# 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, we could 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

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

z0 is 1, of course, as is any value raised to the power of 0. And for normalized biquads, b0 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 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, e0jω 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). 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, ejx = 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 physicists and used j. Computer programming often use j, maybe because i is 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 is 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 use the atan2 function, which takes into account quadrants). Note that this is the same conversion we use for FFT results, as you can see in my article, A gentle introduction to the FFT.

##### $$phase := atan2(H.imag,H.real)$$

For now, I’ll leave you with some Python code, as it’s cleaner and leaner than a C or C++ implementation. It will make it easier to transfer to any language you might want (Python can be quite compact and elegant—I’m going for easy to understand and translate with this code). Here’s the direct evaluation routine corresponding to the summation part of the equation (you’ll also need to “import numpy” to have e available—also available in the math library, but we’ll use numpy later, so we’ll stick with numpy alone):

import numpy as np

# direct evaluation of coefficients at a given angular frequency
def coefsEval(coefs, w):
res = 0
idx = 0
for x in coefs:
res += x * np.e**(-idx * 1j * w)
idx += 1
return res

Again, we call this with the coefficients for each frequency of interest. Once for the numerator coefficients (the a coefficients on this website, corresponding to zeros), once for the denominator coefficients (b, for the poles—and don’t forget that if there is no b0, the case for a normalized filter, insert a 1.0 in its place). Divide the first result by the second. Use use abs (or equivalent) for magnitude and atan2 for phase on the result. Repeat for every frequency of interest.

Here’s a python function that evaluates numerator and denominator coefficients at an arbitrary number of points from 0 to π radians, with equal spacing, returning arrays of magnitude (in dB) and phase (in radian, between +/- π):

# filter response, evaluated at numPoints from 0-pi, inclusive
def filterEval(zeros, poles, numPoints):
magdB = np.empty(0)
phase = np.empty(0)
for jdx in range(0, numPoints):
w = jdx * math.pi / (numPoints - 1)
resZeros = coefsEval(zeros, w)
resPoles = coefsEval(poles, w)

# output magnitude in dB, phase in radians
Hw = resZeros / resPoles
mag = abs(Hw)
if mag == 0:
mag = 0.0000000001  # limit to -200 dB for log
magdB = np.append(magdB, 20 * np.log10(mag))
phase = np.append(phase, math.atan2(Hw.imag, Hw.real))
return (magdB, phase)

Here’s an example of evaluating biquad coefficients at 64 evenly spaced frequencies from 0 Hz to half the sample rate (these coefficients are right out of the biquad calculator on this website—don’t forget to include b0 = 1.0):

zeros = [ 0.2513643668578741, 0.5027287337157482, 0.2513643668578741 ]
poles = [ 1.0, -0.17123074520885395, 0.1766882126403502 ]

(magdB, phase) = filterEval(zeros, poles, 64)

print("\nMagnitude:\n")
for x in magdB:
print(x)

print("\nPhase:\n")
for x in phase:
print(x)

Next up, a javascript widget to plot magnitude and phase of arbitrary filter coefficients.

### Extra credit

The direct evaluation function performs a Fourier analysis at a frequency of interest. For better understanding, reconcile it with the discrete Fourier transform described in A gentle introduction to the FFT. In that article, I describe probing the signal with cosine and sine waves to obtain the response at a given frequency. Look again at Euler’s formula, which shows that e is cosine (real part) and sine (imaginary part), which the article alludes to this under the section “Getting complex”. You should understand that the direct evaluation function presented here could be used to produce a DFT (given complete evaluation of the signals at appropriately spaced frequencies). The main difference is that for this analysis, we need not do a complete and reversible transform—we need only analyze frequency response values that we want to graph.

This entry was posted in Biquads, FFT, Filters, FIR Filters, IIR Filters. Bookmark the permalink.

### 27 Responses to Evaluating filter frequency response

1. Matthieu says:

Nice sum up! This is actually what scipy.signal.freqz does with the polynomial.
I would definitely change the Python code though as you could benefit greatly from vector instructions. The first would be np.exp() instead of np.e**/for coeffs loop and then in your second script, you can use array masking to vectorize the range call.

• Nigel Redmon says:

Thanks for the input, Matthieu, good points. Yes, if I really needed a Python routine, I’d definitely code it differently, but going for easy translation. In fact, I wrote a Javascript widget to display frequency and phase response (will post tonight, probably, after I get a chance to add some text explanation and instructions) that was nearly a direct translation. I love Python, but it’s not so good for translatable examples unless you code “C-like”. 😉

2. Howard says:

Hi Nigel,

I am always amazed at your ability to write easy to understand descriptions of things that are not so easy to understand, if that makes sense. 😉

Really love when you post new articles; they are always useful to me.

Thank you very much!

3. markzzz says:

Hi Nigel,

as usual, amazing job 🙂
Why did you choose nPoints = 64? What does it means?

Somethings near “10 hz”? Or it is just random?

• markzzz says:

I mean: which w must I need to supply for example if I want to catch magnitude at 100 and 2300 hz?

• Nigel Redmon says:

The filterEval function shown specifically divides the frequency range into every spaced points and returns an array of readings. That’s what you’d want to do when plotting linear response, for instance. I was going for minimal code here, to keep the idea clear. But in practice you might want to break the inner loop out to evaluate a single point—to support log frequency spacing, for instance.

You can also look at the JavaScript source code for the grapher widget in your browser; here’s the coefficient evaluation function for a single point (normalized frequency—frequency in Hz divided by sample rate); the function implements just the single summation operation in the equations in the article—the coefsEval Python function—so you’d use it on the numerator (zeros) coefficients, the denominator (poles) coefficients, and divide; also note I’m using math.js—it keeps the code a lot cleaner since it handles the complex math:

// frequency response
function evalCoefs(coefs, freq) {
var len = coefs.length;
var res = 0;
for (var idx = 0; idx < len; idx++)
res = math.add(res, math.multiply(coefs[idx], math.exp(math.complex(0,-idx * freq * 2 * math.pi))));
return res
}

• markzzz says:

Clear! Thank you very much. Ill try to convert it into C++ 😉

• markzzz says:

Not sure why, but I’ve a problem getting amp from freq. Check this code in JavaScript (mostly copy and paste from your code): https://jsfiddle.net/oznbs4kt/
It alerts the amp in dB of the freq 606Hz. I have those coefficients:
a0 = 1.1311308700603495
a1 = -1.9742954081351101
a2 = 0.8512071363732295
b0 = 1.0000000000000000
b1 = -1.9742954081351101
b2 = 0.9823380064335793

If I show them on your tool http://www.earlevel.com/main/2016/12/08/filter-frequency-response-grapher/ , I can clearly see that near the freq 606 its up of +20db more or less. Thats correct, the filter is working weel (even in my C++ plugin).

But when I try to plot from your JS code (the code used in this tutorial), I just get +7db at that Hz. The whole slope is slowed down in gain.

Where am I wrong?

• Nigel Redmon says:

If you take a look at some of your intermediate results, I think you’ll see the problem is that you’re not doing complex math. Note that I’m using “math.abs”, etc., not “Math.abs”—I’m using the js.math library. Otherwise, you need to use Euler’s identity, unroll the complex math (more code).

• markzzz says:

What a mistake! I see “0” as real part and I didn’t figure out that exp will produce value as well 🙂 Thanks

• Nigel Redmon says:

Just an arbitrary number of points to evaluate. Note that the function explains the parameter—but I added a few words to the text to make it clearer.

4. Princeton University says:

It’s surprising to find on earlevel.com a resource so
We will note your page as a benchmark for Evaluating filter frequency response .

We also invite you to link and other web resources for equations like http://equation-solver.org/ or https://en.wikipedia.org/wiki/Equation.
Thank you and good luck!

5. jan16 says:

Thanks for an awesome article! Helped a lot to understand how to plot these.

6. Alan Cooper says:

Thanks,
It was all I needed to check my coefficients by plotting then in a spreadsheet.
Regards AlanC

7. RobBon says:

Here is the simply implementation in C# (without any “magic” Complex computation).

public double filterEvaluate(double freq, double sampleRate, double a0, double a1, double a2, double b1, double b2) {

double w = 2 * Math.PI * freq / sampleRate;

double cos1 = Math.Cos(-1 * w);
double cos2 = Math.Cos(-2 * w);

double sin1 = Math.Sin(-1 * w);
double sin2 = Math.Sin(-2 * w);

double realZeros = a0 + a1 * cos1 + a2 * cos2;
double imagZeros = a1 * sin1 + a2 * sin2;

double realPoles = 1 + b1 * cos1 + b2 * cos2;
double imagPoles = b1 * sin1 + b2 * sin2;

double divider = realPoles * realPoles + imagPoles * imagPoles;

double realHw = (realZeros * realPoles + imagZeros * imagPoles) / divider;
double imagHw = (imagZeros * realPoles - realZeros * imagPoles) / divider;

double magnitude = Math.Sqrt(realHw * realHw + imagHw * imagHw);
double phase = Math.Atan2(imagHw, realHw);

//return magnitude;     //gain in Au
return 20 * Math.Log10(magnitude);  // gain in dB
//return phase;                   //phase
}
• Malcolm Rest says:

Hi,

In this implementation, what does ‘freq’ stand by?
Is it the ‘freq’ value of the specific filter?

i.e if the dialed in filter is:
[F = 1000 [Hz], Q = 1.414, G = 6 [db], Sample rate = 48KHz, shape = bell]

Then ‘freq’ is simply 1000 ?

Isn’t this information ‘folded’ into the 5 (or six) biquads ?

• Malcolm Rest says:

Or is ‘freq’ a single calulated point for the full graph?

Example:
Calculate the full graph in which ‘freq’ = [20: 20 : 20000] ?

• Nigel Redmon says:

Yes, this—the routine determines the filter gain at a given frequency.

8. bart says:

Very well written! What I’m still can’t seem to wrap my head around is: you write:
“To translate that to an analog response, we substitute e^jω for z”.
In the original equation, H(z), you see the coefficients, which are in the time domain. Also z^-1 means “one sample delay in time”. Then you suddenly introduce omega into the equation, where omega is frequency dependent.
This substitution, is it actually executing the fourier transform, ie. time-to-frequency domain translation?

• Nigel Redmon says:

Evaluating the z transform on the unit circle (which is what e does) at an appropriate number of equal intervals is equivalent to the DFT. But of course there is no such requirement in evaluating e, so they aren’t the same—they’re related.

You can search for something like “e^jw substitution for z”, I see a nice discussion on researchgate.

I considered writing this article a different way, avoiding the exponential. For instance, the magnitude at a point on the unit circle is equal to the product of the length of vectors from zeros to that point, divided by the length of vectors of the poles to that point. It might have been more intuitive or interesting for some, but it gets uglier translating that into graphs and code.

9. Bart says:

Thanks! Since I wanted to get a better feel of the relation between the polynomals of the z transform, their roots and the frequency response, I decided to draw a 3d mesh of the magnitude of a H(z) transfer function. A kind of 3d version of a pole/zero plot in the z domain. Intersection with a ‘unit tube’ yields your bode plot, bent over the tube! Something hard to find images of. I now understand why… it is kinda hard to interpret all in 3d. Although I’m still a bit puzzled by the math, it was nevertheless a fun exercise.

10. Bart says:

Here the resulting 3d plot with octave code, if anybody would be interested.

11. Marc says:

You may also want to note that e^(-jw) = cos(w) – j*sin(w) which can be more directly applied. It’s been quite a while since I’ve used complex numbers, so that little thing wasn’t immediately obvious to me.

12. Mik says:

Hi Nigel,
I am using this method to visualize biquad response line in my eq. I would like to make sloopes 24/oct and higher. I have vector of calculated gains from standard biquad and assuming that i should multiply it by itself to make 4th order filter but i don’t know what to do with phase in this situation.

• Nigel Redmon says:

Yes, if you use two identical biquads to get 24 dB/oct, it would be squared fro frequency. Although in the most obvious cases you might want fourth order (highpass and lowpass in particular), you probably don’t want two identical biquads. See Cascading filters. For visualization it’s probably easier to calculate the single higher order filter. I don’t have a higher order calculator on my site, but you can use matlab/octave, or probably find a calculator on the web (in a quick look, digital Butterworth filter calculator, but you need to divide all on the right by by the first number and flip the signs for y coefficients).

• Mik says:

Hi Nigel, thank you for the answer.
So to make 4th order biquad (cascade of two identicall) should i only square complex frequency response?
I checked it and it makes magnitude only two times larger (i had peaking biquad with peak in -9dB, now i have -18dB).
Should i use a scaling factor or something else?

• Nigel Redmon says:

If I understand correct, that sounds right.