filter coefficients
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Just a quick question - are 'filter coefficients' the coefficients of the impulse response, desired transfer function, or actual transfer function?
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Matt Fig
2012 年 11 月 2 日
Just a quick question - are 'filter coefficients' the coefficients of the impulse response, desired transfer function, or actual transfer function?
採用された回答
Wayne King
2012 年 5 月 25 日
That depends. The filter coefficients are the coefficients of the difference equation. If your filter is an FIR filter, then the filter coefficients are the values of the impulse response.
If you have an IIR filter, then the filter coefficients are not the same as the impulse response. Remember in that case the impulse response is infinite.
For example:
b = fir1(10,0.2);
stem(b)
h = impz(b);
stem(h)
isequal(b',h)
but
[b,a] = butter(10,0.2);
h = impz(b,a);
But the ratio of Z-transforms of the numerator coefficients to denominator coefficiens is equal to the Z-transform of the impulse response.
For example - consider the IIR system with the following difference equation
y(n)-0.8*y(n-1) = x(n)
So the filter coefficients are:
A = [1 -0.8]; B =1;
The impulse response is:
h(n) = 0.8^n*u(n) where u(n) is the unit step. But compare:
h1 = impz(B,A);
subplot(211)
stem(h1,'color',[1 0 0]);
subplot(212)
n = 0:length(h1)-1;
h = 0.8.^n;
hold on;
stem(h,'color',[0 0 1]);
その他の回答 (2 件)
Wayne King
2012 年 5 月 28 日
The unit step should be treated in your DSP book.
For a sequence, the unit step sequence is the sequence which is 1 for n>=0 and 0 for all n<0.
As far as your other question, I solved the difference equation. I think if you search the web, you will find plenty of material on solving difference equations. This forum does not lend itself to me walking through that solution here.
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