ECE 3640
Katelyn Johnson
A01984215
April 23, 2019
HW10
1. This problem explores the design a symmetric real-valued linear phase FIR bandpass filter that meets
the following specifications.
• Design this filter using the window technique using the best window from the set (rectangular,
Bartlett, Hann, Hamming, Blackman). Indicate which window shape you used.
• Design this filter using the window technique using the Kaiser window. Indicate the Kaiser window
parameters (β and M).
• Using FDATool in Matlab, design this filter using equiripple design method.
Make a table with one row for each filter you designed (window, Kaiser, equiripple). The table should
have a column for the following parameters: length, order, group delay, type (I, II, III, or IV), and
window parameters.
Answer: The length of the Hann filter was increased to make the filter a type I.
Hann
Kaiser
Equiripple
Length
63
47
37
Order
62
46
36
Group Delay
31
23
18
Type
I
I
I
Window Parameters
N/A
B = 3.3953
N/A
For each filter you design, turn in plots of the impulse response, magnitude response (linear and dB),
and phase response. The frequency domain plots should show response over the interval − 12 ≤ f ≤ 21 .
Also include a pole-zero plot.
Plots for the Hann filter
1
ECE 3640
Plots for the Kaiser filter
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ECE 3640
Plots for the Equiripple filter
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ECE 3640
Matlab Code
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%Hann
L = 63;
M = L−1;
n = [ 0 :M] ’ ;
a = M/ 2 ;
w = hann ( L ) ;
h1 = s i n ( 2 ∗ p i ∗ 0 . 3 5 ∗ ( n−a ) ) . / ( p i ∗ ( n−a ) ) ;
i n d = f i n d ( n−a==0) ;
h1 ( i n d ) = 2 ∗ 0 . 3 5 ;
h2 = s i n ( 2 ∗ p i ∗ . 1 5 ∗ ( n−a ) ) . / ( p i ∗ ( n−a ) ) ;
i n d = f i n d ( n−a==0) ;
h2 ( i n d ) = 2 ∗ 0 . 1 5 ;
h = h1−h2 ;
H = f f t s h i f t ( f f t ( h . ∗ w , N) ) ;
figure (1) ;
stem ( n , h . ∗w) ;
t i t l e ( ’ Impulse Response ’ ) ;
N = 2ˆ16;
f = [−N / 2 : ( N−1) / 2 ] /N ;
p r i n t −dpng 1 h i m p u l s e . png
figure (2) ;
p l o t ( f , 2 0 ∗ l o g 1 0 ( a b s (H) ) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e dB ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gni tu de ’ ) ;
g r i d on ;
p r i n t −dpng 1hmagdB . png
figure (3) ;
p l o t ( f , a b s (H) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gni tu de ’ ) ;
g r i d on ;
p r i n t −dpng 1hmagL . png
figure (4) ;
p l o t ( f , unwrap ( a n g l e (H) ) ) ;
t i t l e ( ’ Phase R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Amplitude ’ ) ;
g r i d on ;
p r i n t −dpng 1 h p h a s e . png
figure (5) ;
z p l a n e ( ( h . ∗w) . ’ , 1 ) ;
a x i s ([ −1 1 −1 1 ] ∗ 2 ) ;
t i t l e ( ’ Zero Pole Plot ’ ) ;
p r i n t −dpng 1 h z e r o . png
%K a i s e r
M = 46;
L = M+1;
n = [ 0 :M] ’ ;
beta = 3 . 3 9 5 3 ;
a = M/ 2 ;
w = k a i s e r (L , beta ) ;
h1 = s i n ( 2 ∗ p i ∗ 0 . 3 5 ∗ ( n−a ) ) . / ( p i ∗ ( n−a ) ) ;
i n d = f i n d ( n−a==0) ;
h1 ( i n d ) = 2 ∗ 0 . 3 5 ;
h2 = s i n ( 2 ∗ p i ∗ . 1 5 ∗ ( n−a ) ) . / ( p i ∗ ( n−a ) ) ;
i n d = f i n d ( n−a==0) ;
h2 ( i n d ) = 2 ∗ 0 . 1 5 ;
h = h1−h2 ;
H = f f t s h i f t ( f f t ( h . ∗ w , N) ) ;
figure (6) ;
stem ( n , h . ∗w) ;
t i t l e ( ’ Impulse Response ’ ) ;
p r i n t −dpng 1 k i m p u l s e . png
figure (7) ;
p l o t ( f , 2 0 ∗ l o g 1 0 ( a b s ( f f t ( h . ∗ w , N) ) ) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e dB ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gni tu de ’ ) ;
g r i d on ;
p r i n t −dpng 1kmagdB . png
figure (8) ;
p l o t ( f , a b s ( f f t ( h . ∗ w , N) ) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gni tu de ’ ) ;
g r i d on ;
p r i n t −dpng 1kmagL . png
figure (9) ;
p l o t ( f , unwrap ( a n g l e ( f f t ( h . ∗ w , N) ) ) ) ;
t i t l e ( ’ Phase R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Amplitude ’ ) ;
g r i d on ;
p r i n t −dpng 1 k p h a s e . png
figure (10) ;
z p l a n e ( ( h . ∗w) . ’ , 1 ) ;
a x i s ([ −1 1 −1 1 ] ∗ 2 ) ;
t i t l e ( ’ Zero Pole Plot ’ ) ;
p r i n t −dpng 1 k z e r o . png
%E q u i r i p p l e
figure (11) ;
stem ( [ 0 : l e n g t h ( e q u i ) −1 ] , e q u i ) ;
ECE 3640
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t i t l e ( ’ Impulse Response ’ ) ;
p r i n t −dpng 1 e i m p u l s e . png
figure (12) ;
p l o t ( f , 2 0 ∗ l o g 1 0 ( a b s ( f f t s h i f t ( f f t ( e q u i , N) ) ) ) ) ;
t i t l e ( ’ Ma gni tu de R e s p o n s e dB ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gn itu de ’ ) ;
g r i d on ;
p r i n t −dpng 1emagdB . png
figure (13) ;
p l o t ( f , a b s ( f f t s h i f t ( f f t ( e q u i , N) ) ) ) ;
t i t l e ( ’ Ma gni tu de R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gn itu de ’ ) ;
g r i d on ;
p r i n t −dpng 1emagL . png
figure (14) ;
%i s t h i s how t h e p h a s e s h o u l d be o r s h o u l d i t
p l o t ( f , unwrap ( a n g l e ( f f t s h i f t ( f f t ( e q u i , N) ) ) ) ) ;
t i t l e ( ’ Phase R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Amplitude ’ ) ;
g r i d on ;
p r i n t −dpng 1 e p h a s e . png
be
the
zero
phase ?
figure (15) ;
zplane (( equi ) ,1) ;
a x i s ([ −1 1 −1 1 ] ∗ 2 ) ;
t i t l e ( ’ Zero Pole Plot ’ ) ;
p r i n t −dpng 1 e z e r o . png
2. This problem explores designing a lowpass filter with a very narrow transition band that achieves the
following specifications: Ap = 0.1 dB ripple in the passband (0 ≤ f ≤ .24) and As = 60 dB suppression
in the stopband (.26 ≤ f ≤ .5).
• Design this filter using the window technique using the best window from the set (rectangular,
Bartlett, Hann, Hamming, Blackman). Indicate which window shape you used.
• Design this filter using the window technique using the Kaiser window. Indicate the Kaiser window
parameters (β and M).
• Using FDATool in Matlab, design this filter using equiripple design method.
Make a table with one row for each filter you designed (window, Kaiser, equiripple). The table should
have a column for the following parameters: length, order, group delay, type (I, II, III, or IV), and
window parameters.
Blackman
Kaiser
Equiripple
Length
275
183
167
Order
274
182
166
Group Delay
137
91
83
Type
I
I
I
Window Parameters
N/A
B = 5.65326
N/A
For each filter you design, turn in plots of the impulse response, magnitude response (linear and dB),
and phase response. The frequency domain plots should show response over the interval − 12 ≤ f ≤ 21 .
Also include a pole-zero plot.
Plots for the Blackman filter
5
ECE 3640
Plots for the Kaiser filter
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ECE 3640
Plots for the Equiripple filter
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ECE 3640
Matlab Code
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%Blackman
L = 275;
M = L−1;
n = [ 0 :M] ’ ;
a = M/ 2 ;
fc = .25
w = blackman ( L ) ;
h = s i n ( 2 ∗ p i ∗ f c ∗ ( n−a ) ) . / ( p i ∗ ( n−a ) ) ;
i n d = f i n d ( n−a==0) ;
h ( i n d ) = 2∗ f c ;
H = f f t s h i f t ( f f t ( h . ∗ w , N) ) ;
figure (1) ;
stem ( n , h . ∗w) ;
t i t l e ( ’ Impulse Response ’ ) ;
N = 2ˆ16;
f = [−N / 2 : ( N−1) / 2 ] /N ;
p r i n t −dpng 2 h i m p u l s e . png
figure (2) ;
p l o t ( f , 2 0 ∗ l o g 1 0 ( a b s (H) ) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e dB ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gni tu de ’ ) ;
g r i d on ;
p r i n t −dpng 2hmagdB . png
figure (3) ;
p l o t ( f , a b s (H) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gni tu de ’ ) ;
g r i d on ;
p r i n t −dpng 2hmagL . png
figure (4) ;
p l o t ( f , unwrap ( a n g l e (H) ) ) ;
t i t l e ( ’ Phase R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Amplitude ’ ) ;
g r i d on ;
p r i n t −dpng 2 h p h a s e . png
figure (5) ;
z p l a n e ( ( h . ∗w) . ’ , 1 ) ;
a x i s ([ −1 1 −1 1 ] ∗ 2 ) ;
t i t l e ( ’ Zero Pole Plot ’ ) ;
p r i n t −dpng 2 h z e r o . png
%K a i s e r
M = 182;
L = M+1;
n = [ 0 :M] ’ ;
beta = 5 . 6 5 3 2 6 ;
a = M/ 2 ;
w = k a i s e r (L , beta ) ;
h = s i n ( 2 ∗ p i ∗ f c ∗ ( n−a ) ) . / ( p i ∗ ( n−a ) ) ;
i n d = f i n d ( n−a==0) ;
h ( i n d ) = 2∗ f c ;
H = f f t s h i f t ( f f t ( h . ∗ w , N) ) ;
figure (6) ;
stem ( n , h . ∗w) ;
t i t l e ( ’ Impulse Response ’ ) ;
p r i n t −dpng 2 k i m p u l s e . png
figure (7) ;
p l o t ( f , 2 0 ∗ l o g 1 0 ( a b s (H) ) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e dB ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gn itu de ’ ) ;
g r i d on ;
p r i n t −dpng 2kmagdB . png
figure (8) ;
p l o t ( f , a b s (H) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gn itu de ’ ) ;
g r i d on ;
p r i n t −dpng 2kmagL . png
figure (9) ;
p l o t ( f , unwrap ( a n g l e (H) ) ) ;
t i t l e ( ’ Phase R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Amplitude ’ ) ;
g r i d on ;
p r i n t −dpng 2 k p h a s e . png
figure (10) ;
z p l a n e ( ( h . ∗w) . ’ , 1 ) ;
a x i s ([ −1 1 −1 1 ] ∗ 2 ) ;
t i t l e ( ’ Zero Pole Plot ’ ) ;
p r i n t −dpng 2 k z e r o . png
%E q u i r i p p l e
figure (11) ;
stem ( [ 0 : l e n g t h ( e q u i ) −1 ] , e q u i ) ;
t i t l e ( ’ Impulse Response ’ ) ;
p r i n t −dpng 2 e i m p u l s e . png
figure (12) ;
p l o t ( f , 2 0 ∗ l o g 1 0 ( a b s ( f f t s h i f t ( f f t ( e q u i , N) ) ) ) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e dB ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
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y l a b e l ( ’ Ma gn itu de ’ ) ;
p r i n t −dpng 2emagdB . png
figure (13) ;
p l o t ( f , a b s ( f f t s h i f t ( f f t ( e q u i , N) ) ) ) ;
t i t l e ( ’ Ma gni tu de R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gn itu de ’ ) ;
g r i d on ;
p r i n t −dpng 2emagL . png
figure (14) ;
p l o t ( f , unwrap ( a n g l e ( f f t s h i f t ( f f t ( e q u i , N) ) ) ) ) ;
t i t l e ( ’ Phase R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Amplitude ’ ) ;
g r i d on ;
p r i n t −dpng 2 e p h a s e . png
figure (15) ;
zplane (( equi ) ,1) ;
a x i s ([ −1 1 −1 1 ] ∗ 2 ) ;
t i t l e ( ’ Zero Pole Plot ’ ) ;
p r i n t −dpng 2 e z e r o . png
3. This problem explores designing a lowpass filter for sample rate conversion using the system depicted in
Fig. 4.29 (b) in the textbook (page 191) in which the upsampling factor is L = 3 and the downsampling
factor is M = 4 . (Note: the L and M parameters in Fig. 4.29 are the up and down sampling factors. We
have also used L and M to denote the length L and order M of FIR filters.) The maximum frequency in
the signal x[n] is fmax,x = 31 . Design the lowpass filter to have Ap = 0.1 dB of ripple in the pass band
and As = 60 dB of attenuation in the stopband. Use your knowledge of upsampling and downsampling
to determine the passband and stopband edge frequencies, fp and fs . What are fp and fs ?
Answer: fp = 19 and fs = 29 .
• Design this filter using the window technique using the best window from the set (rectangular,
Bartlett, Hann, Hamming, Blackman). Indicate which window shape you used.
• Design this filter using the window technique using the Kaiser window. Indicate the Kaiser window
parameters (β and M).
• Using FDATool in Matlab, design this filter using equiripple design method.
Make a table with one row for each filter you designed (window, Kaiser, equiripple). The table should
have a column for the following parameters: length, order, group delay, type (I, II, III, or IV), and
window parameters.
Answer: The length of the Blackman filter was increased to make the filter a type I.
Blackman
Kaiser
Equiripple
Length
51
35
33
Order
50
34
32
Group Delay
25
17
16
Type
I
I
I
Window Parameters
N/A
B = 5.65326
N/A
For each filter you design, turn in plots of the impulse response, magnitude response (linear and dB),
and phase response. The frequency domain plots should show response over the interval − 12 ≤ f ≤ 21 .
Also include a pole-zero plot.
Plots for the Blackman filter
9
ECE 3640
Plots for the Kaiser filter
10
ECE 3640
Plots for the Equiripple filter
11
ECE 3640
Matlab Code
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%Blackman
L = 51;
M = L−1;
n = [ 0 :M] ’ ;
a = M/ 2 ;
fc = 3/18;
w = blackman ( L ) ;
h = s i n ( 2 ∗ p i ∗ f c ∗ ( n−a ) ) . / ( p i ∗ ( n−a ) ) ;
i n d = f i n d ( n−a==0) ;
h ( i n d ) = 2∗ f c ;
H = f f t s h i f t ( f f t ( h . ∗ w , N) ) ;
figure (1) ;
stem ( n , h . ∗w) ;
t i t l e ( ’ Impulse Response ’ ) ;
N = 2ˆ16;
f = [−N / 2 : ( N−1) / 2 ] /N ;
p r i n t −dpng 3 h i m p u l s e . png
figure (2) ;
p l o t ( f , 2 0 ∗ l o g 1 0 ( a b s (H) ) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e dB ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gni tu de ’ ) ;
g r i d on ;
p r i n t −dpng 3hmagdB . png
figure (3) ;
p l o t ( f , a b s (H) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gni tu de ’ ) ;
g r i d on ;
p r i n t −dpng 3hmagL . png
figure (4) ;
p l o t ( f , unwrap ( a n g l e (H) ) ) ;
t i t l e ( ’ Phase R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Amplitude ’ ) ;
g r i d on ;
p r i n t −dpng 3 h p h a s e . png
figure (5) ;
z p l a n e ( ( h . ∗w) . ’ , 1 ) ;
a x i s ([ −1 1 −1 1 ] ∗ 2 ) ;
t i t l e ( ’ Zero Pole Plot ’ ) ;
p r i n t −dpng 3 h z e r o . png
%K a i s e r
M = 34;
L = M+1;
n = [ 0 :M] ’ ;
beta = 5 . 6 5 3 2 6 ;
a = M/ 2 ;
w = k a i s e r (L , beta ) ;
h = s i n ( 2 ∗ p i ∗ f c ∗ ( n−a ) ) . / ( p i ∗ ( n−a ) ) ;
i n d = f i n d ( n−a==0) ;
h ( i n d ) = 2∗ f c ;
H = f f t s h i f t ( f f t ( h . ∗ w , N) ) ;
figure (6) ;
stem ( n , h . ∗w) ;
t i t l e ( ’ Impulse Response ’ ) ;
p r i n t −dpng 3 k i m p u l s e . png
figure (7) ;
p l o t ( f , 2 0 ∗ l o g 1 0 ( a b s (H) ) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e dB ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gn itu de ’ ) ;
g r i d on ;
p r i n t −dpng 3kmagdB . png
figure (8) ;
p l o t ( f , a b s (H) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gn itu de ’ ) ;
g r i d on ;
p r i n t −dpng 3kmagL . png
figure (9) ;
p l o t ( f , unwrap ( a n g l e (H) ) ) ;
t i t l e ( ’ Phase R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Amplitude ’ ) ;
g r i d on ;
p r i n t −dpng 3 k p h a s e . png
figure (10) ;
z p l a n e ( ( h . ∗w) . ’ , 1 ) ;
a x i s ([ −1 1 −1 1 ] ∗ 2 ) ;
t i t l e ( ’ Zero Pole Plot ’ ) ;
p r i n t −dpng 3 k z e r o . png
%E q u i r i p p l e
figure (11) ;
stem ( [ 0 : l e n g t h ( e q u i ) −1 ] , e q u i ) ;
t i t l e ( ’ Impulse Response ’ ) ;
p r i n t −dpng 3 e i m p u l s e . png
figure (12) ;
p l o t ( f , 2 0 ∗ l o g 1 0 ( a b s ( f f t s h i f t ( f f t ( e q u i , N) ) ) ) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e dB ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
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y l a b e l ( ’ Ma gn itu de ’ ) ;
p r i n t −dpng 3emagdB . png
figure (13) ;
p l o t ( f , a b s ( f f t s h i f t ( f f t ( e q u i , N) ) ) ) ;
t i t l e ( ’ Ma gni tu de R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gn itu de ’ ) ;
g r i d on ;
p r i n t −dpng 3emagL . png
figure (14) ;
p l o t ( f , unwrap ( a n g l e ( f f t s h i f t ( f f t ( e q u i , N) ) ) ) ) ;
t i t l e ( ’ Phase R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Amplitude ’ ) ;
g r i d on ;
p r i n t −dpng 3 e p h a s e . png
figure (15) ;
zplane (( equi ) ,1) ;
a x i s ([ −1 1 −1 1 ] ∗ 2 ) ;
t i t l e ( ’ Zero Pole Plot ’ ) ;
p r i n t −dpng 3 e z e r o . png
4. Suppose you would like to boost by a factor of 4 the low frequencies of an audio signal between the
range of 0 ≤ |F | ≤ 150 Hz and leave the scale of frequencies above 300 Hz unchanged. This type of filter
has two passbands and no stopband. It is called a shelving filter. The transition band extends from
150 Hz to 300 Hz. You plan to perform this processing using the system depicted in Fig. 4.10 of the
textbook (page 167) using a sample rate of T1 = Fs = 20k samples/second. Convert this information
to a specification for a discrete-time filter and design a linear phase filter to meet this spec. Require
your filter to limit deviations from the desired response to δ = 0.01 in both pass bands.
• Design this filter using the window technique using the best window from the set (rectangular,
Bartlett, Hann, Hamming, Blackman). Indicate which window shape you used.
• Design this filter using the window technique using the Kaiser window. Indicate the Kaiser window
parameters (β and M).
• Using FDATool in Matlab, design this filter using equiripple design method.
Make a table with one row for each filter you designed (window, Kaiser, equiripple). The table should
have a column for the following parameters: length, order, group delay, type (I, II, III, or IV), and
window parameters.
Answer: The length of the Hann filter was increased to make the filter a type I.
Hann
Kaiser
Equiripple
Length
415
299
281
Order
414
298
280
Group Delay
207
149
140
Type
I
I
I
Window Parameters
N/A
B = 3.39532
N/A
For each filter you design, turn in plots of the impulse response, magnitude response (linear and dB),
and phase response. The frequency domain plots should show response over the interval − 12 ≤ f ≤ 21 .
Also include a pole-zero plot.
Plots for the Blackman filter
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Plots for the Kaiser filter
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Plots for the Equiripple filter
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Matlab Code
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%Hann
L = 415;
M = L−1;
n = [ 0 :M] ’ ;
a = M/ 2 ;
fc = .01125;
w = hann ( L ) ;
h = s i n ( 2 ∗ p i ∗ f c ∗ ( n−a ) ) . / ( p i ∗ ( n−a ) ) ;
i n d = f i n d ( n−a==0) ;
h ( i n d ) = 2∗ f c ;
Hw = 3∗ h . ∗w ;
Hw( a +1) = Hw( a +1) +1;
H = f f t s h i f t ( f f t (Hw, N) ) ;
figure (1) ;
stem ( n , Hw) ;
t i t l e ( ’ Impulse Response ’ ) ;
N = 2ˆ16;
f = [−N / 2 : ( N−1) / 2 ] /N ;
p r i n t −dpng 4 h i m p u l s e . png
figure (2) ;
p l o t ( f , 2 0 ∗ l o g 1 0 ( a b s (H) ) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e dB ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gni tu de ’ ) ;
g r i d on ;
p r i n t −dpng 4hmagdB . png
figure (3) ;
p l o t ( f , a b s (H) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gni tu de ’ ) ;
g r i d on ;
p r i n t −dpng 4hmagL . png
figure (4) ;
p l o t ( f , unwrap ( a n g l e (H) ) ) ;
t i t l e ( ’ Phase R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Amplitude ’ ) ;
g r i d on ;
p r i n t −dpng 4 h p h a s e . png
figure (5) ;
z p l a n e ( (Hw) . ’ , 1 ) ;
a x i s ([ −1 1 −1 1 ] ∗ 2 ) ;
t i t l e ( ’ Zero Pole Plot ’ ) ;
p r i n t −dpng 4 h z e r o . png
%K a i s e r
M = 298;
L = M+1;
n = [ 0 :M] ’ ;
beta = 3 . 3 9 5 3 2 ;
a = M/ 2 ;
w = k a i s e r (L , beta ) ;
h = s i n ( 2 ∗ p i ∗ f c ∗ ( n−a ) ) . / ( p i ∗ ( n−a ) ) ;
i n d = f i n d ( n−a==0) ;
h ( i n d ) = 2∗ f c ;
Hw = 3∗ h . ∗w ;
Hw( a +1) = Hw( a +1) +1;
H = f f t s h i f t ( f f t (Hw, N) ) ;
figure (6) ;
stem ( n , Hw) ;
t i t l e ( ’ Impulse Response ’ ) ;
p r i n t −dpng 4 k i m p u l s e . png
figure (7) ;
p l o t ( f , 2 0 ∗ l o g 1 0 ( a b s (H) ) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e dB ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gn itu de ’ ) ;
g r i d on ;
p r i n t −dpng 4kmagdB . png
figure (8) ;
p l o t ( f , a b s (H) ) ;
t i t l e ( ’ Mag ni tu de R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gn itu de ’ ) ;
g r i d on ;
p r i n t −dpng 4kmagL . png
figure (9) ;
p l o t ( f , unwrap ( a n g l e (H) ) ) ;
t i t l e ( ’ Phase R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Amplitude ’ ) ;
g r i d on ;
p r i n t −dpng 4 k p h a s e . png
figure (10) ;
z p l a n e ( (Hw) . ’ , 1 ) ;
a x i s ([ −1 1 −1 1 ] ∗ 2 ) ;
t i t l e ( ’ Zero Pole Plot ’ ) ;
p r i n t −dpng 4 k z e r o . png
%E q u i r i p p l e
Hw = 3∗ e q u i ;
len = length ( equi ) ;
a = ( l e n −1) / 2 ;
Hw( a +1) = Hw( a +1) +1;
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figure (11) ;
stem ( [ 0 : l e n g t h ( e q u i ) −1 ] ,Hw) ;
t i t l e ( ’ Impulse Response ’ ) ;
p r i n t −dpng 4 e i m p u l s e . png
figure (12) ;
p l o t ( f , 2 0 ∗ l o g 1 0 ( a b s ( f f t s h i f t ( f f t (Hw, N) ) ) ) ) ;
t i t l e ( ’ Ma gni tu de R e s p o n s e dB ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gn itu de ’ ) ;
p r i n t −dpng 4emagdB . png
figure (13) ;
p l o t ( f , a b s ( f f t s h i f t ( f f t (Hw, N) ) ) ) ;
t i t l e ( ’ Ma gni tu de R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Ma gn itu de ’ ) ;
g r i d on ;
p r i n t −dpng 4emagL . png
figure (14) ;
p l o t ( f , unwrap ( a n g l e ( f f t s h i f t ( f f t (Hw, N) ) ) ) ) ;
t i t l e ( ’ Phase R e s p o n s e ’ ) ;
x l a b e l ( ’ F r e q u e n c y [ Hz ] ’ ) ;
y l a b e l ( ’ Amplitude ’ ) ;
g r i d on ;
p r i n t −dpng 4 e p h a s e . png
figure (15) ;
z p l a n e ( (Hw) , 1 ) ;
a x i s ([ −1 1 −1 1 ] ∗ 2 ) ;
t i t l e ( ’ Zero Pole Plot ’ ) ;
p r i n t −dpng 4 e z e r o . png
5. After working the previous 4 problems, summarize your observations about the best filter design
method to use if the goal is to minimize the filter length.
Answer: The equiripple filter is the best method if the goal is to minimize the filter length. the
equiripple instance of each problem had a consistently shorter length.
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