Why FFT function returns amplitude divided by 2 ?
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Hello,
I doesn't understand the amplitude given by the FFT function. Indeed I use it to calculate the DFT of a sum of sine and the amplitude returned is divided by 2 wrt the amplitude of my original function. The frequencies are correct
Does anyone can explain it to me ?
Please find hereunder the code I used :
T = 0:0.01:10;
T = T';
x1 = 0.5*sin(2*pi*2*T);
x2 = 1.2*sin(2*pi*5.4*T);
x3 = 0.7*sin(2*pi*7*T);
y = x1+x2+x3;
N = length(y);
duree = max(T)-min(T);
Delta_T = duree/N;
Fe = N/duree;
Delta_F = 1/duree;
xfft = 1/N*fft(y);
mag = abs(xfft);
freq = 0:Delta_F:(Fe-Delta_F);
freq = freq';
figure(2)
hold all;
plot(freq,mag);
legend('abs');
xlabel('Freq in Hz');
title('FFT');
box on;
set(gca,'Xlim',[0 100]);
grid on;
figure(1)
hold all;
plot(T,y)
grid on
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Réponse acceptée
Wayne King
le 6 Août 2013
Modifié(e) : Wayne King
le 6 Août 2013
Because the discrete Fourier transform matches the input signal with complex exponentials and a cosine is the sum of two complex exponentials divided by 2. The same is true of a sine (except it's divided by 2i)
That is where the factor of 1/2 is coming from. Since you have a real-valued signal, if you are only interested in looking at the magnitude, you can just keep the "positive" frequencies and scale them by 2.
T = 0:0.01:10-0.01;
T = T';
x1 = 0.5*sin(2*pi*2*T);
x2 = 1.2*sin(2*pi*5.4*T);
x3 = 0.7*sin(2*pi*7*T);
y = x1+x2+x3;
N = length(y);
duree = max(T)-min(T);
Delta_T = duree/N;
Fe = N/duree;
Delta_F = 1/duree;
xfft = 1/N*fft(y);
magfft = abs(xfft);
magfft = magfft(1:length(xfft)/2+1);
magfft(2:end-1) = 2*magfft(2:end-1);
freq = 0:100/length(y):100/2;
plot(freq,abs(magfft))
grid on;
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