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Loksewa GK IQ

7.1 MCQs-Laplace Transform

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Laplace Transform MCQs

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Definition and Basic Transforms

1. The Laplace transform of a function f(t)f(t) is defined as:

  1. ∫0∞e−stf(t) dt\int_{0}^{\infty} e^{-st} f(t) \, dt

  2. ∫−∞∞e−stf(t) dt\int_{-\infty}^{\infty} e^{-st} f(t) \, dt

  3. ∫0∞estf(t) dt\int_{0}^{\infty} e^{st} f(t) \, dt

  4. ∫−∞∞estf(t) dt\int_{-\infty}^{\infty} e^{st} f(t) \, dt

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Answer: 1. ∫0∞e−stf(t) dt\int_{0}^{\infty} e^{-st} f(t) \, dt

Explanation:

  • The (unilateral) Laplace transform L{f(t)}=F(s)\mathcal{L}\{f(t)\} = F(s) is defined as: F(s)=∫0∞e−stf(t) dtF(s) = \int_{0}^{\infty} e^{-st} f(t) \, dt

  • The integral is taken from 00 to ∞\infty (one-sided).

  • The parameter ss is a complex variable: s=σ+iωs = \sigma + i\omega.

2. The Laplace transform of the unit step function u(t)u(t) (also denoted as H(t)H(t)) is:

  1. 1s\frac{1}{s} for Re(s)>0\text{Re}(s) > 0

  2. 1s2\frac{1}{s^2}

  3. 1

  4. ss

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Answer: 1. 1s\frac{1}{s} for Re(s)>0\text{Re}(s) > 0

Explanation:

  • u(t)={0,t<01,t≥0u(t) = \begin{cases} 0, & t < 0 \\ 1, & t \ge 0 \end{cases}

  • L{u(t)}=∫0∞e−st⋅1 dt=[e−st−s]0∞\mathcal{L}\{u(t)\} = \int_{0}^{\infty} e^{-st} \cdot 1 \, dt = \left[ \frac{e^{-st}}{-s} \right]_{0}^{\infty}

  • For Re(s)>0\text{Re}(s) > 0, e−s⋅∞→0e^{-s \cdot \infty} \to 0.

  • Therefore, L{u(t)}=0−(1−s)=1s\mathcal{L}\{u(t)\} = 0 - \left( \frac{1}{-s} \right) = \frac{1}{s}.

3. The Laplace transform of f(t)=eatf(t) = e^{at} is:

  1. 1s−a\frac{1}{s-a} for Re(s)>Re(a)\text{Re}(s) > \text{Re}(a)

  2. 1s+a\frac{1}{s+a}

  3. ss2+a2\frac{s}{s^2 + a^2}

  4. as2+a2\frac{a}{s^2 + a^2}

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Answer: 1. 1s−a\frac{1}{s-a} for Re(s)>Re(a)\text{Re}(s) > \text{Re}(a)

Explanation:

  • L{eat}=∫0∞e−steat dt=∫0∞e−(s−a)t dt\mathcal{L}\{e^{at}\} = \int_{0}^{\infty} e^{-st} e^{at} \, dt = \int_{0}^{\infty} e^{-(s-a)t} \, dt

  • This integral converges if Re(s−a)>0\text{Re}(s-a) > 0, i.e., Re(s)>Re(a)\text{Re}(s) > \text{Re}(a).

  • ∫0∞e−(s−a)t dt=[e−(s−a)t−(s−a)]0∞=0−1−(s−a)=1s−a\int_{0}^{\infty} e^{-(s-a)t} \, dt = \left[ \frac{e^{-(s-a)t}}{-(s-a)} \right]_{0}^{\infty} = 0 - \frac{1}{-(s-a)} = \frac{1}{s-a}.

4. The Laplace transform of f(t)=tnf(t) = t^n where n is a positive integer is:

  1. n!sn+1\frac{n!}{s^{n+1}}

  2. n!sn\frac{n!}{s^n}

  3. ss2+n2\frac{s}{s^2 + n^2}

  4. 1sn\frac{1}{s^n}

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Answer: 1. n!sn+1\frac{n!}{s^{n+1}}

Explanation:

  • L{tn}=∫0∞e−sttn dt\mathcal{L}\{t^n\} = \int_{0}^{\infty} e^{-st} t^n \, dt

  • Using integration by parts or the Gamma function: ∫0∞e−sttn dt=Γ(n+1)sn+1\int_{0}^{\infty} e^{-st} t^n \, dt = \frac{\Gamma(n+1)}{s^{n+1}}.

  • Since n is a positive integer, Γ(n+1)=n!\Gamma(n+1) = n!.

  • Therefore, L{tn}=n!sn+1\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} for Re(s)>0\text{Re}(s) > 0.

  • Special cases: L{t}=1s2\mathcal{L}\{t\} = \frac{1}{s^2}, L{t2}=2s3\mathcal{L}\{t^2\} = \frac{2}{s^3}.

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Transforms of Trigonometric Functions

5. The Laplace transform of cos⁡(ωt)\cos(\omega t) is:

  1. ss2+ω2\frac{s}{s^2 + \omega^2}

  2. ωs2+ω2\frac{\omega}{s^2 + \omega^2}

  3. 1s2+ω2\frac{1}{s^2 + \omega^2}

  4. ss2−ω2\frac{s}{s^2 - \omega^2}

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Answer: 1. ss2+ω2\frac{s}{s^2 + \omega^2}

Explanation:

  • Using Euler's formula: cos⁡(ωt)=eiωt+e−iωt2\cos(\omega t) = \frac{e^{i\omega t} + e^{-i\omega t}}{2}.

  • Then L{cos⁡(ωt)}=12(L{eiωt}+L{e−iωt})=12(1s−iω+1s+iω)\mathcal{L}\{\cos(\omega t)\} = \frac{1}{2} \left( \mathcal{L}\{e^{i\omega t}\} + \mathcal{L}\{e^{-i\omega t}\} \right) = \frac{1}{2} \left( \frac{1}{s - i\omega} + \frac{1}{s + i\omega} \right).

  • Simplify: 12⋅(s+iω)+(s−iω)s2+ω2=12⋅2ss2+ω2=ss2+ω2\frac{1}{2} \cdot \frac{(s + i\omega) + (s - i\omega)}{s^2 + \omega^2} = \frac{1}{2} \cdot \frac{2s}{s^2 + \omega^2} = \frac{s}{s^2 + \omega^2}.

6. The Laplace transform of sin⁡(ωt)\sin(\omega t) is:

  1. ss2+ω2\frac{s}{s^2 + \omega^2}

  2. ωs2+ω2\frac{\omega}{s^2 + \omega^2}

  3. 1s2+ω2\frac{1}{s^2 + \omega^2}

  4. ss2−ω2\frac{s}{s^2 - \omega^2}

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Answer: 2. ωs2+ω2\frac{\omega}{s^2 + \omega^2}

Explanation:

  • Using Euler's formula: sin⁡(ωt)=eiωt−e−iωt2i\sin(\omega t) = \frac{e^{i\omega t} - e^{-i\omega t}}{2i}.

  • L{sin⁡(ωt)}=12i(1s−iω−1s+iω)\mathcal{L}\{\sin(\omega t)\} = \frac{1}{2i} \left( \frac{1}{s - i\omega} - \frac{1}{s + i\omega} \right).

  • Simplify: 12i⋅(s+iω)−(s−iω)s2+ω2=12i⋅2iωs2+ω2=ωs2+ω2\frac{1}{2i} \cdot \frac{(s + i\omega) - (s - i\omega)}{s^2 + \omega^2} = \frac{1}{2i} \cdot \frac{2i\omega}{s^2 + \omega^2} = \frac{\omega}{s^2 + \omega^2}.

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Properties of Laplace Transform

7. The linearity property of Laplace transform states that:

  1. L{af(t)+bg(t)}=aL{f(t)}+bL{g(t)}\mathcal{L}\{af(t) + bg(t)\} = a\mathcal{L}\{f(t)\} + b\mathcal{L}\{g(t)\}

  2. L{f(t)g(t)}=L{f(t)}⋅L{g(t)}\mathcal{L}\{f(t)g(t)\} = \mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\}

  3. L{f(t)}=sL{f(t)}−f(0)\mathcal{L}\{f(t)\} = s\mathcal{L}\{f(t)\} - f(0)

  4. L{f(at)}=1aF(sa)\mathcal{L}\{f(at)\} = \frac{1}{a}F\left(\frac{s}{a}\right)

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Answer: 1. L{af(t)+bg(t)}=aL{f(t)}+bL{g(t)}\mathcal{L}\{af(t) + bg(t)\} = a\mathcal{L}\{f(t)\} + b\mathcal{L}\{g(t)\}

Explanation:

  • Linearity is a fundamental property: The Laplace transform of a linear combination is the same linear combination of the transforms.

  • This follows directly from the linearity of the integral: ∫(af+bg)=a∫f+b∫g\int (af + bg) = a\int f + b\int g.

  • This property makes the Laplace transform easy to apply to linear differential equations.

8. The First Shifting Theorem (Frequency Shift) states: If L{f(t)}=F(s)\mathcal{L}\{f(t)\} = F(s), then L{eatf(t)}=\mathcal{L}\{e^{at}f(t)\} =

  1. F(s−a)F(s-a)

  2. F(s+a)F(s+a)

  3. easF(s)e^{as}F(s)

  4. e−asF(s)e^{-as}F(s)

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Answer: 1. F(s−a)F(s-a)

Explanation:

  • The First Shifting Theorem: L{eatf(t)}=F(s−a)\mathcal{L}\{e^{at}f(t)\} = F(s-a).

  • Proof: L{eatf(t)}=∫0∞e−steatf(t) dt=∫0∞e−(s−a)tf(t) dt=F(s−a)\mathcal{L}\{e^{at}f(t)\} = \int_{0}^{\infty} e^{-st} e^{at} f(t) \, dt = \int_{0}^{\infty} e^{-(s-a)t} f(t) \, dt = F(s-a).

  • This is also called the s-shifting property.

  • Example: Since L{cos⁡(ωt)}=ss2+ω2\mathcal{L}\{\cos(\omega t)\} = \frac{s}{s^2+\omega^2}, then L{eatcos⁡(ωt)}=s−a(s−a)2+ω2\mathcal{L}\{e^{at}\cos(\omega t)\} = \frac{s-a}{(s-a)^2+\omega^2}.

9. The Laplace transform of the derivative f′(t)f'(t) is:

  1. sF(s)sF(s)

  2. sF(s)−f(0)sF(s) - f(0)

  3. F(s)s\frac{F(s)}{s}

  4. F(s)−f(0)F(s) - f(0)

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Answer: 2. sF(s)−f(0)sF(s) - f(0)

Explanation:

  • This is one of the most important properties for solving differential equations.

  • L{f′(t)}=sL{f(t)}−f(0)=sF(s)−f(0)\mathcal{L}\{f'(t)\} = s\mathcal{L}\{f(t)\} - f(0) = sF(s) - f(0).

  • For the second derivative: L{f′′(t)}=s2F(s)−sf(0)−f′(0)\mathcal{L}\{f''(t)\} = s^2F(s) - sf(0) - f'(0).

  • This property converts differential equations in t to algebraic equations in s.

10. The Laplace transform of the integral ∫0tf(τ) dτ\int_0^t f(\tau) \, d\tau is:

  1. F(s)s\frac{F(s)}{s}

  2. sF(s)sF(s)

  3. F(s)−f(0)F(s) - f(0)

  4. F(s)s+f(0)s\frac{F(s)}{s} + \frac{f(0)}{s}

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Answer: 1. F(s)s\frac{F(s)}{s}

Explanation:

  • Integration in the time domain corresponds to division by s in the frequency domain.

  • If g(t)=∫0tf(τ) dτg(t) = \int_0^t f(\tau) \, d\tau, then g′(t)=f(t)g'(t) = f(t) and g(0)=0g(0) = 0.

  • Using the derivative property: L{g′(t)}=sL{g(t)}−g(0)\mathcal{L}\{g'(t)\} = s\mathcal{L}\{g(t)\} - g(0).

  • So F(s)=sL{g(t)}−0F(s) = s\mathcal{L}\{g(t)\} - 0, thus L{g(t)}=F(s)s\mathcal{L}\{g(t)\} = \frac{F(s)}{s}.

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Unit Step Function and Time Shifting

11. The Laplace transform of the unit step function shifted by a, i.e., u(t−a)u(t-a) (with a>0a > 0), is:

  1. e−ass\frac{e^{-as}}{s}

  2. eass\frac{e^{as}}{s}

  3. 1s\frac{1}{s}

  4. e−ase^{-as}

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Answer: 1. e−ass\frac{e^{-as}}{s}

Explanation:

  • u(t−a)={0,t<a1,t≥au(t-a) = \begin{cases} 0, & t < a \\ 1, & t \ge a \end{cases}

  • L{u(t−a)}=∫0∞e−stu(t−a) dt=∫a∞e−st dt\mathcal{L}\{u(t-a)\} = \int_{0}^{\infty} e^{-st} u(t-a) \, dt = \int_{a}^{\infty} e^{-st} \, dt

  • Let τ=t−a\tau = t-a, then dt=dτdt = d\tau, and when t=at=a, τ=0\tau=0.

  • ∫a∞e−st dt=∫0∞e−s(τ+a) dτ=e−as∫0∞e−sτ dτ=e−as⋅1s\int_{a}^{\infty} e^{-st} \, dt = \int_{0}^{\infty} e^{-s(\tau+a)} \, d\tau = e^{-as} \int_{0}^{\infty} e^{-s\tau} \, d\tau = e^{-as} \cdot \frac{1}{s}.

12. The Second Shifting Theorem (Time Shift) states: If L{f(t)}=F(s)\mathcal{L}\{f(t)\} = F(s), then L{f(t−a)u(t−a)}=\mathcal{L}\{f(t-a)u(t-a)\} =

  1. e−asF(s)e^{-as}F(s)

  2. easF(s)e^{as}F(s)

  3. F(s−a)F(s-a)

  4. F(s+a)F(s+a)

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Answer: 1. e−asF(s)e^{-as}F(s)

Explanation:

  • Also known as the t-shifting property.

  • The function f(t−a)u(t−a)f(t-a)u(t-a) represents f(t)f(t) shifted to the right by a units.

  • L{f(t−a)u(t−a)}=e−asF(s)\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s).

  • Example: To find L{(t−2)2u(t−2)}\mathcal{L}\{(t-2)^2 u(t-2)\}, we know L{t2}=2s3\mathcal{L}\{t^2\} = \frac{2}{s^3}, so the transform is e−2s⋅2s3e^{-2s} \cdot \frac{2}{s^3}.

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Dirac Delta Function

13. The Laplace transform of the Dirac delta function δ(t)\delta(t) is:

  1. 0

  2. 1

  3. 1s\frac{1}{s}

  4. ss

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Answer: 2. 1

Explanation:

  • The Dirac delta function δ(t)\delta(t) is defined such that ∫−∞∞δ(t) dt=1\int_{-\infty}^{\infty} \delta(t) \, dt = 1 and δ(t)=0\delta(t) = 0 for t≠0t \neq 0.

  • L{δ(t)}=∫0∞e−stδ(t) dt=e−s⋅0=1\mathcal{L}\{\delta(t)\} = \int_{0}^{\infty} e^{-st} \delta(t) \, dt = e^{-s \cdot 0} = 1 (using the sifting property).

  • For a shifted delta: L{δ(t−a)}=e−as\mathcal{L}\{\delta(t-a)\} = e^{-as} for a>0a > 0.

14. The Laplace transform of δ(t−a)\delta(t-a) for a>0a > 0 is:

  1. 1

  2. e−ase^{-as}

  3. ease^{as}

  4. e−ass\frac{e^{-as}}{s}

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Answer: 2. e−ase^{-as}

Explanation:

  • Using the sifting property of the Dirac delta function: ∫0∞e−stδ(t−a) dt=e−sa\int_{0}^{\infty} e^{-st} \delta(t-a) \, dt = e^{-sa} for a>0a > 0.

  • This follows because δ(t−a)\delta(t-a) "picks out" the value of the integrand at t=at = a.

  • Therefore, L{δ(t−a)}=e−as\mathcal{L}\{\delta(t-a)\} = e^{-as}.

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Inverse Laplace Transform

15. The inverse Laplace transform of 1s−a\frac{1}{s-a} is:

  1. eate^{at}

  2. e−ate^{-at}

  3. tnt^n

  4. cos⁡(at)\cos(at)

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Answer: 1. eate^{at}

Explanation:

  • This is the inverse of the basic transform L{eat}=1s−a\mathcal{L}\{e^{at}\} = \frac{1}{s-a}.

  • So, L−1{1s−a}=eat\mathcal{L}^{-1}\left\{ \frac{1}{s-a} \right\} = e^{at} for t≥0t \ge 0.

16. The inverse Laplace transform of ss2+4\frac{s}{s^2 + 4} is:

  1. sin⁡(2t)\sin(2t)

  2. cos⁡(2t)\cos(2t)

  3. e2te^{2t}

  4. e−2te^{-2t}

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Answer: 2. cos⁡(2t)\cos(2t)

Explanation:

  • We know L{cos⁡(ωt)}=ss2+ω2\mathcal{L}\{\cos(\omega t)\} = \frac{s}{s^2 + \omega^2}.

  • Here, ss2+4=ss2+22\frac{s}{s^2 + 4} = \frac{s}{s^2 + 2^2}, so ω=2\omega = 2.

  • Therefore, L−1{ss2+4}=cos⁡(2t)\mathcal{L}^{-1}\left\{ \frac{s}{s^2 + 4} \right\} = \cos(2t).

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Convolution Theorem

17. The Convolution Theorem states: The Laplace transform of the convolution (f∗g)(t)=∫0tf(τ)g(t−τ) dτ(f * g)(t) = \int_0^t f(\tau)g(t-\tau) \, d\tau is:

  1. F(s)+G(s)F(s) + G(s)

  2. F(s)⋅G(s)F(s) \cdot G(s)

  3. F(s)−G(s)F(s) - G(s)

  4. F(s)G(s)\frac{F(s)}{G(s)}

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Answer: 2. F(s)⋅G(s)F(s) \cdot G(s)

Explanation:

  • The Convolution Theorem: L{(f∗g)(t)}=F(s)⋅G(s)\mathcal{L}\{(f * g)(t)\} = F(s) \cdot G(s).

  • Conversely: L−1{F(s)⋅G(s)}=(f∗g)(t)\mathcal{L}^{-1}\{F(s) \cdot G(s)\} = (f * g)(t).

  • This is useful for finding inverse transforms of products and for solving integral equations.

  • Convolution is commutative: f∗g=g∗ff * g = g * f.

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Solving Differential Equations

18. When solving an initial value problem using Laplace transforms, the differential equation in the time domain is converted to:

  1. Another differential equation in s

  2. An integral equation

  3. An algebraic equation in s

  4. A partial differential equation

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Answer: 3. An algebraic equation in s

Explanation:

  • This is the primary advantage of the Laplace transform method for solving linear ordinary differential equations with constant coefficients.

  • Derivatives become algebraic terms: L{y′}=sY(s)−y(0)\mathcal{L}\{y'\} = sY(s) - y(0), L{y′′}=s2Y(s)−sy(0)−y′(0)\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0), etc.

  • The ODE transforms into an algebraic equation for Y(s)Y(s), which is then solved algebraically.

  • Finally, y(t)y(t) is obtained by taking the inverse Laplace transform of Y(s)Y(s).

19. The Laplace transform is particularly useful for solving:

  1. Nonlinear differential equations

  2. Linear differential equations with constant coefficients and initial conditions

  3. Partial differential equations without boundary conditions

  4. Integral equations only

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Answer: 2. Linear differential equations with constant coefficients and initial conditions

Explanation:

  • The Laplace transform method excels at solving linear ODEs with constant coefficients and given initial conditions (initial value problems).

  • It handles discontinuous forcing functions (like step functions) and impulse functions (Dirac delta) naturally.

  • The method automatically incorporates the initial conditions.

  • For boundary value problems or variable coefficient equations, other methods may be more appropriate.

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Final Value Theorem

20. The Final Value Theorem states: If the limits exist, then lim⁡t→∞f(t)=\lim_{t \to \infty} f(t) =

  1. lim⁡s→0sF(s)\lim_{s \to 0} sF(s)

  2. lim⁡s→∞sF(s)\lim_{s \to \infty} sF(s)

  3. lim⁡s→0F(s)\lim_{s \to 0} F(s)

  4. lim⁡s→∞F(s)\lim_{s \to \infty} F(s)

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Answer: 1. lim⁡s→0sF(s)\lim_{s \to 0} sF(s)

Explanation:

  • Final Value Theorem: lim⁡t→∞f(t)=lim⁡s→0sF(s)\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s), provided all poles of sF(s)sF(s) have negative real parts (i.e., the limit exists and is finite).

  • This theorem gives the steady-state value of a time function without needing the inverse transform.

  • Initial Value Theorem (dual): lim⁡t→0+f(t)=lim⁡s→∞sF(s)\lim_{t \to 0^+} f(t) = \lim_{s \to \infty} sF(s).

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