System Analysis Using Laplace

System analysis using the Laplace transform converts a differential equation describing a circuit or control loop into an algebraic equation in s, solves it, then recovers the time-domain response by inverse Laplace. This approach handles arbitrary initial conditions and any input shape in a single unified procedure.

Core Concept

When you take the Laplace transform of a differential equation with non-zero initial conditions, the initial conditions appear as additional source terms in the algebraic equation. The total response Y(s) then has two parts: one driven by the initial conditions and one driven by the input X(s).

The zero-input response (ZIR) is what the system produces from its stored energy alone, with no external input. The zero-state response (ZSR) is what the system produces from the input alone, starting from rest. The ZSR is always H(s)*X(s). The total response is their sum, and linearity guarantees that superposition applies.

After solving for Y(s) as a rational function, the time-domain output y(t) is found by partial fraction expansion and inverse Laplace. The denominator poles of H(s) determine the natural frequency terms. The poles of X(s) determine the forced response terms. If input poles match system poles, resonance or repeated poles occur and the expansion needs special treatment.

Key Formula

For a second-order system a*y' + b*y' + c*y = x(t), the Laplace transform gives Y(s) * [as^2 + bs + c] = X(s) + [a*s*y(0-) + a*y'(0-) + b*y(0-)]. The bracketed term on the right is the initial condition contribution. Rearranging: Y(s) = X(s)/[as^2+bs+c] + [IC terms]/[as^2+bs+c]. The first term is the ZSR = H(s)*X(s). The second is the ZIR. Both share the same denominator because the system's natural frequencies appear in both.

Given: y' + 3y' + 2y = u(t),  y(0-) = 1,  y'(0-) = 0
Find y(t) for t >= 0.

Step 1: Laplace of ODE
  [s^2*Y - s*y(0-) - y'(0-)] + 3[s*Y - y(0-)] + 2Y = 1/s
  [s^2 + 3s + 2]*Y - s*1 - 0 - 3*1 = 1/s
  [s^2 + 3s + 2]*Y = 1/s + s + 3

Step 2: Solve for Y(s)
  Y(s) = 1/[s(s^2+3s+2)] + (s+3)/(s^2+3s+2)
  Denominator factors: (s+1)(s+2)
  Y(s) = 1/[s(s+1)(s+2)] + (s+3)/[(s+1)(s+2)]

Step 3: Partial fractions for first term
  1/[s(s+1)(s+2)] = 1/(2s) - 1/(s+1) + 1/(2(s+2))

Step 4: Partial fractions for second term
  (s+3)/[(s+1)(s+2)] = 2/(s+1) - 1/(s+2)

Step 5: Add and collect
  Y(s) = 1/(2s) + (2-1)/(s+1) + (-1+1/2)/(s+2)
       = 1/(2s) + 1/(s+1) - 1/(2(s+2))

Step 6: Inverse Laplace
  y(t) = [0.5 + e^(-t) - 0.5*e^(-2t)] u(t)

Final Answer: y(t) = (0.5 + e^(-t) - 0.5*e^(-2t))u(t)

Exam Tip: When applying Laplace to a differential equation with initial conditions, include every initial condition term that comes from differentiating. For y', the transform is s^2*Y - s*y(0-) - y'(0-). A very common error is writing s^2*Y - y(0-) and forgetting the s*y(0-) term. Always write out the full transform of each derivative before substituting numbers.

Properties Summary

• Total response Y(s) = ZIR (initial condition driven) + ZSR = H(s)*X(s)
• ZIR: set input X(s)=0, solve from IC terms alone
• ZSR: set all ICs=0, output is H(s)*X(s)
• Natural response: terms from poles of H(s); forced response: terms from poles of X(s)
• For a stable system, natural response decays; forced response matches input shape in steady state
• Resonance occurs when a pole of X(s) coincides with a pole of H(s); produces t*e^(-at) terms
• Stability tested by checking whether all poles of the characteristic polynomial (denominator of H(s)) lie in left half-plane

Quick Revision

• Laplace method handles both initial conditions and any input in one algebraic step
• Initial conditions enter as additive source terms when transforming derivatives
• Total response = ZIR + ZSR; both share the same system denominator
• ZSR is always H(s)*X(s); requires zero initial conditions
• Poles of H(s): natural frequencies; poles of X(s): forced frequencies
• After finding Y(s), use partial fractions and table lookup to get y(t)
• Stability: all characteristic roots must have negative real parts
• Exam trap: missing initial condition terms when transforming second or higher-order derivatives; for y' the full Laplace is s^2Y - s*y(0-) - y'(0-), not just s^2Y