Why FFT function returns amplitude divided by 2 ?

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

Answer 1

[Wayne King]

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;