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As an example, I know that sampling a signal $s$ is modelled by multiplication of s by a dirac comb, which has the effect of convolving the Fourier Transform (FT) of $s$ by the FT of the dirac comb which is another dirac comb but with inverse spacing between peaks.

My question is what is the corresponding process for when one upsamples (e.g., by a factor of 2), , i.e., inserts zeros between samples, and NOT yet applies interpolation since I know for interpolation you need to follow upsampling by a lowpassfilter whicg is a convolution a signal $s$?

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  • $\begingroup$ Maybe I should have posted that in signal processing forum since here you dont even have the upsampling tag as part of the keywords ; maybe someone can move my q there? $\endgroup$ – Machupicchu Sep 18 '20 at 18:40
  • $\begingroup$ Not sure how to move it there, but do you thing of bandlimited or finite dimensional signals? If sooner, then just sample it by the Dirac comb twice more frequently. If it is finite-dimensional, then we sample each entry twice. It is useful if the measurements are noisy or may be missing (frames with erasures). $\endgroup$ – kate-melnykova Sep 18 '20 at 20:12
  • $\begingroup$ @Machupicchu I've provided an answer, have a look at it... $\endgroup$ – Fat32 Sep 19 '20 at 9:05
  • $\begingroup$ Thanks for your answer. So it looks like a convolution of the signal with a dirac comb but with a shift that is a multiple of the upsampling rate L, correct? I was looking for this to try to understand why when you just upsample (without interpolation) the Fourier Transform of a signal and the ifft it, you get your signal replicated L times. (Note that i was upsampling the FT, not the signal , but i imagine the FT of s would be replicated L times the other way round , correct?) $\endgroup$ – Machupicchu Sep 19 '20 at 16:51
  • $\begingroup$ You cannot express the I/O relationship of an expander by a convolution, because the expander is not a linear time-invariant system, therefore it does not have an impulse response h[n] and its output cannot be written as a convolution. All you have is the Eq.3 to describe it as a shifted impulses weighted by samples of x[n]... If you are interested in the frquency domain description of the expander, then you may ask a new question on dsp.se or better read many already existing answers on expanders, upsampler, rate converters... $\endgroup$ – Fat32 Sep 19 '20 at 17:43
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If I understood your question right, you want a mathematical expression for the I/O (input/output) relationship of a signal expander (name of the block that expands (upsamples) -without interpolation filering- an input signal $x[n]$)

Below is a block diagram of signal expander by a factor of $L$:

$$ x[n] \longrightarrow \boxed{ \uparrow L } \longrightarrow x_e[n] \tag{1}$$

An expression for the expanded sequence $x_e[n]$ can be written as : $$ x_e[n] = \begin{cases} { x[\frac{n}{L}] ~~~,~~~ n=m\cdot L ~~~,~~~ m=...,-1,0,1,... \\ ~~~ 0 ~~~~~~,~~ ~ \text{otherwise} } \end{cases} \tag{2} $$

An identical expression is also the following : $$x_e[n] = \sum_{k=-\infty}^{\infty} x[k] \delta[n-L\cdot k] \tag{3}$$

where $\delta[n]$ is a unit sample (discrete-time impulse).

With $L=3$, an input $x[n]=[1,2,3,4]$ becomes output $x_e[n] =[1,0,0,2,0,0,3,0,0,4,0,0]$.

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