# Downsampling of signal

I have two signals, for example, $$x_1[n]$$ and $$x_2[n]$$, which are the same for every $$n$$ except for $$n_0$$ and $$n_1$$. They are both bounded within a frequency range of $$\pi/3$$.

I want to reconstruct $$x_2[n]$$ from $$x_1[n]$$. If $$n_0$$ and $$n_1$$ are divisible by 3, I can simply apply downsampling by 3, upsampling by 3, and a low-pass filter that multiplies by 3 (the height).

However, what will happen if, for example, $$n_0 = 0$$? Downsampling by 3 will not erase it because it will erase the time instances 1, 2, 4, 5, and so on.

How can I perform downsampling that starts from $$n = 0$$ rather than $$n = 1$$?

I thought to do shift timing but it will destroy the frequency

I have two signals, for example, $$x_1[n]$$ and $$x_2[n]$$, which are the same for every $$n$$ except for $$n_0$$ and $$n_1$$. They are both bounded within a frequency range of $$\pi/3$$.

That's a contradiction right there. If they are only different at two instances we can write this as

$$x_2[n] = x_1[n] + a\cdot \delta[n-n_0] +b \cdot \delta[n-n_1] = x_1[n] + d[n]$$

where $$\delta[n]$$ is the Kronecker Delta and $$a$$ and $$b$$ are suitable constants. $$d[n] = a\cdot \delta[n-n_0] + b \cdot \delta[n-n_1]$$ is not bandlimited that means that $$x_2[n]$$ can also not be bandlimited even if $$x_1[n]$$ is.

That means that down-sampling and up-sampling won't work here.

Edit: this is not a facetious answer but a fundamental problem

This is a variant of question that's always been bugging me. Let's say I have a finite length discrete sequence that's bandlimited at 20 kHz and sampled at 44.1kHz. You should be able to resample this at 48kHz and back to 44.1kHz and get the exact same sequence back. However, you can't. The (somewhat frustrating) reason is simply "finite sequences cannot be bandlimited", so the initial premise of the original question is flawed.

This has serious consequences for the sampling theorem as well: the only signals that can be sampled without error are bandlimited which means they MUST be infinitely long. This, of course, doesn't reflect the real world and is highly impractical. In practice, we cannot sample any real world signal without some amount of aliasiang.

• There is an option with downsampling and upsampling for sure... Feb 10 at 14:32
• I think something is missing from this proof... Every DT signal is a sum of weighted deltas. Can any DT signal be bandlimited then?
– MBaz
Feb 10 at 16:40
• @MBaz: not if it's finite in length. Feb 11 at 15:25
• Ya, this has bugged me for a while. I played with the idea of using a POCS (projections onto convex sets approach) to finding ONE signal that this could be done with... and the only ones I could find are constant signals. This image shows the evolving spectrum of going back and forth (filtering, decimation, interpolation) many, many times. Dumb code here.
– Peter K.
Feb 11 at 20:34