LT of Standard Signals

The Laplace transform converts standard time-domain signals into algebraic expressions in the complex frequency variable s. Memorizing the transforms of unit impulse, unit step, ramp, exponential, and sinusoidal signals is the foundation for solving any circuit or control system problem.

Core Concept

The unilateral Laplace transform is defined for causal signals as X(s) = ∫₀^∞ x(t)e^(-st)dt. For standard signals that start at t = 0, this integral has closed-form results. Each result is a rational function of s — a ratio of polynomials whose poles directly encode the signal's decay rates and oscillation frequencies.

The unit impulse δ(t) transforms to 1 because the sifting property of the impulse picks out e^(-s·0) = 1. This makes the impulse the identity element for convolution and means a system's transfer function H(s) is literally the Laplace transform of its impulse response.

The damped sinusoid e^(-at)sin(ωt)u(t) has poles at s = -a±jω. These are complex conjugate poles in the left half-plane. Plotting them tells you the decay rate a and oscillation frequency ω immediately, which is why the pole-zero plot is so useful for understanding transient behavior.

Key Formula

Core pairs: δ(t) ↔ 1; u(t) ↔ 1/s; t^n·u(t) ↔ n!/s^(n+1); e^(-at)u(t) ↔ 1/(s+a); sin(ωt)u(t) ↔ ω/(s²+ω²); cos(ωt)u(t) ↔ s/(s²+ω²); e^(-at)sin(ωt)u(t) ↔ ω/((s+a)²+ω²); e^(-at)cos(ωt)u(t) ↔ (s+a)/((s+a)²+ω²). All unilateral transforms have ROC Re(s) > -a, where -a is the real part of the rightmost pole.

Problem: Find the Laplace transform of x(t) = (3e^(-2t) - 2e^(-3t))u(t).

Step 1 — Split using linearity:
  X(s) = L{3e^(-2t)u(t)} - L{2e^(-3t)u(t)}

Step 2 — Apply standard pair e^(-at)u(t) ↔ 1/(s+a):
  L{3e^(-2t)u(t)} = 3/(s+2)
  L{2e^(-3t)u(t)} = 2/(s+3)

Step 3 — Combine:
  X(s) = 3/(s+2) - 2/(s+3)
       = [3(s+3) - 2(s+2)] / [(s+2)(s+3)]
       = [3s+9-2s-4] / [(s+2)(s+3)]
       = (s+5) / [(s+2)(s+3)]

Step 4 — Identify poles and zeros:
  Zero at s = -5
  Poles at s = -2 and s = -3
  ROC: Re(s) > -2  (rightmost pole)

Final Answer:
  X(s) = (s+5)/[(s+2)(s+3)],  Re(s) > -2

Exam Tip: For the ramp signal t^n u(t), the transform is n!/s^(n+1). For n=1 (ramp), it is 1/s². For n=0 (step), 0!/s^1 = 1/s — the formula covers both. When you encounter a damped sinusoid, complete the square in the denominator: s²+2s+5 = (s+1)²+4 = (s+1)²+2². Then match ω=2, a=1 to the standard pair directly.

Properties Summary

• δ(t) ↔ 1 — impulse transforms to a constant, making it the convolution identity.
• u(t) ↔ 1/s — step has a pole at origin; ROC is Re(s) > 0.
• t^n u(t) ↔ n!/s^(n+1) — each integration in time adds one factor of 1/s.
• e^(-at)u(t) ↔ 1/(s+a) — pole at s=-a; decay rate a sets pole location.
• sin/cos pairs: ω/(s²+ω²) and s/(s²+ω²) — poles at ±jω on imaginary axis.
• Damped pairs: shift s→s+a to get e^(-at)-multiplied versions.
• ROC for all causal signals: Re(s) > real part of rightmost pole.

Quick Revision

• δ(t) → 1; u(t) → 1/s; ramp → 1/s²; t^n → n!/s^(n+1).
• Exponential e^(-at) → pole at s = -a.
• Sinusoids → poles on imaginary axis at ±jω.
• Damped sinusoids → complex poles at -a±jω.
• Complete the square to match damped sinusoid form.
• Linearity splits sums; s-shift handles exponential multipliers.
• All unilateral (causal) transforms: ROC is Re(s) > rightmost pole real part.
• Exam trap: writing the transform of cos(ωt)u(t) as ω/(s²+ω²) instead of s/(s²+ω²) — cosine gives the s in the numerator, not ω.