Pass Band Signal

7/23/2019 Pass Band Signal

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Pass-band Data Transmission

Dr. Teerasit Kasetkasem



Block Diagram

Functional model of pass-band data transmissionsystem.



Signaling

Illustrative aveforms for t!e t!ree basic forms ofsignaling binary information. "a# $mplitude-s!ift keying."b# P!ase-s!ift keying. "c # Fre%uency-s!ift keying it!continuous p!ase.



&!at do e ant to study'

&e are going to study and compare

different modulation tec!ni%ues in terms of

Probability of errors

Poer SpectrumBandidt! efficiency

B

Rb= ρ Bits(s()*



+o!erent PSK

Binary P!ase S!ift Keying "BPSK#

+onsider t!e system it! , basis functions

and

( ) t f T t cb

π φ 2cos

21 =

( ) t f T t cb π φ 2sin

2

2 =



BPSK

If e ant to fi t!at for bot! symbols " and /# t!e

transmitted energies are e%ual0 e !ave

φ/

φ,

s/

&e place s to minimi*e

probability of error

s

s



BPSK

&e found t!at p!ase of s/

and s. are /1

degree difference. &e can rotate s/

and

s.

φ/

φ,

s/

s

2otate



BPSK

&e observe t!at φ, !as not!ing to do it!

signals. )ence0 only one basis function is

sufficient to represent t!e signals

φ,

s/

s

φ/



BPSK

Finally0 e !ave

( ) t f T

E t E t s c

b

bb π φ 2cos

2)(11 ==

( ) t f T

E t E t s c

b

bb π φ 2cos

2)(10 −=−=



BPSK

Signal-space diagram for co!erent binary PSK system.

T!e aveforms depicting t!e transmitted signals s/"t # and

s,"t #0 displayed in t!e inserts0 assume nc = ,.



BPSK

Probability of error calculation. In t!e case of

e%ually likely "Pr"m.#3Pr"m/##0 e !ave

=

=

0

0

erfc

2

1

2erfc

2

1

N

E

N

d P

b

ik e



BPSK

Block diagrams for "a# binary PSK transmitter and "b#

co!erent binary PSK receiver.



4uadrip!ase-S!ift Keying "4PSK#

( ) ( ) T t it f T

E t s ci <≤

−+= 0;

4122cos

2 π π

T is symbol duration

5 is signal energy per symbol

T!ere are 6 symbols for i 3 /0 ,0 70 and 6



4PSK

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) T t t i E t i E

t f T

i E t f T

i E t s cci

<≤

−−

−=

−−

−=

0;4

12sin4

12cos

2sin2

412sin2cos

2

412cos

21 φ π

φ π

π π

π π

&!ic! e can rite in vector format as

( )

( )

−−

−

=4

12sin4

12cos

π

π

i E

i E

is



4PSK

i Input Dibit P!ase of4PSK

signaling

+oordinate of8essage point

si/ si,

/ /

, 7 /

6 //

4/π

4/3π

4/5π

4/7π

2/ E

2/ E

2/ E

2/ E

2/ E −

2/ E −2/ E −

2/ E −



4PSK

φ,

φ/

s6

s/

s7

s,

"/#"#

"/# "//#



4PSK signals



4PSK

Block

diagrams of

"a# 4PSK

transmitter

and "b#co!erent

4PSK

receiver.



4PSK9 5rror Probability 4PSK

+onsider signal

constellation

given in t!e

figure

φ,

φ/

s6

s/

s7

s,

"/#"#

"/# "//#

:6

:/

:7

:,

2/ E

2/ E

2/ E −

2/ E −



4PSK

&e can treat 4PSK as t!e combination

of , independent BPSK over t!e interval

T3,Tb

since t!e first bit is transmitted by φ/ and

t!e second bit is transmitted by φ,.

Probability of error for eac! c!annel isgiven by

=

=′

00

12

2erf

2

1

2erfc

2

1

N

E c

N

d P



4PSK

If symbol is to be received correctly bot! bits

must be received correctly.

)ence0 t!e average probability of correct

decision is given by&!ic! gives t!e probability of errors e%ual to

( ) 21 P P c ′−=

≈

−

=−=

0

0

2

0

2erfc

2

erfc

4

1

2

erfc1

N

E

N

E

N

E P P C e



4PSK

Since one symbol of 4PSK consists of tobits0 e !ave E = ,E b.

T!e above probability is t!e error probabilityper symbol. T!e avg. probability of error per bit

&!ic! is eactly t!e same as BPSK .

( )

≈

0

erfcsymbol per N

E Pe b

( ) ( )

≈= 0erfc2

1

symbol per2

1

bit per N

E

Pe Pe

b



BPSK vs 4PSK

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Power spectrum density of BPSK vs. QPSK

Normalized frequency,fTb

N o r m a l i z e d P S D , S f / 2 E b

BPSK

QPSK



4PSK

+onclusion4PSK is capable of transmitting data tice

as faster as BPSK it! t!e same energy per

bit.&e ill also learn in t!e future t!at 4PSK

!as !alf of t!e bandidt! of BPSK.



;FFS5T 4PSK

φ,

φ/

s6

s/

s7

s,

"/#"#

"/# "//#

< degree s!ift in p!ase

/1 degree s!ift in p!ase



;FFS5T 4PSK



;FFS5T 4PSK

&!enever bot! bits are c!angedsimultaneously0 /1 degree p!ase-s!iftoccurs.

$t /1 p!ase-s!ift0 t!e amplitude of t!etransmitted signal c!anges very rapidlycosting amplitude fluctuation.

T!is signal may be distorted !en ispassed t!roug! t!e filter or nonlinearamplifier.



;FFS5T 4PSK

0 1 2 3 4 5 6 7 8

-2

-1

0

1

2

0 1 2 3 4 5 6 7 8-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

;riginal Signal

Filtered signal



;FFS5T 4PSK

To solve t!e amplitude fluctuation

problem0 e propose t!e offset 4PSK.

;ffset 4PSK delay t!e data in

%uadrature component by T(, seconds

"!alf of symbol#.

=o0 no ay t!at bot! bits can c!ange

at t!e same time.



;FFS5T 4PSK

In t!e offset 4PSK0 t!e p!ase of t!e

signal can c!ange by ±< or degree

only !ile in t!e 4PSK t!e p!ase of t!e

signal can c!ange by ±/1 ±< or degree.



0 1 2 3 4 5 6 7 8

-2

-1

0

1

2

0 1 2 3 4 5 6 7 8

-1

-0.5

0

0.5

1

0 1 2 3 4 5 6 7 8

-1

-0.5

0

0.5

1

;FFS5T 4PSK

0 1 2 3 4 5 6 7 8-2

-1

0

1

2

0 1 2 3 4 5 6 7 8-1

-0.5

0

0.5

1

0 1 2 3 4 5 6 7 8

-1

-0.5

0

0.5

1

/ /

/

/ / /

Inp!ase4PSK

4 p!ase

4PSK

4PSK

/ /

/

/ / /

Inp!ase

;ffset 4PSK

4 p!ase

;ffset 4PSK

;ffset 4PSK



;ffset 4PSK

Possible pat!s for sitc!ing beteen t!e

message points in "a# 4PSK and "b# offset 4PSK.



;FFS5T 4PSK

Bandidt!s of t!e offset 4PSK and t!e

regular 4PSK is t!e same.

From signal constellation e !ave t!at

&!ic! is eactly t!e same as t!e regular4PSK.

≈

02erfc

N

E P e



π(6-s!ifted 4PSK

Try to reduce amplitude fluctuation by

sitc!ing beteen , signal constellation



π(6-s!ifted 4PSK

$s t!e result0

t!e p!ase of t!e

signal can be

c!anged in

order of ±π(6 or±7π(6



π(6-s!ifted 4PSK

Since t!e p!ase of t!e net ill be varied in

order of ±π(6 and ±7π(60 e can designed t!e

differential π(6-s!ifted 4PSK as given belo

>ray-5ncoded Input Data P!ase +!ange in radians

?π(6

/

?7π(6// -7π(6

/ -π(6



π(6-s!ifted 4PSK9///

Step Initialp!ase

Input Dibit P!asec!ange

Transmittedp!ase

/ π(6 π(6 π(,

, π(, / -π(6 π(6

7 π(6 / -π(6

6 / 7π(6 7π(6



π(6-s!ifted 4PSK

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5-2

0

2

QPSK

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5-2

0

2

OFFSET QPSK

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5-1

0

1

D OFFSET QPSK

/ / / /



π(6-s!ifted 4PSK

Since e only measure t!e p!ase

different beteen ad@acent symbols0 no

p!ase information is necessary. )ence0

non-co!erent receiver can be used.



Block diagram of t!e π (6-s!ifted D4PSK

detector.



π(6-s!ifted 4PSK

Illustrating t!e

possibility of p!ase

angles rapping

around t!e positive

real ais.



8-array PSK

$t a moment0 t!ere are 8 possible

symbol values being sent for 8 different

p!ase values0 ( ) M ii /12 π θ −=

( ) ( ) M ii M

t f T

E t s ci ,,2,1,1

22cos

2=

−+= π π



8-array PSK

Signal-space diagram

for octap!ase-s!ift

keying "i.e.0 M = 1#. T!e

decision boundaries are

s!on as das!ed lines.

Signal-space diagram

illustrating t!e

application of t!e union

bound for octap!ase-

s!ift keying.



8-array PSK

Probability of errors

( ) M E d d /sin21812 π ==∴

( ) 4 ;/sinerfc 0 ≥

≈ M M N

E

P e π



8-ary PSK

0 5 10 15 20 25 3010-50

10-40

10-30

10-20

10-10

100

Eb /N0 dB

P r o b a b i l i t y o f S y m b o l e r r o r s

QPSK

8-ary PSK

16-ary PSK



8-array PSK

Poer Spectra "8-array#

83,0 e !ave

( )

( ) M f T M E

Tf E f S

bb

PSK

22

2

2

logsinclog2

sinc2)(

=

=

( ) f T E f S bb BPSK 2sinc2)( =



8-array PSK

Poer spectra of M -ary PSK

signals for 8 = ,0 60 1.

T bf



8-array PSK

Bandidt! efficiency9

&e only consider t!e bandidt! of t!e main lobe

"or null-to-null bandidt!#

Bandidt! efficiency of 8-ary PSK is given by

M R

M T T B b

b 22 log2

log22 ===

M M R

R

B

R

b

bb22 log5.0log

2=== ρ



8-ary 4$8

4$8 3 4uadrature $mplitude 8odulation

Bot! $mplitude and p!ase of carrier

c!ange according to t!e transmitted

symbol0 mi .

!ere ai and bi are integers.( ) ( ) ( ) T t t f bT

E

t f aT

E

t s cicii ≤<−= 0;2sin

2

2cos

2 00

π π



8-ary 4$8

$gain0 e !ave

as t!e basis functions

( ) T t t f T

t c ≤<= 0;2cos2

1 π φ

( ) T t t f T

t cb

≤<= 02sin2

2 π φ



8-ary 4$8

4$8 s%uare +onstellation)aving even number of bits per symbol0

denoted by ,n.

83L L possible valuesDenoting M L =



/A-4$8

[ ]

−−−−−−−−−−−−

−−

−−

=

)3,3()3,1()3,1()3,3(

)1,3()1,1()1,1()1,3()1,3()1,1()1,1()1,3(

)3,3()3,1()3,1()3,3(

, ii ba



/A-4$8

-ary0 6-P$8



/A-4$8

+alculation of Probability of errorsSince bot! basis functions are ort!ogonal0

e can treat t!e /A-4$8 as combination of

to 6-ary P$8 systems.For eac! system0 t!e probability of error is

given by

−=

−=′

0

0

0

11

2

11

N

E erfc

M N

d erfc

L P e



/A-4$8

$ symbol ill be received correctly if data

transmitted on bot! 6-ary P$8 systems are

received correctly. )ence0 e !ave

Probability of symbol error is given by

( ) ( ) 21 ec P symbol P ′−=

( ) ( ) ( )

( ) eee

ece

P P P

P symbol P symbol P

′≈′−′+−=

′−−=−=

2211

111

2

2



/A-4$8

)ence0 e !ave

But because average energy is given by

&e !ave

( )

−=

0

0112

N

E erfc

M symbol P e

( ) ( )

3

1212

22 0

2/

1

20 E M i

L

E E

L

iav

−=

−= ∑

=

( )( )

−

−=

012

3112

N M

E erfc

M symbol P av

e



+o!erent FSK

FSK 3 fre%uency s!ift keying

+o!erent 3 receiver !ave information on

!ere t!e *ero p!ase of carrier.

&e can treat it as non-linear modulation

since information is put into t!e

fre%uency.



Binary FSK

Transmitted signals are

!ere

( ) ( )

≤<

=

elsewere ,0

0,2cos2

bib

b

i

T t t f T

E

t sπ

2,1; =+= iT

in f b

ci



Binary FSK

S/"t# represented symbol C/.

S,"t# represented symbol C.

T!is FSK is also knon as SundeEs FSK.It is continuous p!ase fre%uency-s!ift

keying "+PFSK#.



Binary FSK

T!ere are to basis functions ritten as

$s a result0 t!e signal vectors are

( ) ( )

≤<

=

elsewere ,0

0,2cos2

bibi

T t t f T t

π φ

=

=

b

b

E

E 0 !n"

021 ss



S



BFSK

From t!e figure0 e !ave

In case of Pr"#3Pr"/#0 t!e probability of

error is given by

&e observe t!at at a given value of Pe0t!e BFSK system re%uires tice as muc!

poer as t!e BPSK system.

b E d 212 =

=

022

1

N

E erfc P b

e



T2$=S8ITT52



25+5I52

P S t l d it f BFSK



Poer Spectral density of BFSK

+onsider t!e SundeEs FSK !ere f / and f , are different

by /(Tb. &e can rite

&e observe t!at in-p!ase component does not

depend on mi since

( )

( ) ( )t f T

t

T

E t f

T

t

T

E

T

t t f

T

E t s

cbb

bc

bb

b

bc

b

bi

π π

π π

π π

2sinsin2

2coscos2

2cos2

±−

±=

±=

=

±

bb

b

bb

b

T

t

T

E

T

t

T

E π π cos

2cos

2

P S t l d it f BFSK




&e !ave

For t!e %uadrature component

( )

++

−=

=

bbb

b

bb

b BI

T f

T f

T

E

T

t

T

E F f S

2

1

2

1

2cos

22

δ δ π

)alf of t!e symbol poer

( )

±=

bb

b

T

t

T

E t g π sin

2 ( )

( )2222

2

14

cos8

−=

f T

f T T E S

b

bbb B

π

π

P S t l d it f BFSK




Finally0 e obtain )()()( f S f S f S B BI B +=

P! T f BFSK



P!ase Tree of BFSK

FSK signal is given by

$t t 3 0 e !ave

T!e p!ase of Signal is *ero.

( )

±=

bc

b

b

T

t t f

T

E t s

π π 2cos

2

( ) ( )0cos20

02cos2

0b

b

bc

b

b

T

E

T f

T

E s =

±=

π π

P! T f BFSK



P!ase Tree of BFSK

$t t 3 Tb0 e !ave

&e observe t!at p!ase c!anges by ±π afterone symbol "Tb seconds#. -π for symbol C/

and ?π for symbol C

&e can dra t!e p!ase trellis as

( ) ( )π π

π ±=

±= cos

22cos

2

b

b

b

bbc

b

bb

T

E

T

T T f

T

E T s



8i i S!ift k i "8SK#



8inimum-S!ift keying "8SK#

8SK tries to s!ift t!e p!ase after one

symbol to @ust !alf of SundeEs FSK

system. T!e transmitted signal is given

by

( ) ( )[ ]

( )[ ]

( )[ ]

+

+

=+= #0#for02cos2

#1#for02cos2

2cos2

2

1

θ π

θ π

θ π

t f T

E

t f T

E

t t f T

E t s

b

b

b

b

cb

b

8SK



8SK

&!ere

;bserve t!at

( ) ( ) t T

!t

b

π θ θ ±= 0

bc

bc

T

! f f

T

! f f

2!n"

221 −=+=

( )212

1 f f f c +=

8SK



8SK

h 3 T b"f /-f ,# is called Cdeviation ratio.

For SundeEs FSK0 h 3 /.

For 8SK0 h 3 .G.

h cannot be any smaller because t!eort!ogonality beteen cos",πf /t # and cos",πf ,t #

is still !eld for h H .G.

;rt!ogonality guarantees t!at bot! signal illnot interfere eac! ot!er in detection process.

8SK



8SK

P!ase trellis diagram for 8SK signal ///

8SK



8SK

Signal s"t# of 8SK can be decomposed

into

!ere

( ) ( )[ ]

( )[ ] ( ) ( )[ ] ( )

( ) ( ) ( ) ( )t f t st f t s

t f t T

E t f t

T

E

t t f

T

E t s

cc I

cb

bc

b

b

c

b

b

π π

π θ π θ

θ π

2sin2cos

2sinsin2

2coscos2

2cos2

−=

−=

+=

( ) ( ) bb

T t t T

t ≤<±= 0;2

0 π

θ θ

8SK



8SK

Symbol θ "# θ "Tb#

/

π(,

π -π(,

-π(,

ππ(,

8SK



8SK

For t!e interval T b H t ≤ 0 e !ave

etEs note !ere t!at t!e ± for t!e interval-T bHt ≤ and H t ≤T b may not be t!e same.

&e kno t!at

( ) ( ) 0;2

0 ≤<−±= t T t T

t bb

π θ θ

( ) ( )[ ] ( )[ ]

=

±

bbb T

t

T

t

T

t

2sin0sin

2cos0cos

20cos

π θ

π θ

π θ

8SK



8SK

Since θ"# can be eit!er or π depending ont!e past !istory. &e !ave

C? for θ"# 3 and C- for θ"# 3 π)ence0 e !ave

( ) ( )[ ]

±=

=

±

bbb T

t

T

t

T

t

2

cos

2

cos0cos

2

0cos π π

θ π

θ

bbbb

b I T t T

T

t

T

E t s ≤<−

±= ;

2cos

2)(

π

8SK



8SK

Similarly e can rite

for H t ≤T b and T b H t ≤,T b. =ote t!e C? and C-may be different beteen t!ese intervals.

Furt!ermore0 e !ave t!at θ "T b# can be ±π(,

depending on t!e past !istory.

( ) ( ) ( )bb

b T t T

T t −±=2

π θ θ

8SK



8SK

)ence0 e !ave

e !ave t!at θ "T b# can be ±π(, depending

on t!e past !istory.

( ) ( )

( )[ ] ( )

( )[ ] ( )

( )[ ] ( )[ ]

−±

−=

−±

−=

−±

22sincos

22cossin

2sincos

2cossin

2sin

π π θ π π θ

π θ

π θ

π θ

bb

bb

b

bb

b

bb

b

bb

T t T

T t T

T

T t T

T

T t T

T

T t T

( ) ( )

±=

−±=

−±

bbb

bb

T

t

T

t

T

T t T

2sin

22cos

2sin

π π π π θ

8SK



8SK

)ence0 e !ave

C? for θ"T b# 3 ?π(, and C- for θ"T b# 3 -π(,T!e basis functions c!ange to

bbb

b T t

T

t

T

E t s 20;

2sin

2)( ≤<

±=

π

( ) ( ) bc

bb

T t t f T

t

T t ≤<

= 0;2cos

2cos

21 π

π φ

( ) ( ) bcbb

T t t f T

t

T t ≤<

= 0;2sin

2sin

22 π

π φ

8SK



8SK

&e rite 8SK signal as

&!ere and

( ) ( )[ ] ( ) ( )[ ] ( )

( )[ ] ( ) ( )[ ] ( )

( )[ ] ( )[ ]

)()(

)(sin$)(0cos$

2sin

2

sinsin

2

2cos

2

cos0cos

2

2sinsin2

2coscos2

2211

2 b1 b

t st s

t T t

t f T

t

T T

E

t f T

t

T

E

t f t T

E t f t

T

E t s

b

cb

bb

b

cbb

b

cb

bc

b

b

φ φ

φ θ φ θ

π

π

θ π

π

θ

π θ π θ

+=

−=

−

=

−=

( )[ ]0cos1 θ b E s = ( )[ ]bb T E s θ sin2 −=

8SK



8SK

Symbol θ "# s/ θ "Tb# s,

/

π(,

π -π(,

-π(,

ππ(,

b E

b E

b E

b E

b E −

b E −

b E −

b E −



=

02

1

N

E erfc P b

e



P!ase9 π(, π π π(, -π(,π(,

8SK



8SK

&e observe t!at 8SK is in fact t!e4PSK !aving t!e pulse s!ape

Block diagrams for transmitter and

receiver are given in t!e net to slides.

bT

t

2cos

π



∫bT

dttt )()( φ



∫ −

=bT

dt t t " " )()( 11 φ

∫ =bT

dt t t " "

2

022 )()( φ

4



0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

2.5

3

3.5

Normalized Frequency, fTb

N o r m a l i z e d P S D ,

S ( f ) / E b

MSK

BPSK

QPSK

8SK



8SK

Probability of error of 8SK system is e%ual toBPSK and 4PSK

T!is due to t!e fact t!at 8SK observes t!e

signal for to symbol intervals !ereas FSKonly observes for single signal interval.

Bandidt! of 8SK system is GJ larger t!an

4PSK.

( ) 2

222 11%

2cos32)(

−=

f T

f T E f S

b

bb MSK

π

π

=onco!erent ;rt!ogonal 8odulation




=onco!erent implies t!at p!ase information isnot available to t!e receiver.

$s a result0 *ero p!ase of t!e receiver canmean any p!ase of t!e transmitter.

$ny modulation tec!ni%ues t!at transmitsinformation t!roug! t!e p!ase cannot be usedin nonco!erent receivers.





cos",πft#

sin",πft#

2eceiver

cos",πft#sin",πft#

Transmitter





It is impossible to dra t!e signal constellationsince e do not kno !ere t!e aes are.

)oever0 e can still determine t!e distance

of t!e eac! signal constellation from t!e origin. $s a result0 t!e modulation tec!ni%ues t!at put

information in t!e amplitude can be detected.

FSK uses t!e amplitude of signals in to

different fre%uencies. )ence non-co!erent

receivers can be employed.





+onsider t!e BFSK system !ere tofre%uencies f / and f , are used to represented

to C/ and C.

T!e transmitted signal is given by

Problem is t!at θ is unknon to t!e receiver.For t!e co!erent receiver0 θ is precisely knon

by receiver.

( ) bi T t it f T

E t s ≤<=+= 0,2,1;2cos

2)( θ π





Furt!ermore0 e !ave

To get rid of t!e p!ase information "θ#0 e use

t!e amplitude

( )

( ) ( ) ( ) ( )

)()(

2sinsin2

2coscos2

2cos2

)(

2211 t st s

t f T

E t f

T

E

t f T

E t s

ii

ii

i

φ φ

π θ π θ

θ π

+=

+=

+=

( ) ( ) ( ) E E E s st s ii =+=+= θ θ 2222

21 sincos





&!ere

T!e amplitude of t!e received signal

∫ ∫ =⇒=T T

i dt t t " "dt t t s s

011

011 )()()()( φ φ

∫ ∫ =⇒=

T T

i dt t t " "dt t t s s0

220

22 )()()()( φ φ

( ) ( )

2/12

0

2

0

2sin)(2cos)(

+

= ∫ ∫ T

iT

ii dt t f t "dt t f t "l π π

4uadrature 2eceiver using



g

correlators

4uadrature 2eceiver using 8atc!ed



g

Filter





Decision rule9 et if l i & l k for all k . Foreamples0 decide if l 1 & l 2

T!is decision rule suggests t!at if t!e envelop"amplitude# of t!e received signal described in

term of cos(2π f 1t ) is greater t!an t!e envelop of

t!e received signal described in term of

cos(2π f 2t ), e say s1(t ) w!s sent.

imm ='1' mm =

=onco!erent 8atc!ed Filter



=onco!erent 8atc!ed Filter





+onsider t!e output of matc!ed filter ofcos(2π f it ).

( ) ( )[ ] ( )[ ]∫ +−=−∗=T

ii d t T f "t T f t "t y

0

2cos)(2cos)( τ τ π τ π

[ ] ( )

[ ] ( )

2sin)()(2in

2cos)()(2cos)(

0

0

∫

∫

−

−=

T

ii

T

ii

d f "t T f s

d f "t T f t y

τ τ π τ π

τ τ π τ π





5nvelope at t3T is

&!ic! is eactly t!e same as in

correlator receiver

( ) ( )

2/12

0

2

0

2cos)(2cos)(

+

=

∫ ∫

T

i

T

ii d f "d f "l τ τ π τ τ τ π τ

>enerali*ed binary receiver for nonco!erent ort!ogonal

modulation



modulation.

4uadrature receiver e%uivalent to eit!er one of t!e to

matc!ed filters in part



matc!ed filters in part




g

Probability of 5rrors

−=

02

ep

2

1

N

E P e

=onco!erent9 BFSK



=onco!erent9 BFSK

For BFSK0 e !ave

( ) ( )

≤<=

elsewere ; 0

0;2cos2

bib

bi

T t t f T

E

t s π

=onco!erent9 BFSK



o co e e S

=onco!erent9 BFSK



Probability of 5rrors

−=

02

ep2

1

N

E P b

e

DPSK



Differential PSK Instead of finding t!e p!ase of t!e signal on

t!e interval 0*t ≤T b$ +is receier "etermines te

p!se "ifference between !"-!cent time inter!ls. f /1 is sent, te p!se will rem!in te s!me

f /0 is sent, te p!se will c!nge 180 "egree.

DPSK



;r e !ave

and

( )

( )

≤<

≤<

=

bbcb

b

bcb

b

T t T t f T

E

T t t f T

E

t s

2;2cos2

20;2cos2

)(1

π

π

( )

( )

≤<+

≤<

=

bbcb

b

bc

b

b

T t T t f T

E

T t t f

T

E

t s

2;2cos2

20;2cos

2)(2

π π

π

DPSK



In t!is case0 e !ave T 2T b !n" E 2 E b

ence, te prob!bility of error is gien by

−= 0ep2

1

N

E

P

b

e

DPSK9 Transmitter



11 −− ⊕= k k k k k d bd bd

DPSK



bk L / / / /

d k 1

L/ / / / /

Differentialencoded d k L

/ / / / /

TransmittedP!ase

π π π π π

DPSK9 2eceiver



DPSK9 2eceiver



From t!e block diagram0 e !ave t!at t!edecision rule as

If t!e p!ase of signal is unc!anged "send C/#t!e sign "C? or C-# of bot! "i and " sol" notc!nge. ence, te l (x) sol" be positie.

If t!e p!ase of signal is unc!anged "send C#t!e sign "C? or C-/# of bot! "i and " sol"c!nge. ence, te l (x) sol" be neg!tie.

( ) 0

1s!y

0s!y

1010 <>

+= I I " " " "l x



Pass Band Signal

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