Question

Let x(t)=e⁻³ᵗ u(t) be a signal, let h(t)=e⁻⁵ᵗ u(t) be a linear time invariant system,

Sketch x(t) and h(t)

Answer 1

The functions x(t) = e^(-3t) u(t) and h(t) = e^(-5t) u(t) represent exponentially decaying signals starting from t=0. They decay at different rates, with h(t) having a steeper decline compared to x(t). Both start at the same point and approach zero as time increases.

Step-by-step explanation:

Sketching x(t) and h(t)

To sketch x(t) = e−3t u(t) and h(t) = e−5t u(t), we first recognize that u(t) represents the unit step function which is 0 for t < 0 and 1 for t ≥ 0. Both functions involve exponential decay that begins at t = 0, only differing in their rates of decay due to the exponents -3 and -5, respectively.

The function x(t) will decay more slowly compared to h(t) because it has a smaller exponent in absolute value. At t = 0, both functions start at 1 because e0 is 1. As t increases, the functions decrease towards 0, with h(t) approaching 0 faster than x(t).

A sketch would show both functions starting at the same point (1,0) and decreasing with time, with the curve for h(t) falling below that of x(t) for any t > 0 because of its steeper decay rate.