Question

Respond to all questions, showing all work. Write your answers on a separate sheet, and submit a scan of that sheet.

Convolution in discrete time is given by
x[k] x h[k]=₋[infinity]∑τ₌₋[infinity] x[τ]h[k−τ]
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

The question relates to showing that the convolution of a discrete signal with a Kronecker delta function yields an output that is the sum of the system response and its shifted version.

Step-by-step explanation:

The question involves the subject of convolution in discrete time, which is a fundamental concept in signal processing. The convolution of two discrete signals x[k] and h[k] is defined by the equation x[k] * h[k] = ∞∑τ=∞ x[τ]h[k−τ], where the asterisk denotes convolution, and δ[k] is the Kronecker delta function. Given the signal x[k] = δ[k] + 3δ[k−2], we want to show that the convolution of x[k] with h[k] yields y[k] = h[k] + 3h[k−2]. To show this, we apply the properties of the Kronecker delta function in convolution. The Kronecker delta function, δ[k], has the property that when convolved with another signal h[k], it picks the value of h at the position of the delta.

So, the convolution x[k] * h[k] simplifies to y[k] = δ[k] * h[k] + 3δ[k−2] * h[k] = h[k] + 3h[k−2], as expected.