Inverse Laplace Transform

The inverse Laplace transform recovers a time-domain signal from its s-domain representation, completing the round trip that makes Laplace analysis useful. Circuit designers use it to find the exact voltage or current waveform after solving an algebraic s-domain equation.

Core Concept

Computing the inverse Laplace transform directly from the Bromwich contour integral is rarely done by hand. Instead, the standard approach is partial fraction expansion (PFE). You decompose X(s) into a sum of simple fractions, each of which matches a known transform pair in a table.

The factored denominator of X(s) gives you the poles. Each distinct real pole at s = -a contributes a term A/(s+a) in the expansion, which maps to Ae^(-at)u(t) in time. A pair of complex conjugate poles at s = -sigma +/- j*omega contributes damped sinusoidal terms in time.

If X(s) is improper (numerator degree greater than or equal to denominator degree), perform polynomial long division first. The quotient gives delta functions and their derivatives in the time domain. The remainder over the denominator is proper and can then be expanded by partial fractions.

Key Formula

For simple poles, the residue at pole s = p_k is found by the cover-up method: A_k = (s - p_k)*X(s) evaluated at s = p_k. For a pair of complex poles s^2 + 2*zeta*omega_n*s + omega_n^2 in the denominator, the inverse is a damped sinusoid of the form (A/omega_d)*e^(-sigma*t)*sin(omega_d*t)u(t) where omega_d = omega_n*sqrt(1-zeta^2). The inverse Laplace of delta(t) is 1, of u(t) is 1/s, of e^(-at)u(t) is 1/(s+a), and of t*e^(-at)u(t) is 1/(s+a)^2.

Given: X(s) = (s + 3) / ((s+1)(s+2))
Find x(t).

Step 1: Partial fraction form
  X(s) = A/(s+1) + B/(s+2)

Step 2: Cover-up for A (multiply both sides by (s+1), set s=-1)
  A = (s+3)/(s+2) at s=-1
    = (-1+3)/(-1+2) = 2/1 = 2

Step 3: Cover-up for B (multiply both sides by (s+2), set s=-2)
  B = (s+3)/(s+1) at s=-2
    = (-2+3)/(-2+1) = 1/(-1) = -1

Step 4: X(s) = 2/(s+1) - 1/(s+2)

Step 5: Lookup table
  x(t) = 2*e^(-t)*u(t) - e^(-2t)*u(t)

Final Answer: x(t) = (2e^(-t) - e^(-2t))u(t)

Exam Tip: Always check whether X(s) is proper before starting partial fractions. If degree of numerator is equal to or greater than degree of denominator, divide first. For complex pole pairs, write the numerator of the combined fraction as (As + B) and solve for A and B by comparing coefficients, since the cover-up method does not directly apply to complex poles in real-coefficient form.

Properties Summary

• Inverse Laplace is defined as the Bromwich contour integral; computed in practice via PFE
• Cover-up method: residue A_k = (s-p_k)*X(s) at s=p_k for simple poles
• Improper X(s): perform long division first; quotient maps to impulse terms
• Real distinct poles -> sum of decaying exponentials in time
• Complex conjugate poles -> damped sinusoidal terms; always come in conjugate pairs for real signals
• Repeated poles require differentiation of residue formula; covered separately for completeness
• ROC determines whether the inverse is causal (right-sided) or anti-causal (left-sided)

Quick Revision

• Standard method: partial fraction expansion then lookup table
• Check proper vs improper before starting; do long division if improper
• Cover-up method gives residues for simple real poles quickly
• Complex poles produce damped sine and cosine pairs in time domain
• ROC: right half-plane means causal signal; left half-plane means anti-causal
• Pole at s=0 gives u(t), pole at s=-a gives e^(-at)u(t)
• Repeated poles require higher-order partial fraction terms 1/(s+a)^k
• Exam trap: using cover-up method for repeated poles; that method only works for simple (non-repeated) poles and gives wrong residues otherwise