CODEWITHSUNDEEP
cfilter
patrick
patrick

Reputation: 1042

How can I apply multiple filters at same time?

I need to apply a FIR low pass, high pass, and notch filter at the same time. Each filter has it's own set of coefficients. All use the following algorithm. Can I add or average the output of all three or is there a better way? 

SAMPLE fir_basic(SAMPLE input, int ntaps, const SAMPLE FIRcoefficient[], delay[])       
    {
        int ii;
        SAMPLE accum;

        /* store input at the beginning of the delay line */
        delay[0] = input;

        /* calc FIR */
        accum = 0;
        for (ii = 0; ii < ntaps; ii++) {
            accum += FIRcoefficient[ii] * delay[ii];
        }

        /* shift delay line */
        for (ii = ntaps - 2; ii >= 0; ii--) {
            delay[ii + 1] = delay[ii];
        }

        return accum;
    }

Upvotes: 1

Views: 395

Answers (1)

onitake
onitake

Reputation: 1409

The problem with multiple stacked convolution filters (as your filter routine implementation suggests) is that for each pass, you (theoretically) need to access all the values that had an influence on the previous filter passes.

To illustrate this, consider the following example:

Input:    abcdefghijk
Filter 1: 121
Filter 2: 1210121

For pass 1, you get the following outputs:

a: 2a+b
b: a+2b+c
c: b+2c+d
d: c+2d+e
e: d+2e+f
...

In pass 2, you normally use the data already processed by pass 1 as a whole:

a: b+2c+d
b: a+c+2d+e
c: 2a+b+d+2e+f
d: a+2b+c+e+2f+g

If you combine the two:

a: (a+2b+c)+2(b+2c+d)+(c+2d+e)
b: (2a+b)+(b+2c+d)+2(c+2d+e)+(d+2e+f)
...

So, by combining the two filters, you make the function much more complicated as each pass is dependent on more original inputs than the previous one. You can precalculate the coefficients, but it needs some manual work.

Edit: I didn't take any scaling into consideration. If the scaling factor is always the same for a single filter pass, it can of course be factored out and applied to the final result.

Upvotes: 1

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