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This code gives me a sound vector and Fs scalar from a .wav sample and plots the waveform and power spectrum:
function [sound, Fs] = analyzer(file)
[sound, Fs] = wavread(file) % y is sound data, Fs is sample frequency.
t = (1:length(sound))/Fs; % time
ind = find(t>0.1 & t<0.12); % set time duration for waveform plot
figure; subplot(1,2,1);
plot(t(ind),sound(ind));
axis tight
title(['Waveform of ' file]);
xlabel('time, s');
N = 2^12; % number of points to analyze
c = fft(sound(1:N))/N; % compute fft of sound data
p = 2*abs( c(2:N/2)); % compute power at each frequency
f = (1:N/2-1)*Fs/N; % frequency corresponding to p
subplot(1,2,2);
semilogy(f,p);
axis([0 4000 10^-4 1]);
title(['Power Spectrum of ' file]);
xlabel('frequency, Hz');
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This code is supposed to write and play a .wav file from the "sound" data above with a specified fundamental frequency and duration:
function guitarSynth(file,f,d,sound,Fs)
nbits=8; % frequency and bit rate of wav file
t = linspace(1/Fs, d, d*Fs); % time
y = zeros(1,Fs*d); % initialize sound data
for n=1:length(sound);
y = y + sound(n).*cos(2*pi*n*f*t); % sythesize waveform
end
y = .5*y/max(y); % normalize. Coefficent controls volume.
wavwrite( y, Fs, nbits, file)
wavplay(y,Fs)
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The "too many output arguments" error has been solved. The problem is that MATLAB is busy forever, and I am forced to kill the program. When I hit Ctrl+C to kill the program, MATLAB says there is an error at "y = y + sound(n)*cos(2*pi*n*f*t);". I don't see what the error is. Please help.
Answers (2)
Walter Roberson
on 31 May 2012
1 vote
Unless your second function is nested inside the first, it has no access to the "sound" array you define inside the first function (because you do not pass the array to the second function.) So inside the second function instead of accessing the array it doesn't know about, it looks around and finds the sound() function and in the process of setting up to run it, notices that sound() does not return any outputs but the context requires that sound returns a value...
We warn people about naming their variables the same thing as MATLAB functions...
15 Comments
Jonathan
on 31 May 2012
Walter Roberson
on 31 May 2012
In your duplicate question (which I have just deleted for the sake of sanity), you indicate
function [sound, Fs] = analyzer(file)
This indicates that you return "sound" and "Fs" when you run the routine. So assign those outputs to variables when you run analyzer():
[sound, Fs] = analyzer('handle.wav');
Now you have sound and Fs ready to pass into synth3 as additional arguments:
synth3('TheOutputFile.wav', 12345, 3.14, sound, Fs)
You would have to adjust the "function" line of synth3() to expect these as inputs. And you probably don't want to overwrite Fs within the synth3() routine like you do now...
Jonathan
on 31 May 2012
Jonathan
on 31 May 2012
Ryan
on 1 Jun 2012
what is there error name? If both sound(n) and cos(A*t) are vectors, you'll want to use 'sound(n).*cos(2*pi*n*f*t)'. This way, instead of trying to multiply the two vectors using matrix algebra, it multiplies each matching array pair, e.g. results = [sound(1) x cos(A*1), sound(2) x cos(A*2)... sound(n) x cos(A*n)].
Jonathan
on 1 Jun 2012
Jonathan
on 1 Jun 2012
Oleg Komarov
on 1 Jun 2012
Please post the exact way you're calling the function with example inputs.
Ryan
on 1 Jun 2012
also please copy the red text exactly that matlab spits back at you.
Walter Roberson
on 1 Jun 2012
At the command line command
dbstop if error
then run your program. When it stops it will indicate the error. You can then examine the size of all of the variables and experiment with sub-expressions until you figure out what it is complaining about and why.
Jonathan
on 1 Jun 2012
Jonathan
on 1 Jun 2012
Walter Roberson
on 1 Jun 2012
That tells you that it was executing that line when you interrupted the program. You have asked it to do so much work that it is not finishing before you interrupt it. Consider using waitbar() to monitor the progress.
Oleg Komarov
on 1 Jun 2012
@Jonathan: it's not amusing to play the discovery game. Please supply all the relevant information at once specifying the context. Were you planning to tell us "when I kill the program"? Or just the right combination of questions will unlock that achievement?
Consider my answer from a slightly humouristic point.
Walter Roberson
on 1 Jun 2012
MATLAB responds to control-C by printing out the line that it was executing at the time you interrupted. The "error" it reports for this purpose is the control-C interruption itself, not an error in the line it prints out.
What value is d*Fs*length(sound) ?
Stephen
on 31 May 2012
if the function usually outputs 1 thing and you ask for more, it will give you that error. for example,
function ans = myfunc(x,y)
ans = x + y;
end
will error when I write:
[ans1, ans2] = myfunc(1,1);
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