fft error: Not enough input arguments.
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Someone can help me in this code for know de heart frequency.
function homoFreq(y,Fs,nfft)
% x is input signal
% nfft is number of fft points
Fs = 16000;%sampling rate
% subplot(2,1,1),plot(x);
% title('Waveform'), xlabel('Time (ms)'), ylabel('Amplitude');
[y,Fs]=wavread('*.wav'); %* is for file name
t=((1:length(y))-1)/Fs;
figure();
plot(t,y); %only for check
ht=fft((y.*hamming(length(y))),nfft); %nfft point -----> fft error in this line
freqAxis=Fs/2*length(ht)/Fs;
f=(0:freqAxis)*Fs/length(ht);
% subplot(2,1,2);
spec = 20*log10(abs(ht(1:length(f))));
plot(f,spec);ylim([-50 35]);
title('Spectrum'), xlabel('Frequency'),ylabel('Log Magnitude');
[pks,locs] = findpeaks(spec);
for m = 5:5,
differ(m-4) = locs(m)-locs(m-1);
%finding the difference between 5th...
%and 4th peak of the spectrum
end
pitchFreq = mean(differ)*Fs/nfft;
fprintf('Pitch Frequency in Frequency Domain = %gHz\n', pitchFreq)
2 件のコメント
採用された回答
Walter Roberson
2012 年 1 月 20 日
You need to invoke
homoFreq([], [], NFFT)
where NFFT is the number of points you want to use for the fft.
3 件のコメント
Dr. Seis
2012 年 1 月 20 日
He is simply suggesting you replace "NFFT" with the number of points you want to use. For example:
homeFreq([],[],1024)
The number you replace NFFT with depends on if you want to truncate "y", keep "y" the same length, or pad "y" with zeros until its length is equal to NFFT.
その他の回答 (1 件)
Dr. Seis
2012 年 1 月 20 日
When you run the function as "homFreq" (with no input arguments) from the command line (or from the editor) you aren't defining "nfft" before it is being called at line 12.
0 件のコメント
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