About FFT of sine wave

Dear Forum

i just do the FFt of a sine wave :

t=0:0.01:1; A=sin(2*pi*10*t);

The FFt is:

U=fft(A,200); plot(abs(U));

Question is why the amplitude of fft is 50 and not 1 (convolution betwen sine wave spectrum and rect spectrum 1*1*sincf

Many thanks

Roberto

Answer 1

[Wayne King]

It is because you have 101 points in your signal. Therefore the magnitude of the DFT at the frequency is going to be approximately N/2 where N is the number of points. If your sine wave had an amplitude other than 1, you would see NA/2

To make this exact, let's create your sine wave with 100 points so that the frequency of 10-Hz falls directly in a DFT bin

   t = 0:0.01:1-0.01;
   x = cos(2*pi*10*t);
   xdft = fft(x);
   plot(abs(xdft))

Now you see the magnitude at -100 and 100 Hz is 100/2=50

Change the amplitude of the sine wave:

   x = 3*cos(2*pi*10*t);
   xdft = fft(x);
   plot(abs(xdft))

Now the magnitude at -100 and 100 Hz is (3N)/2=150

This dependence on N comes from the fact that the DFT sums all the element-by-element products of your signal and complex exponentials with frequencies of (kFs)/N where Fs is the sampling frequency and N is the length of the signal and k=0,1,2,3, N-1

When your signal matches the frequency of one of those complex exponentials exactly, as is the case above, you get the coherent sum that results in N times the amplitude. Since your input is a cosine remember that results in TWO complex exponentials each with amplitude 1/2 the amplitude of the cosine.

Answer 2

[Merve Karabakla]

x = sin(2*pi*n*0.01) ( 0 ≤ n ≤ 200 )

What is its Fourier transform? Obtain and plot its magnitude spectrum