Evaluating filter frequency response

A question that pops up for many DSP-ers working with IIR and FIR filters, I think, is how to look at a filter’s frequency and phase response. For many, maybe they’ve calculated filter coefficients with something like the biquad calculator on this site, or maybe they’ve used a MATLAB, Octave, python (with the scipy library) and functions like freqz to compute and plot responses. But what if you want to code your own, perhaps to plot within a plugin written in c++?

You can find methods of calculating biquads, for instance, but here we’ll discuss a general solution. Fortunately, the general solution is easier to understand than starting with an equation that may have been optimized for a specific task, such as plotting biquad response.

Plotting an impulse response

One way we could approach it is to plot the impulse response of the filter. That works for any linear, time-invariant process, and a fixed filter qualifies. One problem is that we don’t know how long the impulse response might be, for an arbitrary filter. IIR (Infinite Impulse Response) filters can have a very long impulse response, as the name implies. We can feed a 1.0 sample followed by 0.0 samples to obtain the impulse response of the filter. While we don’t know how long it will be, wecould take a long impulse response, perhaps windowing it, use an FFT to convert it to the frequency domain, and get a pretty good picture. But it’s not perfect.

For an FIR (Finite Impulse Response) filter, though, the results are precise. And the impulse response is equal to the coefficients themselves. So:

For the FIR, we simply run the coefficients through an FFT, and take the absolute value of the complex result to get the magnitude response.

(The FFT requires a power-of-2 length, so we’d need to append zeros to fill, or use a DFT. But we probably want to append zeros anyway, to get more frequency points out for our graph.)

Plotting the filter precisely

Let’s look for a more precise way to plot an arbitrary filter’s response, which might be IIR. Fortunately, if we have the filter coefficients, we have everything we need, because we have the filter’s transfer function, from which we can calculate a response for any frequency.

The transfer function of an IIR filter is given by

\(H(z)=\frac{a_{0}z^{0}+a_{1}z^{-1}+a_{2}z^{-2}…}{b_{0}z^{0}+b_{1}z^{-1}+b_{2}z^{-2}…}\)

z 0 is 1, of course, as is any value raised to the power of 0. And for normalized biquads, b 0 is always 1, but I’ll leave it here for generality―you’ll see why soon.

To translate that to an analog response, we substitute e jω for z , where ω is 2π*freq, with freq being the normalized frequency, or frequency/samplerate:

\(H(e^{j\omega})=\frac{a_{0}e^{0j\omega}+a_{1}e^{-1j\omega}+a_{2}e^{-2j\omega}…}{b_{0}e^{0j\omega}+b_{1}e^{-1j\omega}+b_{2}e^{-2j\omega}…}\)

Again, e 0jω is simply 1.0, but left so you can see the pattern. Here it is restated using summations of an arbitrary number of poles and zeros:

\(H(e^{j\omega})=\frac{\sum_{n=0}^{N}a_{n}e^{-nj\omega}}{\sum_{m=0}^{M}b_{m}e^{-mj\omega}}\)

For any angular frequency, ω , we can solve H(e jω ) . A normalized frequency of 0.5 is half the sample rate, so we probably want to step it from 0 to 0.5― ω from 0 to π―for however many points we want to evaluate and plot.

Coding it

From that last equation, we can see that a single FOR loop will handle the top or the bottom coefficient sets. Here, we’ll code that into a function that can evaluate either zeros ( a terms) or poles ( b terms). We ’ll refer to this as our direct evaluation function, since it evaluates the coefficients directly (as opposed to evaluating an impulse response).

You’ve probably noticed the j , meaning an imaginary part of a complex number―the output will be complex. That’s OK, the output of an FFT is complex too, and we know how to get magnitude and phase from it already.

Some languages support complex arithmetic, and have no problem evaluating “ e**(-2*j*0.5) ”―either directly, or with an “exp” (exponential) function. It’s pretty easy in Python, for instance. (Something like, coef[idx] * math.e**(-idx * w * 1j) , as the variable idx steps through the coefficients array.)

For languages that don’t, we can use Euler’s formula, e jx = cos(x) + j * sin(x) ; that is, the real part is the cosine of the argument, and the imaginary part is the sine of it.

(Remember, j is the same as i ―electrical engineers already used i to symbolize current, so they diverged from physicist and used j . Computer programming similarly went with j , apparently because i was a commonly used index variable.)

So, we create our function, run it on the numerator coefficients for a given frequency, run it again on the denominator coefficients, and divide the two. The result will be complex―taking the absolute value gives us the magnitude response at that frequency.

Revisiting the FIR

Since we already had a precise method of looking at FIR response via the FFT/DFT, let’s compare the two methods to see how similar they are.

To use our new method for the case of an FIR, we note that the denominator is simply 1, so there is no denominator to evaluate, no need for division. So:

For the FIR, we simply run the coefficients through our evaluation function, and take the absolute value of the complex result to get the magnitude response.

Does that sound familiar? It’s the same process we outlined using the FFT.

And back to IIR

OK, we just showed that our new evaluation function and the FFT are equivalent. (There is a difference―our evaluation function can check the response at an arbitrary frequency, whereas the FFT frequency spacing si defined by the FFT size, but we’ll set that aside for the moment. For a given frequency, the two produce identical results.)

Now, if the direct evaluation function and the FFT give the same results, for the same frequency point, and the numerator and denominator are evaluated by the same function, by extension we could also get a precise evaluation by substituting an FFT process for both the numerator and denominator, and dividing the two as before. Note that we’re no longer talking about the FFT of the impulse response, but the coefficients themselves. That means we no longer have the problem of getting the response of an impulse that can ring out for an unknown time―we have a known number of coefficients to run through the FFT.

Which is better?

In general, the answer is our direct evaluation method. Why? We can decide exactly where we want to evaluate each point. That means that we can just as easily plot with log frequency as we can linear.

But, there may be times that the FFT is more suitable―it is extremely efficient for power-of-2 lengths. (And don’t forget that we can use a real FFT―the upper half of the general FFT results would mirror the lower half and not be needed.)

An implementation We probably want to evaluate ω from 0 to π, corresponding to a range of half the sample rate. So, we’d call the evaluation function with the numerator coefficients and with the denominator coefficients, for every ω that we want to know (spacing can be linear or log), and divide the two. For frequency response, we’d take the absolute value (equivalently, the square root of the sum of the squared real and imaginary parts) of each complex result to obtain magnitude, and arc tangent of the imaginary part divided by the real part (specifically, we