How the input signal get multiplied with the window functions like Hamming, Hanning, Blackman, etc?

I have tried to multiply the input signal of $10\textrm{ MHz}$ sampled at $100\textrm{ MHz}$ using the Blackmann window function. The output graph is enter image description here

  • How is the multiplication perfomed ?

The input signal is $1\times 64$ row matrix, the Blackman window generates its samples in $64\times 1$ column matrix. The multiplication generates a $64\times 64$ matrix i.e one sample of input signal is multiplied to 64 samples of the window function. For a particular point sample there are 64 sample value. I intend to calculate 128-point FFT with padding zeros.

  • So for a particular sample point which value should be considered to calculate FFT?

Here is my code :

N = 64; % signal length (power of 2)
T = 10*(10^-9) ; % sampling period (and rate) is set to 1
A = 1; % sinusoid amplitude
phi = 0; % phase of zero
f = 10*(10^6); % frequency (under Nyquist limit)
nT = [0:N-1]*T; % discrete time axis
w = blackman(N,'periodic');
x1 = A*cos(2*pi*f*nT);
x = w*x1  ; % windowed signal
subplot (311);
stem (x1);
wp = [w;zeros((zpf-1)*N,1)];
subplot (313);
stem (x);
subplot (312);
stem (w);
disp (x1);
disp (w);
disp (x);
  • $\begingroup$ Welcome to SE.DSP! You appear to have done the right thing. What do you think is wrong? Please edit your question with the extra information. $\endgroup$
    – Peter K.
    Jul 25, 2017 at 12:48
  • $\begingroup$ I would say his problem is that he wanted to obtain his original signal weighted by the window, but instead he is obtaining a set of signals corresponding to his window function weighted by each sample of his signal. In my opinion it's just a problem of Matlab code, he did a matrix product instead of a dotted product. $\endgroup$
    – Florent
    Jul 25, 2017 at 23:46
  • $\begingroup$ Husain Asif : you should post your code, I'm pretty sure it's just a simple syntax issue $\endgroup$
    – Florent
    Jul 25, 2017 at 23:46
  • 1
    $\begingroup$ @HusainAsif with a dot... A .* B. As said earlier, post your code if you want more than guesses. source $\endgroup$
    – Florent
    Jul 26, 2017 at 4:42
  • 1
    $\begingroup$ not here, in your question, with proper formatting please. $\endgroup$
    – Florent
    Jul 26, 2017 at 4:51

1 Answer 1


Matlab window functions such as hamming(N) returns the $N$ point Hamming window sequence $w[n]$ in a $N \times 1$ matrix (a column vector).

Therefore if your sequence $x[n]$ is represented in a row vector of size $1 \times N$ then for obtaining the sample-by-sample product $x[n]w[n]$, you should either transpose $w[n]$ or the other $x[n]$, so that their dimensions permit the correct sample-by-sample multiplication in between and produce the output sequence $v[n]=w[n]x[n]$ represented in a matrix of the size $N \times 1$ or $1 \times N$ depending on which one you have transposed.

v0 = x.*w      % x: Nx1, w: Nx1, v0: Nx1
v1 = x .* w'   % x: 1xN, w: Nx1, transpose w => v1: 1xN
v3 = x' .* w   % x: 1xN, w: Nx1, transpose x => v3: Nx1

Note that if the transposed signal is complex valued, then you should use the conjugate transpose instead :

v1 = conj(x').* w  % x[n] was complex valued...

Note also that the sample by sample product is perfomed by the $.*$ operation under Matlab;

  v = x .* w    % is a sample by sample product implementing v[n] = x[n]w[n]

If you omit the period and use the $*$ alone, you get matrix product or the dot product in case of vectors involved.

For this case when you have two vectors, $x$ of size $N \times 1$ and $w$ of size $1 \times N$, then their matrix product would produce a matrix of size $N \times N$

v = x * w    % x: N x 1, w: 1 x N => v: N x N

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