Question

Consider a causal LTI system with impulse response \( h(t) = e^{-bt}u(t) \). Find the output of the system for an input \( x(t) = 3e^{-at}u(t) \).

• A. \( y(t) = e^{-2t}u(t) + e^{-4t}u(t) \)
• B. \( y(t) = e^{-bt}u(t) + e^{-2at}u(t) \)
• C. \( y(t) = e^{-at}u(t) - e^{-2at}u(t) \)
• D. \( y(t) = e^{-at}u(t)-e^{-bt}u(t) \)

Hint:

Convolution of \(Ae^{-at}u(t)\) and \(Be^{-bt}u(t)\) (for \(a \neq b\)) is \(AB \frac{e^{-at}-e^{-bt}}{b-a} u(t)\).
If \(a=b\), the convolution of \(Ae^{-at}u(t)\) and \(Be^{-at}u(t)\) is \(ABte^{-at}u(t)\).

Answer 1

The Correct Option is D

The output \(y(t)\) is the convolution \(y(t) = x(t) * h(t)\).

Given \(x(t) = 3e^{-at}u(t)\) and \(h(t) = e^{-bt}u(t)\). 

For \(a \neq b\), the convolution is:

\(y(t) = \int_{0}^{t} 3e^{-a\tau} e^{-b(t-\tau)} d\tau = 3e^{-bt} \int_{0}^{t} e^{(b-a)\tau} d\tau\)

\(y(t) = 3e^{-bt} \left[ \frac{e^{(b-a)\tau}}{b-a} \right]_{0}^{t} = 3e^{-bt} \left( \frac{e^{(b-a)t} - 1}{b-a} \right)\)

\(y(t) = \frac{3}{b-a} (e^{-at} - e^{-bt})\) for \(t \ge 0\).

So, \(y(t) = \frac{3}{b-a} (e^{-at} - e^{-bt})u(t)\).

If option (d) \(y(t) = (e^{-at} - e^{-bt})u(t)\) is correct, it implies that the pre-factor \(\frac{3}{b-a} = 1\). This means \(b-a=3\). This is a specific condition not stated in the problem but implied if option (d) is the intended answer.

If \(a=b\), then \(y(t) = 3te^{-at}u(t)\), which is not an option.

Assuming the condition \(b-a=3\) for option (d) to hold:

\[ \boxed{y(t) = (e^{-at}-e^{-bt})u(t) \text{ (assuming } b-a=3)} \]