This document discusses digital filters and the design of finite impulse response (FIR) filters. It covers topics such as the design of FIR filters using windows, properties of FIR filters including their linear phase characteristics, and comparisons between FIR and infinite impulse response (IIR) filters. MATLAB code is provided to demonstrate the effects of linear phase characteristics on filtered signals. Linear phase filters are shown to preserve signal shape while non-linear phase filters can distort signals.
The presentation covers sampling theorem, ideal sampling, flat top sampling, natural sampling, reconstruction of signals from samples, aliasing effect, zero order hold, upsampling, downsampling, and discrete time processing of continuous time signals.
It is sometimes desirable to have circuits capable of selectively filtering one frequency or range of frequencies out of a mix of different frequencies in a circuit. A circuit designed to perform this frequency selection is called a filter circuit, or simply a filter. A common need for filter circuits is in high-performance stereo systems, where certain ranges of audio frequencies need to be amplified or suppressed for best sound quality and power efficiency. You may be familiar with equalizers, which allow the amplitudes of several frequency ranges to be adjusted to suit the listener's taste and acoustic properties of the listening area. You may also be familiar with crossover networks, which block certain ranges of frequencies from reaching speakers. A tweeter (high-frequency speaker) is inefficient at reproducing low-frequency signals such as drum beats, so a crossover circuit is connected between the tweeter and the stereo's output terminals to block low-frequency signals, only passing high-frequency signals to the speaker's connection terminals. This gives better audio system efficiency and thus better performance. Both equalizers and crossover networks are examples of filters, designed to accomplish filtering of certain frequencies.
The presentation covers sampling theorem, ideal sampling, flat top sampling, natural sampling, reconstruction of signals from samples, aliasing effect, zero order hold, upsampling, downsampling, and discrete time processing of continuous time signals.
It is sometimes desirable to have circuits capable of selectively filtering one frequency or range of frequencies out of a mix of different frequencies in a circuit. A circuit designed to perform this frequency selection is called a filter circuit, or simply a filter. A common need for filter circuits is in high-performance stereo systems, where certain ranges of audio frequencies need to be amplified or suppressed for best sound quality and power efficiency. You may be familiar with equalizers, which allow the amplitudes of several frequency ranges to be adjusted to suit the listener's taste and acoustic properties of the listening area. You may also be familiar with crossover networks, which block certain ranges of frequencies from reaching speakers. A tweeter (high-frequency speaker) is inefficient at reproducing low-frequency signals such as drum beats, so a crossover circuit is connected between the tweeter and the stereo's output terminals to block low-frequency signals, only passing high-frequency signals to the speaker's connection terminals. This gives better audio system efficiency and thus better performance. Both equalizers and crossover networks are examples of filters, designed to accomplish filtering of certain frequencies.
A signal is a pattern of variation that carry information.
Signals are represented mathematically as a function of one or more independent variable
basic concept of signals
types of signals
system concepts
Using Chebyshev filter design, there are two sub groups,
Type-I Chebyshev Filter
Type-II Chebyshev Filter
The major difference between butterworth and chebyshev filter is that the poles of butterworth filter lie on the circle while the poles of chebyshev filter lie on ellipse.
A signal is a pattern of variation that carry information.
Signals are represented mathematically as a function of one or more independent variable
basic concept of signals
types of signals
system concepts
Using Chebyshev filter design, there are two sub groups,
Type-I Chebyshev Filter
Type-II Chebyshev Filter
The major difference between butterworth and chebyshev filter is that the poles of butterworth filter lie on the circle while the poles of chebyshev filter lie on ellipse.
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Digital Signal Processing Tutorial: Chapt 4 design of digital filters (FIR)
1. 10/21/2009
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Chapt 04
Design of Digital Filters ( FIR)
B.E. Comps, Mumbai Uni
PrePrepared by
IRDC India
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Chapt 04 Design of Digital Filters
•Design of FIR filters
•Design of IIR filters from analog filters
•Frequency transformation
•Design of digital filters based on least-squares method
•Digital filters from analog filters
•Properties of FIR filters
•Design of FIR filters using windows
•Comparison of IIR and FIR filters
•Linear phase filters
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What is filter ?
Basic building block of DSP:
Input is given to filter and output of the system would be
signal obtained from input and filter's impulse response.
Filter
Input Output
[ 1 1]
Input Output
[ 1 -1]
Input Output
Integrator ( Low
pass filter)
Differentiator
(High pass filter)
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Contd..
Different impulse response different characteristics
Characteristic of the filter – Frequency domain characteristics
Frequency
Characteristic
Phase
Characteristic
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Frequency Characteristic
H(w)
Wπ
Frequency characteristic plot between frequencies and magnitude
For given system if particular frequency is passed thorough the system,
its magnitude at output can be obtained from corresponding magnitude
from frequency characteristic
Note: most of the time input wont be single pure sinusoidal/frequency but
would be a set of frequencies
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Phase Characteristic
ϕ(w)
Wπ
Phase characteristic plot between frequencies and phase shift
For given system if particular frequency is passed thorough the system,
its delay/phase shift at output can be obtained from corresponding phse
from phase characteristic
Note: most of the time input wont be single pure sinusoidal/frequency but
would be a set of frequencies
π−
π
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Ideal Characteristics
Frequency Phase
Sharp cutoff
Linear
phase
Sharp cutoff is requirement for
almost all the applications
Non-linear phase characteristic
distorts the signal.
Linear phase characteristic has
constant group delay.
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Effect of Linear Phase Characteristic
Let us pass signal , made up of summing frequency 1 Hz, 2Hz & 3
Hz, through filters with different phase characteristics
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M file used in MATLAB for this simulation
clear all;
close all;
figure,
subplot(4,1,1);
plot(sin(2*pi*1*[0:0.01:3]));
title(' Inoput Signal Composition');
subplot(4,1,2);
plot(sin(2*pi*2*[0:0.01:3]));
subplot(4,1,3);
plot(sin(2*pi*3*[0:0.01:3]));
subplot(4,1,4);
plot(sum([sin(2*pi*1*[0:0.01:3]);sin(2*pi*2*[0:0.01:3]);sin(2*pi*3*[0:0.01:3])]));
figure,
subplot(4,1,1);
plot(sin(2*pi*1*[0:0.01:3]+pi/8));
title(' Filtered Signal with phase characteristic pi/8');
subplot(4,1,2);
plot(sin(2*pi*2*[0:0.01:3]+pi/8));
subplot(4,1,3);
plot(sin(2*pi*3*[0:0.01:3]+pi/8));
subplot(4,1,4);
plot(sum([sin(2*pi*1*[0:0.01:3]+pi/8);sin(2*pi*2*[0:0.01:3]+pi/8);sin(2*pi*3*[0:0.01:3]+pi/8)]));
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Contd..
figure,
subplot(4,1,1);
plot(sin(2*pi*1*[0:0.01:3]+pi/4));
title(' Filtered Signal with phase characteristic pi/4');
subplot(4,1,2);
plot(sin(2*pi*2*[0:0.01:3]+pi/4));
subplot(4,1,3);
plot(sin(2*pi*3*[0:0.01:3]+pi/4));
subplot(4,1,4);
plot(sum([sin(2*pi*1*[0:0.01:3]+pi/4);sin(2*pi*2*[0:0.01:3]+pi/4);sin(2*pi*3*[0:0.01:3]+pi/4)]));
figure,
subplot(4,1,1);
plot(sin(2*pi*1*[0:0.01:3]+pi/2));
title(' Filtered Signal with phase characteristic pi/2');
subplot(4,1,2);
plot(sin(2*pi*2*[0:0.01:3]+pi/2));
subplot(4,1,3);
plot(sin(2*pi*3*[0:0.01:3]+pi/2));
subplot(4,1,4);
plot(sum([sin(2*pi*1*[0:0.01:3]+pi/2);sin(2*pi*2*[0:0.01:3]+pi/2);sin(2*pi*3*[0:0.01:3]+pi/2)]));
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Contd..
figure,
subplot(4,1,1);
plot(sin(2*pi*1*[0:0.01:3]+pi/8));
title(' Filtered Signal with phase characteristic w(linear phase with group delay 1)');
subplot(4,1,2);
plot(sin(2*pi*2*[0:0.01:3]+pi/4));
subplot(4,1,3);
plot(sin(2*pi*3*[0:0.01:3]+pi/2));
subplot(4,1,4);
plot(sum([sin(2*pi*1*[0:0.01:3]+pi/8);sin(2*pi*2*[0:0.01:3]+pi/4);sin(2*pi*3*[0:0.01:3]+pi/2)]));
figure,
subplot(4,1,1);
plot(sin(2*pi*1*[0:0.01:3]+pi/4));
title(' Filtered Signal with phase characteristic 2w(linear phase with group delay 2)');
subplot(4,1,2);
plot(sin(2*pi*2*[0:0.01:3]+pi/2));
subplot(4,1,3);
plot(sin(2*pi*3*[0:0.01:3]+pi/1));
subplot(4,1,4);
plot(sum([sin(2*pi*1*[0:0.01:3]+pi/4);sin(2*pi*2*[0:0.01:3]+pi/2);sin(2*pi*3*[0:0.01:3]+pi/1)]));
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Chapt 04
figure,
subplot(6,1,1);
plot(sum([sin(2*pi*1*[0:0.01:3]);sin(2*pi*2*[0:0.01:3]);sin(2*pi*3*[0:0.01:3])]));
title('I am Input Signal , which of followings looks like me ?');
subplot(6,1,2);
plot(sum([sin(2*pi*1*[0:0.01:3]+pi/8);sin(2*pi*2*[0:0.01:3]+pi/8);sin(2*pi*3*[0:0.01:3]+pi/8)]));
%title(' Filtered Signal with phase characteristic pi/8');
subplot(6,1,3);
plot(sum([sin(2*pi*1*[0:0.01:3]+pi/4);sin(2*pi*2*[0:0.01:3]+pi/4);sin(2*pi*3*[0:0.01:3]+pi/4)]));
%title(' Filtered Signal with phase characteristic pi/4');
subplot(6,1,4);
plot(sum([sin(2*pi*1*[0:0.01:3]+pi/2);sin(2*pi*2*[0:0.01:3]+pi/2);sin(2*pi*3*[0:0.01:3]+pi/2)]));
%title(' Filtered Signal with phase characteristic pi/2');
subplot(6,1,5);
plot(sum([sin(2*pi*1*[0:0.01:3]+pi/8);sin(2*pi*2*[0:0.01:3]+pi/4);sin(2*pi*3*[0:0.01:3]+pi/2)]));
%title(' Filtered Signal with phase characteristic w(linear phase with group delay 1)');
subplot(6,1,6);
plot(sum([sin(2*pi*1*[0:0.01:3]+pi/4);sin(2*pi*2*[0:0.01:3]+pi/2);sin(2*pi*3*[0:0.01:3]+pi/1)]));
%title(' Filtered Signal with phase characteristic 2w(linear phase with group delay 2)');
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Chapt 04
figure,
subplot(6,1,1);
plot(sum([sin(2*pi*1*[0:0.01:3]);sin(2*pi*2*[0:0.01:3]);sin(2*pi*3*[0:0.01:3])]));
title(' Input Signal ');
subplot(6,1,2);
plot(sum([sin(2*pi*1*[0:0.01:3]+pi/8);sin(2*pi*2*[0:0.01:3]+pi/8);sin(2*pi*3*[0:0.01:3]+pi/8)]));
title(' Filtered Signal with phase characteristic pi/8');
subplot(6,1,3);
plot(sum([sin(2*pi*1*[0:0.01:3]+pi/4);sin(2*pi*2*[0:0.01:3]+pi/4);sin(2*pi*3*[0:0.01:3]+pi/4)]));
title(' Filtered Signal with phase characteristic pi/4');
subplot(6,1,4);
plot(sum([sin(2*pi*1*[0:0.01:3]+pi/2);sin(2*pi*2*[0:0.01:3]+pi/2);sin(2*pi*3*[0:0.01:3]+pi/2)]));
title(' Filtered Signal with phase characteristic pi/2');
subplot(6,1,5);
plot(sum([sin(2*pi*1*[0:0.01:3]+pi/8);sin(2*pi*2*[0:0.01:3]+pi/4);sin(2*pi*3*[0:0.01:3]+pi/2)]));
title(' Filtered Signal with phase characteristic w(linear phase with group delay 1)');
subplot(6,1,6);
plot(sum([sin(2*pi*1*[0:0.01:3]+pi/4);sin(2*pi*2*[0:0.01:3]+pi/2);sin(2*pi*3*[0:0.01:3]+pi/1)]));
title(' Filtered Signal with phase characteristic 2w(linear phase with group delay 2)');
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Linear Phase Conclusion
If filter doesn't have linear phase characteristic( constant group
delay) then signal shape gets distorts.
e.g. First 3 filtered outputs
If filter has linear phase characteristic( constant group delay) then
signal shape would be preserved.
e.g. Last 2 filtered outputs
Linear phase can be obtained easily by FIR filter by having
symmetric/anti-symmetric property in its impulse response.
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FIR Filter
FIR filter has finite number of samples in its impulse response
Advantages:
• The question of stability and realizability never arise for FIR filters ( it is
always stable and realizable)
• It gives linear phase relationship with frequency , which can be
achieved by having symmetric or anti-symmetric impulse response of the
filter
Disadvantages :
• Long sequences for h(n) are generally required to adequately
approximate sharp cut-off filters
•Hence , higher computation complexity
•The delay of linear phase FIR filters need not always be an integer
number of samples . This non-integral delays can lead to problems in
some signal processing applications
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Characteristic of FIR filter
Let { h(n)} be a causal finite durations sequence of length N( 0 to N-1).
Its z-transform
∑
−
=
−
=
1
0
)()(
N
n
n
znhzH --------(1)
Its Fourier transform,
∑
−
=
−
=
1
0
)()(
N
n
jwnjw
enheH
Which is periodic in frequency with period of 2π
)()( )2( mjj
eHeH πωω +
= ......3,2,1,0 ±±±=mfor
--------(2)
Consider h(n) be real and its magnitude and phase ,
)(
)()( ωθjjwjw
eeHeH ±=
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Contd..
From eq (2) , since wnjwne jwn
sincos −=−
symmetric
Anti-
symmetric
Magnitude of Fourier transform is symmetric and the phase
is an anti-symmetric function
)()( jwjw
eHeH −
=
πω ≤≤0
)()( ωθωθ −−=
Consider , we have to have linear phase response
i.e. αωωθ −=)(
where α is constant ( phase delay in samples)
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Contd..
We would find , condition/restriction on impulse response such that it
will give linear phase response
Mathematically,
∑
−
=
−
=
1
0
)()(
N
n
jwnjw
enheH
αωjjw
eeH −
±= )( Requirement
∑
−
=
−
=
1
0
)()(
N
n
jwnjw
enheH
Given
∑∑
−
=
−
=
−=
1
0
1
0
)sin()()cos()(
N
n
N
n
wnnhjwnnh
−
=−
∑
∑
−
=
−
=−
1
0
1
01
)cos()(
)sin()(
tan N
n
N
n
wnnh
wnnh
αω
32. Since n=0, sin(wn)=0
Hence , limit starts
from 1
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Contd..
=
∑
∑
−
=
−
=
1
0
1
0
)cos()(
)sin()(
)tan( N
n
N
n
wnnh
wnnh
αω
+
=∴
∑
∑
−
=
−
=
1
1
1
1
)cos()()0(
)sin()(
)tan( N
n
N
n
wnnhh
wnnh
αω
There are two possible solutions
1) α=0 , h(0) can be arbitrary
& h(n)=0 for n≠0
In this case filter will have order 0 and it would be
just amplifier and not a filter. This is not too
useful result
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Contd..
2) α≠0
==∴
∑
∑
−
=
−
=
1
0
1
0
)cos()(
)sin()(
)cos(
)sin(
)tan( N
n
N
n
nnh
nnh
ω
ω
αω
αω
αω
∑∑
−
=
−
=
=∴
1
0
1
0
)cos()sin()()sin()cos()(
N
n
N
n
nnhnnh αωωαωω
0)}cos()sin()sin()){cos((
1
0
=−∴∑
−
=
αωωαωω nnnh
N
n
0)sin()(
1
0
=−∴∑
−
=
nnh
N
n
ωαω
0)(sin)(
1
0
=−∴∑
−
=
nnh
N
n
αω
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Contd..
From this equation , unique solution is obtained for the set of
conditions
1) α= (N-1)/2
i.e. for any value of sequence , there is only one value of phase delay
α , which is the condition to obtain linear phase.
2) h(n)=±h(N-1-n) for 0 ≤ n ≤ N-1
i.e. impulse response sequence must have a special kind of symmetry
for the value of α
For linear phase filter , impulse response should be either symmetric or
anti-symmetric.
As impulse response can be of odd or even length, there are four
possible types of impulse response which will have linear phase
characteristic.
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Linear Phase Filters – Impulse Responses
1) Symmetric and Even N
2) Anti-symmetric and Even N
3) Symmetric and Odd N
4) Anti-symmetric and Odd N
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Symmetric and Even & Odd N
0 1 2 3 4 5 6 7 8 9 10
N=11 ( Odd)
α= 5
Center of
symmetry
0 1 2 3 4 5 6 7 8 9
N=10 ( Even)
α= 4.5
Center of
symmetry
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Anti-Symmetric and Even & Odd N
0 1 2 3 4 5 6 7 8 9 10
N=11 ( Odd)
α= 5
Center of
symmetry
0 1 2 3 4 5 6 7 8 9
N=10 ( Even)
α= 4.5
Center of
symmetry
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Proofs of Linear phase characteristics for all four types
Refer Rabinar and Gold
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Position of zeros in Linear phase filters
∑
−
=
−
=
1
0
)()(
N
n
n
znhzHWe have
)1()2()3(
4321
)0()1()2(
)4()3()2()1()0(
−−−−−−
−−−−
±±±−−−
−−−++++=
NNN
zhzhzh
zhzhzhzhh
→± Symmetry/ anti-symmetry in impulse response
2/)1(
)( −−
= N
zzH [ ]
[ ]
[ ]
}
)2(
)1(
)0({
2/)5(2/)5(
2/)3(2/)3(
2/)1(2/)1(
−−−−−+
±+
±+
±
−−−
−−−
−−−
NN
NN
NN
zzh
zzh
zzh
-----------(1)
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Contd..
If z is replaced by z-1 in eq 1 we get
2/)1(1
)( −−
= N
zzH [ ]
[ ]
[ ]
}
)2(
)1(
)0({
2/)5(2/)5(
2/)3(2/)3(
2/)1(2/)1(
−−−−−+
±+
±+
±
−−−
−−−
−−−
NN
NN
NN
zzh
zzh
zzh
-----------(2)
Comparing eq 1 & 2 we get
)()( )1(1
zHzzH N −−
±=
H(z) and H(z-1) are identical within a
delay of (N-1) samples and multiplier ±1
(r,-Ɵ)
(r,Ɵ)
(1/r,Ɵ)
(1/r,-Ɵ)
43. 10/21/2009
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Ideal Filter
Consider ideal low-pass filter characteristic
H(w)
Wc w
=
0
1
)(ωH
πωω
ωω
≤<
≤
c
c
By taking inverse fourier transform we get impulse response of the
above filter as
=
n
n
c
cc
c
nh
ω
ω
π
ω
π
ω
sin
)(
0
0
≠
=
n
n
44. 10/21/2009
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Why ideal filter is not physically realizable ?
Two observations of above h(n) that makes ideal filter not realizable
•Length of h(n) is infinite
•Impulse response is not causal. To make is causal , shifting
response could be one way, but shifting by infinity amount is not
possible .
48. 10/21/2009
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Design of FIR filter
FIR filters can be designed by using following methods
•Window method ( Fourier Series Method)
•Frequency sampling method
•Optimal filter design
49. 10/21/2009
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Windowing Technique
• Desired frequency response with specifications Hd(w)
•Take inverse Fourier transform of Hd(w) to obtain filter impulse
response hd(n)
•Since , impulse response hd(n) is infinite in duration, window
w(n) is used to truncate it and gives impulse sample response of
the FIR filter.
i.e. hw(n)= hd(n). w(n)
Since window function w(n) will have finite samples (say M ),
we get
=
0
)().(
)(
nwnh
nh d
w
otherwise
Mn 2/,......2,1,0 ±±±=
• Give a shift of M/2 samples to hw(n) to make it causal
)2/()( Mnhnh w −=
50. Name of
Window
Time domain sequence h(n) , 0 ≤
n ≤ M-1
Approx
Transition width
of main lobe
Peak
Sidelobe
(dB)
Rectangular 1 4π/M -13
Barlett (
traingualr)
8π/M -27
Blakman 8π/M -32
Hamming 8π/M -43
Hanning 12π/M -58
Kaiser -- --
10/21/2009
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Different Windows
1
4
cos08.0
1
2
cos5.042.0
−
+
−
−
M
n
M
n ππ
1
2
1
2
1
−
−
−
−
M
M
n
1
2
cos46.054.0
−
−
M
nπ
−
−
1
2
cos1
2
1
M
nπ
−
−
−−
−
2
1
2
1
2
1
22
M
I
M
n
M
I
o
o
α
α
52. 10/21/2009
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Frequency response to impulse response
1) Ideal low pass filter
=
0
1
)(ωdH
otherwise
cωω ≤≤ ||0H(w)
-π -Wc Wc π w
∫−
=
π
π
ω
π ωω deHnh nj
dd )()( 2
1
c
c
c
c
jn
e
de
nj
nj
ω
ω
ω
π
ω
ω
ω
π ω
−−
== ∫ 2
1
2
1
−
=
−
jn
ee njnj cc ωω
π2
1
−
=
−
j
ee
n
njnj cc
2
11 ωω
π
53. 10/21/2009
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Contd..
n
n
d
c
nh π
ωsin
)( = for 0≠n
for 0=n
∫−
=
c
c
denh nj
d
ω
ω
ω
π ω2
1
)(
∫∫ −−
==
c
c
c
c
ddeh j
d
ω
ω
π
ω
ω
ω
π ωω .1)0( 2
10.
2
1
π
ω
π
ω
π
ωω cc
d
cc
h === −−
2
2
)0( 2
)(
=
n
n
nh
c
c
d
π
ω
π
ω
sin
)(
0
0
≠
=
n
n
55. 10/21/2009
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Contd..
=
−−
0
)(
)( 2
1 ω
ω
Nj
d
e
H
otherwise
cωω ≤≤ ||0
If
then
−
−
=
−
−
)(
)(sin
)(
2
1
2
1
N
N
c
c
d
n
n
nh
π
ω
π
ω
otherwise
n N
2
1−
=
56. 10/21/2009
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High Pass Filter
H(w)
-π -Wc Wc π w
=
0
1
)(ωdH
otherwise
c πωω ≤≤ ||
∫−
=
π
π
ω
ωω
π
deHnh nj
dd )(
2
1
)(
+= ∫∫
− π
ω
ω
ω
π
ω
ωω
π c
c
dede njnj
.1.1
2
1
+=
−
−
π
ω
ωω
π
ω
π c
c
jn
e
jn
e njnj
2
1
[ ]njnjnjnj cc
eeee
jn
ωππω
π
−+−= −−
2
1
−
−
−
−=
−−
j
ee
j
j
ee
j
jn
njnjnjnj cc
2
2
2
2
2
1 ππωω
π
57. 10/21/2009
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Contd..
( )nn
n
nh cd πω
π
sinsin
1
)( +
−
=
n
n
nh c
d
π
ωsin
)(
−
= 0sin =nπQ for all integers n
for 0=n
+= ∫∫
− π
ω
ω
ω
π
ω
ωω
π c
c
dedeh jj
d
0.0.
.1.1
2
1
)0(
+= ∫∫
− π
ω
ω
π
ωω
π c
c
dd .1.1
2
1 ( )cc ωππω
π
−++−=
2
1
( )cωπ
π
22
2
1
−=
π
ωc
−=1
−
−
=
n
n
nh
c
c
d
π
ω
π
ω
sin
1
)(
0
0
≠
=
n
n
60. 10/21/2009
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Contd..
( )nn
n
nh ccd 12 sinsin
1
)( ωω
π
−= 0≠n
( )
π
ωω
ωω
π
12
12 )(2
2
1
)( cc
ccd nh
−
=−= 0=n
61. 10/21/2009
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Band-stop filter
( )nn
n
nh ccd 21 sinsin
1
)( ωω
π
−= 0≠n
1)( 21
+
−
=
π
ωω cc
d nh 0=n
Proceed in same way ….
67. 10/21/2009
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Problem
a) Design a low pass , linear phase , FIR filter of length 9 ( order 8) with
cut-off frequency of 5kHz and sampling frequency 20 kHz. Use Hann
window with
Find the delay involved.
a) Find magnitude and phase response of this filter and obtain the plots.
Derive necessary expressions
b) Find the cut-off frequency of a designed filter
−
−=
1
2
cos1
2
1
)(
M
n
nw
π
10 −≤≤ Mn
68. 10/21/2009
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Solution
Ideal low pass filter with cutoff frequency ,fc= 5kHz and sampling frequency
fs=20kHz
=
0
1
)(ωdH
otherwise
cωω ≤≤ ||0H(w)
-π -Wc Wc π w
∫−
=
π
π
ω
π ωω deHnh nj
dd )()( 2
1
c
c
c
c
jn
e
de
nj
nj
ω
ω
ω
π
ω
ω
ω
π ω
−−
== ∫ 2
1
2
1
−
=
−
jn
ee njnj cc ωω
π2
1
−
=
−
j
ee
n
njnj cc
2
11 ωω
π
As fs=20 kHz is equivalent to 2π,
fc=5kHz corresponds to wc=2π*5/20=0.5
π
69. 10/21/2009
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Contd..
n
n
d
c
nh π
ωsin
)( = for 0≠n
for 0=n
∫−
=
c
c
denh nj
d
ω
ω
ω
π ω2
1
)(
∫∫ −−
==
c
c
c
c
ddeh j
d
ω
ω
π
ω
ω
ω
π ωω .1)0( 2
10.
2
1
π
ω
π
ω
π
ωω cc
d
cc
h === −−
2
2
)0( 2
)(
=
n
n
nh
c
c
d
π
ω
π
ω
sin
)(
0
0
≠
=
n
n
70. 10/21/2009
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Contd..
Using above equation , obtain desired filter coefficients from h(-4) to
h(4) for filter with wc=0.5 π and length 9
hd(0)=0.5
hd(1)= 0.31830 =hd (-1)
hd(2)=0 =hd (-2)
hd(3)=0.10615 =hd (-3)
hd(4)=0 =hd (-4) hc(4)=0.5
hc(5)= 0.31830 =hc (3)
hc(6)=0 =hc (2)
hc(7)=0.10615 =hc (1)
hc(8)=0 =hc (0)
Shift by 4 to
make it causal
n’ =n+4
72. 10/21/2009
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Contd..
Multiply filter coefficients now with window function
−
−=
1
2
cos1
2
1
)(
M
n
nw
π
10 −≤≤ Mn
h(4)=0.5*w(4)=0.5
h(3)= 0.31830*w(3)=0.27168 =h (5)
h(2)=0*w(2)=0 =h(6)
h(1)=0.10615*w(1)=0.01554 =h (7)
h(0)=0*w(0)=0 =h (8)
73. 10/21/2009
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Frequency Sampling Technique
In this method, desired frequency response is sampled to obtain DFT
coefficients, which are then passed through IDFT to get impulse response
Desired
Frequency
Response
Hd(ejw)
Sampling
DFT coefficients
H(k)
IDFT Filter coefficients
h(n)
Let us sample desired frequency response Hd(ejw) at N points
wk , k=0 ,1 , 2 ….N-1( N being length of filter)
N
k
wk
π2
= 1,,.........1,0 −= Nk
kww
jw
d eHkH
=
= )()(
~
Sampled desired frequency response
1,,.........1,0 −= Nk
)( /2 Nkj
d eH π
= 1,,.........1,0 −= Nk
74. 10/21/2009
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Contd..
)(
~
kHLet us consider, as a DFT coefficients , then
Nknj
N
k
ekH
N
nh /2
1
0
~
)(
1
)( π
∑
−
=
= 1,,.........1,0 −= Nn
For the implementation of filter , taking z transform of h(n)
n
N
n
Nknj
N
k
zekH
N
−
−
=
−
=
∑ ∑
=
1
0
/2
1
0
~
)(
1 π
∑
−
=
−
=
1
0
)()(
N
n
n
znhzH
By changing the order of summation , we get
∑ ∑
−
=
−
=
−
=
1
0
1
0
/2
~ 1
)()(
N
k
N
k
nNknj
ze
N
kHzH π
75. 10/21/2009
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Contd..
( )∑ ∑
−
=
−
=
−
=
1
0
1
0
1/2
~ 1
)()(
N
k
N
k
nNkj
ze
N
kHzH π
∑
−
=
−
−
−
−
=
1
0
1/2
2~
1
11
)(
N
k
Nkj
kjN
ze
ez
N
kH π
π
∑
−
=
−
−
−
−
=
1
0
1/2
~
1
11
)(
N
k
Nkj
N
ze
z
N
kH π
∑
−
=
−
−
−
−
=
1
0
1/2
~
1
)(1 N
k
Nkj
N
ze
kH
N
z
π
This equation can be
directly used for realization
of FIR filter
77. 10/21/2009
e-TECHNote from IRDC India
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Contd..
Realization Structure
x(n) +
Z-M
N
1
+
Z-1
Nj
e /0.2π
)0(
~
H
+
Z-1
Nj
e /1.2π
)1(
~
H
+
Z-1
NNj
e /)1(2 −π
)1(
~
−NH
+
+
y(n)
78. 10/21/2009
e-TECHNote from IRDC India
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Problem
Design a low pass digital filter with cut-off frequency wc = π/2 using
frequency sampling technique for N=9.
79. 10/21/2009
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Solution
∑
−
=
−
−
−
−
=
1
0
1/2
~
1
)(1
)(
N
k
Nkj
N
ze
kH
N
z
zH πDerive for realization
)(
~
kHCalculate 1,,.........1,0 −= Nk
As given ,
=
0
1
1
)(ωdH
otherwise
πωπ
πω
2||2/3
2/||0
≤≤
≤≤
=
N
k
HkH d
π2
)(
~
80. 10/21/2009
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Contd..
k or
0 0 No 1
1 π/4.5 No 1
2 π/2.25 No 1
3 π/1.5 Yes 0
4 π/1.125 Yes 0
5 π/0.9 Yes 0
6 π/0.75 Yes 0
7 π/0.642 No 1
8 π/0.562 No 1
9
2 kπ 2/π>
)(
~
kH2/3π<