ec3400: introduction to digital signal processing by roberto cristi professor dept. of ece naval...
TRANSCRIPT
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.