Yes there is.

Suppose you want to analyze a block of samples that represents a simple sine wave. You have your block of samples x[n] and you apply the FFT to obtain the spectrum X[k] which is the spectrum of the infinite series that you would get if you periodically repeat x[n]. Problems arise when the block that you analyze does not contain an integer number of periods, causing discontinuities when you would repeat the blocks.

What you actually analyze is a windowed version of the real infinite periodic signal, i.e. you take the fft of w[n].x[n]. Not explicitly applying a window function w[n] means that a rectangular window is used, i.e. w[n] = 1 for n = 1 to N and zero otherwise. In the frequency domain this means that the true spectrum is convolved with a sinc function which causes the leakage.

If the block that you analyze becomes smaller this means that your rectangular window is narrower. This means that its Fourier transform becomes wider and hence you can expect more leakage.

Different window functions have different spectral properties: spectral resolution vs leakage. You should pick one that suits your application.

Yes there is.

Suppose you want to analyze a block of samples that represents a simple sine wave. You have your block of samples x[n] and you apply the FFT to obtain the spectrum X[k] which is the spectrum of the infinite series that you would get if you periodically repeat x[n]. Problems arise when the block that you analyze does not contain an integer number of periods, causing discontinuities when you would repeat the blocks.

What you actually analyze is a windowed version of the real infinite periodic signal, i.e. you take the fft of w[n].x[n]. Not explicitly applying a window function w[n] means that a rectangular window is used, i.e. w[n] = 1 for n = 1 to N and zero otherwise. In the frequency domain this means that the true spectrum is convolved with a sinc function which causes the leakage.

If the block that you analyze becomes smaller this means that your rectangular window is narrower. This means that its Fourier transform becomes wider and hence you can expect more leakage.

Different window functions have different spectral properties: spectral resolution vs leakage. You should pick one that suits your application.

Thank you very much for your reply.

Is it possible to calculate the block size so that it contain an integer number of periods for a given signal frequency and sampling frequency?

For a simple sine wave you can compute how many samples there are in a single period given the frequency of the sine and the sampling frequency.

If your signal consists of multiple sine waves you can compute for each sine individually how many samples per period it contains and then the total block size must be divisible by all those block sizes...

You can already see that the ideal block size will rapidly increase as the hypothetical signal consists of more and more frequency components. Therefore, for realistic signals you will need to observe the signal for an infinite amount of time if you want perfect frequency information and this is of course not possible in practical applications.

(added later) For completeness:

The above reasoning was purely intuitive. You can also state it more mathematically: the computed spectrum from the windowed signal is always the convolution of the true spectrum with the Fourier transform of the window function. Ideally you want to compute the true spectrum so you want the Fourier transform of the window to be a dirac delta and as you may know the dirac delta is the transform of the infinite rectangular window.

Hi Subash,

Beerend has right when he talks about rectangular window. If you calculate fft for N=2^15 but you give only 32000 samples of signal then your window equals [ones(32000,1);zeros(2^15-32000)] (in Matlab script language). So you can easily see what is the spectrum of window. In frequency domain you obtain result of convolution of your signal spectrum and window spectrum.

It is possible that you can obtain better result when you calculate fft with size the same as the signal is - it means if signal has 32000 then you should calculate fft for the same length. To avoid frequency leakage you can apply other window like hamming(), hanning() or gausswin(). Of course in this case you also obtain convolution in frequency domain.

Regads,

Boguslaw

As Kumar said above, if your window corresponds to an integer number of periods you will not notice leakage (as you would sample the spectrum in 'lucky' places,, where the leakage part passes through zero). In general that is not the case as you don't know the frequency of your signal and/or you cannot synchronise the sampling interval to start-finish comprising an interval that corresponds to an integer number of periods. Consult the excellent paper by Fredric Harris on windows and their effect on leakage. The full reference to Fredric's paper is given by Neil Scott below.

The reference may be to "On the use of Windows for harmonic Analysis with the Discrete Fourier Transform", Proc. IEEE, vol 66, pp 51-83, Jan 1978, authourship listed as "FREDRIC HARRIS" which is ironic as "fredric harris" is notorious for his desire to use lower case for his name?

Please note, FFT is just an algorithm to compute DFT. DFT can be thought of as a sampled version of DTFT. There are two questions here. The first one is the spectral leakage of DTFT if using finite window to an infinite sequence. The second question is how we are going to sample this DTFT, i.e. choosing the length of DFT. It is possible that there is leakage in DTFT, but DFT missed sampling the leakage, and the DFT “looks” perfect. Looking the problem in this two stages may be less confusion.

Although this question is a few months old, I want to place one suggestion:

Most FFT algorithms work in block sizes of 2^N because of their recursive structure and calculation of coefficients.

So 32k of samples mean 32768 samples. If you don`t fit the border (in your example you used 32000 instead) then your leakage effect can occur.

Just try to fill the remaining 768 values as zeros and see what happen. The leakage should disappear regardless of any N*lambda periods.

Dear Marc,

I do not think that this is correct. Please refer to this MATLAB script that illustrates how I interpret what you say.

This is also addressed on the following wikipedia page: http://en.wikipedia.org/wiki/DTFT#Sampling_the_DTFT

close all; clear all; clc;

Fs = 1000; % Sampling frequency

L = 1000; % Length of signal

t = (0:L-1)/Fs; % Time vector

F = 50; % 50 Hz sine wave signal

y = sin(2*pi*F*t);

% FFT with number of samples equal to a power of 2 and zero padding

NFFT = 2^nextpow2(L); % Next power of 2 from L

y = [y zeros(1,NFFT-L)]; % zero pad signal

Y = fft(y,NFFT)/L;

f = Fs/2*linspace(0,1,NFFT/2+1);

figure; stem(f,2*abs(Y(1:NFFT/2+1)));

% graph shows leakage

% FFT in L points, but L is chosen to be a multiple of the signal's period

NFFT = L;

Y = fft(y,NFFT)/L;

f = Fs/2*linspace(0,1,NFFT/2+1);

figure;

stem(f,2*abs(Y(1:NFFT/2+1)));

% graph shows a single peak at 50 Hz

Dear Beerend,

thanks for your MATLAB example.

First of all it`s not clear, if the problem to the author occurs using MATLAB. We have no information about the algorithm he used.

Second we have no information, if he uses plain 3MHz continuous sine wave or a real signal with less periodically intervalls.

Using continuous sine wave it`s clear, that best results for FFT will be at N*lambda border or N periods. On a fraction of a full period the signal is hard cutted by a rectangular window (results in new frequencies and so on).

Assuming the author uses MATLAB and a continous sine wave of 3MHz at a sampling frequency of 76,8MHz, block sizes of either 32768 or even 32000 should result in a good FFT because only full periods are sampled (1280 and 1250 periods).

BUT only 32000 samples resulted in a "leaked" FFT.

That`s in opposite to your opinion and your given example.

So the fact, that 2^15 (32768) data result in a good FFT seems to underline my opinion regarding 2^N FFT algorithms.

?

Best regards,

Marc

Yes, there is a relation between number of input samples given to FFT and spectral leakage (and the frequency content of the signal being analysed). The number of samples determines where, in the frequency domain your samples occur (at k/(NT), where k is the frequency conter, N is the number of samples and T is the sampling period). So, if you have one sinusoid at an exact integer fraction of Fs (sampling frequency) and you choose N such that that frequency is sampled then no apparent leakage will show. For example, if Fs = 128 Hz and you are analysing a sinusoid of f=10 Hz with N = 256, then you will 'see' a single peak at k= 20. That does not mean that there is no leakage, it simply means all your frequency samples, except for k=20, are at places that happen to cross zero. See The Fast Fourier Transform by E. Oran Brigham, Prentice Hall, 1974.

I guess the author of this question is not interested any more in the solution because his last answer was on Dec9,2013.

When N=32768, the bin number is k=3 MHz *32768/76.8 MHz = 1280, an integer value. 

For N=32000, k is not  an integer.

For the magical number N= 32768, you forced FFT to calculate the frequency content at exactly 3 MHz. The frequency contents at 1281 and 1279 are zero. This can be understood by noting that you are sampling a sin wave multiplied by a rectangular window of width N/Fs. In this particular case, the Fourier spectrum of the rect is a sinc function with a first zero at  Fs/N= the step in the frequency domain. 

As you convolute the sinc with your sinwave (delta in the frequency domain), you move the sinc by exactly Fs/N; multiply and sum; then move by Fs/N and repeat. The multiplication will always be zero unless the peak of the sinc is at your frequency 3 MHz; for other values a zero of the sinc will appear at 3 MHz.

Dear Mr. Sameh Ahmed Yousry Elnaggar

For N=32000, k = 3 MHz *32000/76.8 MHz  = 1250 

So, why 1250 is not an integer?