EE345S Real-Time Digital Signal Processing Lab Spring 2006

Lecture 15

Quadrature Amplitude Modulation (QAM) Transmitter

Prof. Brian L. Evans

Dept. of Electrical and Computer Engineering

The University of Texas at Austin

15 - 2

Introduction

• Digital Pulse Amplitude Modulation (PAM) modulates digital information onto amplitude of pulse and may be later modulated by sinusoid

• Digital Quadrature Amplitude Modulation (QAM) is two-dimensional extension of digital PAM that requires sinusoidal modulation

• Digital QAM modulates digital information onto pulses that are modulated ontoAmplitudes of a sine and a cosine, or equivalently

Amplitude and phase of single sinusoid

15 - 3

Amplitude Modulation by Cosine

• Example: y(t) = f(t) cos(c t)Assume f(t) is an ideal lowpass signal with bandwidth 1

Assume 1 < c

Y() is real-valued if F() is real-valued

• Demodulation: modulation then lowpass filtering• Similar derivation for modulation with sin(0 t)

0

1

-

F()

0

Y()½

-c - -c + c

c - c + c

½Fc½Fc

Review

15 - 4

Amplitude Modulation by Sine

• Example: y(t) = f(t) sin(c t)

Assume f(t) is an ideal lowpass signal with bandwidth 1

Assume 1 < c

Y() is imaginary-valued if F() is real-valued

• Demodulation: modulation then lowpass filtering

Y()j ½

-c - -c + c

c - c +

c

-j ½Fcj ½Fc

-j ½

0

1

-

F()

Review

15 - 5

Digital QAM Modulator

Serial/ParallelMap to 2-D constellation

Impulse modulator

Impulse modulator

Pulse shaping gT(t)

Local Oscillator

+

90o

Pulse shaping gT(t)

d[n]an

bn

a*(t)

b*(t)

s(t)

1 J

MatchedDelay

Matched delay matches delay through 90o phase shifter

bit stream

15 - 6

Phase Shift by 90 Degrees

• 90o phase shift performed by Hilbert transformer

cosine => sine

sine => – cosine

• Frequency response of idealHilbert transformer:

)(2

1)(

2

1) 2cos( 000 fffftf

)(2

)(2

) 2sin( 000 ffj

ffj

tf

)sgn()( fjfH

0 if1

0 if0

0 if1

)sgn(

x

x

x

x

15 - 7

Hilbert Transformer

• Magnitude responseAll pass except at origin

• For fc > 0

• Phase responsePiecewise constant

• For fc < 0)2sin()

22cos( tftf cc

))(2sin()2

)(2cos(

))2

2(cos()2

2cos(

tftf

tftf

cc

cc

f

)( fH

-90o

90o

f

|)(| fH

)sgn()( fjfH

15 - 8

Hilbert Transformer

• Continuous-time ideal Hilbert transformer

• Discrete-time ideal Hilbert transformer

h(t) =

1/( t) if t 0

0 if t = 0h[n] =

if n0

0 if n=0

nn )2/(sin2 2

Even-indexed samples are zero

t

h(t)

)sgn()( fjfH )sgn()( jH

n

h[n]

15 - 9

Discrete-Time Hilbert Transformer

• Approximate by odd-length linear phase FIR filterTruncate response to 2 L + 1 samples: L samples left of

origin, L samples right of origin, and originShift truncated impulse response by L samples to right to

make it causalL is odd because every other sample of impulse response is 0

• Linear phase FIR filter of length N has same phase response as a delay of length (N-1)/2(N-1)/2 is an integer when N is odd (here N = 2 L + 1)

• How would you make sure that delay from local oscillator to sine modulator is equal to delay from local oscillator to cosine modulator?

15 - 10

Performance Analysis of PAM

• If we sample matched filter output at correct time instances, nTsym, without any ISI, received signal

where the signal component is

v(t) output of matched filter Gr() for input ofchannel additive white Gaussian noise N(0; 2)

Gr() passes frequencies from -sym/2 to sym/2 ,where sym = 2 fsym = 2 / Tsym

• Matched filter has impulse response gr(t)

)()()( symsymsym nTvnTsnTx

dianTs nsym )12()( for i = -M/2+1, …, M/2

v(nT) ~ N(0; 2/Tsym)

4-PAM

d

-d

3 d

-3 d

15 - 11

Performance Analysis of PAM

dnTwgnTv r

)()()(

22 )()()( dnTwgEnTvE r

})()()()({ 212211

ddnTwgnTwgE TT

212121 )}()({)()( ddnTwnTwEgg TT

TdGdg

sym

sym

rr

22/

2/

2222 )(2

1)(

Noise power

T = TsymFiltered noise

2 (1–2)

15 - 12

Performance Analysis of PAM

• Decision errorfor inner points

• Decision errorfor outer points

• Symbol error probability

-7d -5d -3d -d d 3d 5d 7d

O- I I I I I I O+

T

dQdnTvPePI 2))(()(

T

dQdnTvPePO

))(()(

T

dQdnTvPdnTvPePO

))(())(()(

Td

QM

MeP

MeP

MeP

M

MeP OOI

)1(2)(

1)(

1)(

2)(

8-PAM Constellation

15 - 13

Performance Analysis of QAM

• Received QAM signal

• Information signal s(nT)

where i,k { -1, 0, 1, 2 } for 16-QAM

• Noise, vI(nT) and vQ(nT) are independent Gaussian random variables ~ N(0; 2/T)

)()()( nTvnTsnTx

dkjdibjanTs nn )12( )12( )(

)( )()( nTvjnTvnTv QI

15 - 14

Performance Analysis of QAM

• Type 1 correct detection

))(&)(()(1 dnTvdnTvPcP QI

))(())(( dnTvPdnTvP QI

)))((1))()((1( dnTvPdnTvP QI

2))(21( Td

Q

3

3

3

3

2 2

2

2

2 2

2

211

1 1

I

Q

16-QAM

)(2 Td

Q

)(2 Td

Q

15 - 15

Performance Analysis of QAM

• Type 2 correct detection

• Type 3 correct detection

))(&)(()(2 dnTvdnTvPcP QI

))(())(( dnTvPdnTvP QI

))(1))((21( Td

QTd

Q

))(&)(()(3 dnTvdnTvPcP QI

))(())(( dnTvPdnTvP QI

2))(1( Td

Q

3

3

3

3

2 2

2

2

2 2

2

211

1 1

I

Q

16-QAM

15 - 16

Performance Analysis of QAM

• Probability of correct detection

• Symbol error probability

22 ))(1(164

))(21(164

)( Td

QTd

QcP

))(1))((21(168

Td

QTd

Q

)(49

)(31 2 Td

QTd

Q

)(49

)(3)(1)( 2 Td

QTd

QcPeP

15 - 17

Average Power Analysis

• PAM and QAM signals are deterministic• For a deterministic signal p(t), instantaneous

power is |p(t)|2

• 4-PAM constellation points: { -3 d, -d, d, 3 d }– Total power 9 d2 + d2 + d2 + 9 d2 = 20 d2

– Average power per symbol 5 d2

• 4-QAM constellation points: { d + j d, -d + j d,d – j d, -d – j d }– Total power 2 d2 + 2 d2 + 2 d2 + 2 d2 = 8 d2

– Average power per symbol 2 d2