Question

1. Calculate the convolution of the following signal:
x(t)=u(t-3)-u(t-5) and h(t)=e^(-3t) u(t)

Answer 1

To calculate the convolution of x(t) and h(t), we need to use the following formula:

y(t) = integral from 0 to t of {x(tau) h(t - tau) d(tau)}

First, we need to find x(tau) and h(t - tau):

x(tau) = u(tau - 3) - u(tau - 5)

h(t - tau) = e^(-3(t - tau)) u(t - tau)

Substituting these into the convolution formula, we get:

y(t) = integral from 0 to t of {(u(tau - 3) - u(tau - 5)) e^[-3(t - tau)] u(t - tau) d(tau)}

Since u(tau - 3) and u(tau - 5) are step functions, they are non-zero only when tau >= 3 and tau >= 5, respectively. Therefore, the integral can be broken up into two parts:

y(t) = integral from 3 to t of {(u(tau - 3) - u(tau - 5)) e^[-3(t - tau)] d(tau)}

- integral from 5 to t of {(u(tau - 5)) e^[-3(t - tau)] d(tau)}

Simplifying this, we get:

y(t) = (e^(-3t)) [integral from 3 to t of e^(3tau) d(tau) - integral from 5 to t of e^(3tau) d(tau)]

- e^(-15t) integral from 5 to t of e^(3tau) d(tau)

Evaluating the integrals, we get:

y(t) = (1/3) e^(-3t) [e^(3t) - e^(9)] u(t - 3) - (1/3) e^(-3t) [e^(3t) - e^(15)] u(t - 5)

Therefore, the convolution of x(t) and h(t) is:

y(t) = (1/3) e^(-3t) [e^(3t) - e^(9)] u(t - 3) - (1/3) e^(-3t) [e^(3t) - e^(15)] u(t - 5)