Frequency doesn't sound right

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Tom
Tom 2012-2-5
Hey, I'm just doing a tutorial for a course I'm doing and one question asks us to recover 'signal 2', where we're given signal 1 (x), and the combination of signal 1 & 2 (z).
Whether or not my process is right or not, there's something I'd like to ask. I thought I'd listen back to signal 2 (y(t) in this case). When looking at the graph of y(t), it's clear that the period T is 0.0025 sec. The inverse of this gives the frequency of the sound - 400 Hz. When I listen to this however, it sounds significantly lower than a tone at 400 Hz.
Is it something to do with my sampling frequency, or my number of points?
Here are the two .wav files - http://dl.dropbox.com/u/11341635/signal_1.wav http://dl.dropbox.com/u/11341635/signal1and2.wav
And here's my code: -
close all; clear all;
[x,fs] = wavread('signal_1');
x = transpose(x);
[z,fs] = wavread('signal1and2');
z = transpose(z);
N = length(x); %Number of points
i_nyquist = N/2+1; %index of nyquist frequency
dt = 1/fs; %sampling interval or sampling period
t = (([1:N]-1)/fs); %Time axis
posfreq = fs*([1:i_nyquist]-1)/N; %positive frequency axis
negfreq = fs*([i_nyquist:(N-1)]-N)/N;%negative frequency axis
X = (fft(x,N)); %Transform into frequency domain using FFT
Z = (fft(z,N)); %Transform into frequency domain using FFT
Y = (Z./X); %Y in the frequency domain
y = ifft(Y); %Transform into time domain using inverse FFT
sound(y)

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采纳的回答

Daniel Shub
Daniel Shub 2012-2-6
The function sound uses a sample rate of 8192 unless you give it something else. My guess is you want
sound(y, fs);
You code also assumes that x and z have the same lengths and sample rates.
  1 个评论
Tom
Tom 2012-2-7
Thanks Daniel, that sorted it. They did have the same length and sample rates, but yeah I needed to add the 'fs' into the sound command to get it sounding at the correct pitch.
Many thanks.

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