I am studying a course in signal analysis and systems. Currently we are looking at aliiasing, samplig,DTFT and reconstruction. I got stuck on an excercise.

For the in signal $x_{a}(t)=2\cos(2\pi F_{0}t)$ with $F_{0}=600Hz$ and $F_{s}=1000Hz$ (Sample freq) I am to determine the outsignal when the signal goes through the circuit below. enter image description here

The reconstruction is ideal and is has the sampling frequency $2F_{s}$.

Okey so there is a bunch of stuff here i can't sort out.

Right away I can se that some aliasing is going to happen in the sampling.

I figured the sampled signal would be $x(n)=2\cos(2\pi n\frac{2}{5})$ The answer sheet gives this graph for $x(n)$ and $x(t)$ enter image description here

At a glance the frequency and normalized frequency looks to be about right. But should'nt the magnitude be twice as high i both cases? Or what am I missing?

I'm not sure how to go on from here. The "signshifter" and "collector" are in quotation marks because that is the litteral translation of the words used in the material. This is not something mentioned in the course litteratur and not something we spoke of in our lectures. Google did'nt have any answers for me either.

In short I'm not sure what the second and third operation means.

Sorry for long post, the course material does not seem to cover some of the excercises we got. Anyway i am happy for any help i get! Thank you!


1 Answer 1


It's hard to give you the "right" hints, since I have no idea what you have covered in class and what you didn't.

But should'nt the magnitude be twice as high i both cases? Or what am I missing?

The answer graph is pretty bad. The axes are not labelled and there are no units. The continuous spectrum would be $\delta(t)$ distribution so the value and the amplitude would be infinite which is typically indicated by putting an arrow head on top of the line. The discrete spectrum has indeed a finite value but that depends a bit on how exactly you define spectrum and what scaling convention you deploy.


Note that $(-1)^n = e^{j\pi n}$ so you can view this a cosine wave of a certain frequency with a typically cosine spectrum. Multiplication in the time domain is equivalent to convolution in the frequency domain. Alternatively you can just throw this into the definition of the z Transform and see what you get.


This is simply up-sampling by a factor of two. If you have covered up-sampling in class then just apply what you have learned. If you didn't, you need again put this into the definition of the z-transform and simplify.

  • $\begingroup$ Thanks a bunch! I look into it and see if i get any wiser! $\endgroup$
    – Aedrha
    Aug 18, 2021 at 18:00
  • $\begingroup$ I think i made some progresse on the signshifter part just by multiplying and using eulers formula: $\cos(2\pi (1-\frac{1}{10})n)+\cos(2\pi \frac{1}{10}n)+j\sin(2\pi (1-\frac{1}{10})n)+j\sin(2\pi \frac{1}{10}n)=2\cos(2\pi\frac{1}{10})$ The answer sheet has a graph fo z(n) similar to the ones I originally posted. The normalised frequency is once again correct but my magnitude here is once again twice as high. Sorry to bother you! And thanks again! $\endgroup$
    – Aedrha
    Aug 19, 2021 at 10:05

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