Output for x = δ[n] + 2δ[n−1] − δ[n−3]

Given:

• $x[n] = \delta[n] + 2\delta[n-1] - \delta[n-3] = \{\underset{\uparrow}{1}, 2, 0, -1\}$, for $0 \le n \le 3$.
• $h[n] = 2\delta[n+1] + 2\delta[n-1] = \{2, \underset{\uparrow}{0}, 2\}$, for $-1 \le n \le 1$.

Method 1: Using $x[n] * \delta[n-k] = x[n-k]$

Convolution with a shifted impulse just shifts the signal, and convolution is distributive:

$$\begin{aligned} y[n] &= x[n] * \left(2\delta[n+1] + 2\delta[n-1]\right) = 2x[n+1] + 2x[n-1] \end{aligned}$$ $$\begin{aligned} 2x[n+1] &= 2\delta[n+1] + 4\delta[n] - 2\delta[n-2] \\ 2x[n-1] &= 2\delta[n-1] + 4\delta[n-2] - 2\delta[n-4] \end{aligned}$$

Adding:

$$y[n] = 2\delta[n+1] + 4\delta[n] + 2\delta[n-1] + 2\delta[n-2] - 2\delta[n-4]$$

Method 2: Tabular check

Output starts at $0 + (-1) = -1$; length $4 + 3 - 1 = 6$, so $-1 \le n \le 4$.

Both methods agree. Sum check: $(\sum x)(\sum h) = 2 \times 4 = 8 = 2 + 4 + 2 + 2 + 0 - 2$.

Answer:

$$y[n] = \{2,\ \underset{\uparrow}{4},\ 2,\ 2,\ 0,\ -2\}$$

with $y[-1] = 2$, $y[0] = 4$, $y[1] = 2$, $y[2] = 2$, $y[3] = 0$, $y[4] = -2$.

 y[n]
  4 |       o
    |       |
  2 |   o   |   o   o
    |   |   |   |   |
  0 +---+---+---+---+---o---+--> n
    |  -1   0   1   2   3   4
 -2 |                       o