• Prof. Brian L. Evans• Dept. of Electrical and Computer Engineering• The University of Texas at Austin

EE345S Real-Time Digital Signal Processing Lab Spring 2006

Lecture 14

Digital Pulse Amplitude Modulation

14 - 2

Introduction• Modulate M = 2J discrete messages or J bits of

information into amplitude of signal• If amplitude mapping changes at symbol rate of

fsym, then bit rate is J fsym

• Conventional mapping of discretemessages to M uniformly space amplitudes

• Pulse amplitude modulated (PAM) signal)12( −= idai 2

, ... , 0 ..., , 12

MMi +−=

) ( )(*sym

kk Tktats −= �

∞

−∞=

δ No pulses overlap in time:requires infinite bandwidth

fsym = 1 / Tsym

14 - 3

Pulse Shaping• Infinite bandwidth cannot be sent in practice• Limit bandwidth by pulse shaping

filter with impulse response gT(t)

• L samples per symbol durationn is symbol index

m is sample index in a symbol: m = 0, 1, 2,…, L-1.

( )symTk

k Tktgatssym

)(* −= �∞

−∞=

��

���

� −+=��

���

� + �∞

−∞=symsymsymT

kksymsym kTT

Lm

nTgaTLm

nTssym

*

( ) ( ) ( )mknLgakLmnLgamLnssymsym T

kkT

kk +−=−+=+ ��

∞

−∞=

∞

−∞=

)( *

At each time t, k is indexed over number of

overlapping pulses

At indices n & m, k is indexed over number of overlapping pulses

14 - 4

Pulse Shaping Example

2-PAM with Raised Cosine Pulse Shaping

14 - 5

Pulse Shaping Block Diagram

• Upsampling by L denoted as LOutpus input sample followed by L-1 zerosUpsampling by converts symbol rate to sampling rate

• Pulse shaping (FIR) filter gTsym[m]Fills in zero values generated by upsamplerMultiplies by zero most of time (L-1 out of every L times)

D/A TransmitFilter

akgTsym[m]L

symbol rate

sampling rate

sampling rate

analog analog

14 - 6

Digital Interpolation Example

• Upsampling by 4 (denoted by 4)Output input sample followed by 3 zerosFour times the samples on output as inputIncreases sampling rate by factor of 4

• FIR filter performs interpolationLowpass filter with stopband frequency ωstopband ≤ π / 4For fsampling = 176.4 kHz, ω = π / 4 corresponds to 22.05 kHz

Digital 4x Oversampling Filter

16 bits44.1 kHz

28 bits176.4 kHz4 FIR Filter16 bits

176.4 kHz

1 2

Input to Upsampler by 4

n

0

n’

Output of Upsampler by 4

1 2 3 4 5 6 7 80

1 2

Output of FIR Filter

3 4 5 6 7 8n’

0

14 - 7

Pulse Shaping Filter Bank

• Simplify by avoiding multiplication by zeroSplit the long pulse shaping filter into L short polyphase filters

operating at symbol rategTsym,0[n]

gTsym,1[n]

gTsym,L-1[n]

akD/A Transmit

Filter

s(Ln)

s(Ln+1)

s(Ln+(L-1))

Filter Bank Implementation

D/A TransmitFilter

akgTsym[m]L

symbol rate

sampling rate

sampling rate

analog analog

14 - 8

Pulse Shaping Filter Bank Example• Pulse length 24 samples and L = 4 samples/symbol

• Derivation in Tretter's manual,

• Define mth polyphase filter

• Four six-tap polyphase filters (next slide)

��

���

� −+=��

���

� + �+

−=symsymsymT

n

nkksymsym kTT

Lm

nTgaTLm

nTs 3

2

* 1,...,1 ,0 −= Lm

��

���

� += symsymTmT TLm

nTgngsymsym

][,

[ ]kngaTLm

nTs mT

n

nkksymsym sym

−=��

���

� + �+

−=,

3

2

*

( ) ( )mknLgamnLs T

n

nkk +−=+ �

+

−=

)( 3

2

* Six pulses contribute to each output sample

1,...,1 ,0 −= Lm

14 - 9

Pulse Shaping Filter Bank Example

gTsym,0[n]24 samples

in pulse

x marks samples ofpolyphase

filter

4 samples per symbol

Polyphase filter 0 response is the first sample of the

pulse shape plus every fourth sample after that

Polyphase filter 0 has only one non-zero sample.

14 - 10

Pulse Shaping Filter Bank Example24 samples

in pulse

x marks samples ofpolyphase

filter

4 samples per symbol

Polyphase filter 1 response is the second sample of the

pulse shape plus every fourth sample after that

gTsym,1[n]

14 - 11

Pulse Shaping Filter Bank Example24 samples

in pulse

x marks samples ofpolyphase

filter

4 samples per symbol

Polyphase filter 2 response is the third sample of the

pulse shape plus every fourth sample after that

gTsym,2[n]

14 - 12

Pulse Shaping Filter Bank Example24 samples

in pulse

x marks samples ofpolyphase

filter

4 samples per symbol

Polyphase filter 3 response is the fourth sample of the

pulse shape plus every fourth sample after that

gTsym,3[n]

14 - 13

Intersymbol Interference• Eye diagram is empirical measure of quality of

received signal

• Intersymbol interference (ISI):

• Raised cosine filter has zeroISI when correctly sampled

� �∞

−∞=

∞+

≠−∞= �

��

�

�

���

�

� −+=−=

knk

k

symsymknsymsymk g

kTnTgaagTkTnganx

)0(

)( )0() ( )(

��∞

≠−∞=

∞

≠−∞=

−=−

−≤nkk

sym

nkk

symsym

g

kTgdM

g

kTnTgdMD

,, )0(

)( )1(

)0(

)( )1(

14 - 14

Symbol Clock Recovery• Transmitter and receiver normally have different

crystal oscillators• Critical for receiver to sample at correct time

instances to have max signal power and min ISI• Receiver should try to synchronize with

transmitter clock (symbol frequency and phase)First extract clock information from received signalThen either adjust analog-to-digital converter or interpolate

• Next slides develop adjustment to A/D converter• Also, see Handout M in the reader

Optional

14 - 15

Symbol Clock Recovery• g1(t) is impulse response of LTI composite channel

of pulse shaper, noise-free channel, receive filter)( )()()( 11

*sym

kk kTtgatgtstq −=∗= �

∞

−∞=

� �∞

−∞=

∞

−∞=

−−==k m

symsymmk mTtgkTtgaatqtp )( )( )()( 112

)(

)( )( } {)}({

21

2

11

symk

symk

symm

mk

kTtga

mTtgkTtgaaEtpE

−=

−−=

�

� �∞

−∞=

∞

−∞=

∞

−∞=

Optional

ReceiveB(ω)

Squarer BPFH(ω)

PLLx(t)

q(t) q2(t)

p(t)

z(t)

E{ak am} = a2 δδδδ[k-m]

g1(t) is deterministic

s*(t) is transmitted signal

Periodic with period Tsym

14 - 16

Symbol Clock Recovery• Fourier series representation of E{ p(t) }

• In terms of g1(t) and using Parseval’s relation

• Fourier series representation of E{ z(t) }

tkj

kk

symeptpE )}({ ω�

∞

−∞=

= �−= sym sym

T tkj

symk dtetpE

Tp

0

)}({1 ω

where

Optional

( ) ( ) ( )��∞

∞−

∞

∞−

− −== ωωωωπ

ω dkGGT

adtetg

Ta

p symsym

tjk

symk

sym

11

221

2

2

ReceiveB(ω)

Squarer BPFH(ω)

PLLx(t)

q(t) q2(t)

p(t)

z(t)

( ) ( ) ( ) ( )�∞

∞−

−== ωωωωπ

ωω dkGGTa

kHkHpz symsym

symsymkk 11

2

2

14 - 17

Symbol Clock Recovery• With G1(ωωωω) = X(ωωωω) B(ωωωω)

Choose B(ω) to pass ± ½ωsym � pk = 0 except k = -1, 0, 1

Choose H(ω) to pass ±ωsym � Zk = 0 except k = -1, 1

• B(ωωωω) is lowpass filter with ωωωωpassband = ½ωωωωsym

• H(ωωωω) is bandpass filter with center frequency ωωωωsym

Optional

( ){ } )cos(2 teeeZtzE symtjtj

k

tjkk

symsymsym ωωωω =+== −�

( ) ( ) ( ) ( )�∞

∞−

−== ωωωωπ

ωω dkGGT

akHkHpZ sym

symsymsymkk 11

2

2

ReceiveB(ω)

Squarer BPFH(ω)

PLLx(t)

q(t) q2(t)

p(t)

z(t)