250+ TOP MCQs on Properties of the Laplace Transform and Answers

Signals & Systems Multiple Choice Questions on “Properties of the Laplace Transform”.

1. Find the Laplace transform of x(t) = u(t+2) + u(t-2).
A. (frac{cos⁡2s}{s})
B. (frac{cosh⁡2s}{s})
C. (frac{sinh⁡2s}{s})
D. (frac{sin⁡2s}{s})
Answer: B
Clarification: Given x(t) = u(t+2) + u(t-2)
We know that the Laplace transform u(t) ↔ (frac{1}{s})
Time shifting property states that L{x(t±t₀)} = X(s)e^±st₀
L{u(t-2)}=(e^{frac{-2s}{s}}); L{u(t+2)}=(e^{frac{2s}{s}})
∴X(s) = L{u(t+2)+u(t-2)} = (frac{e^{-2s}+e^{-2s}}{s} = frac{cosh⁡2s}{s}).

2. Find the Laplace transform of the signal x(t) = e^-2t cos⁡(200πt)u(t).
A. (frac{s}{s^2+(200π)^2})
B. (frac{s}{s^2-(200π)^2})
C. (frac{s-2}{(s-2)^2+(200π)^2})
D. (frac{s+2}{(s+2)^2+(200π)^2})
Answer: D
Clarification: Given x(t) = e^-2t cos⁡(200πt)u(t)
We know that L{cos⁡ωt u(t)} = (frac{s}{s^2+ω^2})
∴L{cos⁡(200πt)u(t)} = (frac{s}{s^2+(200π)^2})
Frequency shifting property states that L{e^-at x(t)} = X(s+A.
L{e^-2t cos⁡(200πt)u(t)} = L{cos⁡(200πt)u(t)}|_s=s+2 = (Big[frac{s}{s^2+(200π)^2}Big]_{s=s+2} = frac{s+2}{(s+2)^2+(200π)^2}).

3. Find the Laplace transform of the signal x(t) = sin⁡((frac{t}{2}))u((frac{t}{2})).
A. (frac{1}{s^2+1})
B. (frac{s}{s^2+1})
C. (frac{2s}{(2s)^2+1})
D. (frac{2}{(2s)^2+1})
Answer: D
Clarification: We know that sin⁡t u(t) ↔ (frac{1}{s^2+1})
Scaling property states that f(at) ↔ (frac{1}{a} F(frac{s}{a}))
(sin⁡(frac{t}{2})u(frac{t}{2}) leftrightarrow frac{1}{(frac{1}{2})} frac{1}{Big[(frac{s}{1/2})^2+1Big]} leftrightarrow frac{2}{(2s)^2+1}).

4. Find the Laplace transform of the signal x(t) = (frac{dδ(t)}{dt}).
A. 1
B. s
C. (frac{1}{s})
D. s²
Answer: B
Clarification: Given x(t) = (frac{dδ(t)}{dt})
We know that L{δ(t)} = 1
Time differentiation property, L{(frac{dδ(t)}{dt})} = sF(s)
L{(frac{dδ(t)}{dt})} = sL{δ(t)} = s × 1 = s.

5. Find the Laplace transform of the signal x(t) = te^-αt.
A. (frac{1}{s^2})
B. (frac{1}{(s+α)^2})
C. (frac{1}{α})
D. (frac{1}{s+α})
Answer: B
Clarification: We know that L{e^-αt} = (frac{1}{s+α})
Differentiation in s-domain property states that (-t)ⁿ f(t) ↔ (frac{d^n F(s)}{ds^n})
L{te^-αt} = –(frac{d}{ds} [frac{1}{s+α}] = frac{1}{(s+α)^2}).

6. Find the Laplace transform for f(t) = (frac{1}{t}) [e^-2t – e^-3t]u(t).
A. ln(left(frac{s-2}{s-3}right))
B. ln(left(frac{s+2}{s+3}right))
C. ln(left(frac{s-2}{s+3}right))
D. ln(left(frac{s+2}{s-3}right))
Answer: B
Clarification: Given f(t) = (frac{1}{t}) [e^-2t – e^-3t]u(t)
We know that L{e^-2t) u(t)} = (frac{1}{s+2}); L{^-3t u(t)} = (frac{1}{s+3})
Integration in s-domain property states that (int_s^∞ F(s)ds leftrightarrow frac{f(t)}{t})
L{(frac{1}{t}) [e^-2t – e^-3)]u(t)} = (int_s^∞ left(frac{1}{s+2} – frac{1}{s+3}right)ds = [ln⁡(s+2) – ln⁡(s+3)]|_s^∞ = lnleft(frac{s+2}{s+3}right)).

7. Find the initial value of f(t) if F(s) = (frac{s}{(s+A.^2+ω^2}).
A. 0
B. -1
C. ∞
D. 1
Answer: D
Clarification: Given F(s) = (frac{s}{(s+A.^2+ω^2})
The initial value is f(0⁺) = lim_s→∞ sF(s)
= lim_s→∞⁡ s (frac{s}{(s+A.^2+ω^2} = lim_{s→∞} frac{1}{(1+a/s)^2+(ω/s)^2} = 1).

8. Find the final value of the function F(s) given by (frac{(s-1)}{s(s^2-1)}).
A. 1
B. 0
C. -1
D. ∞
Answer: A
Clarification: Given F(s) = (frac{(s-1)}{s(s^2-1)})
The final value is f(∞)=lim_s→0⁡ sF(s)
= lim_s→0⁡ s (frac{(s-1)}{s(s^2-1)} = lim_{s→0} frac{1}{s+1} = 1).

9. Determine the initial value x(0⁺) for the Laplace transform X(s) = (frac{4}{s^2+3s-5}).
A. -1
B. 0
C. 1
D. ∞
Answer: B
Clarification: Given X(s) = (frac{4}{s^2+3s-5})
Initial value, x(0⁺) = lim_s→∞⁡ sX(s) = lim_s→∞⁡ s((frac{4}{s^2+3s-5})) = lim_x→0⁡ (frac{4x}{1+3x-5x^2} = 0) [let s = 1/x].

10. Find x(∞) if X(s) is given by (frac{s-2}{s(s+4)}).
A. 1
B. -1
C. (frac{1}{2})
D. –(frac{1}{2})
Answer: D
Clarification: Given X(s) = (frac{s-2}{s(s+4)})
Final value, x(∞) = lim_s→0 sX(s) = lim_s→0 (frac{(s-2)}{s(s+4)} = -frac{2}{4} = -frac{1}{2}).