Question

Given two signals: \(x(t)=u(t-2)-u(t-3)\), and \(y(t)=e^{-4t}u(t)\).
If \(z(t)=x(t)*y(t)\), then \(Z(s)\) is ______.

• A. \(\dfrac{e^{-3s}\left[e^{s}-1\right]}{s(s+4)}\)
• B. \(\dfrac{e^{-s}\left[e^{-3s}-1\right]}{(s+3)(s+4)}\)
• C. \(\dfrac{e^{-3s}\left[1-e^{s}\right]}{(s+3)(s+4)}\)
• D. \(\dfrac{e^{-3s}\left[1-e^{s}\right]}{s(s+4)}\)

Hint:

Convolution in time becomes multiplication in the Laplace domain: find \(X(s)\) and \(Y(s)\) separately, then multiply.

Answer 1

The Correct Option is A

Step 1: Recall the convolution property of the Laplace transform.
If \(z(t)=x(t)*y(t)\), then in the Laplace domain convolution turns into a plain product:
\[ Z(s)=X(s)\,Y(s) \]
So we only need the Laplace transforms of \(x(t)\) and \(y(t)\) separately, then multiply them.

Step 2: Find \(X(s)\).
\(x(t)=u(t-2)-u(t-3)\) is a rectangular pulse of height \(1\) that turns on at \(t=2\) and turns off at \(t=3\). Using the time-shift property \(\mathcal{L}\{u(t-a)\}=\dfrac{e^{-as}}{s}\),
\[ X(s)=\frac{e^{-2s}}{s}-\frac{e^{-3s}}{s}=\frac{e^{-2s}-e^{-3s}}{s} \]

Step 3: Find \(Y(s)\).
\(y(t)=e^{-4t}u(t)\) is a standard exponential, so
\[ Y(s)=\frac{1}{s+4} \]

Step 4: Multiply to get \(Z(s)\).
\[ Z(s)=X(s)\,Y(s)=\frac{e^{-2s}-e^{-3s}}{s}\cdot\frac{1}{s+4}=\frac{e^{-2s}-e^{-3s}}{s(s+4)} \]

Step 5: Write the numerator in the same style as the options.
Factor \(e^{-3s}\) out of the numerator:
\[ e^{-2s}-e^{-3s}=e^{-3s}\left(e^{s}-1\right) \]
since \(e^{-3s}\cdot e^{s}=e^{-2s}\). So
\[ Z(s)=\frac{e^{-3s}\left[e^{s}-1\right]}{s(s+4)} \]

Step 6: Compare with the options and rule out the rest.
This is exactly option (A). Option (D) has the sign inside the bracket flipped to \(1-e^{s}\), which would come from a sign slip while factoring. Options (B) and (C) both carry \((s+3)(s+4)\) in the denominator, which would only appear if \(x(t)\) itself were an exponential like \(e^{-3t}\) rather than a plain rectangular pulse; since \(x(t)\) has no exponential decay at all, the denominator must stay as \(s(s+4)\), so (B) and (C) are wrong.

Final Answer:
\[ \boxed{Z(s)=\frac{e^{-3s}\left[e^{s}-1\right]}{s(s+4)}} \]