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![]() What is this filter?
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| Author | Topic: What is this filter? | |
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pete Member |
A few weeks ago I found out what people ment when there were talking about FIR and IIR filters, so I've made a low pass filter out of two delay lines (attached) . Can any one tell me if this is a FIR or an IIR or something else. Note that "NoOfTaps" controls the cutoff frequency. Let me know if you need an explaination on how it works or if its self explainitory. IP: Logged | |
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David McClain Member |
Pete, If you are using delay lines with feedback of any sort then you have an IIR filter. Delay lines without feedback implement FIR filters. So in this case, you have an FIR filter going from the input up to the first mixer, and then the comb filter with feedback gives you an IIR filter. So you have built a hybrid.
[This message has been edited by David McClain (edited 25 February 2001).] IP: Logged | |
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pete Member |
Thanks David What made me think though was that although the IIR had a feedback value of 1 , the filter as a whole doesn't have an infinate impulse responce. Is this a trick that filter designers use all the time ? pete. IP: Logged | |
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David McClain Member |
Pete, are you just playing with me here? Your last filter made of a 1 tap feedback delay is an IIR integrator which has its pole at 1 on the real axis of the Z-plane. Ideally, it should have infinite amplitude response for DC... ...but your first filter is actually a kind of differentiator which ensures that the DC component gets flushed away so the second filter can't go unstable.... You apply a subtraction in the mixer by adding the incoming signal and its delayed, scaled, and negated, partner... So in essence you have built a highpass filter on the front to remove DC components and you reconstitute the sound in the second filter by means of integrating the differentiated signal. An IIR filter is only named that because in pure mathematical terms it has an infinite response to an input impulse function. In reality, there is no inifinity, and finite mathematics limits the pureness of our approximations. In this case, the second stage is about as perfect an IIR as you can get... try it out. An input impulse to that stage keeps recirculating the impulse value forever. The impulse doesn't get through the whole system (not for very long) because the first stage delay provides a negative copy some cycles later to cancel the effect of the impulse. So you should see a short rectangular pulse coming out of the system when you feed an impulse. The duration of the pulse should be equal to the number of taps in the first filter divided by your sample rate. Actually, in this case your first filter provides some interesting comb filtering on the input. The slope of the response is controlled mainly by the integrator stage and it has gain that is roughly 1/f. But by varying the number of taps in the first stage, you create comb teeth - the more taps the more comb teeth. Passing white noise through this composite filter makes some interesting sounds when you sweep the number of taps in the first stage. - DM
[This message has been edited by David McClain (edited 25 February 2001).] IP: Logged | |
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David McClain Member |
This is not a very efficient way to do things but it illustrates that Kyma can do just about anything... - DM IP: Logged | |
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David McClain Member |
Attached is the well known solution to that problem... It plants two zeros in the Z-plane, at +/- 1 on the real axis. The end result is to provide a nearly constant amplitude gain as a function of frequency. Also attached is a classic IIR parametric filter with a pair of poles and zeros. The poles and zeros lie along the same radial direction, and when the radius of the pole is greater than that of the zero you have a peaking filter. When the radius of the zero is larger than the pole you have a notching filter. The classic all-pass is very much like this except that the radius of the zero is always at 1/Rp (the reciprocal of the pole radius). And, hence, the zero always lies outside the unit circle. But when you pass white noise through an all-pass filter you can't see anything happening unless to add or subtract this filtered signal from the original. Then you get another parametric filter like before. - DM [This message has been edited by David McClain (edited 26 February 2001).] IP: Logged |
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