Output for h[n] = 2ⁿ{u[n] − u[n−3]}

Given:

• $h[n] = 2^n\{u[n] - u[n-3]\}$: non-zero for $n = 0, 1, 2$, so $h[n] = \{\underset{\uparrow}{1}, 2, 4\}$.
• $x[n] = \delta[n] + \delta[n-1] + \delta[n-3] = \{\underset{\uparrow}{1}, 1, 0, 1\}$.

Method 1: Impulse shifting

Since $\delta[n-k]*h[n] = h[n-k]$ and convolution is distributive:

$$y[n] = h[n] + h[n-1] + h[n-3]$$ $$y[n] = \{\underset{\uparrow}{1},\ 3,\ 6,\ 5,\ 2,\ 4\}$$

i.e. $y[n] = \delta[n] + 3\delta[n-1] + 6\delta[n-2] + 5\delta[n-3] + 2\delta[n-4] + 4\delta[n-5]$.

Verification (convolution sum, $y[n] = \sum_k x[k]h[n-k]$)

Output length $= 4 + 3 - 1 = 6$, from $n = 0$ to $5$:

• $y[0] = x[0]h[0] = 1(1) = 1$
• $y[1] = x[0]h[1] + x[1]h[0] = 2 + 1 = 3$
• $y[2] = x[0]h[2] + x[1]h[1] + x[2]h[0] = 4 + 2 + 0 = 6$
• $y[3] = x[1]h[2] + x[2]h[1] + x[3]h[0] = 4 + 0 + 1 = 5$
• $y[4] = x[2]h[2] + x[3]h[1] = 0 + 2 = 2$
• $y[5] = x[3]h[2] = 4$

Sum check: $(\sum x)(\sum h) = 3 \times 7 = 21$ and $1 + 3 + 6 + 5 + 2 + 4 = 21$.

Both methods give the same result, so the answer is verified.