The document discusses the design of a finite impulse response (FIR) filter using the frequency sampling technique. It describes how to determine the impulse response of an FIR filter of length 7 to meet specific frequency response specifications. Frequency samples are taken at points and the desired response is defined. The discrete-time Fourier transform (DTFT) and inverse DTFT are used to calculate the impulse response coefficients that produce the desired filter frequency response. Equations for the impulse response h(n) are provided.

1. FIR filter design using frequency
sampling technique
-SARANG JOSHI

2. c
M
j
d eH 




||0.............{)(
)
2
1
(
Desired frequency response

3. • Frequency samples are taken at
The sampled Frequency response is denoted by
Frequency Sampling Method
1-0,1,...Mk.........
2

M
k
K


M
kdd HHkH k 

  2|)(|)()(

 
DFT

4. 1,..1,0n....)(
1
)(
1
0
2
 


NekX
N
nx
N
k
N
n
kj 
1,..1,0n....)(
1
)(
1
0
2
 


MekH
M
nh
M
k
M
n
kj 
evenisMwhen....])(Re[2)0(
1
)(
oddisMwhen....])(Re[2)0(
1
)(
1
2
1
2
2
1
1
2




























M
k
M
n
kj
M
k
M
n
kj
ekHH
M
nh
ekHH
M
nh


IDFT
)(*)( kNHkH 

5. Q. Determine the impulse response of FIR filter of length 7 to meet the following
specifications:
otherwise....0
2
0{)( 2
)1( 





foreeH
Mj
j
d
otherwise....0
2
0{)( 3 

 
foreeH jj
d
Solution:
M=7

6. kdHkH   |)()(
M
kdHkH 

 2|)()(


7
2|)()( kdHkH 





7. otherwise....0
27
2
0{)(
)
7
2
(3 

 k
forekH
k
j
otherwise....0
4
7
0{)(
)
7
6
(


kforekH
k
j

275.1
4
7


8. otherwise....0
20{)(
)
7
6
(


kforekH
k
j

oddisMwhen....])(Re[2)0(
1
)(
2
1
1
2










 


M
k
M
n
kj
ekHH
M
nh

]Re[2)0(
7
1
)(
3
1
7
2
7
6






 

k
n
kj
k
j
eeHnh

1)0( 0
 eH

9. ]Re[21
7
1
)(
3
1
)3(
7
2






 

k
n
k
j
enh

)sin()cos( 
je j

])3(
7
2cos[21
7
1
)(
3
1






 k
n
k
nh 

10. ])3(
7
2cos[21
7
1
)(
3
1






 k
n
k
nh 
)])3(
7
3
2cos[)]3(
7
2
2cos[)]3(
7
1
2(cos[21
7
1
)0( 



 h
])
7
18
cos[]
7
12
cos[]
7
6
(cos[21
7
1
)0( 





h
0)0( h

11. )])2(
7
3
2cos[)]2(
7
2
2cos[)]2(
7
1
2(cos[21
7
1
)1( 



 h
0)1( h
)])1(
7
3
2cos[)]1(
7
2
2cos[)]1(
7
1
2(cos[21
7
1
)2( 



 h
0)2( h

12. )])0(
7
3
2cos[)]0(
7
2
2cos[)]0(
7
1
2(cos[21
7
1
)3( 



 h
1)3( h
)])1(
7
3
2cos[)]1(
7
2
2cos[)]1(
7
1
2(cos[21
7
1
)4( 



 h

13. )])1(
7
3
2cos[)]1(
7
2
2cos[)]1(
7
1
2(cos[21
7
1
)4( 



 h
)])1(
7
3
2cos[)]1(
7
2
2cos[)]1(
7
1
2(cos[21
7
1
)2( 



 h
)cos()cos(  
0)0()6(
0)1()5(
0)2()4(



hh
hh
hh

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