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Communication Systems II[KECE322_01]

<2012-2nd Semester>

Lecture #222012. 11. 21

School of Electrical EngineeringKorea University

Prof. Young-Chai Ko

12년 11월 21일 수요일

Prof. Young-Chai Ko Communication System II Korea University

Outline

Approximate symbol error rate of PSK (phase-shift keying) modulation

Binary differential PSK (BDPSK)

Quadrature Amplitude Modulation (QAM)



Transformation of

Joint PDF of

Marginal PDF of

y1 and y2

V =q

y21 + y22 ,

⇥ = tan�1 y2y1

.

V and ⇥

fV,⇥(v, ✓) =v

2⇡�2

y

e�(v2+Es�2

pEsv cos ✓)/2�2

y .

⇥

f⇥

(✓) =

Z 1

0

fV,⇥(v, ✓) dv =1

2⇡e�⇢s sin

2 ✓

Z 1

0

ve�(v�p2⇢s cos ✓)2/2 dv

⇢s =EsN0

where is the SNR per symbol.



PDF of p⇥(✓)

[Fig. 10. 15, Proakis textbook]



Symbol error rate

PM = 1�Z ⇡/M

�⇡/Mf⇥(✓) d✓.

⇡

M

� ⇡

M

decision region for s1

PM = 1�Z ⇡/M

�⇡/Mf⇥

(✓) d✓ = 1�Z ⇡/M

�⇡/M

e�⇢s sin

2 ✓

2⇡

Z 1

0

ve�(v�p2⇢s cos ✓)2/2 dv d✓



BER of BPSK

SER of QPSK

P2 = Q

r2EbN0

!

Pc = (1� P2)2 =

"1�Q

r2EbN0

!#2,

P4 = 1� Pc

⇡ 2Q

r2EbN0

!.P4 = 2Q

r2EbN0

!"1� 1

2Q

r2EbN0

!#



Approximation of SER for large Es/N0

f⇥

(✓) =1

2⇡e�⇢s sin

2 ✓

Z 1

0

ve�(v�p2⇢s cos ✓)2/2 dv

⇡r

⇢s⇡

cos ✓e�⇢s sin2 ✓

1p2⇡

Z 1

�1e�

(v�m)2

2 dv = 11p2⇡

Z 1

�1ve�

(v�m)2

2 dv = m

Note

For high SNR per symbol, p2⇢s cos ✓ >> 1

−5 0 5 10 15 200

0.1

0.2

0.3

0.4

Z 1

�1ve�(v�m)2/2 dv ⇡

Z 1

0ve�(v�m)2/2 dv =

p2⇡m



PM ⇡ 1�Z ⇡/M

�⇡/M

r⇢s⇡

cos ✓e�⇢s sin2 ✓ d✓

⇡ 2p2⇡

Z 1

p2⇢s sin⇡/M

e�u2/2 du

= 2Q⇣p

2⇢s sin⇡

M

⌘

= 2Q

r2k sin2

⇣ ⇡

M

⌘ EbN0

!

⇡ 2Q

0

@s

2⇡2log2 M

M2

EbN0

1

A

k = log2 M

⇢s = k⇢b

sin⇡

M⇡ ⇡

M for large M

u = 2p⇢s sin

2 ✓

change in variable:

where



Approximation of SER for M-PSK

[Fig. 10. 6, Proakis textbook]



Differential coherent modulations are used when tracking the phase is difficult or costly.

These techniques allow the extraction/recovery of the information bits without knowledge of phase.

Let us treat only binary differential phase-shift-keying (BDPSK).

Binary Differential Phase-Shift-Keying (BDPSK)



Block diagram

XBPSK

modulator

+

Delay by T Band-limitedchannel

X

Delay by T

Z T

0(·) dtDecision

↵k = ±1 ak = ±1

n(t)

y(t)

y(t� T )

s(k)

z(t)



Received signal for

After delay by

Key assumption

✦ Assume that is unknown to the receiver but slowly varying such as it is roughly constant over 2 successive bits.

t 2 [kT, (k + 1)T ]

y(t) = ak

r2Eb

Tcos(2⇡fct+ ✓) + n(t)

y(t� T ) = ak�1

r2Eb

Tcos(2⇡fc(t� T ) + ✓) + n(t)

= ak�1

r2Eb

Tcos(2⇡fct+ ✓) + n(t)

fcT = integerAssume

✓(t)



z(t) = y(t)y(t� T )

= akak�12EbT

cos(2⇡fct+ ✓) cos(2⇡fct+ ✓) + ak

r2EbT

cos(2⇡fct+ ✓)n(t)

+ak�1

r2EbT

cos(2⇡fct+ ✓)n(t) + n2(t)

1 + cos(4⇡fct+ 2✓)

2

After integration

z = akak�12EbT

Z (k+1)T

kT

1 + cos(4⇡fct+ 2✓)

2

dt+ noise

= akak�12EbT

T

2

+ noise

= akak�1Eb + noise

noise =

⇢2n1 + n2

n2

n1 =

r2EbN0

Z T

0cos(2⇡fct+ ✓)n(t) dt

n2 =

Z T

0n2(t) dt



Differential encoding and decoding

If the symbol are obtained from the information symbols .

From the differential encoder we have:

ak 2 {�1, 1}

ak = ak�1 · ↵k

1 1 1

1 -1 -1

-1 1 -1

-1 -1 1

ak ak�1 ↵k

z = akak�1Eb + noise

= (ak�1)2↵kEb + noise

= ↵kEb + noise

↵k 2 {�1, 1}



Performance of BDPSK

Pb =1

2e�Eb/N0

0 5 10 1510−6

10−5

10−4

10−3

10−2

10−1

100

BDPSKBPSK



Transmit signal waveforms

Quadrature Amplitude Modulation (QAM)

um(t) = AmcgT (t) cos 2⇡fct+AmsgT (t) sin 2⇡fct, m = 1, 2, . . . ,M

=

pA2

mc +A2msgT (t) cos(2⇡fct+ ✓m)

where

= sm1�1(t) + sm2�2(t)

= <{p

A2mc +A2

msgT (t)ej(2⇡fct+✓m)}

= <{slm(t)e2j⇡fct} slm(t) =pA2

mc +A2msgT (t)e

j✓m

✓m = tan�1 Ams

Amc

�1(t) =

r1

EsgT (t) cos(2⇡fct)

�2(t) =

r1

EsgT (t) sin(2⇡fct)sm =

⇣pEsAmc,

pEsAms

⌘



Example

M = 8

M = 16



Spectral efficiency

Number of bits per symbol

✦ Symbol rate

um(t) =p

A2mc +A2

msgT (t) cos(2⇡fct+ ✓m) = AmgT cos(2⇡fct+ ✓n),

m = 1, 2, . . . ,M1

n = 1, 2, ...,M2

M1 = 2k1 and M2 = 2k2Let

k1 + k2 = log2 M1 + log2 M2 bits/symbol

Rs =Rb

k1 + k2



Rectangular QAM

Signal amplitudes take the set of values

Signal space diagram

M = 4

M = 8

M = 16

M = 32

M = 64

Rectangular QAM

{(2m� 1�M)d, m = 1, 2, . . . ,M}



Average energy per symbol

Distance between two symbols

Eav =1

M

MX

i=1

||si||2.

dmn =p||sm � sn||2.



For rectangular signal constellations in which where is even, the QAM signal constellation is equivalent to two PAM signals on quadrature carriers, each having signal points.

Since the signals in the phase-quadrature components can be perfectly separated at the demodulator, the probability of error for QAM is easily determined from the probability of error for PAM.

Specifically, the probability of a correct decision for the M-ary QAM system is

✦ where is the probability of error of an -ary PAM with one-half the average power in each quadrature signal of the equivalent QAM system.

By appropriately modifying the probability of error for M-ary QAM, we obtain

Probability of Error for QAMM = 2k k p

M = 2k/2

Pc = (1� PpM )2

PpM

pM

PpM = 2

✓1� 1p

M

◆Q

r3

M � 1

Eav

N0

!

where is the SNR per symbol.Eav/N0



−5 0 5 10 15 2010−6

10−5

10−4

10−3

10−2

10−1

100

SNR per bit (dB)

Prob

abilit

y of

a b

it er

ror,

P M

Therefore, the probability of a symbol error for the M-ary QAM is

Note that this result is exact for when is even.

PM = 1� (1� PpM )2

M = 2k k

QAMM=64

QAMM=16QAM

M=4


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