1
$\begingroup$

Let's assume that a linear and time-invariant system is sampled at 2 different frequencies $F_{s}$ and $2F_{s}$ (e.g. 5Hz and 10 Hz). It gives $$Y_{F_{s}}(z) = H_{F_{s}}(z)X_{F_{s}}(z)$$ $$Y_{2F_{s}}(z) = H_{2F_{s}}(z)X_{2F_{s}}(z)$$ where $Y_{F_{s}}$ and $Y_{2F_{s}}$ are respectively the z-transforms of the sampled signals, $H_{F_{s}}$ and $H_{2F_{s}}$ are the impulse responses of the systems and $X_{F_{s}}$ and $X_{2F_{s}}$ are its inputs. The signals $Y$ are known, the impulse response are also known : $$ H_{F_{s}}(z) = \frac{a+bz^{-1}}{1+cz^{-1}+dz^{-2}}$$ $$ H_{2F_{s}}(z) = \frac{a'+b'z^{-1}}{1+c'z^{-1}+d'z^{-2}}$$ and the inputs are unknown.

I am wondering whether it is possible to express $X_{2F_{s}}$ in terms of $X_{F_{s}}$.

Since I am dealing with upsampling, I thought that $X_{2F_{s}}(z)=X_{F_{s}}(z^2)$ and so I had $$Y_{2F_{s}}(z) = H_{2F_{s}}(z)X_{F_{s}}(z^2).$$ But I performed numerical simulations and it seemed to be false. So, where is my mistake and is it possible to express $X_{2F_{s}}$ in terms of $X_{F_{s}}$.

Any suggestions will be appreciated.

$\endgroup$
  • $\begingroup$ Maybe you mean the $H$'s are the $\mathcal Z$-transforms of impulse responses and the $X$'s the $\mathcal Z$-transforms of inputs ? $\endgroup$ – Gilles Jun 22 '16 at 12:19
1
$\begingroup$

I am wondering whether it is possible to express $X_{2F_s}$ in terms of $X_{F_s}$.

In general, no.

This is the sampling theorem: unless you've bandlimited $X_{2F_s}$ first to be alias-free sampled by $F_s$, then the digital signal obtained by sampling twice as fast will contain information that the digital signal sampled at $F_s$ does not contain. And information that is not contained cannot be recovered.

If you, however, did not band limit $X$ to the Nyquist bandwidth dictated by the sampling rate, your $X_{F_s}$ will already contain aliases. Then you can simply look at the "upper" half of $X_{2F_s}$, shift it down to overlay exactly the lower half, which is exactly what's called aliasing, and reconstruct $Y_{F_s}$ from $Y_{2F_s}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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