Question

Convolution in discrete time is given by

[infinity]
x[k] ⋆ h[k] = ∑ x [T]h[k - T].
T= -[infinity]
Let x[k] = δ[k] + 3δ[k - 2], where δ[k] is the kronecker delta function. In this case, show that
y[k] = h[k] + 3h[k - 2].

Answer 1

To show that y[k] = h[k] + 3h[k - 2] using convolution, we need to convolve the input signal x[k] = δ[k] + 3δ[k - 2] with the impulse response h[k]. The final result is y[k] = h[k] + 3h[k - 2].

Step-by-step explanation:

To show that y[k] = h[k] + 3h[k - 2] using convolution, we need to convolve the input signal x[k] = δ[k] + 3δ[k - 2] with the impulse response h[k].

Let's calculate:

[infinity]x[k] ⋆ h[k] = ∑ x[T]h[k - T]

= ∑ (δ[T] + 3δ[T - 2])h[k - T]

Considering T = -[infinity],

= (∑ δ[T]h[k - T]) + 3(∑ δ[T - 2]h[k - T])

= h[k] + 3h[k - 2]

So, y[k] = h[k] + 3h[k - 2].