posted 19 July 2006 06:28            

how do filter co-efficients work, and how to go about generating them from the filter characteristics of an impulse response?

and how to think about co-efficients in Waveshaping? Are they points on an input->output map or what?

there isn't really anything about co-efficients in the book, so a little guide would be useful ot help grasp what is audibly a very powerful and unique way of shaping sound

posted 19 July 2006 08:29            

You can think of it as a multi tap delay line but where all the taps are right next to each other. The co-efficients are like the levels of each tap. So if you had 1 1 1 1 1 it would be like playing the signal 5 times each time delayed by one sample. In this case you would probably over load and clip the signal as a lo frequency signal could become amplified five times. A hi frequency signal would be smeared and the plus and minus would tend to cancel a bit and hence get attenuated. This would be a very basic lo pass filter. If instead you put in {1/5} {1/5} {1/5} {1/5} {1/5} then you would get the same thing but without the clipping. If you now put in 1 {-1/4} {-1/4} {-1/4} {-1/4} you can see that a lo frequency will tend to average out to zero and hence get cut quite a lot, but a hi frequency is more likely to get through. This is a very basic hi pass filter.

This is a very basic description of a huge subject and I'm sure there are plenty of filter design experts that will tell you all the floors and disadvantages of using the figures I've given, but it might give you a taste of something the you could end up reading about for years.

hope this helps

posted 19 July 2006 08:35            

For the Waveshaper, a rule of thumb is that even numbered coefficients control the amplitude of even-numbered harmonics in the distortion and odd-numbered coefficients control the amplitude of odd-numbered harmonics in the distortion.

In this case the coefficients are like faders on increasing powers of the Input signal that are all mixed together to form the output. For example, using ^ to mean raised-to-the-power-of, we have:

(a0 * Input^0) + (a1 * Input^1) + (a2 * Input^2) + .. + (an * Input^n)

posted 19 July 2006 23:05            

I found this on line somewhere ...

"It so happens that this frequency response is just the Fourier transform of the filter coefficients.

The inverse solution to a Fourier transform is well known: it is simply the inverse Fourier transform.

So the coefficients for an FIR filter can be calculated simply by taking the inverse Fourier transform of the desired frequency response.

Here is a recipe for calculating FIR filter coefficients:

decide upon the desired frequency response
calculate the inverse Fourier transform
use the result as the filter coefficients"

posted 19 July 2006 23:27            

instead, trawling the web, I find...
http://www-users.cs.york.ac.uk/~fisher/mkfilter/ - an interactive digital filter designer

and
http://www.dsptutor.freeuk.com/IIRFilterDesign/IIRFilterDesign.html - a Java Applet for generating co-efficients of IIR filters , both Chebyshev and Butterworth

posted 20 July 2006 00:44            

http://www.dspguru.com/info/faqs/fir/basics.htm

[This message has been edited by cristian_vogel (edited 20 July 2006).]

posted 20 July 2006 05:42            

" I'm still keen to find out if an inverse FFT can be used to analyse an impulse response into co-efficients"

No it's actually much simpler than that. If you play a single pulse (one sample wide) into a filter and record the output, that recording is the impulse response without any need for FFT. i.e. the first sample of the recording is the first co-effecient the second sample is the second co-effecient etc.

What they meant by the inverse FFT method is that if you didn't have a filter to test in the first place you could make one with manuel control over the spectrum. Imagine if you was to put this recorded response into an FFT (non inverse), then you could see the spectrum of the filter, well sort of as the levels could be in the real or the imaginary parts and could be positive or negative. But this still represents the filters spectrum. Obviously if you did an inverse FFT on this you would get back the response you had in the first place. So what they are saying is that if you didn't have a filter in the first place you could draw a spectrum by hand, and then do the inverse FFT on that, and the resulting output could be used as the response (stream of co-efficients) for an FIR filter that would match the spectrum you drew.

Did you realize that the crossfilter module is a very big FIR but instead of typing in all the co-effecients, you load them in in the form of the response signal. A one second response is like 44100 co-effecients (if thats the sample rate you are using).

hope it helps

posted 20 July 2006 09:46            

What is normally done is to go back and forth between frequency and time using FFTs, changing the values slightly each time until you end up with an impulse response that contains only real numbers.

Another alternative is to look up the formulas for a standard form filter (like a Butterworth low pass filter).

posted 20 July 2006 11:16            

One of the problems with the inverse-FFT method of FIR filter design is that you can control the frequency response of the filter at the spectrum points in the FFT, but you cannot easily control the response in-between those points. For example, if you make a perfect lowpass filter using this method, the resulting filter will have very large ripples in the frequency response.