how to change the window length???

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raj
raj 2012-1-26
I am implementing an FFT over 2 sin signals which are placed at frequencies 300Hz and 380 Hz as the frequency components are very close to each other,I cant resolve then and i know if i increase the window length tbut i cannot change the window length is it possible to change window length in a normal FFT or should i consider the STFT???

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Wayne King
Wayne King 2012-1-26
Yes in the sense that they will typically change the DFT bin in which they are located, but you have to remember that the frequency axis changes too, which you did not do above.
Fs = 1e3;
t = 0:1/Fs:(100*1/Fs)-1/Fs;
x = cos(2*pi*300*t)+cos(2*pi*310*t);
xdft = fft(x,256);
freq = 0:Fs/256:500;
stem(freq,abs(xdft(1:256/2+1)));
xlabel('Hz'); ylabel('Magnitude');

更多回答（5 个）

Wayne King
Wayne King 2012-1-26
Hi Raj, The frequency resolution depends on the length of the signal and the sampling frequency. The DFT bins are spaced as Fs/N, where Fs is the sampling frequency and N is the length of the input signal. Are you able to increase your signal length?
The STFT is surely going to worsen your frequency resolution because you will be segmenting your signal in order to have some time-frequency analysis.
  2 个评论
raj
raj 2012-1-26
Hallo wayne ,but does the window length has no effect over the frequency resolution and i cant vary my signal length if i have to change the window length should i apply zero padding to my signal.
raj
raj 2012-1-26
IS this the correct way to apply window
x = sin(2*pi*300*t)+sin(2*pi*380*t);
y1 = fft(x,256);
win1 = hamming(256)';
xw1 = win1.*y1;

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Wayne King
Wayne King 2012-1-26
You don't tell us what the size of your t vector is. Applying a window is going to worsen frequency resolution because it broadens the main lobe.
You apply the window in the time domain, not in the frequency domain.
Fs = 100;
t = 0:1/100:1-1/100;
x = cos(2*pi*20*t)+randn(size(t));
x = x'.*hamming(length(x));
xdft = fft(x);
Perhaps your problem in resolving the peaks is because you applied a Hamming window to the DFT coefficients.
  1 个评论
raj
raj 2012-1-26
suppose i consider a rectangular window(or no window) and i have two frequency components(like 300Hz and 350Hz) if i want to distinguish those frequency components then how can I do it.

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Wayne King
Wayne King 2012-1-26
You need to tell me what the sampling frequency is and how many samples you have.
Assume that the sampling is 1 kHz, then if you have even a short signal, let's say 100 samples, you can easily tell them apart.
Fs = 1e3;
t = 0:1/Fs:(100*1/Fs)-1/Fs;
x = cos(2*pi*300*t)+cos(2*pi*350*t);
xdft = fft(x);
freq = 0:Fs/length(x):500;
plot(freq,abs(xdft(1:length(x)/2+1)));
xlabel('Hz'); ylabel('Magnitude');
raj
raj 2012-1-26
clear all
close all
clc
f1 = 2000;
f2 = 1700;
T = 1/f1;
tmin = 0;
tmax = 5*T; % Plot 5 cycles
dt1 = 1/30000;
fs = 1/dt1;
t1 = tmin:dt1:tmax;
X = sin(2*pi*f1*t1)+ sin(2*pi*f2*t1);
nx = length(X);
y = abs(fft(X));
figure;
plot(y,'Linewidth',2)
grid on
shg
this is the corrected one
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raj
raj 2012-1-26
hallo wayne thanks a lot can you tell me an optimal values of sampling frequency and number of values
Wayne King
Wayne King 2012-1-26
There are so many possibilities. You have to decide what kind of frequency resolution you want, e.g. 1 Hz, 2 Hz, 10 Hz, and what the highest frequency you want to resolve is.

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raj
raj 2012-1-26
One last question what are the effects of increasing the FFT length I tried it on the code you gave it to me and i couldnt even recognise the dominant frequencies in it, is the effects of zero padding so bad
clear all
close all
clc
Fs = 1e3;
t = 0:1/Fs:(100*1/Fs)-1/Fs;
x = cos(2*pi*300*t)+cos(2*pi*310*t);
xdft = fft(x,256);
freq = 0:Fs/length(x):500;
stem(freq,abs(xdft(1:length(x)/2+1)));
xlabel('Hz'); ylabel('Magnitude');
  2 个评论
Wayne King
Wayne King 2012-1-26
zero padding does not change the frequency resolution, zero padding only interpolates. You cannot use zero padding to increase your frequency resolution.
raj
raj 2012-1-26
No i dont want to increase my frequency resolution with zero padding but I just wanted to know if there would be change in the placing of the frequency components

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