Designfilt and crossover filters?

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Darsana P M 2017-3-16

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评论： Star Strider 2017-3-24

Can anybody help me with an alternative function for DesignFilt and crossover filters. These functions work only with matlab 2016. it doesn't work in matlab 2013. What is the alternative function?

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Answer 1

Star Strider 2017-3-16

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In R2013a, see the documentation on fdatool (link) and the other functions such as fdesign (link) that appear in the same documentation page.

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Star Strider 2017-3-16

在 MATLAB Online 中打开

I no longer have access to those functions (in R2017a). Since you want to design a bandpass filter, one option is the fdesign.bandpass function.

You can also design it yourself.

Example (using your design criteria):

Fs = 1500;                                      % Sampling Frequency
Fn = Fs/2;                                      % Nyquist Frequency
Wp = [500  560]/Fn;                             % Passband Frequencies (Normalised)
Ws = [495  565]/Fn;                             % Stopband Frequencies (Normalised)
Rp =  5;                                        % Passband Ripple (dB)
Rs = 50;                                        % Stopband Ripple (dB)
[n,Ws] = cheb2ord(Wp,Ws,Rp,Rs);                 % Calculate Filter Order
[z,p,k] = cheby2(n,Rs,Ws);                      % Design Filter
[sos,g] = zp2sos(z,p,k);                        % Convert To Second-Order-Section For Stability
figure(1)
freqz(sos, 2^16, Fs)
set(subplot(2,1,1), 'XLim',[0 700])             % ‘Zoom’ X-Axis
set(subplot(2,1,1), 'YLim',[-50 100])           % ‘Zoom’ Y-Axis (Amplitude)
set(subplot(2,1,2), 'XLim',[0 700])             % ‘Zoom’ X-Axis

Darsana P M 2017-3-17

编辑：Star Strider 2017-3-17

在 MATLAB Online 中打开

Can you please explain it once again. Is this the code that i must use?? But i got an error in the for loop part??

Fs = 8200;                                      % Samping Frequency (Hz)
Fn = Fs/2;                                      % Nyquist Frequency
pf = linspace(20,4000,17);                      % Passband Frequencies
cf = pf(1:end-1)+(pf(2)-pf(1))/2;               % Centre Frequencies
for k1 = 1:length(cf)
    [z(k1,:),p(k1,:),k(k1)] = butter(7, [pf(k1) pf(k1+1)]/Fn);
    [sos{k1},g{k1}] = zp2sos(z(k1,:),p(k1,:),k(k1));
    [h(k1,:),w(k1,:)] = freqz(sos{k1},512,Fs);
end
figure(1)
plot(w([1 16],:), abs(h([1 16],:)))
grid
% axis([0  0.2    ylim])
figure(2)
freqz(sos{1})
hold on
for k1 = 2:16
    freqz(sos{k1})
end
hold off

Star Strider 2017-3-17

You will need to change the code to design the filters that meet your requirements. Use your own sampling frequency, and define your filter passband frequencies in a vector and assign them to the ‘pf’ variable. You might be able to use the linspace call with only minor modifications. The rest of my code should work without any need to change it.

I ran that code just now and it ran for me without error (in R2017a, the code was written in R2015a). What error did you see? Please copy the entire red text of the error from the Command Window and paste it to a comment here.

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Answer 2

Darsana P M 2017-3-17

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Error using vertcat Dimensions of matrices being concatenated are not consistent.

Error in zp2sos>orderp (line 247) new_p = [p_conj;p_real];

Error in zp2sos (line 94) [new_p,p_conj,p_real] = orderp(p_conj,p_real);

Error in band8 (line 7) [sos{k1},g{k1}] = zp2sos(z(k1,:),p(k1,:),k(k1));

This is the error that i obtained. I tried it in matlab 2013a

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Star Strider 2017-3-18

在 MATLAB Online 中打开

‘Wow that really worked!!!!!!!!! Thanks a lot.’

Thank you! My pleasure!

‘Which parameter should i change?’

I am not certain what you want. You cannot have any of the passbands or stopbands exactly at 0 or 1 (actually normalised frequencies of 0 or pi or 0 and your Nyquist frequency) because those are the limits of the frequency space for your sampled signal. You can get close to them (you have to experiment with your filters to determine how close you can get, depending on the other filter parameters) but you cannot by definition include them. (They are included by default in lowpass and highpass filters that do not specifically include them. You cannot specify filter band limits that include them.)

So if you want a filter that goes from (and includes) 0 Hz to the lower passband of your lowest bandpass filter, use a lowpass filter design with the passband approaching the lower passband of your lowest filter. If you want a filter that goes from the upper passband of your highest filter to the Nyquist frequency, use a highpass filter with the passband just above the upper passband of your highest filter. That is a long reply to a short question. I wanted to provide background.

You can definitely filter your audio signal. If it is a (Nx2) stereo matrix, filter it as you would filter any other signal. Both channels will be filtered at the same time, and the output will be a (Nx2) array of your filtered signal.

It is easiest to filter your signal through each filter in a loop similar to the way you created them. Use the filtfilt function to do the actual filtering.

Example code to do the filtering:

for k1 = 1:length(cf)
    filt_sig{k1} = filtfilt(sos{k1}, g{k1}, sig);
end

where ‘sig’ is the signal you are filtering. I use a cell array for ‘filt_sig’ because I do not know anything about your signal. The cell array will work for ‘sig’ being either a vector or a matrix (for example an (Nx2) stereo sound signal). You can use a double array, and assign the output to the appropriate rows or columns, but here, the cell array is easier. It is easy to recover your filtered signal from the cell array. The filtered output signal will be in the cell array.

Darsana P M 2017-3-19

Ok...Thanks a lot

Star Strider 2017-3-19

My pleasure.

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Answer 3

Darsana P M 2017-3-22

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https://ww2.mathworks.cn/matlabcentral/answers/330218-designfilt-and-crossover-filters#answer_259808

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    Fs = 8200;                                      % Samping Frequency (Hz)
Fn = Fs/2;                                      % Nyquist Frequency
pf = [20 100:100:2E+4]                     % Passband Frequencies
cf = pf(1:end-1)+(pf(2)-pf(1))/2;               % Centre Frequencies
for k1 = 1:length(cf)
    [z(k1,:),p(k1,:),k(k1)] = butter(7, [pf(k1) pf(k1+1)]/Fn);
    [sos{k1},g{k1}] = zp2sos(z(k1,:),p(k1,:),k(k1));
    [h(k1,:),w(k1,:)] = freqz(sos{k1},512,Fs);
end
figure(1)
plot(w([1 16],:), abs(h([1 16],:)))
grid
% axis([0  0.2    ylim])
figure(2)
freqz(sos{1})
hold on
for k1 = 2:16
    freqz(sos{k1})
end
hold off
I wrote the pf command as you said. pf=[20 100:100:2E+4]. But i got an error saying
  the cut off frequencies must be within the interval (0,1);   line 57
    [z(k1,:),p(k1,:),k(k1)] = butter(7, [pf(k1) pf(k1+1)]/Fn);  line 7

This was done to get a similar graph as shown in figure?? What will i do?

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Darsana P M 2017-3-24

Thanks a lot

Star Strider 2017-3-24

My pleasure.

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