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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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