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Consider the original signal (just a sine) and the same signal interpolated with zeros:

16-point signal

enter image description here

The Fourier transform of them:

enter image description here

In this example I inserted only one zero between the samples, as if I sampled the original signal with $2f_s$ (where $f_s$ is the sampling for original signal), but added zero after each sample.

Here for original signal the 14th sample in the frequency domain corresponds to the aliased signal, but in case of interpolated signal it is not aliased and corresponds to the $f_s-f_{signal}$, which I don't understand. I just started sampling faster the same signal with the same frequency, but with inserting zeros between the samples (which I think is a key point). It looks like there are two sampling frequencies: $f_s$ and $2f_s$ for interpolated signal, because if I just increase the sampling twice for original signal, I'll not see such effect.

Why does the interpolation with zeros introduce such frequency artefacts?

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    $\begingroup$ That's not an artifact, that's the actual spectrum you've produced! $\endgroup$ Feb 26, 2023 at 19:27
  • $\begingroup$ ok, but why the spectrum is doubled when interpolating with zeors? $\endgroup$
    – Curious
    Feb 26, 2023 at 19:31
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    $\begingroup$ Are you aware of the fact that a discrete signal has a periodic spectrum? Inserting zeros just shows more repetitions of that periodic spectrum. $\endgroup$
    – Matt L.
    Feb 26, 2023 at 19:35
  • $\begingroup$ @MattL., I read about it, but unfortunately have no enough details to get the idea of such periodicity... could you please clarify a bit? for now I have no idea, how inserting zeros may influence the periodicity of the signal... $\endgroup$
    – Curious
    Feb 26, 2023 at 19:38
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    $\begingroup$ Not sure about the downvote here? The OP's question is clearly stated and illustrated, and shows effort at solving before asking, as well as being within the scope of DSP (arguably about the very fundamental aspect of DSP, which is sampling theory...) $\endgroup$
    – Jdip
    Feb 26, 2023 at 21:43

1 Answer 1

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First of all, you should (re-)read the corresponding chapter(s) in your textbook or lecture notes. Understanding these basic properties of discrete-time signals is essential. Second, what you see is not an artifact but it's exactly what one would expect. You see a replica of the original spectrum. If you had inserted more zeros between the samples you would have seen even more periods of the (necessarily periodic) spectrum. This can be shown as follows:

The DFT of a discrete-time signal $x[n]$ of length $N$ is

$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N}nk},\qquad k=0,1,\ldots, N-1\tag{1}$$

Note that if you choose $k$ outside the interval $[0,N-1]$ you get a periodic repetition of the spectrum, i.e., $X[k]=X[k+lN]$ for integer $l$. That's simply the case because $\exp(-j2\pi nk /N)$ is $N$-periodic in $k$ (and in $n$).

Now you upsample the sequence $x[n]$ by inserting one zero after each sample:

$$\tilde{x}[n]=\begin{cases}x[n/2],&n\textrm{ even}\\0,&n\textrm{ odd}\end{cases}$$

The DFT of the upsampled signal (of length $2N$) is

$$\begin{align}\tilde{X}[k]&=\sum_{n=0}^{2N-1}\tilde{x}[n]e^{-j\frac{2\pi}{2N}nk},\qquad k=0,1,\ldots, 2N-1\end{align}$$

Since every other sample of $\tilde{x}[n]$ is zero, this can be rewritten as

$$\begin{align}\tilde{X}[k]&=\sum_{n=0}^{N-1}\tilde{x}[2n]e^{-j\frac{2\pi}{2N}2nk}\\&=\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N}nk}\\&=X[k],\qquad k=0,1,\ldots, 2N-1\tag{2}\end{align}$$

which is just the same as $(1)$, but with $k$ going from zero to $2N-1$. So the original DFT $X[k]$ is just repeated due to the periodicity shown above.

You can show with equal ease that inserting $L-1$ zeros will result in $L$ repetitions of the original spectrum (including the original one).

The job of an interpolation filter is to remove (read: sufficiently attenuate ) those spectral replicas.

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  • $\begingroup$ by replica you mean all the frequencies above $N/2$? $\endgroup$
    – Curious
    Mar 1, 2023 at 11:20
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    $\begingroup$ @Curious: All the copies of the original spectrum. $\endgroup$
    – Matt L.
    Mar 1, 2023 at 12:08

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