fft of a signal

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Mangesh KAle
Mangesh KAle on 27 Sep 2022
Answered: Star Strider on 27 Sep 2022
I would like to know whether the code is right or wrong because I am not getting the frequency plot
clc;
%% load a signal
[y,Fs] = audioread("bodytune.wav"); % Fs= Sampling Frequency
% y = 393216*2 double and Fs= 16000
info = audioinfo("bodytune.wav");
y1= y(:,1);
L = length(y); % Length of a signal
dt= 1/Fs;
t = (0:L-1)*dt; % Time vector
%% plot and label the graph
figure(1)
plot(t,y1);
xlabel('time')
ylabel('Amplitude')
grid on
%% fourier transform
nfft = 2^( nextpow2(length(y1)) );
df = Fs/nfft;
f = 0:df:Fs/2;
X_fft = fft(y1,nfft);
X_fft = X_fft(1:nfft/2+1);
figure(2);
plot(f,abs(X_fft));
xlabel('Frequency')
ylabel('Amplitude')
grid on

Accepted Answer

Star Strider
Star Strider on 27 Sep 2022
You need to normalise the fft result by the length of the signal:
X_fft = fft(y1,nfft)/L;
Otherwise, using my test signal, it appears to be correct —
% clc;
% % load a signal
% [y,Fs] = audioread("bodytune.wav"); % Fs= Sampling Frequency
% y = 393216*2 double and Fs= 16000
% info = audioinfo("bodytune.wav");
Fs = 44100;
t = linspace(0, 5*Fs-1, 5*Fs)/Fs;
y = sum(sin([1; 5000; 10000; 15000]*2*pi*t)).';
y1= y(:,1);
L = length(y); % Length of a signal
dt= 1/Fs;
t = (0:L-1)*dt; % Time vector
%% plot and label the graph
figure(1)
plot(t,y1);
xlabel('time')
ylabel('Amplitude')
grid on
%% fourier transform
nfft = 2^( nextpow2(length(y1)) );
df = Fs/nfft;
f = 0:df:Fs/2;
X_fft = fft(y1,nfft)/L;
X_fft = X_fft(1:nfft/2+1);
figure(2);
plot(f,abs(X_fft));
xlabel('Frequency')
ylabel('Amplitude')
grid on
.

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