Lecture 4: March 13, 2007
Topic:
1. Uniform Frequency-Sampling Methods
(cont.)
1
Lecture 4: March 13, 2007
Topics:
1. Nonrecursive FIR Filter Design by DFT
Applications (cont.)
2. Nonrecursive FIR Filter Design by Direct
Computation of Unit Sample Response
3. Recursive FIR Filter Design
2
4.2.3. Uniform Frequency-Sampling Method B.
Nonrecursive FIR Filter Design by DFT
Applications
3
4.2.3. Uniform Frequency-Sampling Method B.
Nonrecursive FIR Filter Design by DFT
Applications
• specification of the desired frequency response at a set
of equally spaced frequencies,
• solving for the unit sample response h(n) of the linear
phase FIR filter from these equally spaced specification
based on DFT application.
M 1
y ( n)   h( k ) x ( n  k )
k 0
M 1
H (e j )   h(k )e  j k
k 0
4
Suppose that we specify the frequency response of
the filter at the frequencies
2 n
n 
M
Then, we obtain:
H (e
H (e
jn
jn
M 1
H (e j )   h(k )e  jk
k 0
 2 n 
)  H  n   H 
  H (n)
 M 
M 1
)  H ( n)   h( k )e
 jkn
k 0
for n  0,1,2,

M 1
 h ( k )e
 jk
2 n
M
k 0
, M 1
5
It is a simple matter to identify the previous expression
as DFT of h(k):
M 1
H ( n )   h ( k )e
j
2 nk
M
H (n)  DFT  h(k )
k 0
Thus, we obtain
h(k )  DFT
1
h( k ) 
M
M 1
 H ( n)e
j
2 nk
M
1
 H ( n) 
k  0,1,2,
2 n
n 
M
, M 1
n 0
This expression allows to compute h(k) from H(n),
where frequency response is specified at the set of
equally spaced frequencies.
Notice: H(n) has to be specified correctly!
6
Important property of H(n) :
H ( n)  H ( M  n )
H ( M  n)  H ( n)
for n= 1, 2, …, M-1
Proof:
M 1
H ( M  n )   h ( k )e
j
2 k ( M  n )
M
k 0
M 1
  h(k )e
j
2 kM
M
e
j
M 1
  h(k )e
j
2 k ( M  n )
M

k 0
2 k (  n )
M
k 0
M 1
  h(k )e j 2 k e
j
2 kn
M

k 0
M 1
  h(k )e
k 0
j
2 kn
M
 H ( n)
7
Frequency response of the linear phase FIR digital
filters:
M 1
 j n
2
H (e jn )  H r ( n )e
H (e
j n
 2 n 
)  H (n)   1 H r 
e
 M 
n
j
n
M
Proof:
H (e
jn
)  H r (n )e
 jn
M 1
2
 2 n 
 Hr 
e
 M 
j
2 n M 1
M
2

M 1
 nM
n

j

n

j
j
2

n
2

n




M
M
M
 Hr 
e

H
e
e

r


 M 
 M 
 2 n   j n
 Hr 
e e
 M 
j
n
M
 2 n 
  1 H r 
e
 M 
n
j
n
M
8
Uniform Frequency-Sampling Method B. : A Review
1. Specification of H(n) by:
n
j
2

n

 M
H (n)   1 H r 
e
 M 
n
H ( M  n)  H ( n)
for n= 1, 2, …, M-1
2. Computation of h(k) by:
1
1
h(k )  DFT  H (n)  
M
M 1
 H ( n )e
j
2 nk
M
n 0
for k= 0, 1, 2, …, M-1
9
Example:
Determine the coefficients h(k) of a linear phase FIR filter
of length M=15, frequency response of which satisfies
the condition:
 2 n  1 n  0,1,2,3
Hr 

 15  0 n  4,5,6,7
Solution:
 2 n 
H (n)   1 H r 
e
 M 
n
H ( M  n)  H ( n)
j
n
M
 2 n 
  1 H r 
e
 15 
n
j
n
15
for n= 1, 2, …, M-1
10
n
j
2

n

 15
H (n)   1 H r 
e
 15 
n
H (0)  1
 1.2
H (1)   1 H r 
 15
1

e

j
1
15
 e
j

15
H ( M  n)  H ( n)
H (14)  H (15  1)  H (1)  e
 2.2
H (2)   1 H r 
 15
2

e

j
2
15
e
j
 3.2
H (3)   1 H r 
 15

e

j
3
15
 e

15
2
15
H (13)  H (15  2)  H (2)  e
3
j
j
j
2
15
3
15
H (12)  H (15  3)  H (3)  e
j
3
15
11
H (4)  0
H (11)  H (15  4)  H (4)  0
H (5)  0
H (10)  H (15  5)  H (5)  0
H (6)  0
H (9)  H (15  6)  H (6)  0
H (7)  0
H (8)  H (15  7)  H (7)  0
H ( M  n)  H ( n)
 2 n 
H (n)   1 H r 
e
 15 
n
j
n
15
12
151
j
1
1
h(k )  DFT  H (n)    H (n)e
15 n0
2 nk
15
j
1 14
  H ( n)e
15 n0
2 nk
15
for k= 0, 1, 2, …, M-1
13
By using MATLAB (function ifft.m) the following filter
coefficients can be obtained:
h(0)=h(14)= -0.0498
h(1)=h(13)= 0.0412
h(2)=h(12)= 0.0667
h(3)=h(11)= -0.0365
h(4)=h(10)= -0.1079
h(5)=h(9)= 0.0341
h(6)=h(8)= 0.3189
h(7)= 0.4667
14
Example Solution: Magnitude Response
H  e j 

15
4.2.4. Uniform Frequency-Sampling Method C.
Nonrecursive FIR Filter Design by Direct
Computation of Unit Sample Response
16
4.2.4. Uniform Frequency-Sampling Method C.
Nonrecursive FIR Filter Design by Direct
Computation of Unit Sample Response
17
4.2.4. Uniform Frequency-Sampling Method C.
Nonrecursive FIR Filter Design by Direct
Computation of Unit Sample Response
 2 n 
We have shown that H (n)   1 H r 
e
 M 
n
j
n
M
It is more convenient, however, to define the samples
H(n) of the frequency response for the FIR filter as
H ( n )  G ( n )e
j
n
M
where
 2 n 
G (n)   1 H r 

M


n
18
 2 n 
Hr 
 R
 M 
G (k )  R
H ( M  n)  H ( n)
for n= 1, 2, …, M-1
G (n)  G ( M  n)
for n= 1, 2, …, M-1
M : even
M
G
 2

0

19
Proof: H (n)  H ( M  n) and H (n)  G (n)e
H ( n )  G ( M  n )e
 G ( M  n )e
H ( n )  G ( n )e
G ( M  n)e
j
n
M
j
j
 ( M n )
j
M
M
M
e
j
 G ( M  n )e
n
M
j
j
n
M
 ( M n )
 G ( M  n)e
M
j

n
M
n
M
 G ( n )e
j
n
M
G (n)  G ( M  n)
for n= 1, 2, …, M-1
20
In the view of the symmetry for the real-valued frequency
samples G(n), it is a simple matter to determine an
expression for h(k) in terms of G(n). If M is even, we
have:
 kn
 n  nk
M 1
j
j
j
1 M 1
1
h( k ) 
H (n)e M 
G (n)e M e M 


M n 0
M n 0
1

M
M 1
 G(k )e
j
 n (2 k 1)
M

n 0
21
In the view of the symmetry for the real-valued frequency
samples G(n), it is a simple matter to determine an
expression for h(k) in terms of G(n). If M is even, we
have:
 kn
 n  nk
M 1
j
j
j
1 M 1
1
h( k ) 
H (n)e M 
G (n)e M e M 


M n 0
M n 0
1

M
M 1
 G(k )e
j
 n (2 k 1)
M

n 0
G(n)  GM(M
 n) for n 1,2,..., M  1 
1
 n (2 k 1)
 ( M  n )(2 k 1)
2
j
j


1 
M
M

e
G (0)   G (n) e

M
n 1




22
Because:
e
j
 ( M  n )(2 k 1)
M
e
j
e
2 kM
M
j

M
e
( M n )2 k ( M n )
j
2 nk
M
e
j
M
M
e
j
e
n
M
j

M
( M  n )2 k
 e
j
n
M
e
j

M
( M n )

 2 k 1
23
We can get:
 n (2 k 1)
 n (2 k 1)
M


j
j
1


M
M
2
1 
e
e



h( k ) 
G (0)  2  G (n) 


M
2
n 1






M
1


2
1 
n

h( k ) 
G (0)  2  G (n)cos  2k  1 
M
M
n 1



for k= 0, 1, 2, …, M-1
24
Thus we can compute h(k) directly from the specified
frequency samples G(n) or equivalently, H(n)
(method 3). This approach eliminates the need for a
matrix inversion (method 1) or for direct DFT
computation (method 2) for determining the unit
sample response of the linear phase FIR filter from
its frequency-domain specification.
25
Thus we can compute h(k) directly from the specified
frequency samples G(n) or equivalently, H(n)
(method 3). This approach eliminates the need for a
matrix inversion (method 1) or for direct DFT
computation (method 2) for determining the unit
sample response of the linear phase FIR filter from
its frequency-domain specification.
26
Thus we can compute h(k) directly from the specified
frequency samples G(n) or equivalently, H(n)
(method 3). This approach eliminates the need for a
matrix inversion (method 1) or for direct DFT
computation (method 2) for determining the unit
sample response of the linear phase FIR filter from
its frequency-domain specification.
27
Summary
on
the Uniform Frequency-Sampling Method 3.
28
1. Unit Sample Response: Symmetrical,   0
H (k )  G(k )e j k / M , k  0,1,
, M 1
 2 k 
G (k )   1 H r 

M


k
G (k )  G ( M  k )
1
h( n) 
M
U
k


G (0)  2 G (k )cos  2n  1 
M
k 1


M 1
U
for M odd
2
M
U
 1 for M even
2
29
1
2. Unit Sample Response: Symmetrical,  
2
1
1   j / 2 j (2 k 1) / 2 M


H  k    G k  e
e
2
2


 2 
1
1 
k

G  k     1 H r   k   
2
2 

M 
1
1


Gk    G M  k  
2
2


2
h( n) 
M
1  2

G  k   sin

2
M

k 0
U
1 
1

 k   n  
2 
2

30
3. Unit Sample Response: Antisymmetrical,   0
H (k )  G(k )e j / 2e j k / M , k  0,1,
 2 k 
G (k )   1 H r 

M


G (k )  G ( M  k )
k
2
h( n)  
M
( M 1) / 2

k 1
G (k )sin
, M 1
k
M
 2n  1
for M odd
( M / 2) 1

1 
k
n 1
h( n) 
 2n  1
 1 G ( M / 2)  2  G ( k )sin
M
M
k 1

for M even
31
1
4. Unit Sample Response: Antisymmetrical,  
2
1
1  j (2 k 1) / 2 M


H  k    G k  e
2
2


 2 
1
1 
k

G  k     1 H r   k   
2
2 

M 
1
1


M 
G  k    G  M  k   ; G    0 for M even
2
2


 2 
2
h( n) 
M
1
2

G  k   cos

2
M

k 0
V
M 1
V
for M odd
2
1 
1

 k   n  
2 
2

M
V
 1 for M even
32
2
Example:
Determine the coefficients h(n) of a linear phase FIR filter
of length M=15 with symmetrical impulse response,
frequency response of which satisfies the condition:
 2 k  1 k  0,1,2,3
Hr 

 15  0 k  4,5,6,7
33
Solution: Unit Sample Response: Symmetrical,   0
H (k )  G(k )e j k / M , k  0,1,
, M 1
 2 k 
G (k )   1 H r 

M


k
G (k )  G ( M  k )
1
h( n) 
M
U
k


G (0)  2 G (k )cos  2n  1 
M
k 1


M 1
U
for M odd
2
M
U
 1 for M even
2
34
k
 2 k 
 2 k 
G (k )   1 H r 
   1 H r 

 M 
 15 
k
 0.2
G (0)   1 H r 
 15

 1

 1.2
G (1)   1 H r 
 15

  1

 2.2
G (2)   1 H r 
 15
3

 3.2
  1 G (3)   1 H r 

 15

  1

0
2
1
G (4)  G (5)  G (6)  G (7)  0
 2 k  1 k  0,1,2,3
Hr 

 15  0 k  4,5,6,7
35
M  1 15  1 14
U


7
2
2
2
7
1
k

h(n)  G (0)  2 G (k )cos  2n  1 
15 
15
k 1

1
h(n)  G (0)  2 X (n)
15
X (n)  G (1)cos

 2n  1 
15
2
3
G (2)cos  2n  1  G (3)cos  2n  1
36
15
15
n0
1

2
3  

h(0)  1  2   cos  cos
 cos   
15 
15
15
15  

1
 1  2  0.9781  0.9135  0.8090 
15
1
 1  2(0.8736)   0.0498
15
Etc.
h(0)=h(14)= -0.0498
h(4)=h(10)= -0.1079
h(1)=h(13)= 0.0412
h(5)=h(9)= 0.0341
h(2)=h(12)= 0.0412
h(6)=h(8)= 0.3189
h(3)=h(11)= -0.0365
h(7)= 0.4667
End.
37
Example Solution: Magnitude Response
H  e j 

38
4.2.5. Uniform Frequency-Sampling Method D.
Recursive FIR Filter Design
39
4.2.5. Uniform Frequency-Sampling Method D.
Recursive FIR Filter Design
40
4.2.5. Uniform Frequency-Sampling Method D.
Recursive FIR Filter Design
An FIR filter can be uniquely specified by giving either
its impulse response (coefficients h(k)) or, equivalently,
by the DFT coefficients H(n). These sequences are
related by the following expressions:
M 1
H ( n)   h( k ) e
j
2 nk
M
k 0
1
h( k ) 
M
M 1
 H ( n) e
n 0
j
 DFT  h(k ) 
2 nk
M
n=0, 1, …, M-1
 DFT 1  H (n)  k=0, 1, …, M-1
41
It has also been shown that H(n) can be regarded as Ztransform of h(k), evaluated at N points equally spaced
around the unit circle:
H (n)  DFT  h(k )  H ( z )
for z  e
j  2 / N  n
and n  0,1,2,
, N 1
42
Thus, the transfer function of an FIR filter can easily be
expressed in terms of DFT coefficients of h(k):
M 1
H ( z )   h( k ) z  k 
k 0
k  1
z 
k 0
M
M 1
M 1
M 1
n 0
k 0
M 1
1

M
1

M
 H ( n )e
 H ( n)
n 0
2 kn
M
n 0
 H (n)  e
M 1
j
1  e
j 2 n / M
j 2 n / M
1 e



1 k
z  
z
j 2 n / M

1 M
z
1

43
1  e j 2 n z  M
H ( n)


j 2 n / M 1
1 e
z
n 0
1

M
M 1
1

M
M 1
1  zM
H ( n)


j 2 n / M 1
1 e
z
n 0
1  zM

M
M 1
H ( n)

j 2 n / M 1
1

e
z
n 0
H1(z)
1  zM
H ( z) 
M
H2(z)
M 1
H ( n)

j 2 n / M 1
1

e
z
n 0
44
The transfer function H(z) can be expressed in the form:
H ( z )  H1 ( z ) H 2 ( z )
zeros
where
1  zM
H1 ( z ) 
M
H 2 ( z) 
M 1
poles !
H ( n)

j 2 n / M 1
1

e
z
n 0
45
H ( z)
x ( n)
H1 ( z )
H 2 ( z)
y ( n)
H(z) can be interpreted as a cascade of
a comb filter and a parallel bank of single-pole filters
The zeros of the comb filter cancel the poles of the
parallel bank of filters, so that in effect the final transfer
function contains no poles, but it is in fact FIR filter.
46
Alternative form to
1  zM
H ( z) 
M
M 1
H ( n)

j 2 n / M 1
z
n 0 1  e
can be obtained by combining a pair of complexconjugate poles of H(z) or two real-valued poles to
form two-pole filters with real valued coefficients and
taking into account H (n)  H ( M  n) .
Then, we can obtain H 2 ( z ) in the forms:
47
M 1
2
1
H (0)
A(k )  B(k ) z
H 2 ( z) 

1
1
2
1 z
1

2cos(2

k
/
M
)
z

z
k 1
for M odd
M
1
2
H (0) H ( M / 2)
A(k )  B(k ) z 1
H 2 ( z) 


1
1
1
2
1 z
1 z
1

2cos(2

k
/
M
)
z

z
k 1
for M even
A(k) and B(k) are the real-valued coefficients given by
A(k )  H (k )  H ( M  k )
B(k )  H (k )e j 2 k / M  H ( M  k )e j 2 k / M
48
Proof:
H (k )
H (M  k )


j 2 k / M 1
j 2 ( M  k ) / M 1
1 e
z
1 e
z
H (k )
H (M  k )



j 2 k / M 1
j 2 M / M  j 2 k / M 1
1 e
z
1 e
e
z
H (k )
H (M  k )



j 2 k / M 1
 j 2 k / M 1
1 e
z
1 e
z

H (k ) 1  e  j 2 k / M z 1   H ( M  k ) 1  e j 2 k / M z 1 
j 2 k / M 1
 j 2 k / M 1
1

e
z
1

e
z 


C

D
49
H (k )
H (M  k )


j 2 k / M 1
j 2 ( M  k ) / M 1
1 e
z
1 e
z

H (k ) 1  e  j 2 k / M z 1   H ( M  k ) 1  e j 2 k / M z 1 
j 2 k / M 1
 j 2 k / M 1
1

e
z
1

e
z 


C

D
C
 H (k ) 1  e  j 2 k / M z 1   H ( M  k ) 1  e j 2 k / M z 1  
  H (k )  H ( M  k )  
  H (k )e  j 2 k / M  H ( M  k )e j 2 k / M  z 1 
 A(k )  B (k ) z 1
50
H (k )
H (M  k )


j 2 k / M 1
j 2 ( M  k ) / M 1
1 e
z
1 e
z

H (k ) 1  e  j 2 k / M z 1   H ( M  k ) 1  e j 2 k / M z 1 
j 2 k / M 1
 j 2 k / M 1
1

e
z
1

e
z 


C

D
D
 1  e j 2 k / M z 1 1  e  j 2 k / M z 1  
e j 2 k / M  e  j 2 k / M 1 2
 1 2
z z 
2
2 k 1 2
 1  2cos
z z
M
51
Then,
H (k )
H (M  k )
A(k )  B(k ) z 1


j 2 k / M 1
j 2 ( M  k ) / M 1
2 k 1 2
1 e
z
1 e
z
1  2cos
z z
M
52
Comments on A(k) and B(k) Computation:
1. Because, H (n)  H ( M  n) it can be found that,
A(k )  H (k )  H ( M  k )  H (k )  H (k )  2Re  H (k )   R
53
2. Let us assume that k  arg  H (k ) then,
B(k )  H (k )e j 2 k / M  H ( M  k )e j 2 k / M 
 H (k ) e jk e j 2 k / M  H (k ) e j 2 k / M 
 H (k ) e jk e  j 2 k / M  H (k ) e  jk e j 2 k / M 
 2 H (k )
e
jk  2 k / M 
e
2
 jk  2 k / M 
 2 H (k ) cos k  2 k / M   R

54
Summary on A(k) and B(k) Computation:
A(k )  2Re  H (k )
B(k )  2 H (k ) cos k  2 k / M 
55
Comments on Quantization Effects:
There is a potential problem in the frequency sampling
realization of the recursive FIR filters. These realizations
introduce poles and zeros at equally spaced points on the
unit circle. In the ideal situation, the zeros cancel the poles
and, consequently, the actual zeros of H(z) are determined
by the selection of the frequency samples H(k). In a
practical implementation, however, quantization effects
preclude a perfect cancellation of the poles and zeros. In
fact, the locations of poles on the unit circle provide no
damping of the round-off noise. As result, such noise tends
to increase with time and, ultimately, may destroy
computations the normal operation of the filter.
56
Comments on Quantization Effects:
There is a potential problem in the frequency sampling
realization of the recursive FIR filters. These realizations
introduce poles and zeros at equally spaced points on the
unit circle. In the ideal situation, the zeros cancel the poles
and, consequently, the actual zeros of H(z) are determined
by the selection of the frequency samples H(k). In a
practical implementation, however, quantization effects
preclude a perfect cancellation of the poles and zeros. In
fact, the locations of poles on the unit circle provide no
damping of the round-off noise. As result, such noise tends
to increase with time and, ultimately, may destroy
computations the normal operation of the filter.
57
Comments on Quantization Effects:
There is a potential problem in the frequency sampling
realization of the recursive FIR filters. These realizations
introduce poles and zeros at equally spaced points on the
unit circle. In the ideal situation, the zeros cancel the poles
and, consequently, the actual zeros of H(z) are determined
by the selection of the frequency samples H(k). In a
practical implementation, however, quantization effects
preclude a perfect cancellation of the poles and zeros. In
fact, the locations of poles on the unit circle provide no
damping of the round-off noise. As result, such noise tends
to increase with time and, ultimately, may destroy
computations the normal operation of the filter.
58
Comments on Quantization Effects:
There is a potential problem in the frequency sampling
realization of the recursive FIR filters. These realizations
introduce poles and zeros at equally spaced points on the
unit circle. In the ideal situation, the zeros cancel the poles
and, consequently, the actual zeros of H(z) are determined
by the selection of the frequency samples H(k). In a
practical implementation, however, quantization effects
preclude a perfect cancellation of the poles and zeros. In
fact, the locations of poles on the unit circle provide no
damping of the round-off noise. As result, such noise tends
to increase with time and, ultimately, may destroy
computations the normal operation of the filter.
59
To mitigate this problem, we can move both the poles
and zeros from the unit circle just inside the unit
circle, let us say at radius r=1-x, where x is a very small
number. Thus the original system function of the linear
phase FIR filter
 M M 1
1 z
H ( n)
H ( z) 

M n0 1  e j 2 n / M z 1
becomes
1 r z
H ( z) 
M
M
 M M 1
H ( n)

j 2 n / M 1
1

re
z
n 0
The corresponding two-pole filter realization can be
modeled accordingly. The damping provided by
selecting r<1 ensures that round-off noise will be
60
bounded and thus instability is avoided.
Example:
Determine the transfer function of a recursive linear phase
FIR filter for M=15. The real-valued frequency response
of the filter to be designed has to satisfy the condition:
 2 k  1 k  0,1,2,3
Hr 

 15  0 k  4,5,6,7
Solution:
n
n
j
2

n
n

 M
 2 n  j 15
H (n)   1 H r 
 e   1 H r 
e
 M 
 15 
n
61
Step 1: H(n) computation for n= 0, 1, 2, …, (M-1)/2
i.e. n= 0, 1, 2, 3, 4, 5, 6, 7.
 0.2
H (0)   1 H r 
 15

e

 1.2
H (1)   1 H r 
 15

e

0
1
 2.2
H (2)   1 H r 
 15
2
j
j
0.
15
1
15

e

j
 e
2
15
0  0
1
e
j
j

15
2
15

14
1     
15
15
2
2 
15
n
j
2

n
n

 15
H (n)   1 H r 
e
 15  62
Step 1 (cont.):
 3.2
H (3)   1 H r 
 15
3

e

j
3
15
 e
j
3
15
12
3  
15
For n  4,5,6,7
Because
 2 n 
Hr 
  0 for n  4,5,6,7
 15 
H (n)  0 for n  4,5,6,7
n
j
2

n
n

 15
H (n)   1 H r 
e
 15  63
Step 2: A review of basic expressions
H ( z )  H1 ( z ) H 2 ( z )
M 1
2
1  zM
H1 ( z ) 
M
H 2 ( z )  H 20 ( z )   H 2 k ( z )
H (0)
H 20 ( z ) 
1
1 z
A(k )  B(k ) z 1
H 2k ( z) 
1  2cos(2 k / M ) z 1  z 2
k 1
A(k )  2Re  H (k )
B(k )  2 H (k ) cos k  2 k / M 
64
Step 3: A(k) computation
A(k )  2Re  H (k )
A(1)  2Re  H (1)  -1.9563
A(2)  2Re  H (2)  1.8271
A(3)  2Re  H (3)  -1.6180
65
Step 4: B(k) computation
B(k )  2 H (k ) cos k  2 k / M 
B (1)  2cos 1  2 /15   2cos  14 /15  2 /15  
 2cos  16 /15   -1.9563
B (2)  2cos 2  4 /15   2cos  2 /15  4 /15  
 2cos(2 /15)  1.8271
B (3)  2cos 3  6 /15   2cos  12 /15  6 /15  
 2cos  18 /15   -1.6180
66
Step 5: H1 ( z ) and H 20 ( z ) computation (M=15):
1  zM
H1 ( z ) 
M
1  z 15
H1 ( z ) 
15
H (0)
H 20 ( z ) 
1  z 1
1
H 20 ( z ) 
1  z 1
67
Step 6: H 2 k ( z ) computation, (M-1)/2=(15-1)/2=7
A(k )  B(k ) z 1
H 2k ( z) 
1
2
1  2cos(2 k / M ) z  z
-1.9563+1.9563z 1
H 21 ( z ) 
1  1.8271z 1  z 2
1.8271-1.8271z 1
H 22 ( z ) 
1  1.3383z 1  z 2
-1.6180+-1.6180z 1
H 23 ( z ) 
1  0.6180z 1  z 2
H 2 k ( z )  0 for k= 4, 5, 6, 7.
68
Step 7: result H(z)
H 20 ( z )
H 21 ( z )
H1 ( z )
1  z 15  1
-1.9563+1.9563z 1
H ( z) 
.


1
1
2
15 1  z
1  1.8271z  z
H 2 ( z)
1
1
1.8271-1.8271z
-1.6180+1.6180z 


1
2
1  1.3383z  z
1  0.6180z 1  z 2 
H 22 ( z )
H 23 ( z )
The different frequency responses and zero-pole plot of the
particular transfer functions are given in the next figures.
69
Magnitude Responses H 2 x ( z )
H 20  e j 
H 21  e
j



H 22  e
j
H 23  e j 


 70
H1  e j 
Magnitude Responses

H 2  e j 

H1  e j  H 2  e j 

71
Magnitude Response H 2  e j  : zoom
H 2  e j 

72
Magnitude Response
H  e j   H1  e j  H 2  e j 
Nonrecursive FIR Filter
Recursive FIR Filter

73
Zero-Pole Plot
Poles
Zeros
Unit Circle
74