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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.

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  • $\begingroup$ 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
    Commented Jun 22, 2016 at 12:19

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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 $Y_{2F_s}$.

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