Impulse response of the inverse system to the backward difference I'm currently stuck on a problem. I have a DT LTI system $\mathcal{T}$ that depicts the first backward difference: 

$$y[n] = \mathcal{T} \{ {x[n] \} } = x[n] - x[n-1]$$

Its impulse response is therefore: $h[n] = \delta[n] - \delta[n-1]$, where $\delta[n]$ is the DT unit impulse.

I want to compute the impulse response $h'$ of the system $\mathcal{T}'$ that, when cascaded with $\mathcal{T}$, recovers my input: 

$$\mathcal{T}'\{\mathcal{T}\{x[n]\}\} = x[n]$$

I know the answer is that $h'[n]$ is the DT unit step $u[n]$, but I can't quite figure out why.

Here is my idea so far:


I know that the impulse response of the cascaded system is: $$h_{casc}[n] = \mathcal{T}'\{\mathcal{T}\{\delta[n]\}\} = \delta[n]$$
I know that this impulse response is the convolution of the impulse responses of $\mathcal{T}'$ and $\mathcal{T}$, so: $$\delta[n] = h[n] \ast h'[n] = h'[n] \ast (\delta[n] - \delta[n-1]) = h'[n] - h'[n-1]$$
So for $n=0$ I have: $$h'[0] - h'[-1] = 1$$
And for $n \ne 0$ I have: $$h'[n] - h'[n-1] = 0 \Leftrightarrow h'[n] = h'[n-1]$$


Still, this isn't enough to solve for $h'$ if I am not mistaken, as for example the following different signals would satisfy this:


$... 4 \ 4 \ 4 \ 4 \ 4 \ 5 \ 5 \ 5 \ 5 \ 5 \ldots$, Here: $h'[-1] = 4$, $h'[0] = 5$
$... 2 \ 2 \ 2 \ 2 \ 2 \ 3 \ 3 \ 3 \ 3 \ 3 \ldots$, Here: $h'[-1] = 2$, $h'[0] = 3$


Can anyone give me a hint what I am missing here or perhaps point me to an error or in the right direction?
You're on the right track. Let's call the impulse response of the second system (equalizing the first one) $h_2(n)$. The trick is to assume causality of the second system, i.e. $h_2(n)=0$ for $n&lt;0$. By doing so, the solution for $h_2(n)$ becomes unique.

The total impulse response of the cascaded system is

$$h(n)=h_2(n)-h_2(n-1)\tag{1}$$

and this must equal $\delta(n)$. So we get

$$h_2(0)=1\\
h_2(n)=h_2(n-1),\quad n\ge 1$$

Which gives

$$h_2(1)=h_2(0)=1\\
h_2(2)=h_2(1)=1\\
h_2(3)=h_2(2)=1\\\vdots$$

So $h_2(n)=u(n)$, where $u(n)$ is the discrete-time step function.

Note that you can also find an anti-causal filter $h_2(n)$ satisfying the requirement $h(n)=\delta(n)$. This anti-causal solution is $h_2(n)=-u(-n-1)$. You can easily convince yourself that for this choice of $h_2(n)$ the total impulse response $h(n)$ given by (1) also equals $\delta(n)$.

Note also that all the solutions you gave at the end of the question have the form $h_2(n)=u(n)+c$ where $c$ is some constant. It is indeed true that all these systems are solutions to the problem because the frequency response of the first system has a zero at DC, compensating for any DC component of $h_2(n)$. However, all of these systems with $c\neq 0$ are neither causal nor stable.