Short notes

Laplace Transform Short Notes

When you analyse the step response of an RC circuit with R = 10 kΩ and C = 10 µF using the Laplace transform, the output voltage becomes V_out(s) = (1/RC) / (s(s + 1/RC)) = 1/(s(s + 10)) in the s-domain — inverse transforming gives v(t) = (1 − e^(−t/0.1))·u(t), and you immediately read off the time constant τ = RC = 0.1 s. That circuit-analysis shortcut is why Laplace is taught before convolution in most university courses.

EEE, ECE, EI

How it works

The unilateral Laplace transform is X(s) = ∫₀^∞ x(t)e^(−st)dt where s = σ + jω. The region of convergence (ROC) is the set of s values for which the integral converges — for a right-sided signal e^(−at)u(t), the ROC is Re(s) > −a. Key pairs: δ(t) ↔ 1 (ROC: entire s-plane), u(t) ↔ 1/s (ROC: Re(s) > 0), e^(−at)u(t) ↔ 1/(s+a) (ROC: Re(s) > −a), tⁿe^(−at)u(t) ↔ n!/(s+a)^(n+1). The convolution property L{x*h} = X(s)H(s) makes transfer function analysis straightforward — poles of H(s) determine natural frequencies of the system.

Key points to remember

The ROC must be specified with every Laplace transform; different ROCs for the same X(s) correspond to different time-domain signals. For a causal stable system, all poles of H(s) must lie in the left half of the s-plane (Re(s) < 0). A repeated pole at s = −a gives time-domain terms of the form tⁿe^(−at). Partial fraction expansion is mandatory when inverting a rational X(s): factorise the denominator, assign A/(s+p₁) + B/(s+p₂) terms, multiply through and equate coefficients. Initial value theorem: x(0⁺) = lim(s→∞) sX(s); final value theorem: x(∞) = lim(s→0) sX(s), valid only when the limit exists and the system is stable.

Exam tip

The examiner always asks you to state and apply the initial and final value theorems for a given X(s) — always check the final value theorem's stability condition first, because applying it to an unstable X(s) is a guaranteed mark-losing error.

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