lecture 22-26 - frequency sampling method
TRANSCRIPT
DIGITAL FILTER DESIGN:FREQUENCY SAMPLING METHOD
احسان احمد عرساڻيLectures 22-26 26کان 22ليڪچر
Digital Signal Processing
A Practical Aproach
Second Edition
Emmanuel C. Ifeachor
Barrie W. Jervis
Chapter 7: FIR filter Design
Frequency Sampling Method
An alternate approach to Fourier Transform –Window method
The filter coefficients can be calculated based onthe specified magnitudes of the desired filterfrequency response, distributed uniformly infrequency domain
More flexibility
Freq. Samp. Method
Create an ideal specification of the filter in the frequency-domain
Take Inverse Discrete Fourier Transform to get the time-domain impulse response of the filter
Remember, in window method we took IDTFT, here we take the IDFT
Derivation
Suppose H(k) represents the sampled version of theideal magnitude response of the desired filter, so,h[n] i.e. IDFT of H(k), would represent its impulseresponse
Nknj
N
k
ekHN
nh2
1
0
)(1
][
Nkn
Nkn
N
k
jkHN
nh 22
1
0
sincos)(1
][
We have established that for linear phase response, werequire the impulse response to be symmetric (takingN as odd number)
Nkn
Nkn
k
jkHN
nh
N
22
0
sincos)(1
][2
1
Now, there is another issue with this equation
There is a possibility that h[n] may be complex, i.e. itmight include the real component & thequadrature (imaginary) component
To avoid the requirement of quadrature processing intime domain, we shall have to make sure that theimpulse response is real
This can only happen if
a) H(0) is real
b) H(k) = H*(N-k)
This concept is illustrated in the figure on next slide
Graph courtesy DSP by Ifeachor & Jervis, pp. 380
N = 15k = 0, 1, . . . , 14
Filter Design Steps
Given the filter length N, specify the magnitude frequency response for the normalized frequency range from 0 to Fs :
Calculate the FIR filter co-efficients
21,...,1,0 N-k
N
kFf s
k
2
1
1
)(2)0(cos)(2
1][
N
k
N
nkHkH
Nnh
2
1
N
Example: Design a low-pass FIR filter with the following specs using the frequency sampling method,
Pass band : 0-5 KHz
Sampling frequency : 18 KHz
filter length : 9
The ideal frequency response can be easily constructed
We need to sample it at
4,3,2,1,0 , 2 9
18 x kKHzkk
N
kFf s
k4
2
1
N
The pass-band edge frequency is 5 KHz, which would actually fall between frequency bins 2 & 3
k =2 → 4 Hz
k =3 → 6 Hz
We’ll choose k=3, since k=2 would mean that the component at 5 KHz would be stopped
|H(k)|
k10 2 3 4 5 6 7 8
009
422cos22
9
412cos120
9
1][
nH
nHHnh
for n = 0, 1, 2, 3, 4
4,30
2,1,01)(
k
kkH
9
44cos2
9
42cos21
9
1][
nnnh
for n = 0, 1, 2, 3, 4
0725.053.1879.119
1]0[ h
4,30
2,1,01)(
k
kkH
9
44cos2
9
42cos21
9
1][
nnnh
9
404cos2
9
402cos21
9
1]0[
h
9
16cos2
9
8cos21
9
1]0[
h
111.01119
1]1[ h
9
4222cos2
9
4212cos21
9
1]2[
h
059.0879.1347.019
1]2[ h
9
44cos2
9
42cos21
9
1][
nnnh
9
414cos2
9
412cos21
9
1]1[
h
9
12cos2
9
6cos21
9
1]1[
h
9
4322cos2
9
4312cos21
9
1]3[
h
31993.0347.0532.119
1]3[ h
9
4422cos2
9
4412cos21
9
1]4[
h
55.02219
1]4[ h
Using the symmetry property, we can find the remaining coeffs
55.0]4[ h
0723.0]8[]0[ hh
111.0]7[]1[ hh
059.0]6[]2[ hh
31993.0]5[]3[ hh
0 1000 2000 3000 4000 5000 6000 7000 8000 9000-600
-400
-200
0
Frequency (Hz)
Phase (
degre
es)
0 1000 2000 3000 4000 5000 6000 7000 8000 9000-60
-40
-20
0
20
Frequency (Hz)
Magnitude (
dB
)
Example: Design a FIR low-pass filter using the frequency sampling method, with the following spec
Pass band : 0-5 KHz
Sampling frequency : 18 KHz
filter length : 15
Solve it
Optimizing the amplitude resp.
From the graph for the previous example weobserved that there were large ripples in the stopband Gibbs effect, caused due to discontinuity
This is similar to what we observed in the FT-Wmethod when rectangular window was used
In that case we used a window function tosmoothen the discontinuity, trading it off with thetransition width
Graph courtesy DSP by Ifeachor & Jervis, pp. 384
For a lowpass filter, the stop-band attenuation
increases, approximately, by 20 dB for each
transition band bin (citing Rabiner’s work in 1970),
with a corresponding increase in the transition
width
Approx. stop-band attenuation : (25+20M) dB
Approx. transition width : (M+1)Fs/N
Where M = number of bins in transition band & N is the
length of the filter impulse response
The key to using transition width bins is to choose
their values properly, to get the required
attenuation & so that bins are not used
Rabiner, also went ahead and calculated some
optimized values of these transition width bins and
are given on next slide
} stopband in the w{
)()(maxminimize
},...,,{ 21
wHwHW d
TTT M
Graph courtesy DSP by Ifeachor & Jervis, pp. 384
Example: Design a FIR low-pass filter using the frequency sampling method, with the following spec
Pass band : 0-5 KHz
Sampling frequency : 18 KHz
filter length : 15
stop band attenuation : 42 dB
Solve it
Comparison of Filter Design Methods
Hd(ω)
Hd(ω)
or H(k)
Window based method Frequency sampling method
k
sampling
d HH thH d
F
d 1
KnnhkH DFT
0 1
nhnh truncate
d
Hnh DTFT
Hnh DTFT
nhth d
Sampling
d