DSP Question: invfreqs.m
I generate some coefficents for a filter and can inspect the frequency response as following: %%Orginal Data N = 5000; data = cumsum(randn(N,1)); t = 252; a = 2 / (t+1); b = repmat(1-a,1 ,N).^(1:N); %b are your filter coeff b = b ./ sum(b); a = 1; %%Plot the Filter on some example data ma = filter(b, a, data); figure;plot(data); hold all; plot(ma, 'r'); %%Plot the Response figure;freqz(b,1); [h,w] = freqz(b,1); I now explain my problem. I am now in the situation where I have a frequency response (i.e. the vector "h") and know nothing else. I would like to estimate from this my original "b" (the filter coefficents) to allow me to estimate my variable "t". I thought I could use invfreqs.m (or invfreqz.m) to do this, but Im afraid I dont know how. %%Find the impluse response n = 10; % I choose a large number allowing a good approximation m = 0; % I choose 0 here as I have 1 in my orignal filter ==> the output comes out as aNew = 1; [bNew,aNew] = invfreqz(h,w,n,m); %[bNew,aNew] = invfreqs(h,w,n,m); sys = tf(bNew,aNew) %%Plot the filter coeffcients x1 = [0: 1/(size(b,2) -1) : 1]; x2 = [0: 1/(size(bNew,2) -1) : 1]; figure;plot(x1,b); hold all; plot(x2,bNew, 'r'); When I inspect the final plot, I would expect to see the red line (bNew) as a good approximation to b. It is not. not even close. Clearly I am doing something very wrong. Please could someone with experince of how this function works, explain my mistake. many thanks!
John Michell answered .
2025-11-20
invfreqz.m has some odd rules re calling it.
%%Orginal Data
N = 5000;
data = cumsum(randn(N,1));
t = 252;
a = 2 / (t+1);
b = repmat(1-a,1 ,N).^(1:N);
b = b ./ sum(b);
a = 1;
%%Plot the Filter on some example data
ma = filter(b, a, data);
figure;plot(data); hold all; plot(ma, 'r');
%%Plot the Response
figure;freqz(b,1, N);
[h,w] = freqz(b,1,N);
%%Using dflit
% b = 1; a = -1; %Sanity Check. simple difference filter
% Hd = dfilt.df1([b a],1); % num/ denom == a/b
% fvtool(Hd);
%%Find the impluse response
% If n>(N-1) then you get a random answer!
%If less then you get a visually different approximation,
%albiet it one with the same frequency response when plotted using freqz.m
n = N-1;
m = 0; % I choose 0 here as I have 1 in my orignal filter ==> the output comes out as aNew = 1;
wt = ones(size(w));
[bNew,aNew] = invfreqz(h,w,n,m);%, wt, 1000);
%[bNew,aNew] = invfreqs(h,w,n,m);
%sys = tf(bNew,aNew);
%%Plot the filter coeffcients
% These now match if you used n = N-1
x1 = [0: 1/(size(b,2) -1) : 1];
x2 = [0: 1/(size(bNew,2) -1) : 1];
figure;plot(x1,b); hold all; plot(x2,bNew, 'r');
%these always match
figure; freqz(b);
figure; freqz(bNew);
%invfreqz expects the complex-valued frequency response, not just the
%magnitude. If you only have the magnitude then use fdesign.arbmag
% http://www.mathworks.co.uk/products/dsp-system/demos.html?file=/products/demos/shipping/dsp/arbmagdemo.html
M = 1000;
F = w/max(w);
A = abs(h);
d = fdesign.arbmag('N,F,A',M,F,A);
Hd = design(d,'freqsamp');
%fvtool(Hd,'MagnitudeDisplay','Zero-phase');
b3 = Hd.Numerator(501:end); %symmetric so just take half.
x1 = [0: 1/(size(b,2) -1) : 1];
x3 = [0: 1/(size(b3,2) -1) : 1];
figure;plot(x1,b); hold all; plot(x2,bNew, 'r'); plot(x3,b3, 'g');