Questions: Laplace Transform: Fundamentals and Properties

Question 1

A system has a pole at s = -3 + 2j in the s-plane. What does this pole tell you about the system's time-domain behavior?

• A. The system oscillates at 3 rad/s with amplitude growing at rate e^(2t)
• B. The system oscillates at 2 rad/s with amplitude decaying at rate e^(-3t)
• C. The system grows exponentially at rate e^(-3t) with no oscillation
• D. The system is unstable because the pole has a nonzero imaginary part

Answer: (B)

A pole at s = σ + jω corresponds to the time-domain mode e^(σt)·cos(ωt). Here σ = -3 and ω = 2, so the mode is e^(-3t)·cos(2t) — a sinusoidal oscillation at 2 rad/s whose amplitude decays exponentially (since σ < 0, the system is stable). The imaginary part sets the oscillation frequency; the real part determines growth or decay. A common misconception is that any complex pole means instability — it is the sign of the real part that matters.

Question 2

The Fourier transform integral fails to converge for the signal x(t) = e^(2t)u(t). How does the Laplace transform handle this signal?

• A. The Laplace transform also fails — no transform can handle exponentially growing signals
• B. The Laplace transform multiplies by e^(-σt) before integrating; choosing σ > 2 creates a net decaying envelope, making the integral converge
• C. The Laplace transform automatically sets σ = 0, recovering the Fourier transform result
• D. The Laplace transform uses integration from -∞ to ∞ rather than 0 to ∞, which makes the integral converge

Answer: (B)

The Laplace transform X(s) = ∫₀^∞ x(t)e^(-st)dt with s = σ + jω multiplies x(t) by e^(-σt) before integrating. For x(t) = e^(2t), the product is e^(2t)·e^(-σt) = e^((2-σ)t). When σ > 2, the exponent (2-σ) is negative, ensuring the integrand decays and the integral converges. The region of convergence (ROC) for this signal is Re(s) > 2. The Fourier transform uses σ = 0 only — it cannot make this adjustment, which is why it fails for growing signals.

Question 3

A system is stable if and only if all poles of its transfer function lie in the left half of the complex s-plane (i.e., have negative real parts).

• T. True
• F. False

Answer: True

A pole at s = σ₀ + jω₀ corresponds to the time-domain mode e^(σ₀t)·cos(ω₀t). If σ₀ < 0 (left half-plane), the mode decays to zero — stable. If σ₀ > 0 (right half-plane), the mode grows without bound — unstable. Poles on the imaginary axis (σ₀ = 0) produce sustained oscillations (marginally stable). The geometry of poles in the s-plane is the central tool for stability analysis in control systems.

Question 4

The Fourier transform is a generalization of the Laplace transform — the Laplace transform is a special case obtained by restricting to real frequencies.

• T. True
• F. False

Answer: False

This has the relationship backwards. The Laplace transform X(s) = ∫₀^∞ x(t)e^(-st)dt uses complex frequency s = σ + jω. The Fourier transform is the special case where σ = 0 — i.e., evaluating the Laplace transform along the imaginary axis. The Laplace transform is the generalization; the Fourier transform is a restricted slice of it. This matters because it means the Laplace transform can analyze signals and systems for which the Fourier transform fails (those that don't decay fast enough).

Question 5

Explain why the property 'differentiation in time becomes multiplication by s in the Laplace domain' is useful for analyzing circuits and mechanical systems.

Model answer: Circuits and mechanical systems are naturally described by differential equations (e.g., V = L·dI/dt for an inductor, F = m·d²x/dt² for a mass). Taking the Laplace transform converts these into algebraic equations: sI(s), s²X(s), etc. An equation that required solving an ODE becomes a polynomial equation in s, solvable by algebra and partial fractions. Initial conditions appear as explicit additive terms (e.g., sX(s) - x(0⁻)), so non-zero initial states are handled automatically without special techniques.

The power is that algebraic manipulation is far easier than differential equation manipulation. Once a system is represented as a rational function of s — a transfer function — questions about stability, frequency response, and transient behavior all reduce to analysis of that polynomial ratio. The Laplace transform doesn't change the physics; it changes the mathematical language to one that is far easier to work with.