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Determine the following discrete-time convolution sums:
y[n]=(−1)ⁿ∗2nᵘ[−n+2]

1 Answer

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Final Answer:

The discrete-time convolution sum y[n] = (-1)^n * 2^n * u[-n+2] is: y[n] = { ... 0, 0, 0, 0, -2u[0], 2u[1], -4u[2], 4u[3], ... }

Step-by-step explanation:

The discrete-time convolution sum y[n] = (-1)^n * 2^n * u[-n+2] can be determined using the following steps:

Shift the sequence u[n] two positions to the right:

u[-n+2] = { ... 0, 0, u[0], u[1], u[2], u[3], ... }

Multiply the shifted sequence u[-n+2] by the sequence (-1)^n * 2^n:

y[n] = (-1)^n * 2^n * u[-n+2] = { ... 0, 0, 0, 0, -2u[0], 2u[1], -4u[2], 4u[3], ... }

Simplify the expression:

y[n] = { ... 0, 0, 0, 0, -2u[0], 2u[1], -4u[2], 4u[3], ... }

Therefore, the discrete-time convolution sum y[n] = (-1)^n * 2^n * u[-n+2] is:

y[n] = { ... 0, 0, 0, 0, -2u[0], 2u[1], -4u[2], 4u[3], ... }

User Gene Vayngrib
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