± is A2£j . Hence the average energy of the M-ary pulse is
EM =—[2£4 + 2(±3)2+2(±5)
2+ +2[±(M- >)f
E»]
M L
M-2
2£i (2*+o
2
M *«
o
Af2 -1
Bb
Each M-ary pulse conveys the information oflog2 M bits. Hence we require only^MM_ary
pulses/second. The power PM is given by
"idPu ’
log, M “3108, M * 24 logi M 24 log, M
7.7-S Each sample requires 8 bits(256 - 2*) . Hence: 24,000 x 8 - 192,000 bta/*c.
Bt = 30 kHz
R = _2_ Bt = —(30,000) = 50,000 M-ary pulses/sec.
1+r 12
We have available up to 5o!oo0 M-ary pulses/second to transmit 192,000 bits/sec. Hence, each pulse must
( 192,000^ai,
..j
transmit
:
• 3.84 bits.
64
Page 65
7.8-1
=e choose 4 bits/pulse
=> AS = 16 is the smallest acceptable value
(a) Baseband polar signal at a rate of IMbits/sec
and using full width pulses has BW = 1MHz . PSK
doubles theBW to 2MHz.
(b) FSK can be viewed as a sum of2 ASK signals.
Each ASK signal BW = 7 MHz. The first ASK signal
occupies a band fgQ i 1 MHz, and die second ASK
signal occupies a band fe\± 1 MHz. Hence, the
bandwidth is 2 MHz + 100 kHz = 2.1 MHz.
PSD
7.8-2
7.8-3
(a) A baseband polar signal at a rate 1 Mbits/sec using Nyquist criterion pulses at r = 02 has a
BW = llld Rb mH x 106 = 6.0 x 10
5 Hz.
PSK doubles BW to 1.2 MHz.
(b) Similar to Prob. 7.8-1
.
B1PFSK =0.6 MHz + 0.6 MHz +100 kHz
SHfa* = 13 MHz
log2 M * 2 for AT = 4.
We need to transmit only 03 x 1064-ary pulses/sec
(a) BW is reduced by a factor of 2.
BWVsk = 1 MHz
(b) In FSK, there are four center (carrier) frequencies
/cl , fa • fa > *nd/c4 . each seParated by 100 kHz
Since ASK signal occupies bandfc ± 03 MHz, the total
bandwidth is
03 MHz + 0.5 MHz + 100 kHz + 100 kHz + 100 kHz = 13 MHz.
PSD ***%
c, -?c-
Jt 3,00 ij
Fig. S7.8-3
7.9-1
65
Page 66
nt.tO•»
\—‘ft; 'T
w**)
—
*vw
—
Fig. (b)
/
-ft * Uioo cobi^/itc
Fig. S7.9-1
7.9-2
Either figure (a) or (b) yields the same result.
mi (/) has 8400 samples/sec.
ffl2 (t), «j(f). mA (t) each has 2800 samples/sec.
Hence, there are a total of 16,800 samples/sec.
Fira, we combine «,(<). -* ^W wiU,,comm.«W^of™0^^
™
slgoe, it now multiplexed with with .c—or tpeed of2.00 touuiont/tec, „e,dm6 dte ompu.
5600 samples/sec.
66
Page 67
r* M'S ^4-s kbls
Fig S7.9-3
Page 68
Chapter 8 Exercises
8 . 1-1 If a plesiochronous network operates from Cesium beam clock which is accurate to ± 3 parts „in 1012 , how long will it take for a DS3 signal transmitted between two networks to become out :
of sync if a 1/4 bit length time error results in desynchronization?
Solution: A DS3 bit is transmitted in 1/(44.736- 106)= 2.235336-10^ sec. In the worst case,
both network clocks will be out of synchronization by 6 parts in 101.
2.235336- 10*®/(6- 1012
) = 3922.27 sec/bit or 980.57 sec/ Vi bit
8.1-2 For the bit stream 01 1100101001 11 101 1001 draw an AMI waveform.
Solution:
8 . 1-3 For the following waveforms, determine if each is a valid AMI format
for a DS1 signal. If not, explain why not.
Solution: No. 16 0’s violation
Solution: No. bi-polar violation
Solution: No. l’s density violation
Solution: Yes
68
Page 69
8.1-4 a) You have received the following sequence of ESF framipp pattern sequence bits
...00110010110010110...
Is this a legitimate framing bit sequence in order to maintain
synchronization between the T1 transmitter and receiver?
Yes NoIf yes, why? If no, why not?
Solution: No. The bit sequence 001 1 cannot be in an ESF framing bit sequence.
b) The following T1 AMI signal is received:
n
Is this an acceptable T1 signal?
Yes Noa. If yes, explain.
. , .
b. If no, explain why not (what, if any, DS1 standards are violated) and
draw on the figure the AMI waveform which would be transmitted by the DSU
Solution: No. 16 0’s violation. The 16 0’s will be replaced by a pattern of 1 s by the
DSU.
8.1-5 Tie signal 110100000000000000001 is received by die DSUin
8ZS format. Draw the output of the DSU for this signal? The
ihow the bit stream which is substituted by the DSU.
a T1 data stream which uses a
first 1 is already drawn.
8 .1.6 T-l synchronization at two distant locations is controlled by separate crystal controlled
oscillators which differ in freqoency by 125 parts per million If the termmj equipm
<
maintain sync in how many complete D4 superframes will the faster oscillator have generated (at
most) one more time slot (8-bit) than the slower oscillator ? Circle the correct answer.
69
Page 70
a) 5
b) 10
c) 15
e) None of the above - if "none", what is the number of D4 superframes before an extra time slot
&£^)lteflSer oscillator will generate 125 10-61.544 106 = 193 bits per second more
than the slower oscillator. This is one frame/sec = 24.125 time slot^ Hence, a time slot
difference will be generated in 1/24.125 = 0.04164498 frames or 0.0034704 superframes.
8 . 1-7 Two plesiochronous digital networks, A and B, utilize Cesium beam clocks accurate to 3
parts in 1013
. The networks are operated by independent long distance companies and are
synchronized to each other by means of a UTC signal.
a If a company leases a T1 line with D4 framing which is terminated at one end m
network A and at the other end in network B, how often must the networks be resync d to
each other to avoid a framing bit error in the customers T1 signal in the worst case {You
may assume a framing bit error occurs when the two networks are out of sync by > 1/2 o
a T1 "bit time".} . ,.
Solution: ATI bit time is 1/(1.544. 106
) = 6.47668- 10'7 sec/bit. In the worst case, the
two clocks would be off by 2-3 = 6 parts in 1013
or 6- 1013
errored bits per bit transmitted.
Hence, 6.47668 10'7 sec/bit / 6- 1013
errored bits per bit = 1 .07945 10 seconds per errored
bit or 5.39723- 106seconds per errored half-bit.
b. UTC operates via GPS satellites which are approximately 23,000 miles above the Earth.
How long, in terms of T1 bits, will a correction signal take to be transmitted to
the network switches?
Solution: The speed of light is approximately 1 86000 miles/sec.
23000x2 » 46000miles up and down. 46000/186000 = 0.247 sec
0.247x1544000 s 381850 bits
70
Page 71
Chapter 10
10. 1-1/ , JV 13+13 1
(a) /’(red card) = — = -
(b) />(black queen) =— =—
(c) /’(picture card) =
(“> ^-5-5. 20 5
(e) /»(« ^ 5) *
52“73
10.1-2 (a) S = 4 occurs as (1,1,2),(1,2,1),(2,1,1)-There are a total of6 * 6 x 6 = 21 6 outcomes.
Hence,^ = 4)
«
lsO,2,6) (1,3,5), (1,4,4), (1,5,3), (1,6,2), )
;5 3W5%I)’
(3,1,5), (3,2,4), (3,3,3), (3,4,2), (3,5,1), (4,1,4), (4,2,3), (4,3,2), (4,4,1), (5,1,3), (5,2,2),(b) S = 9 occurs as
c(5 ,3 , 1 ), (6, 1 ,2), (6 ,2, 1 )
P(5 =9)«^
(c) S = 1 5 occurs as (3,6,6), (4,5,6), (4,6,5), (5,4,6), (5,5,5), (5,6,4), (6,3,6), (6,4,5), (6,5,4), (6,6,3)
10.1-3 Note: There is a typo in this problem. The probability that the number, appears should be ki not k,
1 = £*/ = * +2* + 3* + 4* + 5* + 6* = 21* =o * *—,=l
P(/) =— (I = 1,2, 3, 4, 5,6)/ 2i
10.1-4 We can draw 2 items out of 5 in 20 ways as follows: 0,0,. 0,0,. 0,P,. 0,Pi. »A, 0A, Wt. 0,OI•
OMpT^O,,“ 0„ PiOj, P,P„ W„ PA. PA, P.Pi All diese outcomes »e equally likely w,0l
^rto, fl U«,fiU.2AUO^U03,UO! fiU flo,Uf1oJ U fl o3U^ 1
Ufi oJUfioJ
Hence, P(E\) =12
20
3
5
(ii) This event £2 * /’i^U/^/’i
Hence, />(£2)
=^ = '^
(Hi) This event £3 = 0j02 UO1O3 (J020i U0203 UO3O1 U0 302
Hence, £(£3) =20
" Jo
(iv) This event £4 * £2 U £3 and bo* E2&E3 an disjoint.
Hence, P(EA )= P{E2 )+ P{E3 ) =^
= 1
71
Page 72
10.1-5
10.1-6
10.1-7
10.1-8
Let x0 be the event that the fust chip is oscillator and be the event that the first chip is PLL. Also,
let jto2and represent events that the second chip drawn is an oscillator and a PLL, respectively. Then
P{ 1 osc and 1 PLL) = p(*0),xh )
+ P(*i\)
=/>(JCo,)/,(x/
i|xo,)+^fiM*o2h)
(BMHHUsing the notation in the solution of Prob. 10.1-5, we find:
(a)
(b) ^o2K) = f
(a) We can have (*§) ways of getting two l*s and eight 0’s in 10 digits
P (two r s and eight 0’ s) *= 45(0J)1
(0.J)' » 45(0.5) -
(b) Piat least four 0’s) = 1 - [/•(exactly one 0)]+[ /(exactlytwo 0’s)]+[/>(exactly three 0’s)]
/>(two 0’s) =(1
2°)(0-5)
,0“^P (three 0’s) -
()(0-5)
10 * -j—
( 5 45 120 'l 849
P(at least four O' s) = 1 - 1— +— +—
J J024
(a) Total ways of drawing 6 balls out of49 are
(4|) = _i£L = 13,983,816V 6
^ 6143!
Hence, Prob(matching all 6 numbers) *^ 3983816
(b) To match exactly 5 number means we pick 5 of the chosen 6 numbers and the last number can be
picked from the remaining 43 number,. We cer, choose 5 number, of our d in ( f)• 6 ways and can choose
number ou. of43 *(<?) - 43 ways. Hence. w, have 43. 6 combirmrions in wh.ch exactly 5 numbem
match. Thus,43x6 5
P (matching exactly 5 numbers) = 139g3gl6= 1-845 x 10
(e) To match exactly 4 numbers means we pick 4 out of the chosen « number in ( t)- 1 5 ways and choose
2 ou, of dr. remaining 43 numbers in(*J) = 903 way,. Thu, there are 15 * 903way, of picking exanly 4
numbers out of6 and. .
15x903 4
P (matching exactly 4 numbers)- ]39g3gl6
- 9 686
72
Page 73
(d) Similarly, we can pick three numbers exactly in(f)(4j)
= 20 * 12341 *246820 ways. Hence,
p (matching exactly 3 numbers)- “ 001 765
10.1*9 (a) Let/ represent the system failure. Then
/>(/) = (l-O.Ol)10 = 0.90438
/>(/) = 1- /»(/) = 0.0956
(b) P(f) « 0.99 and P(f) = 0.01
If the probability of failure ofa subsystem st
is p, then
p[f) - (1 - pf or 0.99 - (1 - pf 0.001004510.1-
10 If/ represents the system failure andfu andfL represent the failure ofthe upper and the lower paths.
respectively, in the system, then:
(a) P{f) - P(fufl) - HfuWl.) =[p(/w)]
2
P(/u) = 1 - P{/«) = l-(l-O.Ol)10 = 0.0956
and
P(f) = (0.0956)2= 0.009143
Reliability is P[f)= 1 - P(f) = 0.9908
(b) />(/) = 0.999
P(f)= 1-0.999 = 0.001
/>(/„)- Voiom - 0.0316
P{fu) (1 - P)W - 1 - 003 16 => P = 0.003206
10.1-
1 1 Let P be the probability of failure of a subsystem (jj or rj)
.
TheSystem ^lils'if the upper and lower branches fail simultaneously. The probability of any branch not
failing is2
(1 _/>)(!_/>) = (l- P)2
. Hence, the probability of any branch failing is 1 - (1 - P) .
Clearly, Pj , the probability of the system failure is Pf * [l “ 0 ~ p? J1 “0 "
]
s 4/>2 P<< '
For the system in Fig. b: .. ..
We may consider this system as a cascade of two subsystems x, and x2 ,where x, is the parallel combmatio
ofjj and r, and x, is the parallel combination ofs2 ands2. Let Pf (x,) be the probability of failure ofx..
Then
r/M-r/M *'7
The system functions if neither s, nors; frils. Hence, the feobability of system not failing
is (l - P2)(l - P2
)•Therefore, the probability of system failing is
Pf = l-(l-P2){l-P2
)= 2P2 -PA s2P2 P«l
Hence the system in Fig. a has twice the probability of failure of the system in Fig. b.
73
Page 74
10.1-12 There are(52
)= 2598960 ways of getting 5 cards out of 52 cards. Number ofways of drawing 5 cards of
the same suit (of 13 cards) isf^3)* 1287. There are 4 suits. Hence there are 4x1287 ways of getting a
flush. Therefore,
/’(flush) =4x1287
2598960= 1.9808 x 10"3
10.1-13 Sum of 4 can be obtained as (1,3), (2,2) and (3,1). The two dice outcomes are independent. Let x, be the
outcome of the regular die and x5be the outcome of irregular die.
?x\xi 0 *3)* ?x\ 0)^*2 0)
=g*3
=jg
^,x, (2J) - (2)^3 (2) - ^« o = 0
Thtrefor, W).Similarly,
W)-P.,.,(ub *U(M)+ p + ^,(4.1)
10.1-14 B=AB\jA eB
P(B) = P{A)P{B\A)+ P(AC)P(E\AC
)
10.1-
15 (a) Two l’s and three 0’s in a sequence of 5 digits can occur in (2)= 10 ways. The probability one such
sequence is
/> = (0.8)2(02)
3 = 0.00512
Since the event can occur in 10 ways, its probability is
10x0.00512 = 0.0512
(b) Three l’s occur with probability (^)(0i)3(02)
2 = 0.2048
Four l’s occur with probability( 4
}(0i)4(0.2)
1 = 0.4096
Five l ’s occur with probability( 5
j(08)S(02)° = 03277
Hence, the probability of at least three l’s occuring is
P = 02048 + 0.4096 + 03277 = 0.942
1
10.1-
16 Prob(no more than 3 error) = P{no error) + /*(l error)+ P{2 error) + P(3 error)
-<- p.r-iH/iii- p.f -(D^o -
s(l-1007‘,)-.K>0/',(l-99P,)*4950P,I(I-98f,,)-.l61700P,
3(l-97/>,)
74
Page 75
10.1-17 Error can occur in 10 ways. Consider case of error over the first link
/^(correct detection over every link) = (1 ~ /*i)0 ” ^t) -0 “ ^lo)
/ȣ =i-pc = i-0-^X 1 -^)
_ i-[i-(/»j + /^+...+/\0 )+ highcr order terms]
sPl+ P2 +...+Plo Pi«'
10.1-
18 *«) - i(!,)P,'(\-P.t‘ - lO'l’ll - rf -‘VO-
;=
3
sl0/»e3(l-/>
«)2
,/>.« 1
10.1-
19 (a) ^(success in 1 trial)»^ = 0.1
(b) /’(success in 5 trials) = 1 - /’(failure in all 5 trials)
•'-WhWSts
Pj - Prob(failure in 1* trial) -9/10
Pj - Prob(failure in 2nd
trial) = 8/9
Similarly, Pfi
=7/8, Pfi
=6/7, and/»/5=5/6
Hence, /’{success in 5 trials) - 1 “(^XfX^X «)“
10.1-10 Lets be the event of drawing the shot straw and the «(*) denote the event that/th person in the sequence
draws the short straw.
Now, /*i(-^)— 0.1
n.M - ProbO" person does not draw the short straw) * Pwb(2- person draws the short straw)
-MmiHS)- 01
"Uneidter I" nor 2“ person draw, the short straw) x ProbO1* person draws the short straw)
-[i-AM-aWjHfiKHSimilarly, /^(x) = Ps{*) *•••“ ^o(x)
~ 01
10.1-21 All digits are generated independently
(a) P(all 10 digits are 0) = (03)‘°
(b) There are ('§) ways of arranging eight l’s and two 0’s. Hence,
/height l’s and two 0’s)-(1
°)(0.7)S(0J)
2
(c) /»(at least five 0’s)=/^five 0’s)+/^six 0’s)+...+/\ten0’s)
= (
1
j)(O.7)S{01)
5+(*6)(0.7)
4(0.3)* + (*7}(8-7)
3(b^)’ + (*1){0^)
2(0J)* + ('o){®.^K®^)
9 + (0-^)10
10.2-1 /y (0)= /xy(l.°) + ^xy(0»°)
= />x(,)
/>y|x(
0l1)+ ^(Oj^yjxW
= 0.6 x 0.1 +0.4[l - /*y |x(l|0)] = 0.06+032 = 038
/y(l) = 1 - /y(°) = 062
75
Page 76
... VMKU) (i -P')Q10.2-
2 (a) ” {\-Q)Pt +{\-Pe)Q
(b) ^x|y(°l1)= 1
- /,xly(
1l1)
(note that Py(l) and Py (0) are derived in Example 10.10)
10.2-
3 (a) P(x * l) = £^xe~xdx =
^
(b) Prob(-l < x <; 2) = £"**X* +lo\
xe-Xdx =
(c) Prob(x 5 -2) - “~f10.2-
4
Fig. S10.2-4
Since this is a half-wave rectifier, y assumes only positive values. So P{y < 0) - 0.
Hence, Fy (y)
= 0 (fory < 0) and P[y < 0+
)
- • Hence, Fy[o
)* ~
10.2-5 x is a gaussian r.v. with mean 4 and <rx - 3
Hence,
(a) P(x ^ 4) * * C(0) = 03
m n* 2 o) -e^) - 1 -e(|) - ' -<m»I7‘ - 0 !l0!3
(c) P(xZ -2) • - 1 - fi(2)'- 1 -0.02275 = 0.9773
Fig. S10.2-5
102-6 (a) From the sketch it is obvious that x is not gaussian. However, it ista unilateral(rectified) version of
Gaussian PDF. Hence, we can use the expression of Gaussian r.v. with a multiplier of .
For a gaussian r.v.
whhffya4
(b) Hence, (i) P(x * l) = 2P{y * 1)= = 05026
(H) p(l < x £ 2) = 2 P{1 < y 5 2) *2^)-12(f)]
=W 856
(c) Ifwe take a Gaussian random variable y
*w-asr^"Fig. SI0.2-6
py(y) **
and rectify y (all negative of y multipled by -1), the resulting variable is the desired random variable x.
76
Page 77
10.2-7 The volume V under Pxy (x,y) must be unity.
A.2
Px(x)=jpxy(x'y)4y
y
But y = -x + 1 and the limits on y are 0 to 1 - x . Therefore,
l~x2dy = 2J
1 *•
0
f2(1—>) Oiyil
Px(x) = lo~
X24' = 2y|
Similarly, py (y) =
2(1-*) 0SxS4
0 otherwise
otherwise
/, X P*y(x>y) 2
Px\y(Ay)- 2(1 -y)
V\-y OsySl
0 otherwise
Fig. S10.2-7
1/1—x OSxSl
Similarly, />y |
x (.y|-*)-
0 otherwise
Clearly x and y are not independent.
10.2-8
—(x2 +>
2y2 . . ,
.
Pxy {x,y) = xye"
' u(x)u{y)
-x2 /2
(a)
Similarly,
and
px (x) = fixye^^ttxfr'= "
u(x)
py {y)
= ye~y 12
’Ky)
(b) From results in (a), it is obvious that x and y are independent.
« X 1 r«4ax2 +by
2-2ay)hM
10.2-9 Px{x) = i^Pxy(x'y)& = 71177l-~e**
1
re-^t2U
J*r-
1s - ' m£y2ftb
/ \1 —y^ /2tf
Similarly we can show thatpy (y)= c
Therefore/
1
\Pxy (
x>y) _ rrcAx
«y)h
‘|y(xW"
px(x) ~1l2xM
, , s Pxy(x>y)
77
Page 78
10.2-10 K
But
jd* = 1
nJ-e^wW
=
^!Ze'x2
[lZ>e~y2-10,
‘fy,
j*. «.4
„ 1 3Hence, a =— .
-V4
,xW . *rjy<‘W)
‘» -^D"W - 'c '/”e'5'
!''' *
. _ |T -3>2/4
Because of symmetry ofPxy (x'>0 w*^ tespect to x and y. py (y) ^^
, , X PxyM _ 1/{* +,y+T
)
Px|y(;tW = ~ r_
/4
PyM V»
and
...'*W'1F 7
Since Pxy (x,y) * px (x)Py(p). * «»d Y are not dependent.
10.2-11 P« = P^P*! 1)* pM°)**(0)
If the optimum threshold iso ,then
j \ / i
__
jP, _ 1 L-{4p-a
)
2/W p^i). e-(^^f/2g2»p
r (0) =0 Fig. SI0.2-11
da 2nan[
Hence, e^f^1
,„(,) . *(.)
?ctlO
And a =2^„
Px(0)
Px (»).
. . 1 -(x-ifna10.3-1 x = 2, <rx =3, P*\x
)-~£j2ne
78
Page 79
10.3-2 PxM = ^M*"W
Because of even symmetry ofpx (x), x = 0 and
x2 = 2^x2
px (x)dx = 2j"x2 ^xe'xdx
=J^°
x3e~xdx = 3! = 6
x2 = a 2 + x
2 -a\ = 3! = 6
10J-3
Fig. S10.3-2
y * J_*ypyb)^- +
J"
- * f® ye~y2,2a*dy = 0399CT
y2 -
= -L- r yV^'V^
+ r-4-ye->W <*2-00
<rv2ff
'e~>1>2°\
5 as2x
a 2 = y2 -(y
)
2 =y _ (°-399^2 = °-3408ct2
x2 = <t«
2
* o«3-4 .x2=
j
0°VpxW*
=
2C x2 /32a=2x42 *
Because * . ^ ?-*.’-<*f -’T'fe)*
10.3-5 The area of the triangle must be 1 . Hence K = - and px(x) *-(x + l) -15x53
^-
i
+ *)*-i(V
+
£
16ff-32
P*C-x. 5
ir«1.2Z 14l.il —«[< 3 4 3J 3
10*3*4 x-Zx,P.(x,).ip)4i(3)4^(4) +^{3)^(«)+^P)4
7.|4/>x(x,).4(4)+4(9)*i(16)+4(25).4(3<)*A(49)+
Fig. S10.3-5
79
Page 80
10 3-7 I" - 1—
=
Fxne~^'1°'dx For n odd, the integrand is an odd function ofx. Therefore x -0.
ax y2/c
For n even, we find from tables
x" =(lX3K5)-(»-«)^ " even
a n odd
10.3-8 Let x, be the outcomes of the rth die. Then,
1 + 2 + 3 + 4 + 5 + 6 _ 2 j= 1,2, .... 101 6 2
—\2 + 21 + i2 + 4
2 + 52 +6 _ 91
xi
=5 6
2 2 /— \2 35<7,
» x.-(x<)
Ifx is a RV representing the sum, then
X*Xj +X2+—+X10 = 1
.2<rx*< +< +" +ax
1o -<a-Tk2 = <r
2 +x2 = —~ + (35)2 = 1254.167
10.4-1 px (x) = ^<*(x) +^{x ~ 3
)
, \1 -~«2 /8
y = x + n
Py (>-) = WVnW = + " 3)]*Fig. S10.4-1
„ * .->»/• + » e^'8
4^» 4V2«-
10.4-2
and
px (x) = 0.4<J(x) + 0.6£(x - 3)
1 -i?ll
, \1 ,-y2 /g +
3c-(y-3)
2/8
\0>[2n
10.4-3 ^(xJ^e^x-O +O-e^x + O. Pn(") = /,<y(w - ,)
+(1-2W ,, + ‘)
pyW=[e^-i) +0-e)^ + 0]*[M3'-i) +0-^(>' +1)]
= (P + Q-2PQ)S(y)+ PQS(y-2)+{\- P){\-Q)S{y + 2)
80
Page 81
10.5-
2 When y = #,x + K2 Hence, y = Kxx + Kj
a\ - K\o\ and trxy - (x-x)(y~F) = (x - x)(#ix + K2- *
1* -*) = #,<rx . Hence,
Px = , Kxo\ / Kx
a\ = 1 if ATj is positive. If #, is negative, * K\°l « negative.
y axay
But o-
x and <xyare both positive. Hence, pxy = -l
10.5-
3 “S 9rf« = 0 Similarly, 5 = 0
. xy .SS .1*3- i
J
0
J'sin2M9><«= 0
2 2
Hence, crxy = x y = 0 and x, y are uncorrelated. But x + y =1.
Hence, x and y are not independent.
10.6-1 In this case
#11 = #22 ~ ^33 =mk *
#12 = #21 #23 “ ^32 “ ^01 = 0325#m
#13 “ #31 = #02 = 0362Pm
#03 = 0308fm
Substituting these values in Eq. (10.86) yields: a, = 1.1025, a2 = -03883, a3
From Eq. (10.87), we obtain
7 = [l- (0.825a, +0362a2 + 0308a3 )]#m = 03753#m
Hence, the SNR improvement is
-0.0779
81
Page 82
Chapter 1
1
11 1-1 This is clearly a non-stationary process. For example,
amplitudes of all sample functions are zero at same
instants (one is shown with a dotted lind). Hence, the
statistics clearly depend on/.
11.1-2 Ensemble statistics varies with I. This can be seen by
finding
^ = AcoiaX + 0) = A J^°°cos(fflr + 0)p(a>)deo
- jL. f
,00Cos(<ar + B\ia . This is a function of /.
"100/ J°
Hence, the process is non-stationary.
Fig.SlU-2
11.1-3 This is clearly a non-stationary process since its
statistics depend on /. For example, at t = 0, the
amplitudes of all sample functions is b. This is not
the case at other values of /.
Fig. SI 1.1-3
11.1-4 x(/) = acos(fflf + 0)
^ = acos(<vf + 0) = acos(«af + 0) = cos(a* + 0) j_Aap»{o)da
=[cos(<b/ + 6)/2A\ j*
Aada = 0
_
*x (,b /2 )- a
2 ccs(/*i+0)cos(a*2+0) - «os(/uf1
+^)cos(/ar2 + *)a2
= CO$(<U/| + d)cos(fflf2 + 0)
2
* ^-cos(aX| + ^)cos(fiJ/2 + 0)
\ _
stA
A A
Fig. SI 1.1-4
t>aco
83
Page 84
(e) The process is not ergodic. Time means of each sample function is different and is not equal to the
ensemble mean.
(f) x2 = *x (0)
= -^
11.2-
1 (a), (d), and (e) are valid PSDs. Others are not valid PSDs. PSD is always a real, non-negative and even
function ofa. Processes in (b), (c), (0, and (g) violate these conditions.11.2-
2 (a) Letx(t) = x, and x(r + r)=x2 Then,
(xi±x2 )
2 =^I + x22 +2x^7s0, x,
2 +x22 *±2^
But, x,x2 = Rx(t) and xj2 * x2
2 - flx (0) Hence, /?x(0)^j/ix (r)|
(b) Rx {r) = x(/)x(/ + r), Hm /?x (r)= Hmx(<)x(< + r)
As t -> oo, x(t) and x(t + r) become independent, so Urn,Rx(r) = x(t)x(t + r) = (x)(x) *11.2-
3 /?x (r)= 0 for r =±— and its Fourier transform Sx (e») is bandlimited to B Hz. Hence, rtx (r)isa
2Bwaveform bandlimited to B Hz and according to Eq. 6.10b
*x(r)= £ «x^sinc(2*Br-/»). Since Rx^j* 0 for all » except « = 0.
Rx (t) - Rx (0) sine (2«Br) andSx (a)= -^rect^j . Hence, x(<) is a white process bandlimited
to B Hz.
11.2-4 ^x( r)= ^x,x2 0»l) + />
XlXj(” 1 “ 1)“ /X|Xj(“ ,» l)“ ^xjxj 0
But because of symmetry of 1 and 0,
/*,x 2 0* 0* fyxjH*-1) *nd P*i*2 (
-1, 0* ^x^O*" 1)
and Rx (t) = 2[/,
X]Xj (I, 1)- Px,X2 (l,-l)]
-2J4, 0)[v, »)-('- JW 1W)] - 2W»-'
Consider the casenT* < |rj < (n+ 1)7* . In this case, there are at least n nodes and a possibility of (n + 1)
nodes Prob[(w + l)nodes] =f
= J~~n
Prob(« nodes) « 1 - Prob[(« + l)nodes] = (n+l)-y*b
k *x
y„/yyy/J//A
^-nTfcH*—nTb
The event (x2 = ijxj = l) can occur if there are N nodes and no state change at any node or state change at
only 2 nodes or state change at only 4 nodes, etc.
Hence, PXJ
j
x,
(l|l) - Prob[(w+ l)nodes] Prob(state change at even number ofnodes) +
Prob(n nodes) Prob(State changes at eveen number of nodes)
The number ofways in which changes at K nodes out ofN nodes occur
(iio-[(r>6)><rWr>‘)Vr'
is(jf). Hence,
+
85
Page 85
[( 0)(0-6)° (0.4)” + (5K°6)
2(0-4)""
2+-
]{
n+l~fb )
aid/?x (r) = 2/>Xl
|X2
(l|l)-l This yields
*x (r) = 1-1.2 \r\<Tb (n = 0)
= -0.44 + 0.24M Tb < |rj < 2Tb (n- 1)
Tb
*0.136-0.048^ 27* <)r)< 37* (« = 2)
Tb
and so on.
Fig. SI 1.2-4
The PSD can be found by differentiating *x (r) twice. The second derivative d2Rx /dr2
is a sequence of
impulses as shown in Fig. SI 1.2-4. From the time-differentiation property.
Page 86
— i?,(r)^(/fl>)2 Sx (ffl) = -®2SxH- Hence, recalling that ^r-rjoe'^.we have
jf2*' ’ ' ’
-ta2Sx (co) = -^-[-2.4 + 1.44(eJ
®7* +e-^)-02SS{^ +e~''2*Tt
)+••••]
= -L[-2.4 +258cosa»r4 -0576cos2fi>rfc+0.1 152cos3wf6 + .]
Tband ,
5x (q,) ,—1— j^2.4- 2£^coi<aTb ~cos2a>7i +^cos3a>r6
-^cos4fl>rA+....Jj
11.2-5 Because Sm(co) is a white process bandlimited to B, /^(r) = *m(°) sinc(2Bt
)811(1
M^) 350’ " = ±1 ’ ±2’ ±3 "‘
This shows that x(f)x^/ +—j = = 0
Thus, all Nyquist sample are uncorrelated. Now, from Eq. 1 1-29,
wRo + lLRm co*na>0Tb
_ n=l
J?n = ikbk+n S 1 and where a * is the klh Nyquist sample.
Rq = a* = x2 = /?m(0)- Hence,
Sy (*>)
= = 2BRm (Q)\p(vy sincere—
11.2-6 For duobinary
Ptk (1)
= PMk (-1) = 025 and Ptk (0)= 05
«.-.I>o)!
r(-o2
r°!
(i)4
K, =a*a* +1 =£ '£*k*k+iP»k*k+\(*k*k+i)
o* «*+t
Because a* and a*+ ,each can take 3 values (0, 1, -1), the double sum on the right-hand side of the above
equation has 9 terms out ofwhich only 4 are nonzero. Thus,
*1 * (W)*W, O*1)+HX-O'W.(-^-O + (1K-0V*+i
OX-0 HX 1)
Because of duobinary rule, the neighboring pulses must have die same polarities. Hence,
*•*»*! 0’1)= R
*k (0V.K *4 (2)
=8
Similarly, /»,*.**,K"0 = £Hence - *1
Also “ 8 *a*+
2
In this case, we have the same four terms as before, buta* anda*+2 are the pulse strengths separated by one
time slot. Hence, by duobinary rule,
*•**2^“ F*b^ />
**+2l#*^ *4(4) 16
Similarly, *a*a*+2H ^
87
Page 87
11.2-7
11.2-8
In a similar way, we can show that Ptk»k+2 O*-1
)= *•*»*1 ( ^ *
16
uILgalTill procedure, we can show that*„=0 for n * 2. Thus, from Eq. ( 1 1 .29) and noting that *„ is
,|2
an even function ofn,we obtain Sy(a>) =———jy
(1 + cos a>7J, )
J= ^
C0S^ 2 J
‘ Mf
)
TFor half-width rectangular pulse P{o)) -£sinc
i7=(0e + (-i)(i-e)=2e-i
/io »5-(1)
2e + (-i)
2(i-e)=i
Because all digits are independent,
Rn = a 4 a*+i - a*at+ i= (20- 1)
2Hence,
SyH =_lfHi2 r -I »
1 + 2(20-1) £cos«ft>7i
V*c l n
Approximate impulses by rectangular pulses each of height A and width e such that he 1 and e -» 0
(Fig.Sll.2-8a)
^x(j‘) = IS xlx2
/>x,x1 (
;rlx2)
Since x, mdx2 can take only two values A and 0, there will only be 4 terms in the summation, out of which
only one is nonzero (corresponding to Xj = A, x2 =* A ). Hence,
*x(r) =A^xjM “ A^x.W^lx,WSince there area pulses/second, pulses occupy Of fraction of time. Hence,
PX] (A) = a* and Rx (r) =aWX2 ,X| (*W *^x,W
Now, consider the range |r| < e. /»X2
j
Xj(/i)A) is Ae -**6*"
Prob (x2 )* K given that xj = A. This means x, lies on one ^
|
of the impulses. Mark off an interval ofr from the edge of ifc- . med—LI
—
this impulse (see fig. SI 1.2-8b). If Xj lies in the hatched
interval, x2 falls on the same pulse.
Hence,
PX2|x
,
(A)A) = Pretax, lie in the hatched region) » - 1 --
and f?x (r)= oA^l-^j
Since Rx {r) is an even function of r, *x( r) = a^
1_ ~j
In the limit ase-kO, Rx {t) becomes an impulse of strength a.
Rx (t) = ccS{t) M = 0.
When r > e, xj and x2 become independent. Hence,
Rx (r) = a2he »a2 |r|>0
Hence, Rx (r) = aS(r) + a2
€-T —< '4
3T
<RxCt) 4
t •
f©..
o
Fig. SI 1.2-8
88
Page 88
11.2-9 In this case the autocorrelation function at r = Oremain same as in Prob 1M ***£ O^henever
x(f), x(r + r) are both nonzero, the product x(/)x(r + r) is equally likely to
*x(r)«0, r * 0 and Rx (r) = aS[r)
Shift. The first-order probability density of the process is p{x\t). It can be shown that isproce
stationary of order 2. Hence, p{x,t) can be expressed as /<*)• We have
Rx (r) - J1£*1*2PxiXl (*i .*2^1*2
(1)
To calculate pXJ(x2 |x,
-x^we observe that in r seconds (interval between x, «idx2 ), thereat two
mutually exclusive possibilities; either there may be no amplitude shift (x2 - xi ),
or there may an
amplitude shift(x2 * x,)- We can therefore express pX2 (x2 |x, = x{)*s
Px2(x2 |x, -*i)-fc2 (*al*, -*i. no amplitude shift)P(no amplitude shift) +
p (x2 |xj = x,, amplitude shift)/^amplitude shift)
t fc-K-'U
*s Flg.Sll.2-10
The number of amplitude shifts are given to have Poisson distribution. The probability of* shifts
in r seconds is given by
„h«« there are on the averege/f shifts pr seeonti. The protabiH^ ofno shifts is obviously»<r) .where
Po(r) *C_/Jr
The probability of amplitude shift * 1 - Po(r) = 1 - er
. Hence
|x, -Wtl-e-^felx.”.. no amplitude shift) +(l -e~^
r)/>*j (ril* I
-x„ amplitude shifty
when there is no shift, x, - x, and the probability density ofn, is eoneentrared at the single value x,.
This is obviously an impulse located at x2 * x j . Thus,
Px2(-X2 |
xi= no amplitude shift) = S(x2 -x,) '
whenever there are one or more shifts involved, in generel, x, . x„ Moreover, we are given that the
amplitudes before and after a shift are independent. Hence,
Px2 (*2lx i= *1 >
atnpltawte shift) = Px2 (X2J = P\x)
89
Page 89
where p%2(x2 )
is the first-order probability density of the process. This is obviously p(x). Substituting
Eqs. (3) and (4) in Eq. (2), we get
Px2(x2 |xj =xx
)*e~fiT8(x2 -*i)+(l-«pt
)Px2 (xz)
* e”^r|<5(x2 - X|)+ [e^ — l)px2 (*2 )|
Substituting this equation in Eq. (1), we get
Rx (t) = e-;?r
j^o J^oX,x2 A>Xl
(x1)|<S(x2 -xj)+(e^
r-l)px
2(x2 )]£&i<*f2
=«"*[££Wx,
(
x,)5(x2 -*i>*,*2+nJlW^r -‘K (
x«)Px2(x2 )rfr,A2
]
* e~P
\\IxfPx^Xl)^ 1 + («* - 0JZo JCl Px
l
(
Xl )^ 1 -CoX2Px2 (
X2)t&2
]
=e-^p +(^-l)x2
]
where x and? are the mean and the mean-square value of the process. For a thermal noise x = 0 and
Eq. (5) becomes __
Rx(r)*x2e~fiT r>0
Since autocorrelation is an even function of r, we have
*x(r) = x2«_/W
and
SXW 2fix2
P2+a>
2
2
11.3-
1 For any real number a, (ax- y) SO
a2x2 +y2 -2axy SO
Therefore the discriminant of the quadratic in a must be non-positive. Hence,
(2xy)2<4x2
y2or (xy)
2< x
2y2
Now, identify x with x(r,)and y with y(/2 ) ,
and the result follows.
1
1.3-
2 *u (r)= u(/)u(f + r) * [x(r) + y(r)Ix(r + r) + y(< + r)]
« /?x (r) + Ry {r) + *xy (r) + Ryx(r) - *x(r) + Ry (r)
since x(r)and y(r) are independent.
Rv (r) = [2x(r) + 3y(/)l2x(/ + r) + 3y(r + r)]
*4/?x (r) + 9rty(r) since *xy(r) = *yx(r) 5=0
Rw {t) = [x(r) + y(t)l2x(/ + r) + 3y(r + rjj = 2Rx (r) + 3Ry (r)
Rm {r) - «u>(-r) - 2/?„(r) + 3/?y(r)
113-3 *xy( r) ABcos{o}qI + 4>) cos(muo(^ + r)+
n
+\
=^ (cosfnio' + »©o(' + f) + (» + 1)*]+ cos[na>0 (' + r) - a>0t + (n-
1)*]}
90
Page 90
11.3-4
11.4-1
CO<^0^ + + 1)#] =^io^ + W = 0
Similarly, cos[»fi>o(, + f)_ ®o, +
(,,“ 1)^]
= 0 8,1(1 *xy(r)*°
x(0 - C0 +£Cn cosmu0 (f- b
)+0n
11=1
00
• C0 + £C„(nfi>of - «a>0b + 0n)
n«l
Since b is a r.v. unifomtl, distributed in the tta,e(0. T„). ».b .^ is . r.v. unifOmtly distribrned in ihe
™big^the^ngument in problem 1 1 .3-3, we observe dint nil haimonics ere incoherent. Hence.be
autocoirelaiion (taction of«.(r) ia the eon, of autottarelation Ibnclion of each tenn. Hence follows
result.
(a) S,(©) * 2KTR{and S2 (f») = 2KTR2 . p„n
Since the two sources are incoherent, the principle of superposition applies to the PS .
IfS^ (o») and 5^ Hare the PSD’s at the output terminals due toS^ru) and S2 {a>) respecivey,
(a>) = $iH and SrfH =|
w2(fl>)fS2H
where
R2 / jaC «2_
Ri+V je£ _ jaRlC + 1 = .
«iW *: ' s2 / tac-” $_ jk.
*2
R-, / icuC " /?2~
J?iO<uR2C+1)+ *2 ja>R\Ric + R
\+ R2
j«
2 :u0
s v0
TV-;
Ri,
c?
*l*i-.. *
/?, + r2 ;
5 v- *
i
• «L_QL__
(a) (b) (c)
Fig. SI 1.4-1
Similarly,
Ri RHiK03)
/?2 (yalR,c + l) + f?i
* jo)R\R2C+Rx+ R2
•f IfTPi ffn t \2KTR7R\“d
2CT*2(*i+*2)5WH =^h +^H -
^2*2*2^ +( *, + Rlf
, „ i/;<uc _ *i+*2(b) #H = ~ =
+ (Rl + *2)
jaC Ry + ^2
(J?i + J?2)2
7KTR\R2 2KTRxR2 {R\ + R2 )
=<u
2C2*,2*2
2+(*, + *2)
2 *> + *2 ’^l«22C2
+(^, + «2)2
which is the same as that found in part (a).
91
Page 91
1 1.4-2 y(f) =£ " a)da
(f) = x(/)y(/ + r) = x(/)£ ^(o)x(/ + r - a)da
= f h(a)x(t)x(t + t -ajrfa = h(a)Rx(r - a)da = h(r)*
R
x (r)wdS^i”) = H{a>)Sx (o)
J-flO
1
. . jeoC 1
in Fig. 11.13. H{») «—
—
=-^71+jwC
and SnVo (©) = 2X77?/ (><aRC + 1)and 7?,^ (r) = 2X77? «- r/*c«(r)
11.4-3 (a) We have found /?x (r)of impulse noise in Prob. 1 1.2-8
R% (r) = a$(r) + ar2 ,and Sx (t») = or +2*a2
$(fl>)
Hence,
Sy(n>) = \H((of[a + 2/ia
2S{o))] = 2;ra
2j//(0)|
2^®)+
and 7?y(r) = ST 1
^®)] = a2\H{of + ah(r)*h(-r)
(b) h(t) = 2-e-,lr
u(t). = l* 1
inJ+ —
Fig. SI 1.4-3
11.5-1 n(/) = ncW C0S<uc, + wsW sina, cf _The PSD of«c (/) and *,(/) are identical. They are shown in Fig. SI 1.5-1- Also, n
1is the area under
5n (o>) , and is given by n2 «
2 y x 104 +“(y 1 = 1-25 x 10
4Ol
n2^orn2
jistheareaunderS„
t(®),andisgivenbyn
2 =n f=2^5(HXUI +yy x5000
j
1-25 10 ^
Fig. SI 1.5-1
92
Page 92
1 1.5-2 We follow a procedure similar to that of the solution of Prob.
11.5-1 except that the center frequencies are different. For the
3 center frequencies S„c(<»)[or (®) ]
8,8 shown in Fig.
SI 1 .5-2. In all die three cases, the area under S„c(<u) is the
same, viz., 125 x 104
cAI. Thus in all 3 cases
I^ = n2 =i25xl04 J(
r . foBk. C*riW “f * ^15 fc- Ctf'H* -£re<j- Qc
Fig. SI 1.5-2
ii.M
Fig. SI 1.5-3
. , \SmH 9 + fiT - ~
—
11.5-4 (a)+ __L_ + 6 &u 2 +60 o> +1
<b)
(c) Tlie time constant is Hence,, retntnmble «loeoftimeslehym<|uimd to entice this filter
Vio
realizable is * 0.949 sec.
>10
(d) Noise power at the output ofthe filter is
. 1 p Sm{a)Sn (eo) p”—$— dco =—^-nrtan"1 =-4-
=2«-J-*Sm (<a) + Sn (<u) ©2 +10 2W10 VIO -« VIO
The signal power at the output and the input are identical
St * S0 *— f*6
g<tto = 1
12ar^-0O 9 + fl»
2
SNR=-^-*— = 1-054
93
Page 93
4
11.5-5 (a)„ ,
, Sm(a) a>2__
H^ {6)) 'Sm(c) +Sn(a>) _L_ +a)2 +4
_a)
2 +64 l[n 5333 1
9a2 +96 eu2 + 10.67J
+ 432
eo2 +64
(b)
(c) The time constant ofthe filter is 0.306 sec.
A reasonable value oftime-delay required to make this filter realizable is 3 x 0306 = 0.91 8 sec
(d) Noise power at the output of die filter is
w _l_f® Sm(fl>)SnH . _ _L p £ -do) = 0544° ' In SmH + S„(a>) 2n ^ + 10.67)
The signal power is
S,=Se =
ia.
No
Fig. SI 1.5-5
94
Page 94
Chapter 12
12.1*1N0
S,
y = 1000= L -=>5, = 0.08
12 .1-2
2xl0"8 x 4000
Also, //c(a>)= 10~3
. Hence, Sr =— = 8xl °4
5r , _L£ [2 x 8000>r] * 8 x 1
0
4 =c 0 * 1
0
£5
— oC ®< (|>*>
Fig. S12.1-2
.2 . _2
aa = 8000/r
5n„H - * 10I0
[ a.2
35 dB = 3 162 = -f- = yT2—: => So " 337 * 10
_x 10-7
3
iq~3
But s0 {t)•—«(0- Hence-
So = 121-«2(/) . 337 x 10'3 => «2
(/)* 215.7 x 10
9
a 2
Also, i? -—f fiU» »^ -mop = 215.7 x 109
2n J~eo x
Hence, 0 = 26.96 x 106 andSm (e>) * 26.96 x 10
6rectQ^j
\ e<x - j / ii2 . 1 f®#
/rJ°
26.96 __-i ©i tan —an a
r io6
]
U2+ «2J
^iCD
26.96
4a: 2.68 X 10“
Sr =I£ SmH^ = ^Joa26*96 x 10
6<to =
2656^0 . 6M5 x 109
12.2-1 (.) 30dB = 1000--^- =y=^ =^ri^^=>S<sl4xl0
‘
(b) From Eq. (12.7), N0 = cAlB = 10" 10(4000) = 4 x 10'7
(c) Sj = |//c(a»)j
2Sr and 10'8Sr = 4 x 10^* => ST = 4 x 10
4
95
Page 95
12 2-2 ft) -^2_ = 1000 =— 7P =5-5/ =4x10'12.2-2 W „ ,wu
jufi io_l°x4000N„
*-7
(b) N0 =JUB = 10',0 x 8000 = 4x10'
(c) S, = |//c(®)|
2Sr = l<T
8Sr = 4 x 10-4
=> 5^ = 4 x 104
12.2-3 Let the signals m,(f) and m2 (f) be transmitted over the same band by carriers of the same
frequency (t»c ), but in pha* quadrature. nte two transmitted signals are ^t)coia> ct +
ft©5 rigp ,^005^
lpf
_lPf
*9
r%
Fig. S12.2-3
The bandpass noise over the channel is nc(/)cosn>er n^sin^t. Hence, the received signal is
y2mx(f) + n c
(f)]cosa>cf + [>/2m2 (t) + n$(')]sm°
c
1
Eliminating the high frequency terms, we get the output ofthe upper lowpass filter as m,(f) +^n c
(r)
Similarly, the output of the lower demodulator is m2 (r) +^ ns(0
ntesearesimilartotheoutputsobtainedforDSB.se on page 535. Hence, we havejf-
= y for both QAM
channels.
12.2-4 (a)Hence, mp >M
Sa m<b) TT-— 2N0 A2 +m2
_H»2
j. y1 *2-V
—£- +mV
2mP
where*-
m2
m2 SIX(c) For tone modulation*
2 =-£r = 2 andfor/r = l, -j*--—r = jmpr
. ST A2 +m2 f»p2 +m2
_,wg^ 1 \{ K2 »\
(d) Ratio -f-=—
ST m2 m m
96
Page 96
2^
^2
12.2-
5 OFromProb. 122-4. For*r-l«dmg, »p •J--£— »
™ o fC T^i
and when// =1, ^ = =^
s„ (os)2
r0,) Whence
12.2-
6 For tone modulation, let m(r) * /*< cos©mf . For BSB-SC,
^XU»(/) * if2pAcosa>mt -co$a>ct
* -^=-[cos(r»c + + cos(<uc - <Wm),j
,, _Zil +Zdi = inland m=-^+^ = V2/*4 Hence, the peak poweri<-
4 4 2P
V2 V2
Sp =(V^f = 2^2
2<2 and^ = r -^ ^ where Si = ^Sp
For SSB-SC
^S$b(0 = m(/)cosoc/ + m*(/)sin/wc/
= pA cos<amf cosojf/ + /*4 sin <amr sin <uct * /*4 cos(<ac - “«)'
2 ,2 2 2 S0 _ Sj p2 A2
_S
f>
Sj = — and mp = /i4. Hence, Sp = m A
N ~ j\b 2JiB 2JiB
For AM^w (r) * /<(l + /«os<»mf)cos<Hcr
c _0,2
_ jl+fiidiS‘~2
+ 2 2 2
mp = .4(1 + //) and Sp * /<2(1 + ^)
2-
Hence,
Sp(2 + /.2
)
Si-4(l + „)
2 “d »* "A2 + m2
' "A2 *{SaH2)
mr =
*A2l2 A. =1f *
21
’
5,(2V)lAIB 1l2 + /r
2J 4(l + /i)
2<^J£
S„ ^pUnder best condition, ie., for// 1,
N<> 1(Ui5
Hence, for a given peak power (givenSp ) DSB-SC has 6dB superiority, and SSB-SC has 9dB sup^onty
over AM. These results are derived for tone modulation and forp = l(the case most favorable for A ).
12.2-7 For 4<r loading, mp •= 4am and the carrier amplitude A = mp = 4<rm (for // = !)• For Gaussian m(r),
m2 s o2m (assuming m-0)
Prob(£ * A) = r-^e-^ l2al dEn = e'^12^ = <
” fT_
Hence, -^i- « - - 4.605 and S(=
2<t?
= 0.01
A2 +m2 \6*l*oh2 2 2
n -n
_ , _S±_ s IZfdZffi.I s —(4.605) = 9.79dBTherefore, r Thresh ^ 2JUB 8 JIB
J8
V
97
Page 97
12 3-1 = 28dB = 63 1. Hence,
Nc -> iw
2(/)
Z£- = 63\ = ‘i02Y— jk r »A m.
= 3(2)27ttV(^m)
,631x9
Therefore, y =—jj— = 973-^3
(.) Also, r = -^L_ => s, = r^«® = 473.25 x 2 x 10" 10 x 1 5000 = 1.4197 x 10
lMB
m B.mJ-Hy^l.igZsL91;,, =20,000*(b) P
2kB 2k
B
30,000*
sa . -ahy„. (10-,
)
I
(20,000*)! - 4/r
2
(c)
12.3-2 mp = S. S 'T “d band',''dth *Y Hen“'
VsMou (2z*2/T0 )2 B7
_3kJ_
(So/»o)FM
=
3(4s/r0 )
2 4
,-3
12.3-3 m(<) = cos3<u0/ and mp = 1
For a maximum:_2 .
m(t) = 0. This yields cos2
<»0/ = 2sin Q>o'
or l-sin2ai 0/ = 2sin
2<u0f=>sin<u0/ = ^, cosav
and
S = |-3<u0 cos2 w0( sinfl>0/|
- 3®0(jj(^J
,2 2
I
(So /No)pM m (
3a>o) mp =
9fi?o _ = 225
&,/**)FM 3m‘>2
<3-’)
12.3-4 m(r) = aiC0S<al/ + fl2 C0Sfi, 2r >
«^=oi +a2
^) = -(a,fi)lsinfi>if + a2 a, 2sina>2 r )'
mP * °l®» + °2®2
(So/^o)^ 3m’p 3{o\0)\ +02^2)
<»2o|KT
3^a22
(
\2
i +^|,
02^2 ;
98
Page 98
_0+*)
2
=3(1 + *y)
2
12.3-5 Error in this problem. There should be 4*2in the denominator (see below).
SA(m) = o>2Sm («>). Hence,
£ -
n
0s lh(2^ -r>v2sm(2w
From Eq. (12.42a)2— i D*M] <*
e”'Js.(VW
These results are true for a waveform m(r)
12.3-6
~2 ^^(///o)2*^ /o3^l + *
2*?Sm “ r 1
/• r —a^l + (///e)
2t oJ^, + x2
‘
t \
fon
“*(§I-/i
2* sin(£)
The definite integrals are found from integral tables.
—2
sin(*/2*) 2 (*l2k ) _ 1 ABm ~ fo
sin(3*/2k)fo
(3/r/2fc) 3°
Ks)
As k -* ®,
f> 1 ledn\tt.
<r a
.Ifl>l -a1 12a1
Hence, the normalized PSD s is--—e2O
,f W = 2«*,then^ =(W - - to2- M«0 » *•
’-41-band-W to W.
la
-W2/2ct
2I
I
,^).2..nd4^ = '-'‘WW -°» = W = 30to' S '°4K'’
/H£^ = 0.95=>W = 2.45<7, B = 0395c
/H= 0.9 =>W = 2.15<t, fl = 0342(7
pH
x = o 99 = 3.06tr2 >W2 =- PM superior
99
Page 99
x = o 95 = 2.00a2 -W2 => PM and FM equal
x = 0 9Wl.
1.54a2 < W2 => FM superior
12.3-8 m(r) = aiCosfl>1f + fl2 cos®2, < 41,(1
m2= |^5m(/y/ = (a? +
)/ 2
a2 + a
2 2
Since B = A,PM is superior to FM if/2 >5 2
a\ +05
)
2__[<fo2
,
4ft]J a
2 + o2 2 2
a?/2 +alfi
=a? +fl2
*
| 1 2
§a- m tfrdL~lp
r . Since mp = 3a, in2 - a2 and 23.4 dB- 218.8 ,
218.8 = =-(2) r
12.3-9 (a)m2 3mP
y =218--~ = 164.1 . Also, r-nn* - 20(^ + 1)
74
„ 164.1,
...So /3Thresh
m_2^
I s7 -21
!
jl = lply = 1(7J21)2(164.1) = 2844 = 3453 dB (40 dB = 10,000)
- ’" 242i
Required increase in y = = 1-479 = 1.7 dB
12.3-10 From Eq. (12.40) p2 * -
(1) Tone modulation
(2) Gaussian with 3a- loading
(3) Gaussian with 4a - loading
where For tone modulation,
[l +(m2
/mj)j
^-KnbW- a47
fl2 1=^ = 0547P
3(l + l/9j
fl2 - if—-—W- 0J6P
3^1 + 1/16j
HL-,05mj,
.
m2 a2
For Gaussian modulation with 3a— loading,2=
/, ,2 "9
'
mp (3o7
100
Page 100
2 _2m o. .. ill u _ 1
For Gaussian modulation with 4cr - loading, -y = 2-
1 <(4c)
12.3-1 1 Let us first analyze the L+R channel. In this case, the demodulator output signal, when passed through the
0-15 kHz(lowpass) filter, is given by (L + R) + n0 (t), where S„0(<a
) = ^-[see Eq. (12.33)].
When this signal is passed through the de-emphasis filterHd(at) -— .*e s'8nal is restored t0 (L
and the output noise power N'a is given by
---3
Let us now consider the (L-R) channel.
Let <yc=2*x 38,000 and =2*x2100.
The received signal is FM demodulated (Fig. 5.19c). The PSD ofthe noise at the output of the FM
demodulator isS„0H = Jla
2/ A2
[see Eq. (12.33)] The output ofthe FM demodulator is separated
into (L + R)' over 0-15 kHz and (L-R)' cos <oct over the band 38 ± 15 or 23 kHz to 53 kHz. Let us consider
the signal over this passband, where the noise can be expressed as nc (r)cos<ucr + n s(/)sino»cf. The signal
is (
L
- R)' cos co ct. Hence, the received signal is [(I - R)' + nc(/)jcosoc + n
s(/)sin©ef. This signal is
multiplied by 2 costu cr and then lowpass-filtered to yield the output (Z. -R) + nc (/). But
S„£(a) = Sn(o + <uc )
+ Sn (<u - a>c )= + °>c)
2+ (® '
“
°>c)2
]
When this signal is passed through de-emphasis filter //„(«) = *e signal is restored to (L-R) and
the output noise power N£ is given by
s«M)da = J \"[(a+o,c)2 +
(a ~”c )
2
]^i^(dco
m2Jia,L f? +
"c-"?. tan"1— W = 2/rx 15,000
jiA2
[<*>\ " 1
.
Hence, the (L-R) channel is noisier than (L+R) channel by factor—2- given by
at: l “i »ij 71
B-f\ tan'
Substituting ZJ - 15,000, fe = 38.000, /, * 2100 in this equation yields:
= 166.16 = 222 dB.
101
Page 101
12.4-1
12.4-2
12.4-3
L = M n =>n = logw L
— = 3Lrm2
(t)
= 3M2n
_2nP
m* 2
Vwp 7
= 55 dB = 316200
For uniform distributionfn
m2m„ J'"1
/-
1
:
3mP
- rA •
(a) 316200 = 3(2)2n m
2
1WP )
o\
Fig. S12.4-2
m i
,2n
Since n must be an integer, choose n = 10 and L = 1024
nil -^2- = 3(2)20 - = 1.048576 * 10
6 = 60.17 dB.
Na 3
Bpcm - * *>MH2 bip0l““f?
1^ „„ . „ men tte new bnndwidih of transmission is
(c) To increase the SNR by 6 dB, increase n by 1 ,that is n - 1 1. l nen tne new
22x45 = 99 MHz.
Sj = 2BnEp ,Ep = 2xlO
-S,
B = 4000, « = 8
Sj = 2x4000x8x2xl0~5 =128
v-^L- 128• = 2i6xl02
\Ali_2x625 xlO
-7 x4000
^JZj-eV32=7i69xl0“9
Bm = nB = 8 x 8000 = 64 kHz (assuming bipolar line code)
1 + 4i(2,6
-l)e(V32)
21 845 = 43.4 dB.
(b) If power is reduced by 10 dB, then y = 25.6, c(V32)- 0(1.79) = 0.0367 and
S0 _N0 l + 4(2
l6-l)e(>/32)
227 = 3.56 dB.
102
Page 102
The table below gives SNR for various values ofn under the reduced power.
(d)
Hence, n = 3 yields the optimum SNR. The bandwidth in this case is Bm = 3 x 8000 * 24 kHz.
12.4-4 i — />£ = p (correct detection over all K links) + smaller order terms
s(l- Ptf~' (1 - Pi) = [!-(*- l)^Jl - Pi]s\- Pi -(K- 1)Pt
So PE = Pi+{K-')P*
(b) y = 25 dB = 3162, y = 23 dB = 199.5
Pt = 2(73162/8) = 2(6287) = 1.6 x 10" 10
Pi * 2(V199i/8 )= CK4-994
)= 3 x 10"7
p£ m 99 X 1.6x 10"10 +3 X 10~7 * 3.16 x 10-7
s /»;
12.4-5 |m| - mpm{m)dm - * y
m2 = = J^m2jfim =y
o-m*m2 -(mf-y_ 2 / 2m/
^[ln(l^)]2(crm
2/«l,
2) + (2H/Aimp ) + (l///
2)
4_
3(
28)
°"2/^ 6383—j ^(In 256)
2 ^,J + 0.0068^“- + L53xl0"
5
m2 255mP (255)
2 mP
12.5-1 As noted on Pg. (570), the optimum filters for DSB-SC and SSB-SC can be obtained from Eqs. (12.83a)
and (12.83b), provided we substitute ^[5m(ty + <w e )+ Sm(to - t»c )]
forS„(a>) in these equations. Let
5m(®)“7[Sm(" + ®e) +5m(®-®c)]
2 [(a> + a;c )
2+o 2
(a>-a>e )
2 +a2
J
a 2(a>
2 +<u2 +a 2
)a»3000»
(a>2+<u
2 + o2
)
— 4<a2a>
2 ai^^xio5
We shall also require the power ofSm (ffl)
.
/ =2
We can simplify the evaluation of this integral by recognizing that the power of the modulated signal
m(/) cos coc t is half the power ofm(f). Hence,
Ilf* , . If* a 2, o _j (o
00 a/=±J_f Sm(twVtu =— f
—; rtf<u=— tan — = —2 2^“°°
m ' ”2jt^ <y
2 +a2 a o 4
We shall use the PDE system shown in Fig. 12.19
103
Page 103
(a) For this system
(n>)/Sm(®)
i „ / \iJ
(3)
(4)
Because tf» and Sn(o) are constants, we have
. ,.,2 SrfilM »o
V
v£H_{hM
"d/qss* -s^)do)
where Sm(o>) is found in Eq. (1). Also from Eg. (12.83b)
ST yli/sm (a>) tfx/JsM
(b) The output signal is Gm(f ) . Hence, S„=G m (0
We have already found the power ofm(/) to be 2(a/4) * 2 . Hence
Gla (lQ-2
)
2(3000ff)
_ 3„So = 2~ ~
2 20
To find the output noise power N„, we observe drat the noise signs! With PSD5,(o>)
-
2xl ° *WSS*S
through the de-emphasis filter in Eq. (4) above. Hence, Sn (flj)the noise PSDat die otdput of
Hd (co) is
SnW-S.H"dH
Also, the output noise power i, nc(,)/V2 and N. * |s« E,. (ITdb)]. where n2 -n2
and
12.5-
2 Similar to Prob. 12.5-1
12.5-
3 The improvement ratio in FM is-
3nJ20
10JoS,(a)da>
B3m2 —,where
m2 = 2j0°°Sm(a>Hf = andJ
0WSn,H# -J ^ 0"^ 2
Hence, the improvement ratio is
B’W-L.lsUdB.H 3
104
Page 104
Chapter 13
13.1-1
Ay '"K
% -t-*
*T>/| *S^D
f & t-
Fig. S13.1-1
Tb *Tb
For the integrate and dump filter (I&D), the output is the integral ofp(t). Hence, at t » Tb , p0{Tb )= ATb .
Ifwe apply S(t) at the input of this filter, the output h(t) = u(t) - u(t - Tb )
.
Hence,
and
and
,2plin) A2T? _ 2
This is exactly the value of
p
2for the matched filter.
13.1-2 The output p0{i) of this R-C filter is I
p0(t)^A[\-e-‘IKC
)
°*‘* Tb ^—= A(\-e-
T>lRC)e*-n)lRC t>Tb
. -f 4 P-t£
The maximum value ofp0 (t) is Ap ,which occurs at Tb .
Ap- Po{Tb)=A(\-e~
T>l*C)
2 _Lffn =
2fr'
2 )-*>\ +o2 R2C2 _ 4RC
.2nP
P ~Z2°n
4A2RC(\-e-T*l*C
)
2
4A2n l'-‘-T>IRC
fJi Tb/RC
K(f>
rh
103
Page 105
We now maximize p2with respect to RC. Letting x = Tb{RC, we have
(>-«-)
JU
and
This gives
and
Hence,
eb2 2xe
~X{'-e-x)-(' -e~
Xf
etc
2xe~x = \-e~x
x = U6
or
or
l+2x = e
1 126
RC* Tb
pin =(0^16)2A2TbJi
Observe that for the matched filter,
22E
p 2A2TbPmtx ~
Ji Ji
The energy of />(/) is 7}, times the power ofp(r)
.
Hence,
A2{l- m2
)„ ,
a2t .AEp Tb +^f-Tb-^-Eb
/rTSimilarly, £, =
^
* = £*
£«-J?*MrfO‘*-J? -^(l-^Jcos^ct + ^Vsin <V
= Tj, + A 2m2 Tb
Hence,
and
4/<2 7*{l-/n
2
)8£*(l-m2
)
maxJU
2£*(l-m2
)
13.2-2 Let Cj be the cost of error when 1 is transmitted, andC0 be the cost of error when 0 is transmitted,
optimum threshold bea0 in Fig. SI 3.2-2. Then:
c, =Qo£ (€
lm “ 0“ Oo Qh?)
Q> = 0)1 P («h °) = Coi 0 — -J
Let the
106
Page 106
The average cost ofan error is
IfPm (l) = JPm {0)= 0J5
C>
C = /m(l) C\ + ^(o) Co(i)
For optimum threshold dC/da0 = 0. Hence, to compute dC/da0 ,we
observe that
and
Hence,
0(x) = 1—i-f* e~y2^dy
i „-*V2—= =—s*edx fin
Hence,
and
But
Hence,
dC _ 1
dlfl0 2<t_V2™
Cio
—3— : 1
^
C,o«2<r" -Q)l*
(Ap+a0 f _(^p-flp)
2
2*2 = 0
2<r» 2<r*
and q„ = ^ In £si
l C10 )
0 2Ap CW .
^J1EP2
and Ap = Ep .
^ln|>l4 L
cioj
13.2-3 We follow the procedure in the solution of Prob. 1 3.2-2. The only difference is Pm (l) and Pm (0) are not
0.5. Hence,
c - pm (i) c, + />„,(<>) c0 = Pm(i) Cio 2 Pm^ C° 1 2\
^\ °n /
and
(Ap~aof (
AP+°o)
\
Hence,
dC1
da0 2<rn -j2n
In
/n(l) Cjq e ^ -Pm(0) <^01 •2d = 0
\pm(0)c0l-
_2a<>Ap O*_ \Pm(0)C0l-
./’mCOQoJPm0)CloJ2
~~ 0 7A<*n
2AP
107
Page 107
But
13.5-1
13.5-2
Hence,
2 cAiEgand Ap = Ep
Pm(0) c01a°"
4ta
[/»m(l)C10 J
/It 1 ~(r+Ep) A®*
A'h)—
W
rVtoJ
a„V2/r
p(,K)—!
—
Jr-bffaon j2it
The thresholds are ± Epj
2
and
2
T“
-4JS'/
/£ Ae-p
Here, />(/)and 9(r)are identified with 3p(t) and p(t), respectively. Hence,
H{a) = [3P(-£u) - /»(-a>)]e'-,0'r* = 2P(-a)e~J<oTt
and
But multiplication of/i(/) by a constant does not affect die performance. Hence we shall choose h(t) to be
p{T^ -i) rather than 2p{7j -/). This will also halve the threshold to a0 =2Ep . This is shown in Fig.
SI 3.5-2. Also,
108
Page 108
and
Pt ^EEEMJiEEE)
9E„ + EpThe energy/bit is Eb
—^ = 5£P Hence,
Pt
13” p"
"
s^
«
msm«sk,» nqufctt . mibimum b»dwi*h 128 kto But^pltode
modulation doubles the bandwidth.
HenceBt = 256 kHz
io-7=et
V- *J
rv—
7
£fc= 2.7x10'
5, = £4£A = 2.7 x 10-7
x 256,000 = 0.069H/
For A/ = 16
o =«WOO = 64kH2
rlog2 16
7 2(15) ( l24EbPtM~Pb\°i2
16-4x10 =J6 21^2550!
This yields Eb= 5.43 x 10
$ = £6£4 = 5.43 x 10-6
x 256,000 = 1391F
For M = 32256^000 =J12kHz
rlog2 32
-7 2(31) J 30gT**i ,082 32*5x10 = ^ Cly
1023jy
This yields £* = 1.719 x 10-5
St * EbRb * 1719 x 10-5
x 256,000 = 4.4IF
-813.5-4 For A/ = 2 and <3! * 2 x 10
This case is identical to MASK for M = 2
l°’7=e
[\
if
L
j^ £6=2 ‘7><10"7
Si = EbRb = 2.7 x 10'7 x 256,000 = 0.0691F
qt =2^6,0<
?2 x 2 = 256 kHz
109
Page 109
For M = 16
PeM * 0°g2 16)^ " Apb a 4 X 107
4xl0"7 s2gj2ir x 4£t
25601. £
fc= 1.67 x 10
-6
In MPSK, the minimum
Hence,
Sj = E„Rb = 1.67 x 10"6 x 256,000 = 0.42751F
bandwidth is equal to the number of M-ary pulses/second.
256 ,000 ^ 641^
' log2 16
For M 32
PtM -0og2 32)Pb = 5xlO-7
5xlO“7 s20|
2*2(5Eb )
102401=o £*=524x10“
S. » EbRb = 524 x 10-6
x 256,000 = 134W
256^000 = 512kHzT
log2 32
110
Page 111
14. 1-3 1) (1,1.0) is -^Lr[l + >/2sinfiv]
2) (2,1, 1) is -^[2-V2sma>o' + V2cos£»o'] 2 + 2cos^a> 0f + 4 j
3) (3’ 2> 4)is
7^3 + 2>/2 sin a>o* --J^costoo1
.
^3+ 2 .9
1
cos(<tfor “ 1040
)]
4)
-I, -1, 1) is ~7^[“|_ ‘^sin<»of + >^ cosa,o,
j
-- + 2cos^o0,+
112
-u
I*
Page 112
14.1-4
< 3,-3, 3, OFig. S14.1-4
b) The energy of each signal is:
c) F3 F4 = (-6-8 + 6+8 + 0) = 0. Hence, /3 (/) and/4 (r) are orthogonal.
14.2-1 Letx(/) = xIi
x(/ + l) = x2 x(f + 2) = x 3
We wish to determine
Fxix2 x 3 (*l»*2 **3 )
Since the process x(/) is Gaussian, x,, x2 , x 3 are jointly Gaussian with identical variance
£ It The covariance matrix is:
ax,
ffX,X2 °*1*30’x
1x 2
* ^xjxi= x
tx2 * x(') x(‘ + 1
)=
O'xjX! a\x2
CTX2XJAlso trX2x3
= crX3x2« = x(* + 0 x(' + 2
)* **(')
^XjXj °*3X2 <
.
x,x3- axjx,
= xi*3 = x(') x(' + 2) = R*&
so
113
Page 113
14.3*1
1 _ 1 IA,i = A33 = 1—j and A 12 = A2 i
= A23 = A32 ~g3~
e6
_ 1
A >3* A31 = 0,
~~* 4
e
And-IIA*V>
<?' J
Fig. S14.3-1
p(C|m]) = Prob^ni < and P(C|mM )= Prob^nj >
2 j
p(C|m2 )= />(CH - = P(ClmM-l) = Prob
(ln
>l < f
)
Hence
. |-[p(m|)f(Cl«i)++p
(mMlacl
mM)]
The signal energies are
114
Page 114
Hence the average pulse energy E is
7_i[£+2d+2d+.....+i*^
2 1" *(M2 -\W-^<2k+,) =
-
£ K -'K"b
log2 M 121og2 M
2(W-1)|
6(log2^)
,>,M “ W S^(W2
-l) Ji
M 4 4
Hence
Which agrees with the result in Eq. (13.52c)
14.3-2 /»(C|m1 )= P{C\mA )
« />(Clm5 )
= />(C|mg )
p(C\m2 )= />(C|m3 )
= />(CK) = ^<>»)
S. S*. *s
<X —:
"t. - • • • •
s5 ; S4 -*7
?(C|wi)*/^n, -e|, n2>y)
/*(cK)* ^Khf* ,!2 > y)
f(0 = ±[j-(ch)+ *ch)] -[
2_3Cfel
P,M .i-f(c).ie(xr)f5- 34ij]]
Fig. S14J-2
115
Page 115
The average pulse energy E is
£* =log2 8
and
«1
This perfonnance is considerably better than MASK in Prob- 14.3-1, which yields
PeM = \.15Q '(““ftlfor tf-8V ^
14.3-3 In this case, constants ak ' s are same for k = 1,2,
in Fig. 14.8 with terms ak 's omitted.
Hence, the optimum receiver is the same as that
We now compare r-j|, r-J2 * r sM
Since r sk= -J~Er cos6k is the angle between t and sk ,
it is clear that we
which r has the smallest angle. In short, the detector is a phase comparator,
at the smallest angle with r.
are to pick that signal sk with
. It chooses that signal which is
14.3-4 Because of symmetry,
= />(C|m2 )
= = P[C\mM )
116
Page 116
and
P(C)
P'M*
Here, M = 2N
Hence
and
14.3-5
Also
Hence
14.3-6
1-22£
JIN
NP{C1«|)
1 - P{C) = 1 -
. Hence, each symbol carries the information log2 M = Nbits
.
Eb = E/Ji
1-2 2
1
£jy
Jl ^(wp),
^ 2d z*(mi) 2
«4ln2 + 42d 2
/>(Cl«,) = P(C\m_,) = Prob.[ni > -(d - /i)] = 1
/’{Cjmo) = Prob(|n,|</i) = l-
/>(C) = 1 /’(Clm0 ) + ^/>(Clm, ) +£
P(Clm_,)
1 Jf-Jilni ] JV +
2 2dJj!/f J \ 2
, 2 d1
E » 0.5(0) + 025d2 + 025d
z *—
'2£-ln2
2/I/jl
'2£
+0+ ln2
2J£/Ji
/»(c)
=^[/’(ct»,,)+
/
>
(c1,«2 )]
= _L]
«JU iJ-«t »"(*/*)
117
Page 117
14.3-7
and
1 7 *,«(*/») e^-ld)l*Abdq2dq\
~nJi h
~q'
_ <j2 +4d2 _5 dj
£=2 '2
d = >/04£ - V°-1£*
FI*. S14.3-7*
Note that
.t/ , - ——A. +
—
5l =--t\~2*2 ’ 229
22
^ .d,
5 . = _— d. +— ^
2
JJ—T#l~r#2 .J4 2 ^ 2
2
118
Page 118
Fig. S14.3-7b
(c) /’(c|m5 )= Prob^noise originating from s5
remains within the square of side
j
w<^)
/>(f)m5 )= 4
0^
d
2V2<t„
_ -| f 4</
2
We also observe that £ ,the average energy is £ « - —
£ 0.4rf2 02</
2
Ji Ji
Therefore F(4”s)s4d
The decision region R2 for is shown in Fig. a and again in Fig. C-l. R2 can be expressed as the first
quadrant (horizontally hatched area in Fig. C-l) -Ax
. Thus
P(Cjm2 )= noise originating from s2 lie in R2
= £(noise lie in 1st quadrant) - /’(noise lie in A,)
- j\ _jj
~ £(noise originating from s2 lie in A\
)
But /’(noise tie in Ax ) = j[/>(noise lie within outer square) - £(noise lie within inner square)] (See
Fig. C-2)
119
Page 119
Moreover, by symmetry
P(e\m2 )= P(fh) = *M"*) = P(*K)
Hence
PiC) = i-l4?(C|mi )= i[?(C|m
1 )+ P(Cl«2 )
+ P(Clm3 ) + P(C|m4 )]
l 6 i=l4
The decision region /?, for m, (see Figure) can be expressed as
= outer square of side d>[2-~ (outer square - inner square of side d)
= I outer square of side djl + 1inner square of side </
4120
Page 120
Now />(Cj«j) = Prob(noise originating from mj lies in R
\
)
= - P(n lie in outer square) + ^P(n lie in inner square)
=Mw<£H fK)
Similarly P2 ,the decision region for m2 (see figure above) can be expressed as
R2 = outer square of side d^2 -— (outer square - inner square of side d)
and
2
= — outer square of side dV2 -— inner square of side d
2 2
P(Cjm2 )= noise originating from m2 lie in R2
f)
i['- 2C(^)] 4[- 2e(^)
m. id
- J'. - -
ft
The decision region P3for «3 can be expressed as
R2 = RA + Rb ~ Rc
and
P(C|m3 )= Prob(noise originating from m3 lie in Ry)
= P(noise in RA )+ P{noise in Rg
)- P(noise in Pc)
" \ n ' '
P{n, >0, hM+^N’ l
n2 l
<^)-^[
;a
(ln
i!- I«»2|<^)-'p(ln il*
a*
•rm4
*4-*i
*
r— ‘ l *' .< 1• • I
V' . ad
h££i i
121
tl|<N
Page 121
The decision region *4 for m4 can be expressed as
and
p(C|oi<) - P(»i > -A "2 ’
For any practical scheme g(-)« I ,»! »•« 'xPress
[l-k&)fs\-2kQ()
Using this approximation, we havex
» 1 -^Ji]-
Hence
P(C) = i[p(clmO + J’tcK)* /’(chh ^cK)]
,3 r/ ^ 1 lof4=V-eUF^I
1'2 el^rJ"4elv^j WJ
Now
Therefore
And
so that
Therefore
Moreover
Hence
£, = d2,E2 - 2d2
, £3 * 4* 2-
£< “^ ’
£ = -(d2 +2d2 +4d2 +W2)*— d
4
'
. _I I* log2 16 4
£L =X = li^ljy
=401 16 ^
And
122
Page 122
Contptoon ofto result wfth tobto* »«.3[E,d457))
approximately 1 .5 times the power of the system m Example 14.3 to ach.eve tne sam ^
14.3-9 If jj is transmitted, we have
bi = E+a + jE n, ,6_,=-£ +a-V£ n,
A2= fl = V£n2 .
b_2=a-Je n2
and
Note that
Hence
Similarly
Hence
bk =a + SE n k ,b-k =a-JEnk
p(C\mi )= Prob.(b
x>6_ h, i_2 ,
— bk , b.k )
bx>b.
ximplies £ + a + 4e n, > -£ +a-V£ ni or «!>->/£
*,>*2 implies £ + a + ^£n, >a + Vfn2orn2 <Vf+ni
6, > 6_2 implies £ + a + V£ n,> a - 41 n2 or > -(Vi + " i
)
A, > &2 and 6_2 implies - (n, +4e)< n2 < (n, + Je
)
bx> bk and 6_* implies - (n, + Je) < n^ < (nj + Je)
cncc
p(c]mi ) * Prob.(Z»i > b.\, b2 , 6-2. 63, 6-3 bk , 6_*)
. Prob.[n, > ~j£, |n2 | <(m VI). |nsl < ("I + &)• '•W < ("'* '^)\
Since n„ n2,-nk «« .11 intlependen, gaussian .toon. variables each wi* variance ^2 .
/.(CH,) -[/>(»,> -41) 4^1 < n, *t/EMM < »l *- HKI ‘ 4*
)]
I" J/JrmWr I dhi
. T e~n'/Ji
THaT-Jj
,-4- J•-"M*
[CSV*'*'*
1-2 dnt
JEni +
,w-«-4W- ,
-g5?jj!^ 4'-^)]” '
Alsolog2 2N
14.4-1 The on-off signal se, to iis—ES££5SZ*
«
K5g£»tS2C?£»*•- - F*
«
is-with half the energy of on-off or orthogonal signals.
123
Page 123
Therefore
5[(/) = V20^2(0
s2 (r) = V? *i(r) s2 =>/5<t>1
jj(0--V5#iW *3 = ~Vs <t>
The vector o = j£s, = j[VlO «D1- y/lO *
,+ ^20 4>a ]
— «>2 •
Hence the minimum energy signal set is given by
i,(0 = = ^20
The optimum receiver - a suitable form - in this case would be that shown in Fig. 14.8a or b.
Fig. SI 4.4-2
1 4.4-3 To find the minimum energy set, we have 0 * j (5
l+ *2 + *3 + *4 )
- ~+l ~h
Hence the new minimum energy set is
lj = V3 + ^2 , s2 = + V3 ^2 , s3 = -V3 = ^i-V3^2
124
Page 124
Note that all the four signals form vertices ofa square because (*i s2 ), (*2 h)> (*3 *4 )- and (s4 *0 are
orthogonal. The distance between these signal pairs is always ifl . This set is shown in Fig. SI 4.4 3a.
J1
1
fien^iJtg„
^ V
:
1
£ rj*ii
> V
. /
1
l
XQ. e-
*1\ - ' ^7 *
iifA
P±-1*3£)—*
—
i> 1
t ~0 V<3 53
UJ, ts.
q. . - -O' Fig. S14.4-3
14.4-4
Observing symmetry we obtain
P(C) * /’(Clm,) = - ^(<>3) * W*)* Z^nj > -V2 and nj > -V2
)
HMPtM = 1 - F(C) - 1 -[1 - 0(3.16)f
s 138 x 10-3
i y /
_ 1
lofislOVg
N ,. . * f iO TRe
s,L$ - 3
~~c
_ it
t-
H1
• 4*3^
Z*\b3
-fc-*
(d)T*E- nin.Kv^ ***&/
Fig. SI 4.4-4
125
Page 125
P(C,.|[2 c(cK) fM-,)] - i[2 -jefro?) * i - 2C(7 0,)
]
= 1-jQ(7.07)
PeM =i-P(C)-^Q(7.07) = l03xlO-12
Also £l= £3“(^)= 4x,0
":>
£ -f-j—f+f—LSI = 2 x 10-3 £ = {(£l + £2+£3)=
3X, °
2
£2‘liovroj iToVToJ
3 3
Mean energy of the minimum energy set:
£min4(2 ’‘ 10'3+0+2 ’“°",*‘ :
5’< '0
"5
14.4-5 Tile use of Eq. (14.76) end siqiial rotation shows th“V*' ^^^“c^atfyMt.^rs'uIeMicssl
that in Prob. 14.4-4. Hence the minimum energy set is as shown in Fig.
to that in Prob. 14.3-5 with * -^ From the results in the solution of Prob. 14.3-5, we have
e =y = 10-3
Also, we are given s„(o>) *^ = 105
* Hence ’ ^ * 2 * 10
(a) From the solution of Prob. 14.3-5
PeM = ~6(7.02) + C(7.12)= 1.09 x 10
- *2
(b) and (c> identical to those in Prob. 14.4-4
14.4^ (a) The center of gravity of the signal set is (», + «a)/2
Hence, the minimum energy signal set is
The minimum energy signals are
*,(,)« 0.5-0.707 sin^-j
x2(0 = 0.707 sin ~ 03
a>0 = 2000a
(b) EX]=°
J
>,
^0i -0,707 sin dt * 0.4984 x 10"5
£ - E = 0.4984 x 10_s
. We are given = 5x 10x2 ^1
fl-dsR?- =2(4.465) = 0.41x10,-5
126
Page 126
(c) We use Gram-Schmidt orthogonalization procedure in appendix C to obtain
yi(*)= Jl(0
.001
JV2sina>0'd*
»(<)-•£ •‘""o'--4—SSi
—
, 8 8
y\ (r) * 1 - cos2a>0/-— sin a>0t +-j
N if
«i(0*y
jj(/) = 7J7^(0 + = O138p2(0+ -o285>’l(0
127
Page 127
Chapter 15
15-1.1 P\ = 0.4, Pj - 03, Pi — 0.2 and P4 - 0.1
H(m) = -{Pi log Pi + />2 log P2 + ^*3 log /*3 + /»4 log ft)
=1 .846 bits (source entropy)
There are 10‘ symbols/s. Hence, the rate of infonnation generation is 1*46 x 10* bits/s.
15.1-2 Infotmation/element = log2 10 = 332 bits.
Information/picture frame - 332 x 300,000 = 9.96 x 105
bits.
15.1-3 lnfoimation/word » log2 10000 * 133 bits.
Informetion content of 1000 words - l^' <“^“'i“2tobt 9 ,6x , 0
5 biK . obviously, It is no,
™sSS“n^rPXcome
pT“c'y% 1000 words, in genentl. Hence, • pieuue is worth 1000 wo,ds
iTvery much an underrating or understating the reality.
15.1-4 (a) Both options are equally likely. Hence,
/ = log(Jj) = lbit
(b) P{2 lanterns) = 0.1
1(2 lanterns) = log2 10= 3322 bits
15.1-5 (a) All 27 symbols equiprobable and P(xt )= •
//,(x) = 27(3^ log2 27) = 4.755 bits / symbol
(b) Using the probability table, we compute
ffw(x) = -I^log «*i) = 4127 bits/symbol
(c) Using Zipf s law, we compute entropy/word Hw(x)-
S727
//w(x)=-iwiogwr=l
8727_ _ £Miog(M) = 9.1353 bits / word.
r=l'
H/letter =1 1/82/5.5-2.14 bits/symbol.
Entropy obtained by Zipf s law is much closer to the real value than W,(x) or H2 (x).
128
Page 128
Redundancy y = (100- rj) * 0%
7
15.2-2 //(m) * -£ log Pj = 2289 bits
i=l
=2-2g— = 1.4442 3 - aty units
log2 3
Message Probability Code Sl
nti 1/3 0 1/3 0 1/3
m2 1/3 1 1/3 1 1/3
m 31/9 20 1/9 20“ 1/3
m4 1/9 21 1/9 21
m51/27 220 9. 1/9 22J
m6 1/27 221
m7 1/27 222
J
0
1
2
i-ifiil -j(')4w+i(2)+
i(2)+3F(3)
3-ary digits
* 1.4442 3 -ary digits
Efficiency 77 =tf(m)
=L4442
xlQ0= 100o/o
L 1.4442
Redundancy y =(1-77)100 = 0%
129
Page 129
1 5.2-3 H(m) = -S Ptlog P
t= 1-69 bits
i=l
Message Probability Code
mim2
m3
nu
0.5
0.3
0.1
0.1
0 0.5
1 0 0.3
1 1 0“(-v 0.2
1 *1
J
0.5
51 oT— 0.
llj
0
1
Efficiency rj = x J00 = -jj- x 100 = 992%
Redundancy y = (1 - >7)100 * 0.8%
For ternary coding, we need one dummy message of probability 0. Thus,
Message Probability Code
mim2
m3
m4
m 5
0.5
0.3
0.1
0.1
0
0 0.5
1 0.3
20*1—r 0.2
sT
0
1
2
L = 03(1) + 03(1) + 0.1(2) + 0.1(2) = 12 3 - ary digits
H{m) = 1.69 bits = -—- = 1.0663 3 -ary units
log2 3
Efficiency „ . SSL , 100 -^ x 100 » S8J6%
Redundancy y = (1 - ^)100 = 1 1.14%
15.2-4
Message
m,
m2
m3
numjm»m7
Probability
1/2
1/4
1/8
1/16
1/32
1/64
1/64
l = ZPiLim T2 3-ary digits
lo
From Problem 15.2-1, Him) =^ bit!
-
1242 3-aiyuniB
Efficiency 7 . * 100- 100- 94«%
Redundancy y - (1 - »7)100 = 5-37%
130
Page 130
m,m2
m3
m5
m«in?
1/9
1/9
1/27
1/27
1/27
Si
"1/3 1
1/3 00
1/9 Oil
1/9 0100
1/27 01010“
0101 lOT-* 2/27 01011
OlOlllJ
63
%1/3 1 1/3
1/3 00 1/3
1/9 Oil f»2/9010 1/3 01J
1/9 0100*1 1 1/9 Oil
1/9 0101
J
l “ £ ^ “ 2 4074 binstfy digits
//(m) = 2289 bits (See Problem 152-2).
Efficiency „ = ^*100 = xlOO = 95.08%
Redundancy y =(1-t?)100 = 4.92%
15.2-6 (a) H(m) = 3(^ log 3)— 1385 bits
(b) Ternary Code
Message Probability Code
m,m 2
m3
1/3
1/3
1/3
0
1
2
I = I(l) + i(l) + -j(l) = l 3 -ary digits
1 <85H(m) * 1.585 bits = - = 1 3-ary un;t
log2 3
Efficiency n - ~~p * 100 = ,00°/#
Redundancy y =(1-^)1 00 = 0%
(c) Binary Code
Message Probability Code Si
m,
m2
m3
1/3
1/3
1/3
0001
2/3
1/3
l = j(l) + (2) j(2)- 1-667 binary digits
Efficiency tj = x 100 = x 1 00 = 95.08%
Redundancy y *=(1-/7)100 = 4.92%
131
Page 131
(d) Second extension - binary code
•
^[(7{03)+(2)
(?)4)
] =
f
= 161 1 binary d5gits
H(m) = 1385 bits
Efficiency tj = —7-^ x 100 = x 100 = 98.39%L 1.6 1
1
Redundancy y * (1 - >7)100 = 1.61%
Message
mimi
ProbTT Code
001 *2/9
S|
01 TT»»
Si
1/9 0000 1/9 001 10
mim) 1/9 0001 1/9 0000 1/9 001
1/9 110 1/9 0001 1/9 0000
lUjinj 1/9 111 1/9 110 1/9 0001
mimi 1/9 100 1/9 111 1/9 no"
ntiiot
IBjIll;
TflylTlj
1/9
1/9
1/9
101
010-1
01 1J
1/9
1/9
100-1
I01_P1/9
2/9
001 f*2/9>1 r»2
I I 1/9 oooonj 1
nj 1/9 0001
15.4-1 (a) The channel matrix can be represented as shown in Fig. S15.4-1
P{y\) • P(y\\x\)P(.xx )+ P(y\\x2 )P(x2 )
(b)
_ 2 jl
+ J_ 2 =13
=3 3
+10 3 45
P{yi)^\- P{y\)-^
H{x) = />(*,) U>g^TT + ^JlogF(x,)
’ F(x2 )
1 2 3= — logj 3 + — logj — = 0.9 1 8 bits
P- 2-r 3
Fig. S15
To compute //(x|y) , we find
FO-,) 13
rto)-u-
PWn)p{yi) 32
F( 2,W)/>0'2 )
64
//(x|y,) = F(x,|yi)»oi + ^(X2l>’l) ,081
^X2 lFl)
10, 13 3 . 13ft_70
:— tog2— +— log2— = 0.779
1382
10 1362
3
H(x\y2 ) = F(xi|>|)log1
+ P(x2 \y2)\og1
P(x2 \y2 )
5, 32 54. 64
132
Page 132
and
Thus,
Also,
H(x|y) « Piyx)H(x\yO+ P(yi)H{x\yi )
- 11(0.779) + 11(0.624) = 0.6687
/(x|y) = H(x) - W(x|y) = 0.91 8 -0.6687 = 024893 bits/ binit
H(y)»Z^) |o8/
1
Piyd
13 . 45 32
45°8
TT « *45
32= 0.8673 bits /symbol
//(y| X) = tf(y)-/(x|y) = 0J8673 - 02493 * 0.618 bits / symbol
-rn. _ .Lo-nal matriv p( 1J ,lr \ IKi. y / P)
l nc buoiui^i uibuia » vjnt -
'10 0
0 p l-p
0 l-p p
(p) *1 • 7\
'
y
)
f<3>j
yAlso, P(yi) = p, P(P2) P(yi) = Q
(<5) x3^- >3 '
P(yj\x,)P(xi) . •
Now we use “ obl,“ Fig. SI 5.4-2
P(x,\y,) matrix as y}
xi10 0
0 p l-p
|_0 1 -p p
1
//(x) = I /*(jc1)log—1— = -/’log P - 2gloge with (20 = 1- P)
syXj
)
P log P + (l-P) log(¥)] n(p)+o-p)
i
//(x|y) = II P(yj)P(xi\yj)'<>& P{x ,
lyj)• j
/> log 1
+^ log-^+ ( 1 - P) log
J
+ ^jd - P) !og— + p log-
O+20£2(p) = (l-P)n(P)
/(x|y) = H(x) - H(x|y) = fl(P)+(l - P)-0 -TO= fi(P) + (l-P)[l-n(P>3
Letting p = 20**0 or £2(p) = \o%P
/(x|y) = n(/’)+(l-/>Xl-log^)
— /(xiy)-O or 4[n</>
> + < l - /’>< 1 - ,oS^ = 0 - 71,151,168,15
4Pv dP
iL[p log P + (1 - P) -0 - P)M1 - P)(1 * log#] = 0
log P - log(l - P) + [1 - »og/?] = 0
133
Page 133
Therefore\og— = -^°lP
PNote: -1 + log2 fi
—^2 2 + l°g2 P 9 lo^ 1
P P_ p = P— and 1 - P - ~—
r
JI7 T =*f + 2 P+2
so
C= MAT/(xjy)to,!
(00)
=
0 - f|« AO-V * ^ _3fifi
fa)
BSta ">k. M M > —y « 'K-lFig. S15.4-3
-ft*-B^CL
*
a;
n-A a' W2 Pi
rU Pi \ -Pi.
Hence, the channel matrix of the cascade is
\l-Px-Pl-2P\Pi P\ + Pi ~2P[Pl
[p, + P2 -2P1P 1-A“^"2^P
This result will prove everything in this problem.
, , with , . bom the above result it follows that the channel matrix *s^**~^
(a) With P\ Pi e>oftwo cascaded channels is M\Mi-
(b) We have already shown thal the c
,with the H* channel.
ret Consider a c^cade of . identica'^h^en^^-JtT^lhs denvcd in (b).
if «... ,s the channel m.mx of the fim * - chenneU ^ ^^ for^ * Wehjye
-*——*7 ‘^.7^ -— « --- «*r«;•
have, for a cascade of 3 channels
\-Pe .Q-Pti* + 3P'2Q-P')
= l-3Pe+ 3?e2 -Pe
3 +3/>r2 -3/»
e3
*1-3P,+*P?-*P?
we
and
/>£ =3Pe -6/>e2 +4/>
e
3
Now
134
Page 134
3 fl -Pe Pt l3 [l-(3Pe -6/»e
2+4/i
3) 3Pe -6Pe
2 +4P,3
M"U J-^J
=[3P# -6Pe
2 +4P<3 l-(3Pe -6Pe
2 +4Pe3).
Clearly
P£ =3P,-6P,2 +4P, 3
which confirms the results in Example 10.7 for * = 3.
^Log-^a-Zilloe^j
(d) From Equation 15.25
C, = l-
wbere PE is the error probability of cascade of k identical channel
We have shown in Example 10.7 that
Pe- 1-
If kPe « 1, PE s kPe
and
C, = 1-\tft log^+0 - ) log
15.4-4 The channel matrix is
>1 >2 >3
X, 9 0 P
Xi0 4 P
<?= 1-P
Let
x, « 0 yi=0
x2 = 1 yi = 1
>3 = £
Also,
’(>1) » P(Xl,>l)+ P(x2 ,>i) - P(x1)P(>i!x,)+ P(x2 )PO-i|x2 )
[y2)‘ P(x,,>2 ) + P(x2 .>2 ) = P(xi)P(>2|x,)+P(x2)PCy2 |x2 )
P(y3 )= 1- P(>i)- PCP2)* 1- 9 * P
«ala)£££i>.i£.i'W,)
/-O',) «/2
'™w)—twT p ^
'!lW) P(«) P 2
135
Page 135
15 .4-5
Therefore,
P{xl ,yi )= P(x\)P(yi\xi) = ^
P(xuy2)= P(x\)P(y2\*\) = °
P(xi,y2 ) = P(x{)P(y-}\x\) = |-
P(x2 >yi) ~ P(.x2 )P(y\\x
‘i )= 0
P{x2 ,yi) = />(*2)/>0'2lx2)=2
P(x2 ,yy) = />(jt2> />0'3lx2) = f
W(x) = -P(*,)l0g />(*!>- /’(*2) ,°8^X2)
~ui-i2 2
W(x|y) = SI /’(*/.>';) log
’
= i(0) + 0 + ^p + 0 +^x0 +^ = P2 2 2 2
/(x|y) = //(x)-Wx|y)
= 1-p bits/ symbols
H(x| z) - W(xly) = 11 P(x, , zk ) log—~ -11 } log
/>(x,|y7 )
Note that for cascaded channel, the output z depends only on y. Therefore, X J
By Bayes’ rule. Fi_ si 5.4-5
P(x,\yj,ik ) - /’(x/l^) r,g> MS-
„ Wry**)//(x|z)- //(x|y) =
/»(jc(|x*
)
r_ .. P(*i\yj»*k)
y x L x
„„ b. shown that the sunonaion ove, , of the «m> Inside the bn*ke, is nonnegative. Hen.., it Mows
**//(x|z)-W(x|y)2 0
From the relationship for /(xly)and /(x|z), it immediately follows that
/(x|y) £ /(x|z)
136
Page 136
15.5-1
15.5-2
We have H(x) = J_^plog-Ur = M ~p\ogpdx and j Mpdx = 1
Thus,
dFF{x,p) = -p\ogp and— = -(1 + log/))
dp
d*4\(x,p) = /) and -r1 = 1
dp
Substituting these quantities in Equation 15.37, we have
-(1 + log /?) + «! =0^ p = ea
'1
and
Hence,
Also,
*r''TS
m* r0,)mTii
We have H(x) = -J" p logpdx, A =J" x/wfr, 1 = J*
dFF(x, /?)
= -/> log /> and — = -(l + log/>)
dp
di\h(x,p) =pxmd— = x
di-, ,
h (x,/>) = pand— = 1
Substituting these quantities in Equation 15.37, we have
-(l + log/») + aiX + a2 =0
p = ea =(e
ar2- 1
)eflr
»jr
Substituting this relationship in earlier constraints, we get
,a2 -l
or
Hence,
ea2~ l = -a.
and
Hence,
so
i/Kx) = -a,ea|Jt
a\ -—- and e®2”
1 = -«i = A
p(x) =1/^ x £ 0
0 x< 0
To obtain //(x)
137
Page 137
H(x) = -J" p(x) logp(x)dx = -J*
/jj^-log^-- loge] dx
= log /if® P(x)& + ^ifj*xp(x)A
* log >< + loge = log(e/l)
15 5.3 Information pm picture f*m« - 9.96 x 10s
bits. (Sec Problem 15 . 1 ). For 30 pientre frames per second, we
need a channel with capacity C given by
C = 30x9.96* 10s = 2.988 x 10
7bits/sec.
But for a white Gaussian noise
C = fllog
We are given — = 50 db = 100,000 (Note: 100,000 = 50 db)
N
K)
Hence,B = 1.8 MHz
15.5-1 Consider a narrowband *4 where Af -+ 0 so that we may consider both signal noise power density to be
constant (bandlimited white) over the interval A/" . The signal and noise power Sand N respectively are
given by
S-2Ss(ia)£4 and N = 2S„(<w)A/
rs + A,1.Anc»S,(a)) + Sn (fli)
[tJ Sn (*>) J
The capacity of the channel over the entire band (/i,/2 ) is given by
n(») .
We now wish to maximize C where the constraint is that the signal power is constant.
f/2
Vi[
5n (fl
<4
2$ Ss(a>) df -S (a constant)
Using Equation 15.37, we obtain
or
Thus,
A log[^Uf^oas, [
s„ Jas.
s + S„ = -— (a constant)s a
Ss(<u) + S„(a>) = ~^
This shows that to attain the maximum channel capacity, the signal power density + noise power density
must be a constant (white). Under this condition,
C= f/2 l0g[W^ni£0l# = |/2|og y-rWC
Vi8
[Sn (ffl) J
j/lL
aSn (tu)J
= (/2 -/i)log (-')'^ log
138
Page 138
15.5-5
* Blog [5s(a) + Sn(®)]-J/f ,06 [5n(©)]#
In this problem, we use the results of Problem 15.5-4. Under the best possible conditions,
C = Blog [St(a>) + Sn (<u)] - \}}\
log [Sn (<w)]
constant
We shall now show that the integral § log [Sn (a>)] df is maximum when $„(*) = constant if the noise
is constrained to have a given mean square value (power). Thus, we wish to maximize
$ log [S„(®)]tf
under the constraint
2jflog [Sn (<u)]4f
= N (a constant)
Using Equation 15.37, we have
-r“(l°B‘Sn ) + flf—- = 0dS, 3Sn
or
and
— + a = 0
S„(<o) *— (a constant)
a
Thus, we have shown that for a noise with a given power, the integral
log [Sn(co)]df
is maximized when the noise is white. This shows that white Guassian noise is the worst possible kind of
noise.
139
Page 139
I Chapter 16
(?)-(?)+(?)(?)*(?)J=0
2048*1 + 23 + 23x11 + 23x77 = 204816.1-
2 (a) There are(})
ways in which j positions can be chosen from n . But for a temery code, a digit can
be mistaken for two other digits. Hence the number of possible errors in j places is
("V3-1)' or 3" * 3* I ("ft*- 3"-* * f (j)2
y
'' j*0 j*
0
(b) (1 1,6) code for t = 2
35
^(y) + (!
1
)2+(y)22 = 1 + 22+220 = 243
This is satisfied exactly.
16.1-
3 For (1 8,7) code to correct up to 3 errors
Jm0
« 1 + 1®: + -I®-- + = 1 + 18 + 153 + 816 = 98817! 2! 16! 3! 15!
211 =2048
Hence
7=0
Clearly, there exists a possibility of 3 error correcting (18,7) code. Since the Hamming bound is
oversatisfied, this code could correct some 4 error patterns in addition to all patterns with up to 3 errors.
[>
16.2-
1 GH t = [/* P]L/m.
= p©p-0
16.2-
2 c — dG where d is a single digit (0 or l)
.
For d = 0
c = 0 [l 1 1] = [000]
For d = 1
c = l[l 1 1]-[1 1 1]
140
Page 140
16.2-3
16.2-4
16.2-5
c = dG where d is a single digit (0 or l)
.
For d = 0
c = 0 [l 1 1 1 1] = [0 0 0 0 0]
For d = 1
c = 1 [1 1 1 1 1] = [1 1 1 11]
Hence in this code a digit repeats 5 times,
detection.
Such a code can correct up to two errors using majority rule for
0 is transmitted by [0 0 0] and 1 is transmitted by [l 11]
(a) This is clearly a systematic code with
10 0- • • 0 f T0 10- • 0 1
p =1
000 •
_jn
i
/*
Note that m = 1
(c) This is a parity check rode, if a single am,, ocean »ywhara in ihe coda woni. ihc pariiy is violated.
Therefore this code can detect a single error.
T(d) Equation (16.9a) in the text shows that cH = 0
.
Nowr~c@e
and
rH T =*(c®e)HT =cH t 9eHT *eH T
If there is no error e * 0 and
rH T = eH T =0Also
-E]-But since m * 1, lm - [l]
and
1
1
HT =
1
1
141
Page 141
ll<km i, a single error in the received word r, e has a single 1 elenrnn. wid, all other elemenrs being 0.
Hence
rH T =eH T = 1 (for single error)
16.2-6
Data word Code word
0 0 0 o 0 0 0 0 0
0 0 1 1 1 0 0 0 1
0 1 o 1110 100 1 1 0 0 10 111 0 0 0 1110 1
1 0 1 10 110 0
1 1 o 10 0 1111 1 1 0 10 110
From this code we see the. die dismnee between any two code words is « least 3. Hence = 3
16.2-7
16 .2-8
Data word Code word
0 0 0 "0 6 0 0 o o
0 0 1 0 0 1 1 1 0
0 1 o 0 1 o 1 o 1
0 1 1 0 110 1110 0 1 0 0 0 1 1
1 0 1 10 110 1
1 1 o 110 1101 1 1 1110 0 0
Observe that dm
m
= 3
T *
H T « a 1 5 x 4 matrix with all distinct rows. One possible H is:
HT
ini 1
1110
110 1
1 1 oo
ion10 10
1001
00 1 1
0 111
ono0 101
1000
0 1 00
00 10
000 1
142
Page 142
G = [/*J*] =
'l 00000000001 1 1 1
010000000001 110
001000000001 101
000100000001 100
0000 1000000 1 01 1
0000010000010100000001 000010010000000 1000001
1
0000000010001 1
1
000000000 100110
000000000010101
For 4 = 10111010101c = dG = [10111010101] <7-101110101011110
16.2-9 (a)
1 1 r
1 1 0100 111
T 1 0 1
0 10 110 & H t =1 00
00 1 1 0 1
Ik P.
0 1 0
00 1
(b)
Data word 1Code word
0 0 0 0 0 0 0 0 0
0 0 1 0 0 1 1 0 1
0 1 0 0 1 0 1 1 0
0 1 1 0 1 1 0 1 1
1 0 0 1 0 0 1 1 1
1 0 1 1 0 1 0 1 0
1 1 0 1 1 0 0 0 1
1 1 1 1 1 1 1 0 0
(c) The minimum distance between any two code words is 3. Hence, this is a single error correcting code.
Since there are 6 single errors and 7 syndromes, we can correct all single errors and one double error.
(d)
e
1 0 0
0 1 0
0 0 1
0 0 0
0 0 0
0 0 0
1 0 0
0000 0 0
0 0 0
1 0 0
0 1 0
0 0 1
1 0 0
s
1 1 1
1 1 o
1 0 1
1 0 0
0 1 0
0 0 1
0 1 1
s = eHT
143
Page 143
(e)
r s *
101100
000110101010
110
110
000
c d
010000 111100 111
010000 010110 010
000000 101010 101
16.2-10 (a) done in Prob. 16.2-7
(b) HT =
0 1 f
1 0 1
1 1 0
100
0 1 0
00 1
six single errors
1 double error
e s
100000 Oil
010000 101
001000 110
000100 100
000010 010
000001 001
100100 111
16.2-11
fl 000 1 0 l
<?=['* p]=0 100111
00 100 11
jo 0 0 1 1 1 o
c “ dG
144
Page 144
d c
0000 0000000
0001 00011100010 0010011
0011 0011101
0100 0100111 101
0101 0101001 1 1
1
0110 0110100 0 1 1
0111 0111010 ht = 1 1 0
1000 1000101 100
1001 1001011 010
1010 1010110 00 1
1011 1011000
1 100 1100010
1101 1101100
1110 1 1 10001
nil 1111111
s = eH T
e 5
0000001 00 1
0000010 010
0000100 100
0001000 1 1 0
0010000 Oil
0100000 1 1
1
1000000 10!
s-rH T where r = received code
c = r®,
c = corrected code
16.2-12 We observe that the syndrome for all the three 2-error patterns 100010. 010100 or 001001 have the same
syndrome namely 111 Since the decoding table specifies , = 111 for , = 100010 whenever^ 100010
occurs, it will be corrected. The other two patients will not be corrected. If for exampte, -0101
occurs ,**111 and we shall read from the decoding table e = 100010 and the error is not corrected.
Ifwe wish to correct the 2-error pattern 010100 (along with six single error patterns), the new decoding
table is identical to that in Table 16.3 except for the last entry which should be
e s
010100 111
16.2-13 From Eq. on P.737, for a simple error correcting code
2""**n + l or 2""*i« + l -*• n-85>log2(«+l)
This is satisfied for « ;> 12 . Choose n = 12 . This gives a (12, 8) code. H Tis chosen to have 12 distinct
rows of four elements with the last 4 rows forming an identity matrix. Hence,
145
Page 145
90 1 1,
0 10 1
0 1100 1111 00 1
101010 11
1 10010000 1 0000 10000 1
G = [I*P]
G -
1000000000 1
1
01000000010100 10000001 1
0
00010000011100001000100100000100101000000010101100000001 1 100
Tile number ofnon-zero syndromes =16-1 = 15. mere ere 12 single error paneros. Hence we may be
able to correct 3 double-error patterns.
£ -
0000 oooooooooooo
0011 100000000000
0101 01 0000000000
0110 001000000000
0111 000100000000
1001 000010000000
1010 000001000000
1011 0000001 00000
1100 000000010000
1000 000000001000
0100 000000000100
0010 000000000010
0001 000000000001
nil 100000010000
1110 001000001000
1101 000000010001
16.2-14
Data word Code word
00 00000001 01101110 1011101 1 110101
The minimum distance between any two code words is dmi„ = 4 . Therefore, it can
patterns. Since the code oversatisfies Hamming bound it can also correct some 2-error
3-error patterns.
correct all 1 -error
and possibly some
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90 1 1,
0 10 1
0 1100 1111 00 1
101010 11
1 10010000 1 0000 10000 1
G = [I*P]
G -
1000000000 1
1
01000000010100 10000001 1
0
00010000011100001000100100000100101000000010101100000001 1 100
Tile number ofnon-zero syndromes =16-1 = 15. mere ere 12 single error paneros. Hence we may be
able to correct 3 double-error patterns.
£ -
0000 oooooooooooo
0011 100000000000
0101 01 0000000000
0110 001000000000
0111 000100000000
1001 000010000000
1010 000001000000
1011 0000001 00000
1100 000000010000
1000 000000001000
0100 000000000100
0010 000000000010
0001 000000000001
nil 100000010000
1110 001000001000
1101 000000010001
16.2-14
Data word Code word
00 00000001 01101110 1011101 1 110101
The minimum distance between any two code words is dmi„ = 4 . Therefore, it can
patterns. Since the code oversatisfies Hamming bound it can also correct some 2-error
3-error patterns.
correct all 1 -error
and possibly some
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(b)
i/r =
1 1 1 0‘
1011
1000
0100
00 1 0
0001.
and s = eH T
6 single error patterns
7 double-error patterns
2 triple-error patterns {
e s
100000 1110010000 1011001000 1000000100 0100000010 0010000001 00011 1 0000 0101101000 0110100100 1010100010 1 1 00100001 1 1 1
1
011000 0011010010 10010001 1 1 0111001101 1101
16.3-1 Systematic (7, 4) cyclic code
g(x) = x3 +x + l
For data 1111 d{x) = x3 + x
2 + x + l
x3(x
3 + X2 + X + l) = X
6 + xs +x4 + x
3
x3 +x2 +
1
X3 + x + l)x
6 +x5 +x4 +x3~~
' - 4 1
+ x +X
x3 +x
2
2
X3 +X + l
x2 + X + 1
c(x) = (x3 +x + l)(x
3 + x + l) = x6 + x
5 + x4 + x
3 +x2 +x + l
The code word is 11111111
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For data 1110 </(*) = x3 + x
2 +x
x3 + x2
x3 + x + ljx6 +xs +x4
r6 +X4 +X 3
+ XJ
+ x 3 + x2
The code word is 1110100 nA similar procedure is used to find the remaining codes (see Table )
(b) From Table 1 it can be seen that the minimum distance between any two codes is 3.
single error correcting code.
d c
1 1 1 1 11111111110 11101001101 1 1010011100 1 1000101011 10 110001010 10100111001 10011101000 10001010111 01110100110 01 100010101 01011000100 01001110011 00111010010 00101100001 000101 1
0000 0000000
Table 1
(c) There are seven possible non-zero syndromes.
x3 + x + l
for * = 1000000 x3 + x + ljx*
x6 + x4 +x3
x4 + x3
X4 +x2
x3 +x2
+ x
+ x
+ X + 1
+ 1
5-101
Hence this is a
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The remaining syndromes are shown in Table 2.
e s
1000000 1010100000 1110010000 1100001000 Oil0000100 1000000010 0100000001 001
Table 2
(d) The received data is 1101100
r(x) = x6 + x
5 +x3 +x 2
x 3 + x + l]x6 + x
3 + x3 + x
2
+ x4 + x
3
Xs + x
4 + x/
+ x3 + x2
x4 +x 3
+ x2 +x
X3 +x2 +x
s(x) = x2 + 1 X3
S = 1 0 1
From Table 2
,=1000000c=r® *= 1101100© 1000000- 0101100
Hence d - 0 1 0 1
16.3-2 g(x) = x" +x9 +x 7 +x6 +x 5 +x + l
c(x) = d(x)g(x)
x + 1
+ 1
and
and
<f, = 00001 11 10000, </,(x) = x7 +x6 +x5 + x
4
d2 = 101010101010,</2 (x) =
x
11 +x9 +x7 +x5 +x3 +x
C,(x) = rf,(x)*(x) = x18 +x 17 +X 13 +x 12 +Xn +x9 +X8 +x7 +X4
C =00001100011101110010000
c2 (x) = d2(x)g(x)= x32 + x
l' + x ' 7 +x l5 + x ,3 + x* + x5 + x4 + x3 + x2
c2=10001101010000100111110
+ x
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16.3-3
16.3-4
16.3-4
x2 +1
x + l')jc3+Jt
2 +x + l
x + 1
x + 1
0
Hence x3 + x
2 + x + 1 = (x + l)(x2 +
1)- (* + ty* + + 1M* +
Hence xs + x
4 +x2 + 1 = (x + l)(x4 +x + l)
x2 + 1
x + 1
x + 1
0
Now dividing s4 +x+l by x+ 1 , we jet a remainder 1. Hence (*+t) is not a factor of (x4 +Jt+ *)'
Hie 2"d-order prime factors nol divisible by a. lure x! andi
2 +a + l Since (x4 * x +
1)is not divisible
by x’.w, tty dividing by (x’-x-l). TO also yields, remainder I . Hence x4+ xtl doesnothave
either a first or a second order factor. This means it cannot have a third order factor erther. Hence
X5 + X
4 + X2 + 1 “ (x + l)(x
4 + X + l)
rx4 +x + l
x + ljx5 +x4 +x2 +l
Xs +X4
Try dividing x7 + l by x + 1
: +1
JV2 +
1
x6 +xS +x4 +x 3 +x2 +x + l
xi+xi.
x6 + l
x6 + x
5
X5 + l
xs + x4
x4 + l
x4 +x 3
x3 + 1
x3 +xi
x2 + l
x2 +x
x + 1
X + 1
0
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Therefore (x7 +
1)= (x + l)(x
6 + xs + x4 + x3 + x2 + x +1)
Now try dividing (x6 + x 5 + x
4 + x3 + x
2 x+1)
by (x + 1) . It does not divide. So try dividing by
(x2 + 1) . It does not divide. Try dividing by (x
2 + x + 1) . It does not divide. Next try dividing by
(x3 + lj . It does not divide either. Now try dividing by (x
3 + x + 1) . It divides. We find
(x6 + X5 + X
4 + x3 + x2 + X +1)= (x
3 + x + iXx3 + X
2 + l)
Since (x3 + x2 +
1)is not divisible by x or x + 1 (the only two first-order prime factors), it must be a
third-order prime factor. Hence
x7 + 1 - (x + l)(x
2 + X + l\x2 + X
2 + l)
16.3-6 For a single error correcting (7, 4) cyclic code with a generator polynomial
g(x) = x3 +x2 +1
A=4 n =
7
Hence
'l 1 0 1 0 0 O'
_0110100
c “ 00110100001101
Each code word is found by matrix multiplication c » dG’
c * [0 0 0 0]
c = [0 0 0 l]
1 101000'
011010000110100001101
1101000'
011010000110100001101
0000000
0001101
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The remaining codes are found in a similar manner. See table below.
d c
0000 00000000001 00011010010 00110100011 00101110100 01101000101 01110010110 01011100111 01000011000 11010001001 11001011010 11100101011 11111111 100 10111001101 10100011110 10001 10
1111 1001011
16.3-7 g(x)^x 3 +x2 *
1
The desired form is
G’ =
1 000 • • 0hn h2 \hi i •*m,
0 100- • o *12 *22 *32• •
*FB2
0010- • • 0 *13 *23 *33'
0000 •
_J *u *2* *3k •
/* r
(kxk) [kxm)
The code is found by using c * dG
Proceeding with matrix multiplication, and noting that
0 + 0-0, 0 + l = l + 0 = l, l + l-o and 0x0-0, Oxl-lxO
we get
c15 =[1111]
10001100100011ooioi 1
1
0001 101
=[1111111]
C14 =[U10]C=[1110010]
and so on.
i, 1x1 = 1
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d c
1111 11111111110 11100101101 11010001100 11001011011 10111001010 10100011001 10010111000 10001 1 o
0111 01110010110 01101000101 01011100100 0100011001 1 00110100010 00101110001 00011010000 0000000
These results agree with those of Table 16.5
16.3-8 (a)
*101 1 0 0 O'
0101100G -
0010110000 1 0 11
(b) The code is found by matrix multiplication. dG'
In general gix)= S\x
" *+ 82x
" * 1+ ’"8»-*+i
For this case gj = 1. g2 * •. S3= 8* ~ 1
Since A,* -ft, ^ =g3 . *» -ft. the fourth row is immediately found. Thus, so far we have
(7 =
0001 1 0 1
Next, to get row 3, use row 4 with one left shift.
00110100001 101
The i is eliminated by adding row 4 to row 3.
00101110001101
Next for row 2, use row 3 with 1 left shift.
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01011100010111000 1 10 1
The 1 is eliminated by adding row 4 to row 2.
010001
1
00101110001101
Next for row 1, use row 2 with 1 left shift.
’10 0 01 10*
0 1000 1 1
0010111000 1101
This is the desired form.
c d
0000 00000000001 00010110010 0010110001 1 00111010100 01011000101 01001110110 01110100111 01100011000 10110001001 10100111010 10011101011 10001011 100 11101001101 11111111110 11000101 1 1
1
1101001
(c) All code words are at a minimum distance of 3 units. Hence this is a single error correcting code.
16.3-9 g(x) = x3 + x + 1 . Hence row 4 is 0 0 0 1 0 1 1.
G'
'1 0 1 1000*
010110000101100001011
Row 4 is ok.
Row 3 is left shift ofrow 4.
For row 2, left shift row 3.
And add row 1 to obtain row 2.
For row 1 , left shift row 2.
And add row 1 to obtain row 1
.
0 0 0 1 0 1 1 «- row 4
0 0 10110 «- row 3
01011000 100111 4- row 2
10011101 000 1 0 1 4- row 1
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1 0 0 0 10 1
0 100111
00101100001011
16.4-1
16.5-1
The burst (of length 5) detection ability is obvious. The single error correcting i
as follows^ If in any segment of b digits a single error occurs, it will violate the parity in that segment.
H^ewe'l<>«M*«8«^*n,w*.ere,^ e,Tor
augmented segment. By checking which bit in the augmented segment violates die parity, we can
die wrong bit position exactly.
The code can correct any 3 bursts of length 10 or less. It can also correct any 3 random errors in each code
word.
16.7-1 PEu = kQ(pEb /Ji) = 120^2x9.12) = 9525 x 10"6
4-4
»(23)[e(V9lT65)] = 9.872 xlO"9
To achieve a value 9.872 x 10-9
for PEu .we need new value Eb/Ji say E'b !^ Thcn
This means Eb /Ji must be increased from 9.12 to 18.18 (nearly doubled).
Pec =(2kE±
nJ!
155