Im thinking whether any convolutional operation can output a unit impulse, an example to further explain: where a convolution between system $h[n]$ and unknown system $g[n]$ would output $\delta[n]$.

conv(h[n], g[n]) = δ[n]


$$h[n] = \delta[n] + \delta[n-1]/2 + \delta[n-2]/4 + \delta[n-3]/8 + \delta[n-4]/16$$

In regards to Z-transform - it can be concluded that

$$H[z] = 1 + 1/2z^{-1} + 1/4z^{-2} + 1/8z^{-3} + 1/16z^{-4}$$

Moreover, it is know that in z domain conv becomes multiplication therefore:

$$H[z] G[z] = 1$$

And in conclusion $G[z] = 1/H[z]$

$$G[z] = \frac{1}{1 + 1/2z^{-1} + 1/4z^{-2} + 1/8z^{-3} + 1/16z^{-4}}$$

If the above is somehow correct, then when trying to achieve the inverse z-transform we will get:

$$g[n] = \delta[n] + 2/\delta[n-1] + 4/\delta[n-2] + 8/\delta[n-3] + 16/\delta[n-4]$$

We get $1/\delta[n]$, which does not bode well for anybody.

Im I in the wrong? Can it be possible?


This problem is related to deconvolution and equalization. You are basically undoing the effect of a filter by another filter, such that the total system has a flat response, i.e., has a unit impulse as its impulse response.

From $$(h\star g)[n]=\delta[n]\tag{1}$$ it follows that $$H(z)G(z)=1\tag{2}$$ must be satisfied. So the solution to the problem is $$G(z)=\frac{1}{H(z)}\tag{3}$$

as you've already found out.

However, the problem with $(3)$ is that the corresponding filter might not be causal and stable. Note that the poles of $G(z)$ are the zeros of $H(z)$, and the pole locations determine whether $G(z)$ is causal and stable. The inverse filter $G(z)$ is stable if all its poles are inside the unit circle. I.e., all zeros of $H(z)$ must be inside the unit circle for the inverse filter $G(z)$ to be causal and stable. Such a filter $H(z)$ with all its zeros inside the unit circle is called minimum-phase.

As a simple example take $h[n]=\delta[n]+\frac12\delta[n-1]$. We have $H(z)=1+\frac12z^{-1}$, and the zero of $H(z)$ is $z_0=-\frac12$. We can conclude that the inverse filter $(3)$ is causal and stable, because its pole is inside the unit circle:


The inverse $\mathcal{Z}$-transform of $G(z)$ is


It's straightforward to show that the sequences $h[n]$ and $g[n]$ indeed satisfy $(1)$.

In practice, we usually assume that an exact inversion $(3)$ is not possible, so we try to approximately invert a given system. And we usually allow a certain delay, so $(1)$ is changed to $$(h\star g)[n]\approx\delta[n-N]\tag{6}$$ with some appropriately chosen $N>0$.

As a final remark, you have to work on your $\mathcal{Z}$-transform skills. Your attempt at inverting $G(z)$ was not successful.

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  • $\begingroup$ Very nice explanation Matt, thank you very much for your time. $\endgroup$ – MadFlax Jun 15 at 9:27
  • $\begingroup$ @MadFlax: Thx, you're welcome. $\endgroup$ – Matt L. Jun 15 at 10:19

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