ECE6602: Digital Communications

Lecture 10: QAM, PAM, and PSK

Digital ModulationDifferent Modulations: Linear & Non-linear, Memoryless & With MemoryMapping from Data to Symbol: Gray MappingLinear Modulation:

Baud Rate: Number of Symbols Transmitted Per second =Bit Rate: Number of Symbols Transmitted Per second = Power Spectrum:

T1

( ){ }∑ π−−π−=

=

∑ −=

∞

−∞=

π

∞

−∞=

kcksckc

tf2jBP

kkB

)tf2sin()kTt(pa)tf2cos()kTt(paetsRe)t(s

)kTt(pa)t(s

c

TMlog 2

( )

[ ] { } ( ) [ ]∑φ=Φ=φ

∫=

Φ=Φ

∞+

−∞=π−

+

∞+

∞−

π−

mmf̂2j

aaaa*nmnaa

ft2j

2aaB

emf̂ ,aaEm

,dte)t(pP(f)

,)f(PTfT1)f(

Amplitude Shift Keying (ASK)Amplitude shift keying (ASK) signals transmit information in the amplitude of the signals.There are two types of ASK signals

� Pulse amplitude modulation (PAM)� Quadrature amplitude modulation (QAM)

Pulse Amplitude Modulation

The Signal Energy is

)1Tf ,dt)t(pE( E2

aE T0 c

2pp

2m

m ∫ >>==

)tf2cos()t(pa)t(s}A)1M(,...,A3,A{a

cmm

mπ=−±±±∈

∑ π−−π−=

∑ −=∞

−∞=

∞

−∞=

kcksckcP

kkB

)tf2sin()kTt(pa)tf2cos()kTt(pa)t(s

)kTt(pa)t(s

PAM – Average Energy

Assuming equally likely symbols, the average symbol energy is

Use the identities

∑ +++−=

∑ −−=

∑=

∑=

=

=

=

=

M

1m22p

2

M

1m2p

M

1m2m

p

M

1mmav

))1M(m)1M(4m4(2EA

M1

}A)M1m2{(2

EM1

a2

EM1

EM1E

6)1n2)(1n(nk

2)1n(nk

n

1k

2

n

1k

++=∑

+=∑

=

=

PAM — Average Energy

The average symbol energy is

The average energy per bit is

The average transmitted power P is given by

6)1M(EA

3)1M(M

2EA

M1E

2p

2

2p

2

av

−=

−=

MlogE

kEE

2

avavav b ==

PTEav =

PAM — Base Function and Signal Vector

The PAM signals can be expressed in terms of signal vectors.Since all the are linearly dependent (they just differ in a scale factor) there is just one basis function. Using

We have

Then

Hence,

tf2cos)t(Ap)t(s c1 π=

)t(sm

( ) tf2cos)t(pE2

2EA

tf2cos)t(Apts)t(s)t( c

pp2

c

1

10 π=π==ϕ

m0p

m a)t(2

E)t(s ϕ=

mp

mm a2

Es )t(s =↔ �

PAM — Signal Space Diagram

Signal space diagram for 8-PAM signals.

The minimum distance is

The normalized distance is

000 001 011 010 110 111 101 100

-7 -5 -3 -1 +1 +3 +5 +7 A2

Ep×

With M-ary PAM, the take one of M possible valus}.A)1M(,...,A3,A{a k −±±±∈

ka

1ME12AE2d2

avpmin −

==

1M12dd21Eminmin

av −==

=

Quadrature amplitude modulated signals can be thought of as independent amplitude modulation on the inphase and quadrature carrier components. The QAM signal has the form

where

With QAM, the take on discrete values from the set

( )kskckk

tf2j k

kcs kcc k

jaaa }e)kTt(paRe{

)tf2sin)kTt(patf2cos)kTt(pa()t(s

c +=∑ −=

∑ π−−π−=∞

−∞=π

∞

−∞=

sequencen informatio quadrature }a{sequencen informatioinphase }a{

s k

c k

==

s kc k a and a

}.A)1M̂(,...,A3,A{a ,a skck −±±±∈

Quadrature Amplitude Modulation (QAM)

QAM — Base Function

The transmitted QAM bandpass waveforms are

The complex envelopes, or baseband signal

QAM signals can be expressed in terms of signal vectors. Since the functionsare orthogonal, we have two basis

functions

M1,...,m , tf2sin)t(patf2cos)t(pa)t(s csmccmm =π−π=

( ) M1,...,m p(t), jaa)t(pa)t(s msmcmm =+==

1Tf with ,t f2sin and tf2cos ccc >>ππ

tf2sin)t(pE2)t(

tf2cos)t(pE2)t(

cp

2

cp

1

π−=ϕ

π=ϕ

QAM — Signal Vector

Then

Hence

For the case when the resulting signal space diagram has a “square constellation”. In this case the QAM signal can be thought of as 2 PAM signals in quadrature with one-half the average power in each of the quadrature components.

Tt0 , M1,...,m , )t(a2

E)t(a

2E

)t(s 2s mp

1c mp

m ≤≤=ϕ+ϕ=

)a,a(2

Es )t(s s mc m

pmm =↔ �

,M̂M 2=

QAM — Signal Constellation

Example signal constellation for 16-QAM (M=16, =4).The minimum distance between QAM signals is

M̂

AE2d pmin =

0000•

0100•

1100•

1000•

0001•

0101•

1101•

1001•

0011•

0111•

1111•

1011•

0010•

0110•

1110•

1010•

QAM — Minimum Distance

The energy in the waveform is

The average energy is

The minimum distance and the normalized minimum distance between signals in terms of the average energy is

)aa(2EA

E 2s m

2c m

p2

m +=)t(sm

( )2p2

2p2

M̂

1m2

s mM̂

1m2

c mp

2M

1mmav

M̂M 3

)1M(EA

3)1M̂(M̂M̂2

2EA

M1

)aM̂aM̂(2EA

M1E

M1E

=−

=

−=

∑+∑=∑====

1M6dd ,

1ME6AE2d 1Eminmin

avpmin av −

==−

== =

Phase Shift Keying (PSK)Phase shift keyed (PSK) signals transmit information in the phase of the signals.The transmitted band pass waveform is

The baseband signal/complex envolope

During any given baud interval, the waveform s(t) can take on one of M possible values, i.e. ,

The PSK signals all have equal energy

∑ θ+π−=∞

−∞=kkcP )tf2cos()kTt(pA)t(s

M,...,1m , tf2sinsin)t(Aptf2coscos)t(Ap M,...,1m ,)tf2cos()t(Ap)t(s

cmcm

mcm

=πθ−πθ==θ+π=

}M1,...,m ,M

1m2{k =−π∈θ

2EA

E m allfor 2EA

E p2

avp

2

m =⇒=

∑ −=∞

−∞=θ

kj

B )kTt(peA)t(s k

PSK — Base Function and Vector Form

As in the case of QAM, we can express the in terms of the orthogonal basis functions

and

ms

tf2sin)t(pE2)t(

tf2cos)t(pE2)t(

cp

2

cp

1

π−=ϕ

π=ϕ

)t(sinA2

E)t(cosA

2E

)t(s 2mp

1mp

m ϕθ+ϕθ=

)sin,(cosA2

Es mm

pm θθ=�

PSK — Minimum DistanceExample of 8-PSK signal constellation (M=8).The minimum distance between any two signals is

,M

sin2d ,M

sinE2M

sinA2

E2d minavP

minπ=π=π=

•

•

•

••

• •

•

A2

Ep

000

001011

010

110

111101

100

)t(2φ

)t(1φ

Summary of PAM, QAM, and PSK

Baseband and Passband Signals

� PAM

� PSK

� QAM

∑ π−−π−=

∑ −=∞

−∞=

∞

−∞=

kcksckcP

kkB

)tf2sin()kTt(pa)tf2cos()kTt(pa)t(s

)kTt(pa)t(s

0a andaa is, that real, is a ks ,kkck ==

QAM of cases special arePSK and PAM random, are a and a kskc

)sin(a)cos(a

kkskkc

θ=θ=

Summary of PAM, QAM, and PSKWhich is which? PAM, PSK, or QAM?