I'm doing some self-study for an upcoming exam and came across the following question: enter image description here

My first idea to solve it was using the bilinear transform to get some approximation of $H(Z)$ (or just using the plain sampled version of the $H(s)$) then $V(Z)$ would be $H^{-1}(Z)$ to invert the effect of $H$.

Is this the correct approach? If yes, is there any other approach to this type of problem, like not using inverses?

Thanks for the help!

  • $\begingroup$ some systems are not invertible. if $H(s)$ has poles with non-negative real parts, then there is no stable inverse. now, if $V(z)$ has a sufficiently long impulse response and if a known and constant delay $\tau>0$ is no problem, then to within some error you can approximate $b[n] \approx x(n - \tau)$ $\endgroup$ Commented Oct 9, 2020 at 1:18

1 Answer 1


Can't be done. Next question?

First, the author does not define $x[n]$. Let's assume, then, that $x[n] = x(nT_s)$, where $T_s$ is the sampling interval; i.e. $f_s = 1/T_s$.

So assume that $x_1(t) = x_1[n]\ \delta \left(t - \left(n + \epsilon\right) T_s \right)$, where $\epsilon$ is any number where $0 < |\epsilon| < 1$, and $x_2(t) = x_2[n] \left( u \left(t - T_s \left(n - \frac{1}{2}\right) \right) - u \left(t - T_s \left(n + \frac{1}{2}\right) \right)\right)$.

Basically, $x_1(t)$ is a string of Dirac delta functionals that do not land on $t = n T_s$, where $x_2(t)$ is a string of rectangular pulses (or a "stair-step wave") where the changes don't land on $t = n T_s$. I contend without proof* that for any given signal $b[n]$, both an $x_1[n]$ and an $x_2[n]$ could be found that would cause the system to generate $b[n]$. These $x_1[n]$ and $x_2[n]$ would have the following properties:

  • In general, $x_1[n] \ne x_2[n]$.
  • In the case where $x(t) = x_1(t)$, $x[n] = x(nT_s)$ would be zero for all $n$
  • In the case where $x(t) = x_2(t)$ (if I've dotted my 'i's and crossed my 't's right), $x[n] = x(nT_s) = x_2[n]$ would be true for all $n$.

Since you can have more than one version of $x(t)$ that leads to the same $b[n]$, there is no inverse. You can define some constraints on the form of $x(t)$ that depend on some $x[n]$, as I did above, and if you do it properly you can then find an inverse system from $b[n]$ to $x[n]$ -- but because the sampling process loses information, you can never reconstruct some generic $x(t)$.

It could be that the author was just implicitly thinking (or stated explicitly, and you missed it), that $x(t)$ was constrained to be bandlimited to $f_s/2$ (i.e., the Nyquist criteria). If that's the case, and if the relationship between $x(t)$ and $x[n]$ were defined then you could invert the system.

* Proof is left an an exercise for the reader, heh heh heh.


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