What kind of filter would have a sawtooth frequency response?

Question

I want a complex FIR or IIR filter whose response is approximately a sawtooth in the frequency domain. How should this be formally described, and what formula or algorithm can I use to calculate taps for one?

Here is a sketch of the magnitude response I want (not analysis of an actual filter). I think the phase response doesn't matter.

Answer 1

A windowed design is easy. Starting with an ideal zero-phase frequency response:

$$\frac{\omega}{2\pi} + \frac{1}{2}$$

The inverse Fourier transform of that gives the ideal impulse response:

$$\int^\pi_{-\pi}\left(\frac{\omega}{2\pi}+\frac{1}{2}\right)e^{i\omega k}d\omega = \frac{\sin(\pi k)}{k} + i\left(\frac{\sin(\pi k)}{\pi k^2} - \frac{\cos(\pi k)}{k}\right).$$

Similar to the $\text{sinc}$ function, the real part $\frac{\sin(\pi k)}{k}$ has an illegal division by zero at $k=0$ where the limiting value $\pi$ should be used instead. For other integer $k$, the real part is zero. The imaginary part has similar problems at $k=0$, where its limit is 0. Drawing from Rick's answer, at integer $k$, $\sin(\pi k)$ is zero and $\cos(\pi k)$ is +1 or -1. This simplifies the ideal impulse response at integer $k$ to:

$$\begin{cases}\pi&\text{if }k=0, \\i\frac{-(-1)^k}{k}&\text{otherwise.}\end{cases}$$

If the filter is implemented as a sum of a purely real and a purely imaginary filter, the purely real filter requires no work other than to multiply the input by a constant. The imaginary part of the impulse response is antisymmetrical, which may enable savings.

To continue with the windowed design, choose a window function and its length (equals the number of filter taps) and multiply the ideal impulse response by it. That gives the impulse response of the filter. The window function should be centered at $k = 0$.

Answer 2

For general filter designs, a least squares method is usually a good start. The window method suggested by Olli Niemitalo is also useful, but it is a bit less flexible because you can't define "don't care regions" (which is important for dealing with discontinuities in the desired response). Furthermore, with least squares you can add a weighting function which marks certain frequency regions as more important than others, resulting in a smaller approximation error in those regions. And finally, with least squares you don't need to analytically compute an inverse DTFT, which is not always possible in practice.

A simple Matlab/Octave script for a least squares design (without weighting) looks like this (note that I assume a linear desired phase response):

N = 101;                                 % choose some filter length
w = 2*pi*linspace(-.45,.45,200);         % frequency grid with don't care regions
d = (.5+w/(2*pi)).*exp(-1i*w*(N-1)/2);   % desired response (linear phase)
A = exp(-1i*w(:)*(0:N-1));               % matrix of overdetermined system A*h=d
h = A\d(:);                              % least squares solution
% plot designed frequency response
wp = linspace(-pi,pi,512);
H = freqz(h,1,wp);
plot(wp/2/pi, abs(H)); axis([-.5,.5,0,1])

Answer 3

What you have is basically the magnitude response of a sampled differentiator.

$y(t|t=nT) = \frac {d}{dt} x(t)$

This is a basic first order high pass filter with a +6dB/octave slope.

The simplest possible approximation would be a one zero filter.

$y[n] = a_0 * (x[n] - x[n-1])$

You could add a pole to that for a better low frequency match, or you could go all in with a windowed FIR design like Olli Niemitalo (what is the markdown syntax for proper @-attribution?) mentioned above.

https://ccrma.stanford.edu/~jos/fp/One_Zero.html

Answer 4

Figuring out the Fourier series for a sawtooth wave in the time domain is a common exercise. That same series will be the time domain impulse response for a sawtooth in the frequency domain. Since you want odd symmetry, all imaginary (but conjugate symmetric if you want a real result).

Answer 5

I plugged a frequency response of $\omega$ into the inverse discrete-time Fourier transform (inverse DTFT) equation and ended up with:

$$h[k] = \frac{\sin(k\omega_c)}{k^2\pi}-\frac{(\omega_c)\cos(k\omega_c)}{k\pi}$$

where $\omega_c$ is the cutoff freq measured in radians. Index $k$ is ...-3,-2,-1,0,1,2,3.... When $k = 0$ set $h[0] = 0$. (You may want to multiply $h[k]$ by a window sequence to reduce the frequency magnitude response ripples.) If we set $\omega_c = \pi$ we end up with:

$$h[k] = \frac{-(-1)^k}{k}$$

Try out the above $h[k]$'s. Who knows, they might work for you. (The $h[k]$ equations are supplied "as is." No warranty of any kind, either expressed or implied.)