Laplace Transform Practice Exam Solutions

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You already took the 30-item practice exam on the Laplace Transform Series Index Page. This post gives you the full solution for every item, written the way you should show your work on exam day. No theory recap. Just the tool being tested, the steps, the answer, and the specific trap to watch for on each item. If you got something wrong, find it below and read slowly. Then close this page and solve a similar problem from scratch.

Part A — Definition, Basic Transforms and Properties (Items 1 to 5)

Full lesson: Part 1 — Definition, Basic Transforms and Properties

Item 1. What is $\mathcal{L}\{7\}$ ?

Choices: (A) $7$ (B) $7/s$ (C) $7s$ (D) $1/(7s)$

Tool tested: Basic pair — $\mathcal{L}\{1\} = 1/s$

Given: $f(t) = 7$

Find: $\mathcal{L}\{f(t)\}$

Solution:

Step 1: A bare constant $c$ is $c \cdot 1$ , so it goes through the $\mathcal{L}\{1\}$ pair.

$\mathcal{L}\{7\} = 7 \cdot \mathcal{L}\{1\} = 7 \cdot \dfrac{1}{s}$

✓ ANSWER: B — $\mathcal{L}\{7\} = 7/s$

Examiner note: Choice A treats the constant as not needing transformation at all — the “bare constant” trap covered in Part 1. Every constant still passes through $\mathcal{L}\{1\}=1/s$ , it does not stay as-is.

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Item 2. Find $\mathcal{L}\{t^3\}$ .

Choices: (A) $3/s^4$ (B) $6/s^3$ (C) $6/s^4$ (D) $3!/s^3$

Tool tested: Basic pair — $\mathcal{L}\{t^n\} = n!/s^{n+1}$

Given: $f(t) = t^3$

Find: $\mathcal{L}\{f(t)\}$

Solution:

Step 1: Apply the power function pair with $n = 3$ .

$\mathcal{L}\{t^3\} = \dfrac{3!}{s^{3+1}} = \dfrac{6}{s^4}$

✓ ANSWER: C — $\mathcal{L}\{t^3\} = 6/s^4$

Examiner note: Choice B has the correct numerator but the wrong power of $s$ — the exponent is $n+1$ , not $n$ . Choice D stops halfway, leaving $3!$ unevaluated and the denominator power wrong. Always write out the factorial fully and double-check the exponent is $n+1$ .

Item 3. Using linearity, find $\mathcal{L}\{2e^{3t} - 4\}$ .

Choices: (A) $\dfrac{2}{s-3} - \dfrac{4}{s}$ (B) $\dfrac{2}{s-3} + \dfrac{4}{s}$ (C) $\dfrac{2}{s+3} - \dfrac{4}{s}$ (D) $\dfrac{2s}{s-3} - 4$

Tool tested: Linearity + basic exponential pair

Given: $f(t) = 2e^{3t} - 4$

Find: $\mathcal{L}\{f(t)\}$

Solution:

Step 1: Split with linearity, keeping the original signs of each term.

$\mathcal{L}\{f(t)\} = 2\mathcal{L}\{e^{3t}\} - 4\mathcal{L}\{1\} = 2\left(\dfrac{1}{s-3}\right) - 4\left(\dfrac{1}{s}\right)$

✓ ANSWER: A — $\dfrac{2}{s-3} - \dfrac{4}{s}$

Examiner note: Choice B flips the sign on the constant term — the original function subtracts 4, so the transform must subtract $4/s$ too. Choice C uses $s+3$ instead of $s-3$ , mixing up the sign convention in the exponential pair.

Item 4. Apply the first shifting theorem to find $\mathcal{L}\{e^{-2t}\sin(3t)\}$ .

Choices: (A) $\dfrac{3}{(s-2)^2+9}$ (B) $\dfrac{3}{(s+2)^2+9}$ (C) $\dfrac{3}{(s+2)^2-9}$ (D) $\dfrac{s+2}{(s+2)^2+9}$

Tool tested: First shifting theorem

Given: $f(t) = e^{-2t}\sin(3t)$

Find: $\mathcal{L}\{f(t)\}$

Solution:

Step 1: Identify $a = -2$ , $g(t) = \sin(3t)$ , so $G(s) = 3/(s^2+9)$ .

Step 2: Shift by replacing $s$ with $s - (-2) = s+2$ .

$F(s) = G(s+2) = \dfrac{3}{(s+2)^2+9}$

✓ ANSWER: B — $\dfrac{3}{(s+2)^2+9}$

Examiner note: Choice A is the wrong-sign shifting trap — using $s-2$ instead of $s+2$ when $a$ is negative. Choice C mixes the circular sine pair’s denominator with the hyperbolic sign. Choice D swaps in the cosine pair’s numerator by mistake.

Item 5. For what values of $s$ does $\mathcal{L}\{e^{5t}\}$ converge?

Choices: (A) $s > 0$ (B) $s > 5$ (C) $s < 5$ (D) $s \geq 0$

Tool tested: Convergence condition on the exponential pair

Given: $f(t) = e^{5t}$

Find: Range of $s$ for convergence

Solution:

Step 1: The exponential pair $\mathcal{L}\{e^{at}\} = 1/(s-a)$ converges for $s > a$ .

$a = 5 \implies s > 5$

✓ ANSWER: B — $s > 5$

Examiner note: Choice C reverses the inequality direction entirely. Choice A confuses this with the convergence condition for $\mathcal{L}\{1\}$ , which is $s>0$ — a different pair with a different bound.

Part B — Inverse Laplace Transform and Partial Fractions (Items 6 to 12)

Full lesson: Part 2 — Inverse Laplace Transform and Partial Fractions

Item 6. Which is the correct partial fraction setup for $\dfrac{2s+3}{(s-1)(s+4)}$ ?

Choices: (A) $\dfrac{A}{s-1}+\dfrac{B}{s+4}$ (B) $\dfrac{A}{(s-1)^2}+\dfrac{B}{s+4}$ (C) $\dfrac{As+B}{(s-1)(s+4)}$ (D) $\dfrac{A}{s+1}+\dfrac{B}{s-4}$

Tool tested: Case 1 — distinct linear factors

Given: $F(s) = \dfrac{2s+3}{(s-1)(s+4)}$

Find: Correct decomposition structure

Solution:

Step 1: Both factors are linear and distinct — no repeats, no quadratics. This is Case 1: one constant per factor.

$\dfrac{2s+3}{(s-1)(s+4)} = \dfrac{A}{s-1}+\dfrac{B}{s+4}$

✓ ANSWER: A — $\dfrac{A}{s-1}+\dfrac{B}{s+4}$

Examiner note: Choice D reverses the signs inside each factor — a common transcription slip when copying from the original denominator. Choice B applies a Case 2 structure to a non-repeated factor.

Item 7. Find $\mathcal{L}^{-1}\left\{\dfrac{4}{(s+2)^2}\right\}$ .

Choices: (A) $4e^{-2t}$ (B) $4te^{-2t}$ (C) $2te^{-2t}$ (D) $te^{-2t}/4$

Tool tested: Case 2 — repeated linear factor inversion

Given: $F(s) = \dfrac{4}{(s+2)^2}$

Find: $f(t) = \mathcal{L}^{-1}\{F(s)\}$

Solution:

Step 1: Recall $\mathcal{L}^{-1}\{1/(s-a)^2\} = te^{at}$ , with $a = -2$ here.

$f(t) = 4 \cdot te^{-2t} = 4te^{-2t}$

✓ ANSWER: B — $4te^{-2t}$

Examiner note: Choice A is the $te^{at}$ vs. $e^{at}$ trap covered repeatedly in Part 2 — a squared denominator always introduces a factor of $t$ , not just the bare exponential. Dropping the $t$ is the single most common repeated-factor error in the series.

Item 8. Find $\mathcal{L}^{-1}\left\{\dfrac{3s+2}{s^2+4}\right\}$ .

Choices: (A) $3\cos(2t)+2\sin(2t)$ (B) $3\cos(2t)+\sin(2t)$ (C) $3\sin(2t)+2\cos(2t)$ (D) $3\cos(2t)-\sin(2t)$

Tool tested: Case 3 — irreducible quadratic factor inversion

Given: $F(s) = \dfrac{3s+2}{s^2+4}$

Find: $f(t) = \mathcal{L}^{-1}\{F(s)\}$

Solution:

Step 1: Split the numerator to match the cosine and sine pair shapes, with $k = 2$ .

$\dfrac{3s+2}{s^2+4} = 3\left(\dfrac{s}{s^2+4}\right) + \dfrac{2}{s^2+4}$

Step 2: Since $k=2$ matches the numerator exactly, $2/(s^2+4)$ is already the sine pair — no rescaling needed.

✓ ANSWER: B — $3\cos(2t)+\sin(2t)$

Examiner note: Choice A assumes the sine numerator always needs rescaling, but here $k=2$ already matches — forcing an unnecessary $1/k$ factor gives the wrong coefficient. Always check whether the numerator already equals $k$ before rescaling.

Item 9. How many unknown constants are needed to decompose $\dfrac{5}{s(s+1)^2}$ ?

Choices: (A) $2$ (B) $3$ (C) $4$ (D) $1$

Tool tested: Mixed Case 1 + Case 2 structure recognition

Given: $F(s) = \dfrac{5}{s(s+1)^2}$

Find: Number of unknown constants in the decomposition

Solution:

Step 1: The factor $s$ is simple (Case 1) — needs one constant. The factor $(s+1)^2$ is repeated (Case 2) — needs two constants, one per power.

$\dfrac{5}{s(s+1)^2} = \dfrac{A}{s} + \dfrac{B}{s+1} + \dfrac{C}{(s+1)^2}$

✓ ANSWER: B — 3 constants

Examiner note: Choice A undercounts by writing only one term for the repeated factor $(s+1)^2$ , the exact structural error the Case 2 warnings in Part 2 call out. Every power from 1 to $n$ needs its own term.

Item 10. Given $\dfrac{5s-2}{(s-1)(s+2)} = \dfrac{A}{s-1}+\dfrac{B}{s+2}$ , find $A$ .

Choices: (A) $1$ (B) $-1$ (C) $3$ (D) $2$

Tool tested: Root substitution method

Given: Partial fraction identity as stated

Find: $A$

Solution:

Step 1: Clear denominators: $5s-2 = A(s+2)+B(s-1)$ .

Step 2: Substitute $s=1$ to zero out the $B$ term.

$5(1)-2 = A(1+2) \implies 3 = 3A \implies A = 1$

✓ ANSWER: A — $A = 1$

Examiner note: Choice C, $3$ , is the unsimplified left side before dividing by the coefficient of $A$ — a common stopping-too-early error. Always divide through to isolate the constant fully.

Item 11. Before decomposing $\dfrac{s^2+2s+3}{s^2+2s}$ , what must be done first?

Choices: (A) Apply Case 1 directly (B) Perform polynomial long division, since the degrees are equal (C) Apply Case 3 (D) Multiply both sides by $s$

Tool tested: Improper fraction recognition

Given: $F(s) = \dfrac{s^2+2s+3}{s^2+2s}$

Find: The required first step

Solution:

Step 1: Compare degrees. Numerator is degree 2, denominator is degree 2 — equal degree means the fraction is improper.

Step 2: Partial fractions only applies to proper rational functions, so long division comes first.

✓ ANSWER: B — Perform polynomial long division first

Examiner note: Choice A skips straight to decomposition, which is the exact improper-fraction trap flagged in Part 2 — the constants found without long division first will not check out when recombined.

Item 12. Which decomposition cases apply to the denominator $(s+1)(s^2+9)$ ?

Choices: (A) Two Case 1 factors (B) Case 1 and Case 2 (C) Case 1 and Case 3 (D) Case 2 and Case 3

Tool tested: Case identification

Given: Denominator $(s+1)(s^2+9)$

Find: Applicable decomposition cases

Solution:

Step 1: $(s+1)$ is a single linear factor — Case 1. $(s^2+9)$ has discriminant $0-36<0$ , no real roots — Case 3.

✓ ANSWER: C — Case 1 and Case 3

Examiner note: Choice A misreads $(s^2+9)$ as two linear factors, but it does not factor over the reals — checking the discriminant first prevents this misclassification.

Part C — Derivatives, Integrals, and Solving ODEs (Items 13 to 18)

Full lesson: Part 3 — Derivatives, Integrals, and Solving ODEs

Item 13. Given $F(s) = 2/(s+1)$ and $f(0) = 3$ , find $\mathcal{L}\{f'(t)\}$ .

Choices: (A) $\dfrac{2s}{s+1} - 3$ (B) $\dfrac{2s}{s+1} + 3$ (C) $\dfrac{2}{s+1} - 3s$ (D) $s\left(\dfrac{2}{s+1}\right) + 3$

Tool tested: First derivative transform formula

Given: $F(s) = 2/(s+1)$ , $f(0)=3$

Find: $\mathcal{L}\{f'(t)\}$

Solution:

Step 1: Apply $\mathcal{L}\{f'(t)\} = sF(s) - f(0)$ .

$= s\left(\dfrac{2}{s+1}\right) - 3 = \dfrac{2s}{s+1} - 3$

✓ ANSWER: A — $\dfrac{2s}{s+1} - 3$

Examiner note: Choice B flips the sign on $f(0)$ — the formula always subtracts the initial condition, never adds it.

Item 14. Given $y(0) = 0$ and $y'(0) = 5$ , write $\mathcal{L}\{y''(t)\}$ in terms of $Y(s)$ .

Choices: (A) $s^2Y(s) - 5$ (B) $s^2Y(s) + 5$ (C) $s^2Y(s) - 5s$ (D) $sY(s) - 5$

Tool tested: Second derivative transform formula

Given: $y(0)=0$ , $y'(0)=5$

Find: $\mathcal{L}\{y''(t)\}$

Solution:

Step 1: Apply $\mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0)$ .

$= s^2Y(s) - s(0) - 5 = s^2Y(s) - 5$

✓ ANSWER: A — $s^2Y(s) - 5$

Examiner note: Choice C keeps a stray $s$ multiplied onto $y'(0)$ — only the $y(0)$ term gets multiplied by $s$ , not $y'(0)$ .

Item 15. Solve $y' - 2y = 0$ with $y(0) = 3$ . Find $y(t)$ .

Choices: (A) $3e^{2t}$ (B) $3e^{-2t}$ (C) $2e^{3t}$ (D) $3e^{2t} - 3$

Tool tested: Four-step IVP method — first order

Given: $y'-2y=0$ , $y(0)=3$

Find: $y(t)$

Solution:

Step 1: Transform: $sY - 3 - 2Y = 0 \implies Y(s-2) = 3$ .

Step 2: Solve and invert: $Y = 3/(s-2) \implies y(t) = 3e^{2t}$ .

✓ ANSWER: A — $y(t) = 3e^{2t}$

Examiner note: Choice B uses the wrong sign on the exponent — the ODE coefficient is $-2y$ , which produces a root at $s=+2$ , not $s=-2$ .

Item 16. Solve $y'' - 5y' + 6y = 0$ with $y(0) = 1$ , $y'(0) = 0$ . Find $y(t)$ .

Choices: (A) $3e^{2t} - 2e^{3t}$ (B) $2e^{2t} - 3e^{3t}$ (C) $3e^{2t} + 2e^{3t}$ (D) $-3e^{2t} + 2e^{3t}$

Tool tested: Four-step IVP method — second order, real roots

Given: $y''-5y'+6y=0$ , $y(0)=1$ , $y'(0)=0$

Find: $y(t)$

Solution:

Step 1: Transform and substitute: $(s^2-5s+6)Y = s-5$ .

Step 2: Factor: $(s-2)(s-3)Y = s-5$ . Decompose using root substitution.

$s=2: A = (2-5)/(2-3) = 3 \qquad s=3: B = (3-5)/(3-2) = -2$

Step 3: Invert: $y(t) = 3e^{2t} - 2e^{3t}$ .

✓ ANSWER: A — $y(t) = 3e^{2t} - 2e^{3t}$

Examiner note: Choice D flips both signs — a full-decomposition sign error that recombining and checking against $y(0)=1$ would have caught: $3-2=1$ checks out, $-3+2=-1$ does not.

Item 17. Find $\mathcal{L}\{(t-3)\,u(t-3)\}$ .

Choices: (A) $\dfrac{e^{-3s}}{s^2}$ (B) $\dfrac{e^{-3s}}{s}$ (C) $\dfrac{3e^{-3s}}{s^2}$ (D) $\dfrac{e^{3s}}{s^2}$

Tool tested: Second shifting theorem

Given: $f(t) = (t-3)u(t-3)$

Find: $\mathcal{L}\{f(t)\}$

Solution:

Step 1: Match the pattern $g(t-a)u(t-a)$ with $a=3$ , $g(t)=t$ , $G(s)=1/s^2$ .

$\mathcal{L}\{f(t)\} = e^{-3s}G(s) = \dfrac{e^{-3s}}{s^2}$

✓ ANSWER: A — $\dfrac{e^{-3s}}{s^2}$

Examiner note: Choice D has the sign flipped in the exponential — the second shifting theorem always uses $e^{-as}$ for a delay, never $e^{+as}$ .

Item 18. Given $\mathcal{L}\{f(t)\} = 1/(s+4)$ , find $\mathcal{L}\left\{\displaystyle\int_0^t f(\tau)\,d\tau\right\}$ .

Choices: (A) $\dfrac{1}{s(s+4)}$ (B) $\dfrac{s}{s+4}$ (C) $\dfrac{1}{(s+4)^2}$ (D) $\dfrac{s+4}{s}$

Tool tested: Integration property

Given: $F(s) = 1/(s+4)$

Find: Transform of the running integral

Solution:

Step 1: Apply $\mathcal{L}\{\int_0^t f(\tau)d\tau\} = F(s)/s$ .

$= \dfrac{1}{s(s+4)}$

✓ ANSWER: A — $\dfrac{1}{s(s+4)}$

Examiner note: Choice B multiplies by $s$ instead of dividing — that would be the derivative property, the opposite operation from what integration does to a transform.

Part D — Impulse, Periodic Functions, and Convolution (Items 19 to 23)

Full lesson: Part 4 — Impulse, Periodic Functions, and Convolution

Item 19. Find $\mathcal{L}\{\delta(t-5)\}$ .

Choices: (A) $e^{-5s}$ (B) $e^{5s}$ (C) $5$ (D) $1/5$

Tool tested: Dirac delta transform

Given: $f(t) = \delta(t-5)$

Find: $\mathcal{L}\{f(t)\}$

Solution:

Step 1: Apply $\mathcal{L}\{\delta(t-a)\} = e^{-as}$ with $a=5$ .

✓ ANSWER: A — $e^{-5s}$

Examiner note: Choice C confuses this with a scaled impulse $5\delta(t)$ , which transforms to the constant $5$ — a different function from $\delta(t-5)$ .

Item 20. Find $\mathcal{L}\{3\delta(t)\}$ .

Choices: (A) $3$ (B) $3/s$ (C) $3s$ (D) $e^{-3s}$

Tool tested: Dirac delta at the origin, scaled

Given: $f(t) = 3\delta(t)$

Find: $\mathcal{L}\{f(t)\}$

Solution:

Step 1: $\mathcal{L}\{\delta(t)\}=1$ , scaled by linearity gives $3 \cdot 1 = 3$ .

✓ ANSWER: A — $3$

Examiner note: Choice B is the $\delta(t)$ vs. $u(t)$ confusion flagged throughout Part 4 — $\mathcal{L}\{u(t)\}=1/s$ , but $\mathcal{L}\{\delta(t)\}=1$ . These are two different functions with two different transforms.

Item 21. Given $F(s) = 1/(s+1)$ and $G(s) = 1/(s+3)$ , find $\mathcal{L}\{f*g\}$ .

Choices: (A) $\dfrac{1}{(s+1)(s+3)}$ (B) $\dfrac{1}{s+1}+\dfrac{1}{s+3}$ (C) $(s+1)(s+3)$ (D) $\dfrac{1}{s+4}$

Tool tested: Convolution theorem, forward direction

Given: $F(s)=1/(s+1)$ , $G(s)=1/(s+3)$

Find: $\mathcal{L}\{f*g\}$

Solution:

Step 1: The convolution theorem replaces the integral with direct multiplication.

$\mathcal{L}\{f*g\} = F(s)\cdot G(s) = \dfrac{1}{(s+1)(s+3)}$

✓ ANSWER: A — $\dfrac{1}{(s+1)(s+3)}$

Examiner note: Choice B adds the transforms instead of multiplying — that would be the transform of $f(t)+g(t)$ , not the convolution $f*g$ .

Item 22. A periodic function has period $T = 3$ . What are the integration limits in the periodic transform formula?

Choices: (A) $0$ to $\infty$ (B) $0$ to $3$ (C) $-3$ to $3$ (D) $3$ to $\infty$

Tool tested: Periodic function formula setup

Given: $T = 3$

Find: Correct integration limits

Solution:

Step 1: The periodic formula only ever integrates across one period, $0$ to $T$ .

✓ ANSWER: B — $0$

to $3$

Examiner note: Choice A is the trap of trying to integrate the entire infinite waveform directly — the $1/(1-e^{-sT})$ factor in the formula already accounts for the repetition, so only one period is ever needed.

Item 23. Express $\mathcal{L}^{-1}\left\{\dfrac{1}{(s+1)(s+2)}\right\}$ as a convolution integral.

Choices: (A) $\displaystyle\int_0^t e^{-\tau}e^{-2(t-\tau)}\,d\tau$ (B) $\displaystyle\int_0^t e^{-\tau}e^{2(t-\tau)}\,d\tau$ (C) $e^{-t} \cdot e^{-2t}$ , no integral (D) $\displaystyle\int_0^{\infty} e^{-\tau}e^{-2(t-\tau)}\,d\tau$

Tool tested: Convolution theorem, reverse direction

Given: $F(s)G(s) = 1/[(s+1)(s+2)]$

Find: Convolution integral form

Solution:

Step 1: Identify $f_1(t)=e^{-t}$ , $f_2(t)=e^{-2t}$ from each factor.

Step 2: For a causal system, the convolution limits run from $0$ to $t$ .

✓ ANSWER: A — $\displaystyle\int_0^t e^{-\tau}e^{-2(t-\tau)}\,d\tau$

Examiner note: Choice C drops the integral entirely — the inverse of a product is a convolution integral, not a bare product of the two time-domain functions. Choice D uses the wrong (non-causal) limits.

Part E — Engineering Applications (Items 24 to 30)

Full lesson: Part 5 — Engineering Applications

Item 24. A series RLC circuit has $R = 5\,\Omega$ , $L = 1\,\text{H}$ , $C = 0.5\,\text{F}$ . Find $Z(s)$ .

Choices: (A) $\dfrac{s^2+5s+2}{s}$ (B) $\dfrac{s^2+5s+2}{2s}$ (C) $\dfrac{s^2+5s+2}{5s}$ (D) $\dfrac{2s^2+5s+1}{s}$

Given: $R=5$ , $L=1$ , $C=0.5$

Find: $Z(s)$

Solution:

Step 1: $Z(s) = Ls+R+1/(Cs) = s+5+2/s$ .

Step 2: Combine over a common denominator.

$Z(s) = \dfrac{s^2+5s+2}{s}$

✓ ANSWER: A — $\dfrac{s^2+5s+2}{s}$

Verification: At $s=1$ (arbitrary check): $Z(1) = 1+5+2=8$ , matches $(1+5+2)/1=8$ from the combined form. ✓

Examiner note: Choice B divides by 2 unnecessarily — a leftover from misreading $1/C=2$ as a scaling factor on the whole expression rather than just the capacitor term.

Item 25. For a transfer function $G(s) = Y(s)/X(s)$ , what condition must hold?

Choices: (A) Zero initial conditions (B) Nonzero initial conditions (C) $X(s) = 1$ (D) $Y(s) = 0$

Given: Definition of $G(s)$

Find: Required condition

Solution:

Step 1: The transfer function definition assumes the system starts entirely at rest.

✓ ANSWER: A — Zero initial conditions

Examiner note: Choice B is a direct contradiction of the definition — if initial conditions are nonzero, $G(s)=Y(s)/X(s)$ no longer describes the system’s response alone; the full IVP method from Part 3 is required instead.

Item 26. A spring-mass-damper system has $m = 1$ , $c = 6$ , $k = 9$ . Classify the damping.

Choices: (A) Underdamped (B) Critically damped (C) Overdamped (D) Undamped

Given: $m=1$ , $c=6$ , $k=9$

Find: Damping classification

Solution:

Step 1: Check the discriminant $c^2-4mk = 36-36=0$ .

✓ ANSWER: B — Critically damped

Verification: $(s+3)^2 = s^2+6s+9$ matches the characteristic polynomial exactly — repeated root at $s=-3$ . ✓

Examiner note: Choice A is the trap for anyone who does not compute the discriminant and instead guesses based on nonzero damping being present — always compute $c^2-4mk$ before classifying, never assume from the presence of a damping term alone.

Item 27. A series RL circuit has $R = 2\,\Omega$ , $L = 1\,\text{H}$ , a step voltage of $6\,\text{V}$ , $i(0)=0$ . Find the steady-state current.

Choices: (A) $2\,\text{A}$ (B) $3\,\text{A}$ (C) $6\,\text{A}$ (D) $1\,\text{A}$

Given: $R=2$ , $V=6$

Find: $i_{ss}$

Solution:

Step 1: For a step input, the steady-state current is $V/R$ .

$i_{ss} = 6/2 = 3\,\text{A}$

✓ ANSWER: B — $3\,\text{A}$

Verification: This matches setting $s\to 0$ in $I(s) = V/[s(R+Ls)]$ : $\lim_{s\to0} sI(s) = V/R = 3$ (final value theorem). ✓

Examiner note: Choice C forgets to divide by $R$ at all, mistaking the applied voltage itself for the steady-state current.

Item 28. A series RLC circuit has $L=1$ , $R=4$ , $C=0.25$ , with an $8\,\text{V}$ step applied and $q(0)=0$ , $q'(0)=0$ . Find the characteristic equation in $q$ .

Choices: (A) $s^2+4s+4=0$ (B) $s^2+4s+8=0$ (C) $s^2+2s+4=0$ (D) $s^2+4s+2=0$

Given: $L=1$ , $R=4$ , $C=0.25$

Find: Characteristic equation

Solution:

Step 1: KVL: $L\ddot{q}+R\dot{q}+(1/C)q=v(t) \implies \ddot{q}+4\dot{q}+4q=8$ .

Step 2: The characteristic equation comes from the left side only.

$s^2+4s+4=0$

✓ ANSWER: A — $s^2+4s+4=0$

Verification: $1/C = 1/0.25 = 4$ , confirmed. This is the same critically damped structure as Item 26 — $(s+2)^2$ . ✓

Examiner note: Choice B mistakenly folds the right-hand side forcing term (8) into the characteristic equation — the characteristic equation only comes from the homogeneous (left) side of the ODE.

Item 29. A control system has $G(s) = 2/(s+4)$ . Find $Y(s)$ for a unit step input.

Choices: (A) $\dfrac{2}{s(s+4)}$ (B) $\dfrac{2}{s+4}$ (C) $\dfrac{2s}{s+4}$ (D) $\dfrac{1}{s(s+4)}$

Given: $G(s)=2/(s+4)$ , $x(t)=u(t)$

Find: $Y(s)$

Solution:

Step 1: $X(s) = 1/s$ . Multiply: $Y(s)=G(s)X(s)$ .

$Y(s) = \dfrac{2}{s+4}\cdot\dfrac{1}{s} = \dfrac{2}{s(s+4)}$

✓ ANSWER: A — $\dfrac{2}{s(s+4)}$

Verification: Steady state check via final value theorem: $\lim_{s\to0}sY(s) = 2/4 = 0.5$ , consistent with $G(0)=2/4$ times a unit step. ✓

Examiner note: Choice D drops the gain of 2 from $G(s)$ entirely — always carry the numerator of $G(s)$ through the multiplication, not just its denominator structure.

Item 30. For $m\ddot{x}+c\dot{x}+kx=F(t)$ with zero initial conditions, what is the correct s-domain equation?

Choices: (A) $(ms^2+cs+k)X(s)=F(s)$ (B) $(ms+c+k)X(s)=F(s)$ (C) $(ms^2+cs+k)X(s)=F(s)+mx(0)$ (D) $msX(s)+cX(s)+kX(s)=F(s)$

Given: $m\ddot{x}+c\dot{x}+kx=F(t)$ , zero initial conditions

Find: Correct transformed equation

Solution:

Step 1: Transform each term with zero initial conditions, so all $-sx(0)$ and $-x'(0)$ terms vanish.

$ms^2X(s)+csX(s)+kX(s)=F(s) \implies (ms^2+cs+k)X(s)=F(s)$

✓ ANSWER: A — $(ms^2+cs+k)X(s)=F(s)$

Verification: Dividing both sides by $F(s)$ recovers the transfer function form from Item 25: $X(s)/F(s)=1/(ms^2+cs+k)$ . ✓

Examiner note: Choice C reintroduces an initial condition term despite the problem explicitly stating zero initial conditions — read the given conditions carefully before writing the transformed equation.

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What to Do With Your Score

Go back to the Series Index Page and check your raw score against the interpretation table there. If you missed three or more items in any single part, that is the part that needs another pass — not this solutions page, but the full post with all 10 worked problems. Work through those problems without looking at the solutions first. That is the only kind of practice that actually moves your score on exam day.

Pay closest attention to misses in Part B and Part E. Part B — partial fractions — carries the most items on this exam because it is the highest point-loss skill in the entire topic, and every later part depends on getting it right. Part E — engineering applications — carries the second-most items because it is where every other part gets combined and tested together; a miss there often traces back to a gap in Part 1, 2, or 3 rather than the applications post itself.

If you are reviewing for EE or ECE, this entire series is high-priority — RLC circuits and transfer functions appear on nearly every exam cycle. If you are reviewing for ME, prioritize Part 3 and Part 5 specifically, since vibration analysis and damping classification are the items most likely to appear on your board.

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Part A — Definition, Basic Transforms and Properties (Items 1 to 5)

Part B — Inverse Laplace Transform and Partial Fractions (Items 6 to 12)

Part C — Derivatives, Integrals, and Solving ODEs (Items 13 to 18)

Part D — Impulse, Periodic Functions, and Convolution (Items 19 to 23)

Part E — Engineering Applications (Items 24 to 30)

What to Do With Your Score

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