0
$\begingroup$

I want to design a filter with a custom phase delay related to frequency. As frequency increases, phase delay should increase.

The time delay as a function of frequency can be expressed as: $t_d = \frac{L}{C_{ph}} - \frac{1}{f}$

$L$ and $C_{ph}$ are constants of $0.0017$ and $2628$ respectively.

The range of $f$ is $500\mathrm{kHz}$ to $1 \mathrm{MHz}$, the sampling frequency, $f_s$ is $78.39\mathrm{MHz}$.

Thus, when the frequency is $500 \mathrm{kHz}$, the delay should be $1.37\mathrm{\mu s}$ or $107$ samples.

For the pass band, $f$, I have used a filter response of $e^{-i2\pi fd}$, where $d$ is the delay in samples required.

I have designed stop bands with a filter response of zero between $0 \mathrm{Hz}-100\mathrm{kHz}$ and $1.4 \mathrm{MHz}-f_s$.

In MATLAB my code looks like this:

n = 50;
fs = double(fs); 
L = double(L);

Cph = 2628;

f1 = linspace(500e3/fs,1e6/fs,100);
f1d = L/Cph - (1./(f1*fs));
%Our array is backwards though innit
f1d = f1d * -1;
f1dz = f1d * fs;
h1 = exp(-1i*2*pi*f1.*f1dz);

fstop1 = linspace(0,100e3/fs, 10);
hstop1 = zeros(size(fstop1));

fstop2 = linspace(1.4e6/fs, 1, 10);
hstop2 = zeros(size(fstop2));

d=fdesign.arbmagnphase('n,b,f,h', n, 3, fstop1, hstop1, f1, h1, fstop2, hstop2);
%d=fdesign.arbmagnphase('n,b,f,h', n, 1, f1, h1);
D = design(d,'equiripple');

fvtool(D,'Analysis','phasedelay');

This is what I get:

enter image description here

Markers are at 500k and 1M.

What am I doing wrong?

$\endgroup$
2
$\begingroup$

I notice two problems in your code:

  • at $f=500\,\text{kHz}$ you require a phase delay of $107$ samples, but your filter length is only $50$ taps; you can't have a delay that's greater than the filter length of an FIR filter.
  • it looks like you normalize all frequencies by the sampling frequency; however, in Matlab a normalized frequency $f=1$ corresponds to the Nyquist frequency, i.e., $f_s/2$, not to $f_s$.
$\endgroup$
  • $\begingroup$ Your second point explains lots of things! Thanks man! :) $\endgroup$ – Chris Adams Jul 15 '16 at 13:53
1
$\begingroup$

I went down the ifft route. Based off this post: What effect does a delay in the time domain have in the frequency domain?

My code looks like this:

N = length(h1_data(:,1));
%N = (2^nextpow2(N));

L = h1_cord(18,1) - h1_cord(2,1);
Cph = 2628;

%Generate real frequency vectors (negative and positive 
freal = linspace(nyq, -nyq, N);
fd = (L/Cph) - (1./freal);
fd = fs * fd;
fd = fd * -1;
%fd(1) = fd(2);
f = exp((-1i*2*pi*(1:N).*fd)/N);
%f(N/2+1:end) = fliplr(f(1:N/2));
%f = [f fliplr(f)];
%Overwrite with freqs we don't care about
fs = 1/(h1_time(2) - h1_time(1));
nyq = fs/2;
bin = nyq/(N/2);

f0bin = round(250e3/bin);
f1bin = round(1.25e6/bin);
f(1:f0bin) = 1;
f(f1bin:N/2) = 1;
f(N/2+1:f1bin+N/2) = 1;
f(N-f0bin:end) = 1;

f(1) = 1;
g = real(ifft(f));

b = [g(N/2+1:end) g(1:N/2)];
b = b .* hamming(N)';

fvtool(b,1);
$\endgroup$
  • $\begingroup$ Chris, if this answers your question, please give it the check mark when the system lets you! Thanks for taking the time to give an update. $\endgroup$ – Peter K. Jul 15 '16 at 15:04
0
$\begingroup$

A possible alternative to designing a FIR filter constrained to both frequency and phase response relationships using IFFT methods is to design an IIR filter using FDLS (Frequency-Domain Least Squares). See: Berchin's FDLS arbitrary filter design algorithm and http://robotics.itee.uq.edu.au/~elec3004/2014/lectures/Precise%20Filter%20Design%20(chapter).pdf

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.