EC3400: Introduction to Digital Signal Processing by Roberto Cristi Professor Dept. of ECE

EC3400: Introduction to Digital Signal Processing

by

Roberto Cristi

Professor

Dept. of ECE

Naval Postgraduate School

Monterey, CA 93943

Week 1 Topics:

• Introduction

• Fourier Transform (Review)

• Sampling

• Reconstruction

• Digital Filtering

• Example: a Digital Notch Filter

Introduction

Objectives In this course we introduce techniques to process signals by digital computers.

DSP

HardwareSoftware

sonar

audio

radar

video...

filtered signal:

• reject disturbances.

transformed signal:

• detection

• compression

A signal can come from a number of different sources:

x t( )

x n[ ]ADC

y n[ ]DSP DACLPF

antialiasing

LPF

reconstruction

LPF ADC DSP

DACLPF

)(ty

A Digital Filter

x t( )LPF

Fs

s t( ) s n[ ]

DAC LPF][nw )(ty

s n[ ] ][nw)(zH

We review the relations between the spectra of the signals in the following operations:

Sampling:

Digital Filtering:

Reconstruction:

Structure of a Digital Filter

x t( ) x n[ ] y n[ ] y t( )

H z( )

ADC DAC

ZOH

clock

Ts TsTs

Problem: determine the continuous time frequency response.

LPF LPF

anti-aliasingfilter

reconstructionfilter

continuous time discrete time continuous time

X x n e DTFT x n

x n X e d

j n

n

j n

( ) [ ] [ ] ;

[ ] ( ) .

1

2

X F x t e dt FT x t

x t X F e dF FT X F

j Ft

j Ft

( ) ( ) ( ) ;

( ) ( ) ( )

2

2 1

Recall:

• the Fourier Transform of a continuous time signal

• the Discrete Time Fourier Transform of a discrete time signal

x t( ) x n[ ]ADC

Ts

x t( )

s t t nTs

n

( ) ( )

x t t nTsn

( ) ( )

mathematical model of the sampler: it appends a to each sample

)(tx

t

)(][ snTxnx

t

Ts

Sampling of a continuous time signal:

x t t nT x n t nTsn

sn

( ) ( ) [ ] ( )

FT

X F S F

X F F F nFs sn

( ) ( )

( ) ( )

F X F nFs sn

( )

since

FT

x n e j nFT

n

s[ ]

2

since

snFTjs enTtFT 2)(

We can write the same expression in two different ways:

As a consequence:

X F X F nFFT s s

ns

( ) ( )

2

FFs

2

Fs

2

X F( )

FFs

2

Fs

2

X ( )

Fs Fs

2 2

0

0

Particular case: if the signal is bandlimited as

X F for F Fs( ) | | / 0 2

FFs

2

Fs

2

X F( )

then

F X F Xs F Fs( ) ( )

/

2

x t( ) x n[ ]

Ts

X ( )

Notice: F is in Hz (1/sec),

is in radians/sample (no dimension).

LPF

y n[ ] y t( )DAC

ZOH

Ts

Reconstruction: the Zero Order Hold

y t y n g t nTsn

( ) [ ] ( )

where g(t) is the pulse associated to each sample.

Then, its FT is computed as:

tTs

1g t( )

Y F y n G F e G F Yj nFT

nF F

s

s( ) [ ] ( ) ( ) ( )

/

22

where G(F)=FT[g(t)] is given by

)(sinc

)sin()]([)(

ssFTj

s

ss

FTj

FTTe

FT

FTTetgFTFG

s

s

kHzFFG s 10 with |)(|

sF

Finally, put everything together and assume ideal analog filters:

x t( ) x n[ ] y n[ ] y t( )

H z( )

ADC DAC

ZOH

clock

Ts TsTs

LPF LPF

anti-aliasingfilter

reconstructionfilter

sF

F

s

s

sFF

X

sFF

FG

s

FFjs FXFH

F

FeT

XHFG

YFGFY

2

/2

)(

/2

)(

/ )()(sinc

)()()(

)()()(

(radians)

| ( )|Y

| ( )|Y F reconstruction filter

| ( )|G F

(radians)

(radians)

)(HzF2/sF sF2/sFsF

)(HzF2/sF sF2/sFsF

Example: suppose we design a notch discrete time filter with transfer function

H z Kz z z z

z p z p( )

( )( )

( )( )

1 2

1 2

z-plane

with zeros and poles

z e p ej j1 2

41 2

40 90,/

,/; .

and sampling frequency . Determine the magnitude of its frequency response in the continuous time domain.

F kHzs 10

Solution: from what we have seen the frequency response is given by

Y F

X FH G F

F Fs

( )

( )( ) ( )

/

2

| ( )|H

5 5 0 125. 1 25. rad sample/

F kHz

| ( )|G F

23 92

. dB

F kHz

Y F

X F

( )

( )

125. 1 25.