Frequency Sampling Structures
Specifying Frequency Response
Set the frequency response of an FIR filter at M equally-spaced
points Ω k = 2π k / M , k = 0,1, 2,", M − 1 and find the impulse
response h [ n ].
Frequency response and impulse response are related by
H (e
jΩ
M −1
− jΩn
h
=
n
e
) ∑ []
n =0
and it follows that
(
H e
jΩ k
)
= H (e
j 2π k / M
M −1
) = ∑ h [ n] e
n =0
This is the DFT of h [ n ].
− j 2π kn / M
, k = 0,1, 2,", M − 1
Specifying Frequency Response
Since H ( e j 2π k / M ) is the DFT of h [ n ] , h [ n ] must be the inverse
DFT of H ( e j 2π k / M ) .
1
h [ n] =
M
M −1
j 2π k / M
j 2π kn / M
H
e
e
, n = 0,1, 2,", M − 1
)
∑ (
k =0
The transfer function and impulse response are related by
M −1
1 M −1 M −1
j 2π k / M
j 2π kn / M − n
H ( z ) = ∑ h [ n] z =
H
e
e
z .
(
)
∑
∑
M n =0 k =0
n =0
Exchanging the order of summation,
1
H(z) =
M
−n
M −1
∑ H (e
k =0
j 2π k / M
M −1
)∑e
n =0
j 2π kn / M
z −n
Specifying Frequency Response
Using the formula for summing a geometric series,
1
H( z) =
M
M −1
∑ H (e
k =0
j 2π k / M
−M
j 2π
− M M −1
1− z e
1− z
) 1 − z −1e j 2π k / M = M
H ( e j 2π k / M )
∑ 1− z
k =0
−1 j 2π k / M
e
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
H2 ( z ) = ∑
k =0
H ( e j 2π k / M )
−1 j 2π k / M
1− z e
.
Specifying Frequency Response
H1 ( z ) has zeros at
H1 (z )
1 − z−M = 0 ⇒ z−M = 1
z = e j 2π q / M , q = 0,1, 2,", 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 2π k / M = 0 ⇒ z −1e j 2π k / M = 1 ⇒ z = e j 2π k / M , k = 0,1, 2,", M − 1
So the transfer function can be written as
M −1
M −1 H e j 2π k / M
⎡
⎤
(
)
1
−1 j 2 π q / M
H(z) =
⎢ ∏1 − z e
⎥∑
−1 j 2 π k / M
1
−
M ⎣ q =0
z
e k
0
=
⎦
H1( z )
H2 ( z )
or
M −1
H (e
j 2π p / M
1
)= M
1− e
M −1 ∏
∑
k =0
− j 2π ( p − q ) / M
q =0
1− e
− j 2π ( p − k ) / M
H ( e j 2π 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 2π p / M ) =
H ( e j 2π p / M ) M −1
M
∏1 − e
− j 2π ( p − q ) / M
, 0≤ p<M
q =0
q≠ p
This result implies that
M −1
∏1 − e
q =0
q≠ p
− j 2π ( p − q ) / M
=M , 0≤ p<M
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 2π k / M ) = H * ( e − j 2π k / M ) = H * e
j 2π ( M − k ) / M
)
Specifying Frequency Response
Taking advantage of the complex-conjugate symmetry, for M even,
M −1
H2 ( z ) = ∑
k =0
H2 ( z ) =
H (e
j 2π k / M
)
−1 j 2π k / M
1− z e
H (e j0 )
1− z
−1
+
=
H (e
j0
1− z
)+
−1
H (e
M / 2−1
∑ 1− z
k =1
j 2π k / M
)
+
−1 j 2 π k / M
e
(
H e
1 − z −1e
A ( k ) − B ( k ) z −1
M / 2−1
∑ 1 − 2cos ( 2π k / M ) z
k =1
(
where A ( k ) = H ( e j 2π k / M ) + H e
j 2π ( M − k ) / M
)
−1
(
+ z −2
and B ( k ) = e − j 2π k / M H ( e j 2π k / M ) + e j 2π k / M H e j 2π ( M −k ) / M
)
Similarly, for M odd,
H2 ( z ) =
H (e j0 )
1− z
−1
+
H ( e jπ )
1+ z
−1
+
( M −1) / 2
∑
k =1
A ( k ) − B ( k ) z −1
j 2π ( M − k ) / M
1 − 2cos ( 2π k / M ) z −1 + z −2
)
j 2π ( M − k ) / M
Example
Design a filter whose frequency response goes through these points.
k
0 1 2 3 4 5 6 7
(
H e j 2π k / M
)
j 0 0 0 −j 1
0 1
− M M −1
1− z
H(z) =
M
1 − z −8
H(z) =
8
H ( e j 2π k / M )
∑ 1− z
k =0
−1 j 2π k / M
e
−j
1
j
1
⎡
⎤
+
+
+
⎢⎣1 − z −1e j 2π / 8 1 − z −1e j 4π / 8 1 − z −1e j12π / 8 1 − z −1e j14π / 8 ⎥⎦
Example
Taking advantage of the periodicity of e j 2π k / 8
1 − z −8
H(z) =
8
or
1
1
j
−j
⎡
⎤
+
+
+
⎢⎣1 − z −1e j 2π / 8 1 − z −1e − j 2π / 8 1 − z −1e j 4π / 8 1 − z −1e − j 4π / 8 ⎥⎦
⎡ 2 − 2 z −1 cos (π / 4 )
⎤
2 z −1 sin (π / 2 )
−
⎢
−1
−2
−1
−2 ⎥
1 − 2 z cos (π / 2 ) + z ⎦
⎣1 − 2 z cos (π / 4 ) + z
and the frequency response is
1 − z −8
H(z) =
8
⎡
⎤
2 − 2e − jΩ cos (π / 4 )
2e − jΩ sin (π / 2 )
−
⎢
− jΩ
− j 2Ω
− jΩ
− j 2Ω ⎥
1 − 2e cos (π / 2 ) + e
⎣1 − 2e cos (π / 4 ) + e
⎦
The impulse response is
− j 8Ω
e
−
1
H ( e jΩ ) =
8
⎧⎪0.25, −0.0732,0,0.0732, ⎫⎪ DFT
j 2π k / M
h [ n] = ⎨ ↑
) = {0,1, j,0,0,0, − j,1}
⎬ ←⎯⎯→ H ( e
⎪⎩−0.25, −0.4268,0,0.4268⎪⎭