2-1 Chapter 2 2.1-1 0 0 0 /2 2 ( ) /2 0 sinc( ) 0 otherwise j j T j m n ft j n T Ae n m Ae c e dt Ae m n T φ φ π φ − − ⎧ = = = − = ⎨ ⎩ ∫ 2.1-2 0 0 0 0 /4 /2 0 /4 0 0 0 () 0 2 2 2 2 cos ( ) cos sin 2 T T n T c vt nt nt A n c A dt A dt T T T n π π π π = = + − = ∫ ∫ n 0 1 2 3 4 5 6 7 n c 0 2 / A π 0 2 /3 A π 0 2 /5 A π 0 2 /7 A π arg n c 0 180 ± ° 0 180 ± ° 2.1-3 0 0 /2 2 0 0 0 0 () /2 2 2 2 cos sin (cos 1) ( ) T n c vt A At nt A A c A dt n n T T T n n π π π π π = = ⎛ ⎞ = − = − − ⎜ ⎟ ⎝ ⎠ ∫ n 0 1 2 3 4 5 6 n c 0.5A 0.2A 0 0.02A 0 0.01A 0 arg n c 0 0 0 0 2.1-4 0 /2 0 0 0 0 2 2 cos 0 T t c A T T π = = ∫
23
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sm-ch02 - testbankcollege.eutestbankcollege.eu/sample/Solution-Manual-Communication-Systems-5...2-2 ()() [] 0 0 /2 /2 00 0 0000 0 00 222 2sin 2 / sin 2 / cos cos 4( )/ 4( )/ /2 1 sinc(1
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Square wave has odd harmonics, and the amplitudes are given in Table T.2 as 2
nj Ac
nπ= − and the coefficients trigonometric Fourier series is 2 nc . The power in a
periodic sine wave is 2
sinewave 2mAP = then the sum of the powers of the odd harmonics are
2 2
harmonics 2 20 odd
2 16 1 1 11 ...2 2 9 25
Nn
n
c APnπ=
⎛ ⎞= = + + +⎜ ⎟
⎝ ⎠∑
2
2 2oddodd
2 2 2total odd
16 1 1 11 ...2 9 25 8 1 1 190% 0.9 1 ...
9 25
AnP
P A nπ
π
⎛ ⎞+ + +⎜ ⎟ ⎛ ⎞⎝ ⎠⇒ ⇒ = = = + + +⎜ ⎟
⎝ ⎠
3n⇒ = i.e. having the first and 3rd harmonic will represent 90% of the signal’s power and 27n = will represent 99% (98.553) of the signal’s power. 2.1-9
y t v w t v w t AB a b e e AB a b e eB j b j B j b j
π ππ
π π
− −− −
− −− −
= = = − = +
= ∗ + ∗ = − − + − −= = − = − =
2.4-10
( ) ( ) ( ) let
( ) ( ) ( ) ( ) ( )
v w t v w t d t
v t w d w v t d w v t
λ λ λ μ λ
μ μ μ μ μ μ
∞
−∞
−∞ ∞
∞ −∞
∗ = − = −
= − − = − = ∗
∫∫ ∫
2.4-11
Let ( ) ( ) ( ) where ( ) ( ), ( ) ( )y t v w t d v t v t w t w tλ λ λ∞
−∞= − − = − =∫
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
y t v w t d v w t d
v w t d v w t d y t
λ λ λ λ λ λ
μ μ μ μ μ μ
∞ ∞
−∞ −∞
∞ ∞
−∞ −∞
− = − − = +
= − − − = − =
∫ ∫∫ ∫
2.4-12
Let ( ) ( ) ( ) where ( ) ( ), ( ) ( )y t v w t d v t v t w t w tλ λ λ∞
−∞= − − = − − = −∫
2-16
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
y t v w t d v w t d
v w t d v w t d y t
λ λ λ λ λ λ
μ μ μ μ μ μ
∞ ∞
−∞ −∞
∞ ∞
−∞ −∞
− = − − = − +
= − − = − =
∫ ∫∫ ∫
2.4-13
0 / 2 2 2
/ 2 0
2
/ 2
Let ( ) ( ) ( / )3( ) ( ) ( ) 0 / 24
1 3( ) / 2 3 / 22 2
t
t
t
w t v v t t
v w t d d t t
d t t
τ
τ
τ
τ
τ τ
τ λ λ τ λ λ τ τ
τ λ λ τ τ τ
+
−
−
= ∗ = Λ
∗ = + + − = − ≤ <
⎛ ⎞= − = − ≤ <⎜ ⎟⎝ ⎠
∫ ∫
∫
2 2
2
3 / 24
1 3Thus ( ) / 2 3 / 22 2
0 3 / 2
t t
v v v t t t
t
τ τ
τ τ τ
τ
⎧ − <⎪⎪⎪ ⎛ ⎞∗ ∗ = − ≤ <⎨ ⎜ ⎟
⎝ ⎠⎪⎪ ≥⎪⎩
2.4-14 { } [ ] [ ]( ) [ ( ) ( )] ( ) ( ) ( ) ( ) ( ) ( )v t w t z t V f W f Z f V f W f Z f∗ ∗ = =F
{ }1so ( ) [ ( ) ( )] [ ( ) ( )] ( ) [ ( ) ( )] ( )v t w t z t V f W f Z f v t w t z t−∗ ∗ = = ∗ ∗F 2.4-15
1( ) ( ) 4 (2 )4 4
( ) ( ) ( ) (2 ) ( ) (1/ 2)sinc( / 2)
fV f W f f
Y f V f W f f y t t
⎛ ⎞= Π = Π⎜ ⎟⎝ ⎠
= = Π ↔ =
2-17
2.5-1
( ) cos ( ) sinc( ) sinc( )2 2
As 0 the cosine pulse ( ) gets narrower and narrower while maintaining height A.This is not the same as an impulse since the area under the cur
c c ct A Az t A t Z f f f f f
z t
τ τω τ ττ
τ
⎛ ⎞= Π = − + +⎜ ⎟⎝ ⎠
→ve is also getting smaller.
As 0 the main lobe and side lobes of the spectrum ( ) get wider and wider, however the height gets smaller and smaller. Eventually the spectrum will cover all frequenci
Z fτ →es with
almost zero energy at each frequency. Again this is different from what happens in the caseof an impulse. 2.5-2
0 0
2 20 0
2 20 0 0 0
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
d d
d d
j ft j ftv
n
j nf t j nf tv w v
n
W f v f e c nf f nf e
c nf e f nf c nf c nf e
π π
π π
δ
δ
− −
− −
⎡ ⎤= = −⎢ ⎥⎣ ⎦⎡ ⎤= − ⇒ =⎣ ⎦
∑
∑
2.5-3
[ ]0 0 0 0 0
0 0 0
( ) 2 ( ) 2 ( ) ( ) 2 ( ) ( )
( ) 2 ( )
v vn n
w v
W f j fV f j f c nf f nf j nf c nf f nf
c nf j nf c nf
π π δ π δ
π
⎡ ⎤= = − = −⎢ ⎥⎣ ⎦⇒ =
∑ ∑
2.5-4
[ ]
{ }
{ }
0 0 0 0 0 0 0
0 0 0 0
0 0 0
0 0 0
1 1( ) ( ) ( ) ( ) ( ) ( )2 2
1 [( ) ] ( ) [( ) ] ( )2
1 [( ) ] [( ) ] ( )2
1so ( ) [( ) ] [( ) ]2
v vn n
v vk k
v vn
w v v
W f V f mf c nf f kf mf c nf f kf mf
c k m f f kf c k m f f kf
c n m f c n m f f nf
c nf c n m f c n m f
δ δ
δ δ
δ
⎡ ⎤= − = − − + − +⎢ ⎥⎣ ⎦
⎡ ⎤= − − + + −⎢ ⎥⎣ ⎦
= − + + −
= − + +
∑ ∑
∑ ∑
∑
2-18
2.5-5
( )
4
4 0
4 2
2
( ) ( ) ( 2 )
1 1 1 1( ) ( ) ( )2 2 2 2
But ( ) ( ), so
( ) 1 2 sinc 22
Agrees with ( ) 2 sinc 22
j f
j ft j
j f j f
j f
v t Au t Au t
V f A f f ej f j f
f e e fAV f e A f e
j ftv t A f e
π τ
π
π τ π τ
π τ
τ
δ δπ π
δ δ
τ τπ
τ τ ττ
−
− −
− −
−
= − −
⎧ ⎫⎡ ⎤= + − +⎨ ⎬⎢ ⎥
⎣ ⎦⎩ ⎭=
= − =
−⎛ ⎞= Π ↔⎜ ⎟⎝ ⎠
2.5-6
( )
2 2
2 2 0
2 2
( ) ( ) ( )
1 1 1 1( ) ( ) ( ) ( )2 2 2 2
But ( ) ( ) ( ), so
1( ) ( ) ( ) 2 sinc 22
Agrees with ( )
j f j f
j f j f j
j f j f
v t A Au t Au t
V f A f f e f ej f j f
f e f e e f
V f A f e e A f A fj f
v t A
π τ π τ
π τ π τ
π τ π τ
τ τ
δ δ δπ π
δ δ δ
δ δ τ τπ
−
−
−
= − + + −
⎧ ⎫⎡ ⎤ ⎡ ⎤= − + − +⎨ ⎬⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦⎩ ⎭= =
⎡ ⎤= − − = −⎢ ⎥
⎣ ⎦= ( / 2 ) ( ) 2 sinc 2A t A f A fτ δ τ τ− Π ↔ −
2.5-7
( )
2 2
2 2 0
2 2
( ) ( ) ( )
1 1 1 1( ) ( ) ( ) ( )2 2 2 2
But ( ) ( ) ( ) ( ), so
( ) cos 22
If 0, ( ) sgn ( )
j fT j fT
j fT j fT j
j fT j fT
v t A Au t T Au t T
V f A f f e f ej f j f
f e f e e f fA AV f e e fT
j f j fAT v t A t V f
j
π π
π π
π π
δ δ δπ π
δ δ δ δ
ππ π
π
−
−
−
= − + − −
⎧ ⎫⎡ ⎤ ⎡ ⎤= − + − +⎨ ⎬⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦⎩ ⎭= = =
− −= + =
−→ = − ↔ = , which agrees with Eq. (17)
f
2-19
2.5-8
( ) sinc and (0) 1, sosinc 1( ) ( )
2 21 1If 0, ( ) ( ) and ( ) ( ), which agrees with Eq. (18)
2 2
V f f VfW f f
j f
w t u t W f fj f
εε δ
π
ε δπ
= =
= +
→ = = +
2.5-9
1/( ) and (0) 1, so1/ 2
1/ 1( ) ( )( 2 )(1/ 2 ) 2
1 1If 0, ( ) ( ) and ( ) ( ), which agrees with Eq. (18)2 2