1 DSP dfgFundamentals
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Fundamentals of Digital Signal Processing
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Fourier Transform of continuous time signals
dtetxtxFTFX Ftj 2)()()(
dFeFXFXIFTtx Ftj 2)()()(
with t in sec and F in Hz (1/sec).
Examples:
0 0 0sinct
TFT rect T FT
02 0 FFeFT tFj
021
021
02cos FFeFFetFFT jj
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Discrete Time Fourier Transform of sampled signals
n
fnjenxnxDTFTfX 2][][)(
21
21
2)()(][ dfefXfXIDTFTnx fnj
with f the digital frequency (no dimensions).
Example:
020( )j f n
k
DTFT e f f k
since, using the Fourier Series,
2( ) j nt
k n
t k e
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Property of DTFT
• f is the digital frequency and has no dimensions
• is periodic with period f = 1. )1()( fXfX
21
21 f
f11
2
1
21
)( fX
• we only define it on one period
f1
2
)( fX
1
2
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tFjetx 02)( nj
ssF
F
enTxnx02
)(][
ss TF /1
Sampled Complex Exponential: no aliasing
0F F
)(FX
f
( )X f
2
1
1
2
1. No Aliasing20
sFF
2sF
2
sF0f
sF
Ff 0
0 digital frequency
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tFjetx 02)( nj
ssF
F
enTxnx02
)(][
ss TF /1
Sampled Complex Exponential: aliasing
0FF
)(FX
f
( )X f
2
1
1
2
2. Aliasing 0 2sF
F
2sF
2
sF0f
0 00
s s
F Ff round
F F
digital frequency
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tFjetx 02)( 02[ ] ( ) j f nsx n x nT e
ss TF /1
Mapping between Analog and Digital Frequency
ss F
Fround
F
Ff 00
0
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Example
tjetx 10002)( kHzFs 3
Then:
• analog frequency
• FT:
• digital frequency
•DTFT: for
)1000()( FFX FT
HzF 10000
0 0 1 1 10 3 3 3s s
F FF Ff round round
31)( ffX DTFT 2
1|| f
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Example
tjetx 20002)( kHzFs 3
Then:
• analog frequency
• FT:
• digital frequency
•DTFT: for
)2000()( FFX FT
HzF 20000
31)( ffX DTFT 2
1|| f
0 0 2 2 10 3 3 3s s
F FF Ff round round
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Example
tjjtjj eeeettx 80001.02180001.0
21)1.08000cos()(
kHzFs 3
Then:
• analog frequencies
• FT:
• digital frequencies
•DTFT
)4000()4000()( 1.0211.0
21 FeFeFX jj
FT
0 14000 , 4000F Hz F Hz
21|| f 3
11.021
311.0
21)( fefefX jj
DTFT
4 4 4 10 3 3 3 31f round
4 4 4 11 3 3 3 3( 1)f round
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Linear Time Invariant (LTI) Systems and z-Transform
][nx ][ny][nh
If the system is LTI we compute the output with the convolution:
m
mnxmhnxnhny ][][][*][][
If the impulse response has a finite duration, the system is called FIR (Finite Impulse Response):
][][...]1[]1[][]0[][ NnxNhnxhnxhny
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n
nznxnxZzX ][][)(
Z-Transform
Facts:
)()()( zXzHzY
][nx ][ny)(zH
Frequency Response of a filter:
fjezzHfH 2)()(
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Digital Filters
)(zH][nx ][ny
Ideal Low Pass Filter
f21
21
)( fH
f21
21
)( fH
Pf
Pf
passband
constant magnitude in passband…
… and linear phase
A
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Impulse Response of Ideal LPF
P
P
f
f
fnjfnjideal dfAedfefHnh 222
1
21
)(][
Assume zero phase shift,
nfsincAfnh PPideal 22][
This has Infinite Impulse Response, non recursive and it is non-causal. Therefore it cannot be realized.
-50 -40 -30 -20 -10 0 10 20 30 40 50-0.05
0
0.05
0.1
0.15
0.2
n
fp=0.1
[ ]h n
n
0.1
1Pf
A
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Non Ideal Ideal LPF
The good news is that for the Ideal LPF
0][lim
nhidealn
n
][nh
L L
nL2L
][nh
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Frequency Response of the Non Ideal LPF
Pf STOPf
passstop stop
f
transition region
attenuation
ripple
|)(| fH11
11
2
LPF specified by:
• passband frequency
• passband ripple or
• stopband frequency
• stopband attenuation or
Pf
1 dB RP 1
1
11
10log20
STOPf
2 dB R 2S 10log20
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Best Design tool for FIR Filters: the Equiripple algorithm (or Remez). It minimizes the maximum error between the frequency responses of the ideal and actual filter.
1f 2f 21
attenuation
ripple
|)(| fH11
11
2
1 2 3 3 1 2, 0, , , / , 1,1,0,0 , ,h firpm N f f f f w w
impulse response ][],...,0[ Nhhh
1f 2f 21
3 f0
Linear Interpolation
11/ w
2/ w
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The total impulse response length N+1 depends on:
• transition region
• attenuation in the stopband
1f 2f
|)(| fH
2
12 fff
N
f1
~ 22)(log20 210
Example:
we want
Passband: 3kHz
Stopband: 3.5kHz
Attenuation: 60dB
Sampling Freq: 15 kHz
Then: from the specs
We determine the order the filter
301
0.150.35.3 f
8230~ 2260 N
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Frequency response
0 0.1 0.2 0.3 0.4 0.5-100
-80
-60
-40
-20
0
20magnitude
digital frequency
dB
0 0.1 0.2 0.3 0.4 0.5-120
-100
-80
-60
-40
-20
0
20magnitude
digital frequency
dB
N=82
N=98
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Example: Low Pass Filter
0 0.1 0.2 0.3 0.4 0.5-120
-100
-80
-60
-40
-20
0
20magnitude
digital frequency
dB
Passband f = 0.2
Stopband f = 0.25 with attenuation 40dB
Choose order N=40/(22*(0.25-0.20))=37
| ( ) |H f
f
Almost 40dB!!!
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Example: Low Pass Filter
Passband f = 0.2
Stopband f = 0.25 with attenuation 40dB
Choose order N=40 > 37
| ( ) |H f
f0 0.1 0.2 0.3 0.4 0.5
-80
-70
-60
-50
-40
-30
-20
-10
0
10magnitude
digital frequency
dB OK!!!
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General FIR Filter of arbitrary Frequency Response
],...,,[
],...,,,0[
10
21
M
M
HHHH
ffff
01f 2f 3f 1Mf 2
1Mf
0H1H
2H
3H 1MH MH
Weights for Error:
1w2w
2/)1( Mw
],...,,[ 2/)1(21 Mwwww
Then apply:
, / , ,Mh firpm N f f H w
… and always check frequency response if it is what you expect!
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Example: ( ) 1/ sinc( )H f f for 0 0.2f
( ) 0H f 0.25 0.5f
0 0.2 0.25 0.5 f
fp=0:0.01:0.2; % vector of passband frequencies
fs=[0.25,0.5]; % stopband frequencies
M=[1./sinc(fp), 0, 0]; % desired magnitudesDf=0.25-0.2; % transition regionN=ceil(A/(22*Df)); % first guess of orderh=firpm(N, [ fp, fs]/0.5,M); % impulse response
40A dB
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0 0.1 0.2 0.3 0.4 0.50
0.2
0.4
0.6
0.8
1
1.2
1.4magnitude
digital frequency
0 0.1 0.2 0.3 0.4 0.5-90
-80
-70
-60
-50
-40
-30
-20
-10
0
10
not very good here!
dB
37N
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To improve it:
1. Increase order
2. Add weights
0 0.2 0.25 0.5 f
40A dB
1w 0.2w
w=[1*ones(1,length(fp)/2), 0.2*ones(1, length(fs)/2)];
h=firpm(N, [fp, fs]/0.5,M,w);
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0 0.1 0.2 0.3 0.4 0.50
0.2
0.4
0.6
0.8
1
1.2
1.4magnitude
digital frequency
0 0.1 0.2 0.3 0.4 0.5-160
-140
-120
-100
-80
-60
-40
-20
0
20
100N dB