0
$\begingroup$

Assume $M=3$ and $L=2$ for the multirate system shown below. Furthermore, assume an arbitrary fourier transform $X(e^{j\Omega})$. This one lies between $-\pi$ and $\pi$. The fourier transform after the filter $H_{LP}(z)$, which is an ideal low-pass filter with a cutoff-frequency of $\Omega=\frac{\pi}{3}$, lies between $-\frac{\pi}{3}$ and $\frac{\pi}{3}$, since we sample up by a factor of three. If I then sample down by two, the resulting fourier transform lies between $-\frac{2\pi}{3}$ and $\frac{2\pi}{3}$. Now the spectra of the input and output will clearly look different, therefore also the signal in time. Is it right to say that with this specific multirate system there will be information loss, although no aliasing effects occur?

Note: In case my problem description is confusing, I can provide an example. enter image description here

$\endgroup$

1 Answer 1

0
$\begingroup$

Since in your system there is no aliasing and you do not filter away important parts of the signal (because LP is having the correct cutoff-frequency), there is no information loss.

When a system changes the signal shape, it does not necessarily mean that information is lost. As long, as the operation is reversible, the same amount of information is contained in $y[l]$ and $x[n]$. However, even though the same information is contained, it can be differently difficult, how to extract the useful information from a given signal, depending on its shape.

However, in reality this ideal low-pass filter does not exist, and it might lose some information.

$\endgroup$
2
  • $\begingroup$ I see, so if I understood corretly, the argument is: Although the signals $x[n]$ and $y[l]$ look different, I can still apply another downsampling operation on $y[n]$ in a way that the fourier transform $Y(e^{j\Omega})$ is scaled and then looks exactly like $X(e^{j\Omega})$, therefore no information is actually lost. Right? $\endgroup$ Nov 24, 2016 at 18:45
  • $\begingroup$ In an ideal world (i.e. mathematically), yes. $\endgroup$ Nov 24, 2016 at 18:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.