Evaluate the step response of the LTI system represented by the following impulse response :
(a) ℎ(𝑛)=𝛿(𝑛)−𝛿(𝑛−2)
(b) ℎ(𝑛)=𝑛𝑢(𝑛)
y(n)=h(n)∗h(n) ⟹ (δ(n)−δ(n−2))∗(δ(n)−δ(n−2))y(n)=h(n)*h(n) \implies (\delta (n)-\delta (n-2))*(\delta (n)-\delta (n-2))y(n)=h(n)∗h(n)⟹(δ(n)−δ(n−2))∗(δ(n)−δ(n−2))
⟹ δ(n)∗δ(n)−δ(n−2)∗δ(n−δ(n)∗δ(n−2))+δ(n−2)∗δ(n−2)\implies \delta (n)*\delta (n)-\delta (n-2)*\delta(n-\delta(n)* \delta(n-2))+\delta(n-2)*\delta(n-2)⟹δ(n)∗δ(n)−δ(n−2)∗δ(n−δ(n)∗δ(n−2))+δ(n−2)∗δ(n−2)
By using the above property
⟹ δ(n)−δ(n−2)−δ(n−2)+δ(n−4)\implies \delta(n)- \delta(n-2)- \delta(n-2)+\delta(n-4)⟹δ(n)−δ(n−2)−δ(n−2)+δ(n−4)
⟹ δ(n)−2δ(n−2)+δ(n−4)\implies \delta(n)- 2\delta(n-2)+\delta(n-4)⟹δ(n)−2δ(n−2)+δ(n−4)
y(n)=δ(n)−2δ(n−2)+δ(n−4)y(n)=\delta(n)-2 \delta(n-2)+\delta(n-4)y(n)=δ(n)−2δ(n−2)+δ(n−4)
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