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?


2 Answers 2


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 $n<0$. By doing so, the solution for $h_2(n)$ becomes unique.

The total impulse response of the cascaded system is


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.

  • $\begingroup$ thank you for the quick response and the great explanation. I definitely see how the solution is obtained when assuming causality for the system described by $ h_2 $. Is this strictly necessary though? The overall system is obviously causal, also the first backward difference (i.e. $ h_1 $) is causal, however that need not imply that $ h_2 $ is causal, am I correct? Since the system for $ h_2 $ is LTI shouldn't there only be one impulse response, regardless of whether the system is causal or not? $\endgroup$
    – YeahShibby
    Jul 2, 2014 at 12:25
  • $\begingroup$ @YeahShibby: You're right that causality is not necessary. I added a paragraph to the end of my answer showing that also the anti-causal impulse response $h_2(n)=u(-n-1)$ is a solution. I'm not sure I understand your last question, but it is the case that the solution is not unique if you don't require causality, i.e. there are several different systems the can be cascaded with the first one to give a total response $h(n)=\delta(n)$. Yet there's only one causal one. $\endgroup$
    – Matt L.
    Jul 2, 2014 at 12:50
  • $\begingroup$ Okay thanks, I was able to verify this. I always thought that the impulse response of an LTI system is unique. So can an LTI system have several different impulse responses, all of which are not impulse responses of other LTI systems? $\endgroup$
    – YeahShibby
    Jul 2, 2014 at 15:45
  • $\begingroup$ @YeahShibby: The impulse response of an LTI system is indeed unique. The thing is that several different LTI systems satisfy the specified condition on the total impulse response. As mentioned before, if you require causality, then the solution becomes indeed unique and there's only 1 causal LTI system resulting in $h(n)=1$. $\endgroup$
    – Matt L.
    Jul 3, 2014 at 7:11
  • $\begingroup$ Perfect, got it :) Thanks for the great help! $\endgroup$
    – YeahShibby
    Jul 3, 2014 at 11:35

But this is first difference system . If you consider constant input, output is always 0. So it should be non-invertible right? I dont think you can always solve like this mathematically by taking input as discrete delta function.

  • $\begingroup$ Thank you for your contribution. It looks more like a comment to me that to an answer. Could you please elaborate a little? $\endgroup$ Jul 18, 2017 at 12:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.