Questions: Laplace Transform Properties and Inverse Transform

Question 1

Applying the Laplace differentiation property to the ODE d²y/dt² + 3(dy/dt) + 2y = x(t) with zero initial conditions converts it into:

• A. A first-order differential equation in the complex variable s that can be reduced further
• B. An algebraic equation: (s² + 3s + 2)Y(s) = X(s), solvable with ordinary algebra
• C. A Fourier series representation relating the frequency content of input and output
• D. A convolution integral: y(t) = ∫₀ᵗ h(τ)x(t−τ)dτ in the time domain

Answer: (B)

The differentiation property L{f'(t)} = sF(s) replaces each time-derivative with multiplication by s. The second-order ODE becomes s²Y(s) + 3sY(s) + 2Y(s) = X(s), which factors as (s² + 3s + 2)Y(s) = X(s) — a simple algebraic equation solvable for Y(s) = X(s)/(s² + 3s + 2). This is the entire point of the Laplace transform: differential equations become polynomial equations. The convolution integral (option D) is the time-domain form of the same relationship, but it's what you want to avoid computing.

Question 2

A system has poles at s = −2 ± 3j. What does this tell you about the system's impulse response?

• A. The response decays to zero at a rate determined by the imaginary part (3 rad/s) while oscillating at a rate determined by the real part
• B. The response grows exponentially because the pole values are complex rather than purely real
• C. The response is a damped sinusoid: it oscillates at 3 rad/s while the amplitude decays with time constant 1/2 second
• D. The response is a pure undamped sinusoid, since the poles are not on the real axis

Answer: (C)

Complex conjugate poles at s = −σ ± jω produce a damped sinusoidal time-domain response of the form e^{−σt}cos(ωt + φ). Here σ = 2 (the negative real part) controls exponential decay — the time constant is 1/σ = 0.5 s — and ω = 3 rad/s controls oscillation frequency. The real part determines stability and decay speed; the imaginary part determines the oscillation frequency. Option A reverses these roles, the most common confusion. Poles with σ > 0 (negative real part) always decay; poles in the right half-plane (σ < 0, positive real part) would grow.

Question 3

A system whose transfer function has all poles located in the left half of the complex s-plane produces impulse responses that decay to zero over time — indicating a stable system.

• T. True
• F. False

Answer: True

Each pole in the left half-plane (negative real part) contributes a decaying exponential or damped sinusoid to the impulse response. Since all such terms vanish as t→∞, the response decays to zero — the definition of bounded-input bounded-output stability. A pole at s = 0 gives a constant (marginally stable); poles in the right half-plane give growing exponentials (unstable). This is why pole-zero plots are the primary stability analysis tool: stability is immediately visible from whether all poles lie left of the imaginary axis.

Question 4

Computing the inverse Laplace transform generally requires evaluating the complex contour integral definition directly.

• T. True
• F. False

Answer: False

The contour integral (Bromwich integral) is the formal definition of the inverse transform, but in practice it is almost never computed directly — it requires residue calculus and is reserved for cases not covered by tables. The practical method is partial fraction decomposition: factor the denominator of F(s), expand into a sum of simple first-order (and second-order for complex poles) fractions, then read off the inverse transform of each term from a table. Each term A/(s−p) inverts to Ae^{pt}u(t). This is why transform tables and partial fractions are the core computational skills, not contour integration.

Question 5

Why does the differentiation property L{f'(t)} = sF(s) − f(0) make the Laplace transform so useful for solving linear ordinary differential equations?

Model answer: Differentiation in the time domain becomes multiplication by s in the frequency domain — an algebraic operation. A linear ODE with terms like y'', y', y becomes a polynomial equation in s: (a₂s² + a₁s + a₀)Y(s) = X(s) + initial condition terms. Solving for Y(s) requires only algebra (dividing polynomials), whereas solving the ODE directly requires finding a particular solution and applying boundary conditions through integration. Once Y(s) is found, partial fraction decomposition and a table recover y(t). The transform converts a calculus problem into an algebra problem, then converts back.

Initial conditions are handled automatically — the f(0) term in the differentiation property incorporates them into the algebraic equation rather than requiring them to be applied after finding the general solution. This is especially useful for systems where initial conditions are the primary driver of the response.