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* The focus is now on user-friendliness and easy of use,
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ï¿ Press down arrow keys to change the mode
ï¿ Press up arrow keys to change the speed
ï¿ Press ESC to exit the menu
ï¿ Press F11 to activate the menu
ï¿ Mouse wheel to scroll through menu items
ï¿ Press Enter to select item
ï¿ Press Esc to quit the program
ï¿ Press Z to rotate the screenQ:
Fourier coefficients in filter design
Suppose I have a signal $x(t)$ that I want to decimate. Then I can define the decimated signal $\hat{x}(t)$ by $x(n)=\hat{x}(nT)$. The Fourier coefficient of the decimated signal is given by
$$\hat{x}_k=\frac{1}{N}\sum_{n=0}^{N-1} x(n) \exp(-i2\pi k n/N).$$
If I want to design a filter $h(t)$ with $\hat{h}(k)$ as the Fourier coefficient of the filter I can do so by defining the filter as
$$h(t)=\frac{1}{N}\sum_{n=0}^{N-1} x(n) \exp(-i2\pi k n/N)$$
My question is this: is it necessary that $h(t)$ be decimated in order to have $\hat{h}(k)$ as the Fourier coefficient of the filter $h(t)$? If $h(t)$ is not decimated, can I still get the coefficients by decimating $h(t)$?
A:
It is possible to keep the time samples of a signal $x(t)$ the same, but at the cost of reduced frequency resolution. The reason is that the discrete-time Fourier transform of $x(t)$ at frequency $k$ is given by $\hat x(k) = \sum_{n=0}^{N-1} x(n) \exp(-i 2 \pi k n / N)$.
Suppose you want $N$ samples of $\hat x(k)$. Then you would sample $x(t)$ exactly $N$ times, and you would get $N$ samples of $x(n)$ that you
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May 31, 2022