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Convolution of δ-sums with h[n] = 2δ[n] + 2δ[n−1]

Short answer

Convolution yields y[n]=2δ[n]+6δ[n-1]+4δ[n-2]-2δ[n-3]-2δ[n-4], or sample values {2,6,4,-2,-2} for n=0..4. Use y[n]=2x[n]+2x[n-1].

Find the convolution sum for a given signal x[n] = δ[n] + 2δ[n−1] − δ[n−3] and h[n] = 2δ[n] + 2δ[n−1]

Write the sequences as sample values:

x[n]=δ[n]+2δ[n−1]−δ[n−3] ⇒ x[n]={1, 2, 0, −1}, n=0..3x[n]=\delta[n]+2\delta[n-1]-\delta[n-3]\ \Rightarrow\ x[n]=\{1,\ 2,\ 0,\ -1\},\ n=0..3 h[n]=2δ[n]+2δ[n−1] ⇒ h[n]={2, 2}, n=0,1h[n]=2\delta[n]+2\delta[n-1]\ \Rightarrow\ h[n]=\{2,\ 2\},\ n=0,1

Method 1: using x[n]∗δ[n−n0]=x[n−n0]x[n]*\delta[n-n_0]=x[n-n_0]

y[n]=x[n]∗(2δ[n]+2δ[n−1])=2x[n]+2x[n−1]=2δ[n]+4δ[n−1]−2δ[n−3]+2δ[n−1]+4δ[n−2]−2δ[n−4]=2δ[n]+6δ[n−1]+4δ[n−2]−2δ[n−3]−2δ[n−4]\begin{aligned} y[n]&=x[n]*\big(2\delta[n]+2\delta[n-1]\big)=2x[n]+2x[n-1]\\ &=2\delta[n]+4\delta[n-1]-2\delta[n-3]+2\delta[n-1]+4\delta[n-2]-2\delta[n-4]\\ &=2\delta[n]+6\delta[n-1]+4\delta[n-2]-2\delta[n-3]-2\delta[n-4] \end{aligned}

Method 2: tabular check

n=0n=01234
2x[n]2x[n]240−2
2x[n−1]2x[n-1]240−2
y[n]y[n]264−2−2

Sum check: ∑y=8=(∑x)(∑h)=2×4\sum y=8=(\sum x)(\sum h)=2\times4.

Answer: y[n]={2, 6, 4, −2, −2}y[n]=\{2,\ 6,\ 4,\ -2,\ -2\} for n=0,…,4n=0,\dots,4, i.e. y[n]=2δ[n]+6δ[n−1]+4δ[n−2]−2δ[n−3]−2δ[n−4]y[n]=2\delta[n]+6\delta[n-1]+4\delta[n-2]-2\delta[n-3]-2\delta[n-4].

✓ Answer checked by Claude Opus 5.5 (high effort)

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