Digital Signal Processing Tutorial: Chapt 4 design of digital filters (FIR)

10/21/2009
e-TECHNote from IRDC India
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Chapt 04
Design of Digital Filters ( FIR)
B.E. Comps, Mumbai Uni
PrePrepared by
IRDC India

10/21/2009
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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Copyright© with Authors. All right reserved
For education purpose.
Commercialization of this material is
strictly not allowed without permission
from author.

10/21/2009
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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from author.

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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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Contd..

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from author.

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Contd..

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Challenge …..

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from author.

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Verify ….

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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));
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);
figure,
subplot(4,1,1);
plot(sin(2*pi*1*[0:0.01:3]+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);

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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);
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);

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from author.

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Chapt 04
figure,
subplot(6,1,1);
title('I am Input Signal , which of followings looks like me ?');
subplot(6,1,2);
%title(' Filtered Signal with phase characteristic pi/8');
subplot(6,1,3);
subplot(6,1,4);
subplot(6,1,5);
%title(' Filtered Signal with phase characteristic w(linear phase with group delay 1)');
subplot(6,1,6);
%title(' Filtered Signal with phase characteristic 2w(linear phase with group delay 2)');

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Chapt 04
figure,
subplot(6,1,1);
title(' Input Signal ');
subplot(6,1,2);
subplot(6,1,3);
subplot(6,1,4);
subplot(6,1,5);
title(' Filtered Signal with phase characteristic w(linear phase with group delay 1)');
subplot(6,1,6);
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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from author.

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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

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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from author.

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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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from author.

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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)

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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

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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 .

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Filter Specifications

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Contd..

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from author.

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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

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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 −=

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 -- --
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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

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from author.

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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

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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

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from author.

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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−
=

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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

10/21/2009
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

-π -Wc2 -Wc1 Wc1 Wc2 π w
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Band-Pass Filter
H(w)



=
0
1
)(ωdH
otherwise
cc 21 || ωωω ≤≤
∫−
=
π
π
ω
ωω
π
deHnh nj
dd )(
2
1
)(








+= ∫∫
−
−
2
1
1
2
.1.1
2
1 c
c
c
c
dede njnj
ω
ω
ω
ω
ω
ω
ωω
π








+=
−
−
2
1
1
2
2
1
c
c
c
c
jn
e
jn
e njnj ω
ω
ωω
ω
ω
π
[ ]njnjnjnj cccc
eeee
jn
1221
2
1 ωωωω
π
−+−= −−











 −
−




 −
−=
−−
j
ee
j
j
ee
j
jn
njnjnjnj cccc
2
2
2
2
2
1 1122 ωωωω
π

10/21/2009
from author.

10/21/2009
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

10/21/2009
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 ….

10/21/2009
Hilbert Transform
-π π w
H(w)
j
-j


−
=
j
j
Hd )(ω
0
0
≤≤−
≤≤
ωπ
πω
∫−
=
π
π
ω
ωω
π
deHnh nj
dd )(
2
1
)(






−= ∫∫−
π
ω
π
ω
ωω
π 0
0
..
2
1
dejdej njnj








−=
−
πω
π
ω
π 0
0
2 jn
e
jn
ej njnj
[ ]11
2
1
+−−= − njnj
ee
jn
ππ
π











 +
−=
−
2
22
2
1 njnj
ee
n
ππ
π

10/21/2009
Contd..
( )n
n
nhd π
π
cos1
1
)( −=
0)( =nhd
0=n
( ))cos(1
1
22 nn
n
ππ
π
+−= ( ))(sin)(cos1
1
2
2
2
2
nn
n
ππ
π
+−=
( ))(sin)(sin
1
2
2
2
2
nn
n
ππ
π
+=
n
n
nhd
π
π
)(sin2
)( 2
2
= 0≠n

10/21/2009
FIR differentiator
πωπ ≤≤−jwHd =)(ω
∫∫ −−
==
π
π
ω
π
π
ω
ωω
π
ωω
π
dejdeHnh njnj
dd .
2
1
)(
2
1
)(
π
π
ωω
ω
π −






−= 22
.
2
1
nj
e
jn
e
j
njnj






−+−=
−
−
jn
e
e
jn
e
e
jn
j nj
nj
nj
nj
π
π
π
π
ππ
π
..
2
( )





−−




 +
= −
−
njnj
njnj
ee
jn
ee
n
ππ
ππ
π
π
1
2
2
2
1

10/21/2009
Contd..











 −
−




 +
=
−−
j
ee
n
ee
n
nh
njnjnjnj
d
2
2
2
2
2
1
)(
ππππ
π
π




−= n
n
n
n
πππ
π
sin
2
cos2
2
1
[ ]0cos2
2
1
−= n
n
ππ
π
n
n
nhd
πcos
)( =
0)( =nhd
0=n
0≠n

10/21/2009
from author.

10/21/2009
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

10/21/2009
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

10/21/2009
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

10/21/2009
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

10/21/2009
from author.

10/21/2009
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)

10/21/2009
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

10/21/2009
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

10/21/2009
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

10/21/2009
from author.

10/21/2009
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)

10/21/2009
Problem
Design a low pass digital filter with cut-off frequency wc = π/2 using
frequency sampling technique for N=9.

10/21/2009
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
)(
~

10/21/2009
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<

10/21/2009
from author.

10/21/2009
Contd..
Realization
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)

10/21/2009
End of Chapter 04
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Digital Signal Processing Tutorial: Chapt 4 design of digital filters (FIR)

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