Question

A causal system is given the input x1(t) = 5 u(t) and the output is y1(t) = e⁽⁻²ᵗ⁾u(t). Let y2(t) be the response of the same system to x2(t) = 5 * 3t * u(t - 1). What is y2(t) for t < -1?

Answer 1

The value of y2(t) is 0 for t < -1.

Step-by-step explanation:

A causal system only "remembers" the inputs it has received up to the current time. Its output at any time t depends only on the input at or before that time. For t < -1, the input x2(t) is zero, as it is multiplied by the unit step function u(t-1) which is 0 for t < -1. Therefore, the output y2(t) will also be zero for t < -1.

Here's a mathematical explanation:

The input-output relationship of a causal system can be described by convolution:

y(t) = x(t) * h(t)

where:

y(t) is the output

x(t) is the input

h(t) is the impulse response of the system

In this case, we know that x2(t) = 5 * 3t * u(t-1) and y1(t) = e^(-2t)u(t). Since the system is causal, its impulse response h(t) is also causal. Therefore, h(t) = 0 for t < 0.

Now, let's calculate y2(t) for t < -1:

y2(t) = x2(t) * h(t)

= (5 * 3t * u(t-1)) * (0 * u(t))

= 0 * 0 * u(t-1) u(t)

= 0

Therefore, y2(t) = 0 for t < -1.