Frequency Sampling Structures
Specifying Frequency Response
Set the frequency response of an FIR filter at M equally-spaced
points  k  2 k / M , k  0,1,2,L , M  1 and find the impulse
response h n .
Frequency response and impulse response are related by
 
M 1
H e j   h n e  jn
n0
and it follows that
  

M 1
H e jk  H e j 2 k / M   h n e  j 2 kn / M , k  0,1,2,L , M  1
n0
This is the DFT of h n .
Specifying Frequency Response


Since H e j 2 k / M is the DFT of h n , h n  must be the inverse


DFT of H e j 2 k / M .
1 M 1
j 2 k / M
j 2  kn / M
h n  
H
e
e
, n  0,1,2,L , M  1

M k0


The transfer function and impulse response are related by
M 1
M 1 M 1
1
H z    h n z  n 
H e j 2 k / M e j 2 kn / M z  n .


M n0 k 0
n0
Exchanging the order of summation,
1
H z  
M

M 1

j 2 k / M
H
e

k 0
M 1
 e
n0

j 2  kn / M  n
z
Specifying Frequency Response
Using the formula for summing a geometric series,
 M j 2
M
1 M 1
1

z
e
1

z
H z  
H e j 2 k / M


1 j 2  k / M
M k0
1 z e
M


M 1

H e j 2 k / M
 1 z
1 j 2 k / M
k0
We can think of this as the cascade of two filters, one with
transfer function
1  zM
H1 z  
M
and one with transfer function
M 1
H 2 z   
k0

H e j 2 k / M

1 j 2  k / M
1 z e
.

e
Specifying Frequency Response
H1 z  has zeros at
H1 z 
1  zM  0  zM  1
z  e j 2 q / M , q  0,1,2,L , M  1.
Specifying Frequency Response
H 2 z  is the sum of the transfer functions of M subsystems, each of
which has a pole where
1  z 1e j 2 k / M  0  z 1e j 2 k / M  1  z  e j 2 k / M , k  0,1,2,L , M  1
So the transfer function can be written as


M 1 H e j 2  k / M
M 1


1
H z     1  z 1e j 2 q / M  
1 j 2  k / M
M  q0
1

z
e
k0

1 4 4 42 4 4 43 1 4 42 4 43
H 2 z 
H1 z 
or
M 1

H e
j 2  p/ M

1

M
 j 2   pq / M
1

e
M 1 

k 0
q0
1 e
 j 2   p k / M

H e j 2 k / M
,
0 pM
Specifying Frequency Response
The poles of H 2 z  are at the same locations as the zeros of
H1 z . So for any particular zero of H1 z , all the terms in
the summation are zero except the one for which the pole and
zero cancel. Therefore

H e
j 2 p/ M



H e j 2 p/ M
M

M 1
 j 2  pq / M
1

e
, 0 pM

q0
q p
This result implies that
M 1
 j 2   pq / M
1

e
 M , 0 pM

q0
q p
Specifying Frequency Response
The poles and zeros could theoretically be anwhere on the unit
circle, but for practical designs they must occur in complex
conjugate pairs and





H e j 2 k / M  H * e  j 2 k / M  H * e j 2 M  k / M

Specifying Frequency Response
Taking advantage of the complex-conjugate symmetry, for M even,
M 1
H 2 z   

H e j 2 k / M
  H e 
j0
1 j 2  k / M
k0 1  z e
H 2 z  
 
H e j0
1  z 1
M /21

k1

1  z 1
M /21

H e j 2 k / M
 1 z
k1
  H e
1 j 2  k / M
e

H e
where A k   H e j 2 k / M  H e j 2 M k / M
and B k   e  j 2 k / M
j 2 k / M
j 2 k / M
j 2  M  k / M

Similarly, for M odd,
H 2 z  
  H e  
H e j0
1 z
1
j
1 z
1
M 1/2

k1

1  z 1e j 2 M  k / M
A k   B k z 1
1  2 cos 2 k / M z 1  z 2
 
H e
 e
j 2  M  k / M
A k   B k z 1
1  2 cos 2 k / M z 1  z 2
Example
Design a filter whose frequency response goes through these points.
k
0 1 2 3 4 5 6 7

H e j 2 k / M

j 0 0 0 j 1
0 1
 M M 1
1 z
H z  
M

H e j 2 k / M
 1 z
k0

1 j 2 k / M
e
1 z 8 
1
j
j
1

H z  



8  1  z 1e j 2 /8 1  z 1e j 4  /8 1  z 1e j12 /8 1  z 1e j14 /8 
Example
Taking advantage of the periodicity of e j 2 k /8
1  z 8
H z  
8
or
1
1
j
j





 1  z 1e j 2 /8 1  z 1e  j 2 /8 1  z 1e j 4  /8 1  z 1e  j 4  /8 
 2  2z 1 cos  / 4 

2z 1 sin  / 2 


1
2
1
2 
1

2z
cos

/
4

z
1

2z
cos

/
2

z






and the frequency response is
1  z 8
H z  
8
 j
 j


2

2e
cos

/
4
2e
sin  / 2 


j
H e


 j
 j 2
 j
 j 2 
1

2e
cos

/
4

e
1

2e
cos

/
2

e






The impulse response is
1  e  j8

8
 
0.25,0.0732,0,0.0732,  D F T
j 2 k / M
h n    
 0,1, j,0,0,0, j,1
  H e
0.25,0.4268,0,0.4268 

