Question

The Laplace Transform of the signal \(x(t)=u(t-2)*(t\,u(t))\) is given by which of the following expressions? ["\(*\)" represents the convolution operator]

• A. \(\dfrac{e^{-2s}}{s^{2}(s-2)}\)
• B. \(\dfrac{e^{-2(s-2)}}{s^{3}}\)
• C. \(\dfrac{se^{-2s}}{(s-2)^{2}}\)
• D. \(\dfrac{e^{-2s}}{s^{3}}\)

Hint:

Convolution in time becomes multiplication in the Laplace domain; transform u(t-2) and t.u(t) separately, then multiply the results.

Answer 1

The Correct Option is D

Step 1: Recall the convolution property of the Laplace Transform.
If \(x(t)=x_1(t)*x_2(t)\), then convolution in time becomes plain multiplication in the Laplace domain: \(X(s)=X_1(s)X_2(s)\). Here \(x_1(t)=u(t-2)\) and \(x_2(t)=t\,u(t)\), so we just need to transform each one separately and multiply.

Step 2: Transform the shifted step \(u(t-2)\).
The Laplace Transform of \(u(t)\) is \(1/s\). Using the time-shift property, \(\mathcal{L}\{u(t-2)\}=e^{-2s}\cdot\dfrac{1}{s}=\dfrac{e^{-2s}}{s}\).

Step 3: Transform the ramp \(t\,u(t)\).
A standard Laplace pair is \(\mathcal{L}\{t\,u(t)\}=\dfrac{1}{s^{2}}\).

Step 4: Multiply the two transforms.
\[ X(s)=\frac{e^{-2s}}{s}\cdot\frac{1}{s^{2}}=\frac{e^{-2s}}{s^{3}} \]

Step 5: Check why the other options are wrong.
Option (A) has an extra factor \((s-2)\) in the denominator, as if a frequency-shift property was mixed in with the time-shift by mistake. Option (B) alters the exponent to \(e^{-2(s-2)}\), which would only appear if the signal carried an extra factor \(e^{2t}\), which it does not. Option (C) has an extra \(s\) in the numerator and a squared \((s-2)\) term, neither of which arises from a plain time delay and a ramp. Since \(u(t-2)\) is a pure time delay with no exponential attached to it, no \((s-2)\) term should appear anywhere.

Step 6: Final conclusion.
\[ \boxed{\dfrac{e^{-2s}}{s^{3}}} \]