---
title: "Inverse of accumulator: u[n] and δ[n] − δ[n−1]"
subject: "Signals and Systems"
chapter: 6
key: "semester_4/signals_and_systems:invertibility-step-difference"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/invertibility-step-difference
site: BCT Engineering
---

# Inverse of accumulator: u[n] and δ[n] − δ[n−1]

Signals and Systems · Chapter 6: Discrete-Time Systems

- Asked in: 2076 Bhadra, 3+4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/invertibility-step-difference
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6

## Question

Define invertibility of LTI system with suitable example if impulse response h[n] = u[n] and inverse system h′[n] = δ[n] − δ[n−1]. Prove that their convolution must be δ[n].

## Short answer

An LTI system is invertible if an inverse h'[n] exists such that h[n]*h'[n] = δ[n]. Example: accumulator h[n] = u[n] is inverted by h'[n] = δ[n] - δ[n-1]; their convolution yields u[n] - u[n-1] = δ[n].

## Answer

#### Invertibility
A system is **invertible** if distinct inputs give distinct outputs, so the input can be recovered from the output. For an LTI system with impulse response $h[n]$, an **inverse system** $h'[n]$ exists such that the cascade gives back the input:

$$
h[n]*h'[n]=\delta[n]
$$

```text
 x[n] +------+  y[n] +-------+  x[n]
 ---->| h[n] |------>| h'[n] |------>
      +------+       +-------+
   overall response h*h' = delta[n]
```

**Example:** $h[n]=u[n]$ is the **accumulator**, $y[n]=\sum_{k=-\infty}^{n}x[k]$. Its inverse is the **first difference**, $h'[n]=\delta[n]-\delta[n-1]$, i.e. $w[n]=y[n]-y[n-1]$, which recovers $x[n]$. By contrast, $y[n]=0$ and $y[n]=x^2[n]$ are not invertible.

#### Proof that $u[n]*h'[n]=\delta[n]$
Using the shifting property of the impulse, $f[n]*\delta[n-n_0]=f[n-n_0]$:

$$
\begin{aligned}
h[n]*h'[n] &= u[n]*\big(\delta[n]-\delta[n-1]\big)\\
&= u[n]*\delta[n]-u[n]*\delta[n-1]\\
&= u[n]-u[n-1]
\end{aligned}
$$

Now evaluate $u[n]-u[n-1]$:

| $n$ | $u[n]$ | $u[n-1]$ | difference |
|---|---|---|---|
| $<0$ | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 |
| $\ge1$ | 1 | 1 | 0 |

So it is 1 only at $n=0$:

$$
u[n]-u[n-1]=\delta[n]
$$

The convolution sum gives the same result: $\sum_k u[k]h'[n-k]=\sum_k u[k](\delta[n-k]-\delta[n-1-k])=u[n]-u[n-1]=\delta[n]$.

Hence the cascade of the accumulator and the first difference is the identity system, and $h'[n]=\delta[n]-\delta[n-1]$ is the inverse of $h[n]=u[n]$. Check in the z-domain: $H(z)H'(z)=\frac{1}{1-z^{-1}}\cdot(1-z^{-1})=1$ (for $|z|>1$), which is the transform of $\delta[n]$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For an LTI system h[n] to have an inverse h'[n], what must the cascade satisfy?
   A. h[n]*h'[n]=delta[n]
   B. h[n]*h'[n]=u[n]
   C. h[n]+h'[n]=delta[n]
   D. h[n]*h'[n]=h[n]
   Answer: A. The cascade must act as the identity system, whose impulse response is delta[n]. (Key point: Definition of inverse)

2. Which system has impulse response h[n]=u[n]?
   A. The first difference y[n]=x[n]-x[n-1]
   B. The accumulator y[n]= sum of x[k] for k up to n
   C. The unit delay y[n]=x[n-1]
   D. The squarer y[n]=x^2[n]
   Answer: B. Summing the input up to time n gives u[n] as the impulse response; the first difference is its inverse. (Key point: Accumulator)

3. In the proof, what does u[n]*(delta[n]-delta[n-1]) reduce to using the shifting property?
   A. u[n]+u[n-1]
   B. u[n]*u[n-1]
   C. u[n]-u[n-1]
   D. delta[n]-u[n-1]
   Answer: C. Convolving with delta[n-n0] just shifts, so the result is u[n]-u[n-1]. (Key point: Proof step)

4. Why does u[n]-u[n-1] equal delta[n]?
   A. It is 1 for all n>=0
   B. It is -1 at n=0 and 0 elsewhere
   C. It is 1 for n<0 and 0 for n>=0
   D. It is 1 only at n=0 and 0 for all other n
   Answer: D. At n=0, u[0]=1 and u[-1]=0, giving 1; for n>=1 both are 1 and for n<0 both are 0. (Key point: Final result)

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