why FFT results in high frequency range are not correct?

Hello All.

I am doing FFT with matlab.the time period i am doing the fft on it is 1 second and it consist of 50000 equlay spaces samples. I want to test the FFT results. so I have given an input as below which is a complex of sinosuidal waves (and I have samples it by sampling frequency of 50 KHZ)and I expect to have the frequency magnitude results as I have given in the input. the results are ok for low frequency ranges but for the higher frequency (5752 Hz and 7993 Hz) results are 5.742 and 6.423 respectively. what Is the origin of this big mistake ? how can I improve my results ?!

wave = 100*sin(2*pi*50*t)+1*sin (2*pi*123*t)+2*sin (2*pi*203*t)+3*sin(2*pi*223*t)+4*sin(2*pi*331*t)+5*sin(2*pi*2812*t)+6*sin(2*pi*5752*t)+7*sin(2*pi*7993*t);

Answer 1

[Geoff Hayes]

Ali - what is the line (or lines) of code that you are using to do the FFT? What is the block size, N?

If I follow the examples from fft, then

 % define the function
 func = @(t)100*sin(2*pi*50*t)   + 1*sin (2*pi*123*t) + 2*sin (2*pi*203*t) + ...
              3*sin(2*pi*223*t)  + 4*sin(2*pi*331*t)  + 5*sin(2*pi*2812*t) + ...
              6*sin(2*pi*5752*t) + 7*sin(2*pi*7993*t);

 % set the sampling rate
 Fs = 50000;

 % set the block size
 N  = 50000;

 % create the time vector
 t = linspace(0,1-1/Fs,Fs);

 % create the time domain data
 y = func(t);

 % do the fft
 Y = fft(y,N);

 % determine the frequencies
 f = Fs/2*linspace(0,1,N/2+1);

 % plot single-sided amplitude spectrum.
 plot(f,abs(2*Y(1:N/2+1))) 
 title('Single-Sided Amplitude Spectrum of y(t)')
 xlabel('Frequency (Hz)')
 ylabel('|Y(f)|')

A figure is created that clearly describes the frequency components of the input. To get the amplitudes of each frequency component, just do

 % get Ym which is just the magnitude of the first N/2+1 elements
 Ym     = abs(Y(1:N/2+1))';

 % find those whose magnitude is greater than one
 [idcs] = find(Ym>1);

 % display "table" of frequency vs amplitude (subtract one since
 % frequencies start at 0; to get amplitude, divide by N/2 since real input)
 [idcs-1 Ym(idcs)/(N/2)]

 ans =
    50     100
    123    0.99999999999999
    203    2
    223    2.99999999999994
    331    4.00000000000001
    2812   5
    5752   6.00000000000004
    7993   6.99999999999998

The above is near identical to the input amplitudes.

[Ali]

thaks a lot, my problem is in definition of the sampling times . it is now working !