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I am taking a DSP course, and I want to understand the motivation behind a definition:

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$x(t)$ is sampled by C/D at time intervals of $T_s$. We get the sampled signal $x[n]$ and it is fed into a discrete-time LTI system $G_d(e^{j\omega})$ which outputs $y[n]$.

I wanted to understand why we define the output of the D\Ch (discrete to an impulse chain/train converter) the way it appears in the photo above ($y_s(t)=\sum_{n=-\infty}^\infty y[n]\delta(t-nT_s)$). I understand that D\Ch's purpose is to create an impulse continuous impulse chain, so we have to give each delta its appropriate "height" $y[n]$. I also didn't understand why we take $T_s$ to multiply $n$ in $\delta(t-nT_s)$, not just $n$, for example, but my reasoning was that we must place the deltas at the corresponding times to the sample time of the signal that enters the D/Ch. So then I asked myself, what is the sample time of $y[n]$?

If there were no $G_d(e^{j\omega})$ it is clear that the sample time remains the same, and in order to construct a continuous impulse chain from $x[n]$, we have to take $x[n]$ and multiply it by the delta that "sits" in the corresponding time which is $nTs$ because the sampling rate hasn't changed.

So my question is - how do we know that the sample rate of $y[n]$ remains the same as $x[n]$'s? I think that it is the same question as asking why an LTI system does not change the sample rate of a signal that is passing through it.

Is it because there is no action is carried out on the $n$ in the LTI system? If we look at an Interpolator L then its output for an input $x[n]$ is the same as if we had sampled the signal $x(t)$ at times $T_s/L$. But here there is nothing of this sort - the LTI system doesn't change the "n axis".

If we denote an LTI system $H(e^{j\omega})$, The output of that system is $y(e^{j\omega})=x(e^{j\omega})H(e^{j\omega})$. So we see that the LTI doesn't move the signal or stretch/squeeze (in the omega axis), it can only change the amplitude or phase - an LTi system can't move a certain frequency to another one, only change the amplitude and phase at that $\omega$.

In contrast, a decimator/interpolator stretches the signal in the omega axis/moves the signal in the $\omega$ domain.

Is this the explanation?

I would be grateful if someone could help me clarify this point. I know it is very basic, but I find that by asking these silly kinds of questions, I see whether I understand every step in the way...

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    $\begingroup$ Can you share the source of the diagram in your post? $\endgroup$ Commented Aug 2 at 16:12
  • $\begingroup$ This is from my professor's notes, but there is a very similar one in Oppenheim's book Digital Signal Processing, 3rd edition, page 174, fig 7a $\endgroup$ Commented Aug 2 at 17:44
  • $\begingroup$ A synchronous or asynchronous sample rate converter is LTI. $\endgroup$ Commented Aug 2 at 19:37
  • $\begingroup$ @robertbristow-johnson so my answer is nok ? $\endgroup$ Commented Aug 2 at 20:43
  • $\begingroup$ Well, there was a term I saw in an early DSP course that used the original O&S text. We called them "Linear Shift-Invariant" systems. What is different between a Linear Shift-Invariant system and a Linear Time-Invariant system? (The latter can apply to totally analog circuits.) So now hang an ADC on the input to your SRC and a DAC on the output of the SRC and put it into a black box. How might that differ from an LPF with a constant delay in it? $\endgroup$ Commented Aug 2 at 22:40

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I think you lost track of actually the one relevant thing to describing what is "time-variant", and what is not. It's the definition of time-variance; I'm using that of Oppenheim's Signals and Systems book:

Given a system with a time-dependent output function ⁠$y(t)⁠$, and a time-dependent input function ⁠ $x(t)$⁠, the system will be considered time-invariant if a time-delay on the input ⁠ $x(t+\delta )$ directly equates to a time-delay of the output ⁠ $y(t+\delta )$ function.

While this is noted in time-continuous manner, considering time-discrete systems (notationally: replacing $(t)$ with $[n]$) is the same here.

So, a discrete-time (sampled) system can only be time-invariant (and thus, linear and time-invariant, LTI) when the statement "I delay the input by 1 sample, that means the output is delayed by exactly 1 sample" holds.

Any system that changes sampling rate must have a a different delay, unless it's not linear¹. Therefore, a system that changes sample rate cannot be LTI. End of proof – really just spelling out the definition is enough to say that. Your excourses into transfer functions and frequency-domain representations only serve as "red herring" – a mental decoy.

¹ Exactly one exception: the system that takes anything and outputs zeros is linear, and you can't really say whether it's a resampler or not, because: all zeros.

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  • $\begingroup$ Thank you! I agree with you, I got confused over nothing. I find that looking at things both in time and frequency domains helps me, so after agreeing on the time domain, I want to make sure that my view of the explanation in the frequency domain is correct. As I have said in the question, if we take a system and say that it changes the sampling rate, let's say by L (T'=Ts*L), the output of that system will be $\frac{L}{T_s} \sum_k X\left(j \frac{\omega - 2\pi k}{T_s / L}\right)$, which can't be achieved with an LTI system as $y(e^{j\omega})=x(e^{j\omega})H(e^{j\omega})$. Is this correct? $\endgroup$ Commented Aug 3 at 13:19
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    $\begingroup$ I can't follow you on this. TI is a constraint on time domain. Your like for looking at it in th e frequency domain is not something that's universally useful. It isn't here, far as I can tell. $\endgroup$ Commented Aug 3 at 13:21
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In basic discrete-time signal processing theory, all sample-rate conversion systems are realised/implemented in some form using either or both of these two exclusive blocks: the signal expander and the signal compressor (decimator). These two blocks are linear but time-varying systems, hence non LTI. Therefore any systems using them (except in inverse-cancelling forms) will be non LTI.

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  • $\begingroup$ Thank you, I understand what you have said but still want to understand why an LTI system can't change a signal's sample rate. I thought that since the relation between the input and output of an LTI system in discrete time is $y(e^{j\omega})=x(e^{j\omega})H(e^{j\omega})$, then $H(e^{j\omega})$ can only change the amplitude or phase of the signal, but it can't squeeze or stretch the signal in the $\omega$ domain which is what changing the sampling rate does - unlike the copressor and expander who, like you said, change the sampling rate by stretching/squeezing their input in $\omega$. $\endgroup$ Commented Aug 3 at 3:44
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    $\begingroup$ @R24698 it's quite simple to come up with an example that shows that a interpolating system can't be time-invariant. Say you have an interpolation by 2. You shift the input by 1 sample in time. If the system was TI, then the output would have to be shifted by 1 sample as well. But that's wrong, it shifts by 2. End of proof, really: for no other input/output rate ratio than 1 can LTI be fulfilled; it's a direct consequence of writing down the definition of TI. $\endgroup$ Commented Aug 3 at 12:20

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