Why is u[n]-u[n-1] zero for n>=1?

Both are 1 there, so they cancel

2082 Bhadra (new course) · 4 marks

Convolve δ-sum with u[n] − u[n−1]

Short answer

h[n] = u[n] - u[n-1] simplifies to δ[n]. Convolution with δ[n] is the identity operation, so y[n] = x[n] = δ[n+2] + δ[n+1] + 2δ[n] - δ[n-1] + δ[n-2]. The sequence values are {1, 1, 2, -1, 1} for n = -2 to 2, zero elsewhere.

Find the convolution between the signal x[n] = δ[n−2] − δ[n−1] + 2δ[n] + δ[n+1] + δ[n+2] and h[n] = u[n] − u[n−1]

First simplify $h[n]$ :

h[n]=u[n]-u[n-1]

$u[n]-u[n-1]$ is 1 only at $n=0$ (both are 1 for $n\ge1$ and both are 0 for $n<0$ ), so

h[n]=\delta[n]

Convolution with the unit impulse leaves a signal unchanged ( $x[n]*\delta[n]=x[n]$ , the identity property). Hence

\begin{aligned} y[n]&=x[n]*h[n]=x[n]*\delta[n]=x[n]\\ &=\delta[n+2]+\delta[n+1]+2\delta[n]-\delta[n-1]+\delta[n-2] \end{aligned}

$n$	−2	−1	0	1	2
$y[n]$	1	1	2	−1	1

Answer: $y[n]=\{1,\ 1,\ 2,\ -1,\ 1\}$ for $n=-2,\dots,2$ (↑ at 2), and zero elsewhere.

✓ Answer checked by Claude Opus 5.5 (high effort)

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