# Convolution that outputs a unit impulse

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]

where

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

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 $$G(z)=\frac{1}{H(z)}\tag{3}$$

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:

$$G(z)=\frac{1}{1+\frac12z^{-1}}\tag{4}$$

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

$$g[n]=\left(-\frac12\right)^nu[n]\tag{5}$$

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.

• Very nice explanation Matt, thank you very much for your time. – MadFlax Jun 15 '20 at 9:27
• @MadFlax: Thx, you're welcome. – Matt L. Jun 15 '20 at 10:19