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I am using DFT with windows. The way I understand how a window makes the DFT "look" better, is that multiplication in time domain is convolution in frequency domain. Therefore a window with following FT (Hann window), will suppress the side lobes found in a signal FT (second picture) :

enter image description here

enter image description here

But I dont understand how are the values |F($\omega$)| related to suppressing the signals side lobes ... e.g. Tukey window plotted as |F($\omega$)|

enter image description here

How is the width and the pace of decreasing sidelobes (of the above plot) related to getting rid of sidelobes of signals FT ? Is there any intuitive way to explain ?

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    $\begingroup$ the DFT of length $N$ of the signal $x_n = e^{2 i \pi f n}, n \in \{0,\ldots N-1\}$ is nice only when $f = k /N $ for some integer $k$, but when it isn't there are the side lobes of $\frac{\sin(\pi f x)}{\sin \pi x}$. now the DFT of $x_n h(n)$ where $h$ is the Hanning window is much nicer. hence if the model of your signal is $x_n = \sum_{m=1}^M C_m e^{2 i \pi f n}$ for some frequencies $f$, you'll have a much nicer with the window since its DFT will be (by linearity and frequency shift) $\sum_{m=1}^M C_m H(k/N- f)$ where $H(x)$ is the Fourier transform of your discretized (Hanning) window. $\endgroup$
    – reuns
    May 22, 2016 at 23:06
  • $\begingroup$ in one word : do you know the Fourier series, the Fourier transform, and the Fourier transform of distributions ? $\endgroup$
    – reuns
    May 22, 2016 at 23:08
  • $\begingroup$ I understand why there are side lobes in the DFT of the signal ... I agree that DFT precisely defines frequency only for $f = k/N$ ... what I am looking for is the link between the abs(window_DFT) (what is the actual name for it by the way ? in books they just present it like this) ... and why wider main lobe and descending side lobes enhance the signal $\endgroup$
    – Martin G
    May 22, 2016 at 23:18
  • $\begingroup$ everything is there : the DFT of a windowed sum of $M$ (complex) sines of amplitudes $C_m$ and frequencies $f_m$ (forgot the indice for $f$) will be $\sum_{m=1}^M C_m H(k/N-f_m)$ where $H$ is the FT of the discretized window $\endgroup$
    – reuns
    May 22, 2016 at 23:21
  • $\begingroup$ now if the model of your signal isn't sinusoidal, you won't like the result of the windowing, for example if your signal is percussive, you'll prefer the (no window) rectangular window case $\endgroup$
    – reuns
    May 22, 2016 at 23:22

2 Answers 2

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the DFT (of size $N$, with a window $h_n$) of $$x_n = \sum_{m=1}^M C_m e^{2 i \pi f_m n} \qquad \qquad n \in \{0,\ldots,N-1\}$$ (a mixture of (complex) sines) is $$X_k = \sum_{n=0}^{N-1} h_n x_n e^{-2 i \pi k n /N} = \sum_{m=1}^M C_m H(k/N-f_m)$$ where $$H(\xi) = \int_{-\infty}^\infty \sum_{n=0}^{N-1} h_n \delta(x-n) e^{-2 i \pi \xi x} dx$$ is the Fourier transform of $\sum_{n=0}^{N-1} h_n \delta(x-n)$ i.e. of the (discretized) window.

Proving it is just a matter of the frequency shift theorem for the Fourier transform (of distributions). with $$G(\xi) = FT[g(x)](\xi) = \int_{-\infty}^\infty g(x) e^{-2 i \pi \xi x} dx$$ then the Fourier transform of $g(x) e^{2 i \pi f x}$ is $$FT[g(x) e^{2 i \pi f x}] (\xi) = \int_{-\infty}^\infty g(x) e^{2 i \pi f x} e^{-2 i \pi \xi x} dx = G(\xi-f)$$

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  • $\begingroup$ Alright, I cant disagree with that ... you showed me how to get from FT(signal) to FT(signal*window) ... thanks ... but I need to repeat my question, how does the width of main lobe (3rd picture) and decreasing side lobes affects the signal ? ... or is the $X_k = \sum_{n=0}^{N-1} h_n x_ne^{-2i{\pi}kn/N} = ...$ the only way to get there ? $\endgroup$
    – Martin G
    May 23, 2016 at 7:59
  • $\begingroup$ @MartinG : $X_k = \sum_{m=1}^M C_m H(k/N-f_m)$ obviously tells you how the shape of $H$ impacts the DFT : just draw it. and I recommend you to try very thin windows, such as $h_n = \left(\frac{1+\cos(\frac{2 \pi n}{N})}{2}\right)^a$ with $a = 4$ or $a= 8$ to see how it changes dramatically the DFT compared to $a=1$ (Hanning) or $a=0$ (rectangular window) $\endgroup$
    – reuns
    May 23, 2016 at 8:23
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Alright, here we come. Even though the above answer is great and actually explains quite a lot, here is an answer using crayons.

compare the following, left side Rectangular window, right side Tukey window with ${\alpha} = 1$ (generated by Matlab).

enter image description here

My question was, how does the lower pair of pictures tells how the window works ? Well, it cant be stated directly from those pictures, or at least it is not as obvious as from the upper pair, where we can see the real part going to positive and negative values.

As mentioned in my question, $$Time Domain(multiplication) = Frequency Domain(Convolution)$$

For Rectangular Window:

No change on the signal DFT. The discrete window have DFT values $c_0 = 1$ and the rest of the $c_k = 0$

For Tukey Window:

The Tukey DFT has 2 spikes. 1 positive and 1 negative. Therefore when it passes the signal DFT during convolution, these 2 spikes flatten everything except big spikes (the side lobes or "skirts" of the signal DFT are flattened, while the main spikes of signal DFT remain)

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