How to correctly make FFT of sound set?

During 7 sec 7 tones plays with frequences (all in Hz), one tone -- one sec. Code:
Fs=44100;f=[261.63 293.67 329.63 349.23 392.00 440 493.88];
duration=7;octava=2;f=f/octava;Df=duration*Fs;FF=2*pi/Fs*f;
len = length(f);tau=Df/len;
n=1:Df;y=zeros(1,Df);
for i = 1:len
amplitude=(n>(i-1)*tau)&(n<i*tau);
y = y + sin(FF(i)*n).*amplitude;
end
soundsc(y, Fs);
This question rather for mathematicians. Now I have to see Fourier tranformation. Perhaps it will not accurate vertical Dirac bars but what?
Code:
Y1=fft(y);plot(abs(Y1))
Two bar on edges? How to make beauty picture? And what frequencies will be?

 Akzeptierte Antwort

Rick Rosson
Rick Rosson am 17 Jun. 2011

0 Stimmen

Maybe this will help:
%%Parameters:
playAudio = false;
%%Time domain:
Fs = 44100;
dt = 1/Fs;
StartTime = 0;
StopTime = 1;
t = (StartTime:dt:StopTime-dt)';
%%Cosine waves:
Fc = [261.63 293.67 329.63 349.23 392.00 440 493.88];
y = cos(2*pi*t*Fc);
N = size(y,2);
%%Reshape signal to a single column vector:
y = y(:);
%%Reformulate time domain:
StopTime = N*StopTime;
t = (StartTime:dt:StopTime-dt)';
M = size(t,1);
%%Frequency domain:
dF = Fs/M;
f = -Fs/2:dF:Fs/2-dF;
Y = (N/M)*fftshift(fft(y));
%%Play the music:
if playAudio
sound(y,Fs);
end
%%Plot time domain:
figure;
plot(t,y);
%%Plot frequency domain:
figure;
plot(f,abs(Y));
HTH.
Rick

4 Kommentare

Igor
Igor am 17 Jun. 2011
The best -- y=y(:) :-)
At zoom on figure I see 7 peaks symmetrically..
And I have to understand this result, and I shell push "Accept"-button!
Rick Rosson
Rick Rosson am 17 Jun. 2011
If you have any questions about this code or why it works, please let me know.
Rick Rosson
Rick Rosson am 17 Jun. 2011
Also, there are a few things about this script that are not quite optimal. Can you figure out what they are and how to fix them?
Danilo Persico
Danilo Persico am 9 Jul. 2021
Hello Mr. Rosson, I would like if you could give an explaination of the code, thank you!

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Weitere Antworten (2)

Daniel Shub
Daniel Shub am 15 Jun. 2011

0 Stimmen

The fft assumes that the signal is cyclical. You need to match up your edges (or zero pad) to get delta functions. The fft also returns both positive and negative frequencies. I would read the documentation about fft and fftshift.
Rick Rosson
Rick Rosson am 16 Jun. 2011

0 Stimmen

Please try the following:
dt = 1/Fs;
len = duration*Fs;
n = dt*(0:len-1);
Df = Fs/len;
f = -Fs/2:Df:Fs/2-Df;
Also:
Y1 = fftshift(fft(y));
plot(abs(Y1));
HTH.

1 Kommentar

Igor
Igor am 17 Jun. 2011
Not clear...
1st part -- replace my 2nd row? but what is the sense of *f*?
how does the 2nd part refer to 1st part?
(because of, y-array is unchanged data)

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