Question:medium
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]
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]
Show Hint
Convolution in time becomes multiplication in the Laplace domain; transform u(t-2) and t.u(t) separately, then multiply the results.
Updated On: Jul 20, 2026
- \(\dfrac{e^{-2s}}{s^{2}(s-2)}\)
- \(\dfrac{e^{-2(s-2)}}{s^{3}}\)
- \(\dfrac{se^{-2s}}{(s-2)^{2}}\)
- \(\dfrac{e^{-2s}}{s^{3}}\)
Show Solution
The Correct Option is D
Solution and Explanation
Step 1: Do the convolution directly in the time domain.
By definition, $x(t)=u(t-2)*t\,u(t)=\displaystyle\int_{-\infty}^{\infty}u(\tau-2)\,(t-\tau)u(t-\tau)\,d\tau$. The factor $u(\tau-2)$ restricts $\tau\ge2$, and $u(t-\tau)$ restricts $\tau\le t$, so the integral only survives for $t\ge2$, running from $\tau=2$ to $\tau=t$.
Step 2: Evaluate the integral.
\[ x(t)=\int_{2}^{t}(t-\tau)\,d\tau,\qquad t\ge2 \] Let $\sigma=t-\tau$; as $\tau$ runs from $2$ to $t$, $\sigma$ runs from $t-2$ down to $0$. Then
\[ x(t)=\int_{0}^{t-2}\sigma\,d\sigma=\frac{(t-2)^2}{2},\qquad t\ge2 \] So $x(t)=\dfrac{(t-2)^2}{2}u(t-2)$.
Step 3: Take the Laplace Transform of this time-domain result.
Since $\mathcal{L}\{t^2u(t)\}=\dfrac{2}{s^3}$, we get $\mathcal{L}\left\{\dfrac{t^2}{2}u(t)\right\}=\dfrac{1}{s^3}$. Applying the time-shift property for the delay by $2$,
\[ X(s)=e^{-2s}\cdot\frac{1}{s^3}=\frac{e^{-2s}}{s^3} \]
Step 4: Match with the options.
This confirms option (D) directly, without ever needing the frequency-domain convolution shortcut.
\[ \boxed{\dfrac{e^{-2s}}{s^3}} \]
By definition, $x(t)=u(t-2)*t\,u(t)=\displaystyle\int_{-\infty}^{\infty}u(\tau-2)\,(t-\tau)u(t-\tau)\,d\tau$. The factor $u(\tau-2)$ restricts $\tau\ge2$, and $u(t-\tau)$ restricts $\tau\le t$, so the integral only survives for $t\ge2$, running from $\tau=2$ to $\tau=t$.
Step 2: Evaluate the integral.
\[ x(t)=\int_{2}^{t}(t-\tau)\,d\tau,\qquad t\ge2 \] Let $\sigma=t-\tau$; as $\tau$ runs from $2$ to $t$, $\sigma$ runs from $t-2$ down to $0$. Then
\[ x(t)=\int_{0}^{t-2}\sigma\,d\sigma=\frac{(t-2)^2}{2},\qquad t\ge2 \] So $x(t)=\dfrac{(t-2)^2}{2}u(t-2)$.
Step 3: Take the Laplace Transform of this time-domain result.
Since $\mathcal{L}\{t^2u(t)\}=\dfrac{2}{s^3}$, we get $\mathcal{L}\left\{\dfrac{t^2}{2}u(t)\right\}=\dfrac{1}{s^3}$. Applying the time-shift property for the delay by $2$,
\[ X(s)=e^{-2s}\cdot\frac{1}{s^3}=\frac{e^{-2s}}{s^3} \]
Step 4: Match with the options.
This confirms option (D) directly, without ever needing the frequency-domain convolution shortcut.
\[ \boxed{\dfrac{e^{-2s}}{s^3}} \]
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