FFT of a single sinusoid showing noise ?!

Hi,

I have a signal which I am sampling so at least 10 wavelengths or 10 time periods can be plotted. It then works out how many samples are required based on the sample frequency.

I have done a FFT quite a few times but this is somehow showing unwanted frequency components on a pure sinusoidal signal...

clear all

f = 30  %Carrier freqency

Fs = 20* f; %Sample frequency atleast 2*f
Ts = 1/Fs; %Sample time

%Number of points must be a factor of 2^n and be greater than the number of
%points required for the wanted number of time periods on the plot

plot_periods_of_f = 10;
min_num_samples = (plot_periods_of_f * (1/f)) / (Ts)

%Make sure the number of samples, Ns, is greater than min_num_samples
a=1
while 2^a < min_num_samples 
    a = a+1;
    Ns = 2^a;
end

n = (0:Ns-1)*Ts; %Time vector

sig = cos(2*pi*f*n );

%Setup FFT
N=Ns;
freq_domain = (0:N/2); %Show positive frequency only
freq_domain = freq_domain * Fs / N;

%FFT
ft_raw = fft(sig)/N;
ft = 2*abs(ft_raw); % 2* to compensate for negative frequency energy
ft = ft(1:N/2+1); %Show positive frequency only so it matches the setup of frequency domain

%plot
figure(1);cla;clf
subplot(2,1,1)
plot(n,sig)
subplot(2,1,2);
bar(ft)

For some odd reason my fft is showing noise and I cant seem to work out why... what have I done incorrectly? I thought its all okay..

[Adam]

It looks fairly normal to me. A discrete FFT will always have some leakage.

[Nathan Kennedy]

Please see my comment to Matt J, it shows another script that has no leakage when two frequencies are multipled together and also shows the individual signals being only composed of 1 frequency - why is this leakage occuring when only 1 sinusoid is present...

[Adam]

The fft assumes an infinite signal. You would only get a pure dirac in the frequency domain if you could run it on an infinite sine wave, which you can't. The truncation means that you are running it on an approximation instead, which will be zero to plus/minus infinity.

Your script below also has leakage, it is just a lot smaller because your sine wave has a lot more cycles so is closer to an infinite sine wave (though obviously still not one).

[Nathan Kennedy]

Do you mean truncated in that when the discrete sinusoid is placed as infinite signal in the fft, the start and end are not seen as continuous because the my signal starts at 1 and finishes at a value other than 1?

If so, then I see what you are getting at. Is there any neat tricks of setting up a discrete wave so ensure its looks sinusoidal when placed end to end ie, start and finish at same value and also I can specify how many wavelengths I want?

Answer 1

[Matt J]

Looks normal to me, too. Aside from the fact that the sinusoid is discrete, it is also a truncated, non-periodic sampling, so there's no reason to expect only a single frequency component.

[Nathan Kennedy]

I am confused why you say this. Why does the FFT give several components when its only one sinusoid?

I setup another type of fft here that picks out the correct frequencies from two frequencies being multiplied together. There is no noise on the single component plots and the multiplied frequency plot...

clear all

Fs =2^8 %Sample Frequency - dictates the length of the positive and negative frequency spectrum bandwidth   
Ts = 1/Fs %Sample time period
Ns = (2^12) %Number of Samples 
%Require more samples than your sampling frequency for this time vector
n = (0:Ns-1)*Ts; %Sample array or time vector

f1 = 30;
f2 = 29;

signal_1 = cos(2*pi*f1*n );
signal_2 = cos(2*pi*f2*n );
signal_3 = signal_1 .* signal_2;

%Setup FFT
N=Ns;
freq_domain = (0:N/2); %Show positive frequency only
freq_domain = freq_domain * Fs / N;

%FFT
ft_raw_1 = fft(signal_1)/N;
ft_1 = 2*abs(ft_raw_1); % 2* to compensate for negative frequency energy
ft_1 = ft_1(1:N/2+1); %Show positive frequency only so it matches the setup of frequency domain

ft_raw_2 = fft(signal_2)/N;
ft_2 = 2*abs(ft_raw_2); % 2* to compensate for negative frequency energy
ft_2 = ft_2(1:N/2+1); %Show positive frequency only so it matches the setup of frequency domain

ft_raw_3 = fft(signal_3)/N;
ft_3 = 2*abs(ft_raw_3); % 2* to compensate for negative frequency energy
ft_3 = ft_3(1:N/2+1); %Show positive frequency only so it matches the setup of frequency domain

figure(1);cla;clf;
subplot(6,1,1);
plot(n,signal_1);
subplot(6,1,2);
bar(freq_domain,ft_1);
subplot(6,1,3);
plot(n,signal_2);
subplot(6,1,4);
bar(freq_domain,ft_2);
subplot(6,1,5);
plot(n,signal_3);
subplot(6,1,6);
bar(freq_domain,ft_3);

[Matt J]

In this case, the sinusoid was not truncated. You sampled an integer number of periods of the wave. So, it can be represented by only a single complex exponential basis element.

[Nathan Kennedy]

Do you mean truncated in that when the discrete sinusoid is placed as infinite signal in the fft, the start and end are not seen as continuous because the my signal starts at 1 and finishes at a value other than 1?

If so, then I see what you are getting at. Is there any neat tricks of setting up a discrete wave so ensure its looks sinusoidal when placed end to end ie, start and finish at same value and also I can specify how many wavelengths I want?

[Matt J]

Do you mean truncated in that when the discrete sinusoid is placed as infinite signal in the fft, the start and end are not seen as continuous because the my signal starts at 1 and finishes at a value other than 1?

Pretty much. Except that you don't want signal(N)=1. You want it to be the value just prior to 1 in the periodic sequence.

Is there any neat tricks of setting up a discrete wave so ensure its looks sinusoidal when placed end to end.

Isn't that what you've already done with signal_1?

[Nathan Kennedy]

Yea I am not sure how I done that though, i read somewhere to get the number of samples as a factor of 2^n where can take n=1,2,3,4.... etc. What does that mean?

[Matt J]

If T is the period of your signal, you want to make sure in cos(2*pi/T*n) that n=(0:m*T-1)*k. Here, T , m, and k must all be integers with 0<=k<=T-1.

[Matt J]

No, powers of 2 are irrelevant.

k and m are integer parameters that you are free to choose.

[Nathan Kennedy]

Fs =2^8   
Ts = 1/Fs 
Ns = (2^12) 
n = (0:Ns-1)*Ts;

I have been using this previously and it works for all frequencies if I dont care how many wavelengths are on the plot and aslong as I sample above nyquist. Is there a general approach to this?

[Matt J]

I have been using this previously and it works for all frequencies

It doesn't. For example,

Fs =2^8; 
Ts = 1/Fs; 
Ns = (2^12); 
n = (0:Ns-1)*Ts; 

s=cos(2*pi*(Fs/567)*n);

figure(1);
 plot(n,s);
figure(2);
  F=abs(fft(s));
 bar(F/max(F))
 xlim([0,100]);
 ylim([0,.3])
 title 'Normalized Amplitude Response'

[Matt J]

Maybe this is what you're looking for

    %%User-selected parameters

   f=30; %Frequency of sinusoid
   M=5;  %number of periods of the sinusoid
   Fs=2^8; %Number of samples per period 

    %%Generate signal

    T=M/f; %Total duration of signal - M periods

    n=linspace(0,T,M*Fs+1); %sample times

    n(end)=[];

    Ts=n(2)-n(1);

    s = cos(2*pi*f*n);

   %%Plot

    figure(1);
     plot(n,s);
    figure(2);
      F=abs(fft(s));
     bar(F/max(F))
     xlim([0,100]);
     ylim([0,.3])
     title 'Normalized Amplitude Response'
    shg

Answer 2

[Bruno Luong]

FFT/DFT returns components of the sins/cosine function that are exactly periodic with respect to the number of input data points. You can see as circular spectrum.

If you feed FFT with an input sig with a mono frequency but that is not exactly one of those periodic frequencies, meaning your signal when wrap around has a jump (~ sig(end)-sig(1)), the spectrum will stretch out to accommodate to the jump (similar to a continuous transform of the heaviside fct).

This is not "noise", but miss-understanding of the fundamental of DFT (by too many people unfortunately).