fourier transform - Does sampling in the frequency domain cause time-domain aliasing?

Let's say I have an impulse response $h[n]$.

I analyze the power spectrum of that impulse response similar to fourier transformed $h[n]$ corresponding to roughly $H[f]$.

Now I compare $H[f]$ with some target $H_{0}[f]$ by subtracting $H_{0}[f]-H[f]$.

I create a compensation filter based on this via inverse fourier transform: $\mathrm{ifft}(H_{0}[f]-H[f])$.

But this is inadequate because time-domain aliasing is observed: Why?

What I want to understand is, just as sampling in the time-domain cause aliasing in the frequency domain, is this operation of creating a filter via ifft akin to sampling in the frequency domain and likewise causing aliasing in the time domain?

Or is this because the $H[f]$ was not sufficiently padded and the result is a circular convolution that causes ringing (but is this time-domain aliasing?)

Is this a direct result of the duality of the DFT? In that, just as a sinusoid of a given frequency will show up as a frequency-axis shifted impulse in the frequency domain, an envelope of some frequency will show up as a time-axis shifted impulse in the time domain?

Confusedly yours, @panthyon

Answer

The answer is: yes, sampling in the frequency domain causes aliasing in the time domain, exactly like the dual case: sampling in the time domain causes aliasing in the frequency domain.

There are many ways to see this. One standard way is to sample the discrete-time Fourier transform (DTFT) of a discrete-time signal by multiplying it with a Dirac comb and see what this does to the time domain signal. Here, however, I would like to show it in another way, namely by taking samples of the DTFT and asking which periodic signal has these samples as its discrete Fourier transform (DFT).

Let $Y(e^{j\omega})$ be the DTFT of the sequence $y[n]$ (note that $y[n]$ is not necessarily of finite length):

$$Y(e^{j\omega})=\sum_{n=-\infty}^{\infty}y[n]e^{-jn\omega}\tag{1}$$

Now we take samples of $Y(e^{j\omega})$ at $\omega_k=2\pi k/N$, $k=0,1,\ldots,N-1$, and derive the time domain sequence with a DFT equal to these samples:

$$\begin{align}Y_k\doteq Y(e^{j2\pi k/N})&=\sum_{n=-\infty}^{\infty}y[n]e^{-j2\pi nk/N}\\&=\sum_{l=-\infty}^{\infty}\sum_{n=0}^{N-1}y[n+lN]e^{-j2\pi nk/N}\\&= \sum_{n=0}^{N-1}\sum_{l=-\infty}^{\infty}y[n+lN]e^{-j2\pi nk/N}\\&=\sum_{n=0}^{N-1}\tilde{y}[n]e^{-j2\pi nk/N}\tag{2}\end{align}$$

where $\tilde{y}[n]$ is an aliased version of the original sequence $y[n]$:

$$\tilde{y}[n]=\sum_{l=-\infty}^{\infty}y[n+lN]\tag{3}$$

Note that the right-hand side of $(2)$ is the DFT of $\tilde{y}[n]$. Eq. $(2)$ shows that the samples $Y_k$ of the DTFT $Y(e^{j\omega})$ are the DFT coefficients of $\tilde{y}[n]$, which is an aliased version of $y[n]$. Consequently, sampling in the frequency domain causes time-domain aliasing, which is obviously the dual of the well-known aliasing in the frequency domain caused by sampling in the time domain.