Inverse LT of Repeated Poles

Repeated poles in the Laplace transform arise when a system has multiple identical time constants, such as a critically damped second-order circuit. Finding the inverse Laplace transform in this case requires a modified partial fraction technique that produces polynomial-times-exponential terms in the time domain.

Core Concept

When a pole of X(s) is repeated r times, the denominator has a factor (s+a)^r. A simple partial fraction with a single residue is not enough. Instead, you need r terms: one for each power from 1 up to r. This is because the repeated pole contributes r linearly independent time functions.

For a double pole at s = -a, the expansion has terms A1/(s+a) and A2/(s+a)^2. In the time domain, 1/(s+a) maps to e^(-at)u(t) and 1/(s+a)^2 maps to t*e^(-at)u(t). The t*e^(-at) function starts at zero, rises to a peak at t=1/a, then decays to zero. This waveform shape appears in critically damped LC circuits.

The residue for the highest-power term A_r is found by the standard cover-up method. Residues for lower-power terms require differentiation. Specifically, A_{r-k} is found by differentiating (s+a)^r * X(s) k times with respect to s and evaluating at s=-a, then dividing by k factorial.

Key Formula

For a pole of order r at s = -a, the partial fraction expansion contains: sum from k=1 to r of A_k/(s+a)^k. The residue formula is A_k = (1/(r-k)!) * (d^(r-k)/ds^(r-k))[(s+a)^r * X(s)] evaluated at s = -a. In the time domain, L^{-1}{1/(s+a)^k} = t^(k-1)*e^(-at)/(k-1)! * u(t). For a triple pole, k goes from 1 to 3 and you get e^(-at), t*e^(-at), and (t^2/2)*e^(-at) terms.

Given: X(s) = (s + 5) / ((s+1)^2 * (s+3))
Find x(t).

Step 1: Write partial fraction form
  X(s) = A1/(s+1) + A2/(s+1)^2 + B/(s+3)

Step 2: Find A2 (highest power of repeated pole, cover-up at s=-1)
  A2 = (s+5)/(s+3) at s=-1
     = (-1+5)/(-1+3) = 4/2 = 2

Step 3: Find B (simple pole at s=-3, cover-up)
  B = (s+5)/(s+1)^2 at s=-3
    = (-3+5)/(-3+1)^2 = 2/4 = 1/2

Step 4: Find A1 (differentiate [(s+1)^2 * X(s)] w.r.t. s, evaluate at s=-1)
  (s+1)^2 * X(s) = (s+5)/(s+3)
  d/ds[(s+5)/(s+3)] = [(s+3)-(s+5)]/(s+3)^2 = -2/(s+3)^2
  At s=-1: -2/(-1+3)^2 = -2/4 = -1/2
  So A1 = -1/2

Step 5: Combine
  X(s) = (-1/2)/(s+1) + 2/(s+1)^2 + (1/2)/(s+3)

Step 6: Inverse Laplace
  x(t) = (-1/2)e^(-t)*u(t) + 2t*e^(-t)*u(t) + (1/2)e^(-3t)*u(t)

Final Answer: x(t) = [(-0.5 + 2t)e^(-t) + 0.5*e^(-3t)]u(t)

Exam Tip: The most common error with repeated poles is applying the cover-up method to find A1 for a double pole. Cover-up gives the residue of the highest-power term only. For all lower-power terms you must differentiate. In a 2-mark GATE question, always write out the full expansion form first before computing residues, to avoid missing terms.

Properties Summary

• Repeated pole of order r at s=-a requires r partial fraction terms: A1/(s+a) through Ar/(s+a)^r
• Highest-order residue Ar: use cover-up, multiply by (s+a)^r and set s=-a
• Lower residues: differentiate (s+a)^r*X(s) with respect to s, divide by factorial, evaluate at s=-a
• Time pair: L^{-1}{1/(s+a)^k} = t^(k-1)*e^(-at)/(k-1)! for causal signals
• Double pole produces t*e^(-at) which peaks at t=1/a then decays; associated with critical damping
• Triple pole produces t^2*e^(-at)/2; appears in higher-order critically damped systems

Quick Revision

• Repeated poles require one partial fraction term per order of repetition
• Cover-up method works only for the highest-power repeated term
• Lower residues require successive differentiation of the cleared expression
• 1/(s+a)^2 maps to t*e^(-at)u(t) in time domain
• 1/(s+a)^3 maps to (t^2/2)*e^(-at)u(t)
• Critically damped second-order system always has a double pole
• Always check the denominator for repeated factors before starting expansion
• Exam trap: writing only one term A/(s+a)^2 for a double pole and ignoring the A1/(s+a) term; the expansion is incomplete and the recovered time function is wrong