# Signals & Systems Questions and Answers – Properties of the Laplace Transform

This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses 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) $$\frac{sin⁡2s}{s}$$

Explanation: 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) $$\frac{s+2}{(s+2)^2+(200π)^2}$$

Explanation: 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) $$\frac{2}{(2s)^2+1}$$

Explanation: We know that sin⁡t u(t) ↔ $$\frac{1}{s^2+1}$$
Scaling property states that f(at) ↔ $$\frac{1}{a} F(\frac{s}{a})$$
$$sin⁡(\frac{t}{2})u(\frac{t}{2}) \leftrightarrow \frac{1}{(\frac{1}{2})} \frac{1}{\Big[(\frac{s}{1/2})^2+1\Big]} \leftrightarrow \frac{2}{(2s)^2+1}$$.

4. Find the Laplace transform of the signal x(t) = $$\frac{dδ(t)}{dt}$$.
a) 1
b) s
c) $$\frac{1}{s}$$
d) s2

Explanation: 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) $$\frac{1}{s+α}$$

Explanation: 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} = –$$\frac{d}{ds} [\frac{1}{s+α}] = \frac{1}{(s+α)^2}$$.
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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$$\left(\frac{s+2}{s-3}\right)$$

Explanation: 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)} = $$\int_s^∞ \left(\frac{1}{s+2} – \frac{1}{s+3}\right)ds = [ln⁡(s+2) – ln⁡(s+3)]|_s^∞ = ln\left(\frac{s+2}{s+3}\right)$$.

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

Explanation: Given F(s) = $$\frac{s}{(s+a)^2+ω^2}$$
The initial value is f(0+) = lims→∞ sF(s)
= lims→∞⁡ 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) ∞

Explanation: Given F(s) = $$\frac{(s-1)}{s(s^2-1)}$$
The final value is f(∞)=lims→0⁡ sF(s)
= lims→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) ∞

Explanation: Given X(s) = $$\frac{4}{s^2+3s-5}$$
Initial value, x(0+) = lims→∞⁡ sX(s) = lims→∞⁡ s($$\frac{4}{s^2+3s-5}$$) = limx→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) –$$\frac{1}{2}$$

Explanation: Given X(s) = $$\frac{s-2}{s(s+4)}$$
Final value, x(∞) = lims→0 sX(s) = lims→0 $$\frac{(s-2)}{s(s+4)} = -\frac{2}{4} = -\frac{1}{2}$$.

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