New answers tagged infinite-impulse-response
1
This is a common problem, and has an easy solution. The following assumes you are using a cascade of second-order direct-form-1 sections.
For a Butterworth highpass filter with a small ratio of Fcutoff to Fs, the numerator coefficients are close to [1 -2 1], and therefore the frequency response of the numerator by itself has a large attenuation at low ...
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An easy way to deal with this would be to actually let the sine run through your filter for a while, and then just at the right phase save the state of the IIR.
However, chances are: if the impulse response in this situation is a problem to you, your IIR isn't the right design.
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"integrator is unstable since it has a pole on the unit circle" -- not true. This Z-1 is done all the time in CIC resamplers. The pole is --on-- the unit circle exactly, and is therefore stable. Fun fact, if you use floating point math this falls apart and the integrator blows up bc the value is not exactly 1, its 1.000000000001 (or something larger than 1)
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There is no real shortcut for computing the impulse response. Partial fractions is the standard way to do it. However, in the case of a second-order transfer function as in your example, the result can be found in tables of $\mathcal{Z}$-transform pairs. E.g., if you use the last two correspondence of this table, you can pretty quickly write down the result.
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HINT:
Find the impulse response corresponding to
$$\tilde{H}(z)=\frac{1}{1+\alpha z^{-1}}\tag{1}$$
and then try to figure out what happens to the impulse response if you replace $z$ by $z^N$. (Note that $H(z)=\tilde{H}(z^N)$).
HINT 2:
Answer the following questions for yourself: what is the impulse response corresponding to $H(z)=1+z^{-1}$? And what is ...
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