Solving Linear LTI Systems with Laplace — Partial-Fraction Inversion

Transform, solve algebraically, invert by partial fractions

Signals & SystemsElectrical Engineering Year 2
⏱️ About 18 min

What if solving a differential equation were as simple as solving for x in a polynomial? The Laplace method makes it exactly that.

💡
The big idea: The Laplace transform reduces linear constant-coefficient ODEs to algebra: transform both sides, solve for Y(s), then invert using partial-fraction decomposition. Each simple pole in Y(s) corresponds to a known exponential mode in time, so inversion is a lookup once the fractions are decomposed.
🎯 By the end, you'll be able to
  • Apply the three-step Laplace method: transform, solve algebraically, invert
  • Perform partial-fraction decomposition using the cover-up method
  • Invert standard Laplace terms back to the time domain using known transform pairs
  • Use the Final Value Theorem as a steady-state sanity check
📎 Helpful to know first

The Three-Step Laplace Method

In the previous lesson, we saw that the Laplace transform's differentiation property L{dy/dt}=sY(s)−y(0⁻) converts derivatives into algebraic multiplications. This unlocks a powerful three-step recipe for solving linear constant-coefficient ODEs:

Step 1 — Transform: Apply the Laplace transform to both sides of the ODE. Each derivative becomes a multiplication by s (plus initial-condition terms), turning the differential equation into an algebraic equation in s.

Step 2 — Solve: Rearrange algebraically to isolate Y(s). This is just polynomial manipulation — no integration, no guessing homogeneous or particular solutions.

Step 3 — Invert: Decompose Y(s) into a sum of simple terms using partial fractions, then look up each term's inverse from a table of known transform pairs.

\[ \mathcal{L}\!\left\{\frac{dy}{dt}\right\} = s\,Y(s) - y(0^-) \]
Recap: differentiation in time becomes multiplication by s, with the initial condition y(0⁻) entering explicitly. Higher derivatives follow the same pattern: L{d²y/dt²} = s²Y(s) − sy(0⁻) − y′(0⁻).

Steps 1–2: Transform and Solve

Consider a first-order system: dy/dt+ay(t)=f(t), with initial condition y(0⁻)=y₀. Taking the Laplace transform of both sides gives sY(s)−y(0⁻)+aY(s)=F(s).

Factor out Y(s): Y(s)(s+a)=F(s)+y(0⁻). Solving for Y(s): Y(s)=[F(s)+y(0⁻)]/(s+a).

That's it — the ODE is solved in the s-domain. The denominator (s+a) is the system's characteristic polynomial; its root s=−a is the pole that determines the natural response mode e−at.

Step 3: Partial-Fraction Decomposition

Y(s) is typically a rational function N(s)/D(s). To invert it, we decompose it into a sum of simple fractions — one per pole — each of which matches a known transform pair.

If Y(s)=N(s)/D(s) and D(s) has distinct roots p₁,p₂,...,pₙ, then Y(s)=A₁/(s−p₁)+A₂/(s−p₂)+⋯+Aₙ/(s−pₙ). Each term Aᵢ/(s−pᵢ) inverts to Aᵢ·epᵢt·u(t). The constants A₁,A₂,... are called residues, and there's an elegant trick to find them: the cover-up method.

\[ Y(s) = \frac{N(s)}{D(s)} = \frac{A_1}{s - p_1} + \frac{A_2}{s - p_2} + \cdots + \frac{A_n}{s - p_n} \]
Partial-fraction expansion of a rational Y(s) with distinct poles p₁, p₂, …, pₙ. Each term is a simple first-order fraction that inverts to an exponential mode.

The Cover-Up Method

To find the residue Aᵢ associated with pole pᵢ, multiply both sides of the partial-fraction equation by (s−pᵢ), which cancels that denominator, then evaluate at s=pᵢ. All other terms vanish, leaving Aᵢ=[(s−pᵢ)·N(s)/D(s)] evaluated at s=pᵢ.

This is the cover-up method: you mentally 'cover up' the factor (s−pᵢ) in D(s) and evaluate the remaining expression at s=pᵢ. It's fast, intuitive, and works for any distinct (non-repeated) pole.

✨ Each pole is a natural mode

When you decompose Y(s) into partial fractions, you're factoring the system's response into its natural exponential modes. A pole at s=−a produces a term e−atu(t); a pole at s=jω₀ produces a sustained oscillation. The residues tell you how strongly each mode is excited. This is the same pole-zero intuition you'll use throughout signals and systems — and it generalizes directly to the Z-transform for discrete-time systems.

Sanity Check: The Final Value Theorem

Before doing the full inversion, you can quickly check what the steady-state value y(∞) will be — no partial fractions required. The Final Value Theorem from the previous lesson states that y(∞)=lim(s→0)s·Y(s), valid when the ROC of sY(s) includes the origin — practically, when all poles of Y(s) are in the left half-plane (stable system). After you complete the full inversion, verify that lim(t→∞)y(t) matches the FVT prediction. If they disagree, you've made an error somewhere.

\[ \lim_{t \to \infty} y(t) = \lim_{s \to 0} s\,Y(s) \]
The Final Value Theorem: the steady-state value of y(t) equals sY(s) evaluated at s → 0, valid when the ROC of sY(s) includes the origin.
⚠️ When the FVT fails

The Final Value Theorem is invalid if any pole of Y(s) lies on or to the right of the jω-axis (e.g., a pole at s=0 that isn't cancelled, or a pole with positive real part). In such cases y(t) does not settle to a finite value, and the limit lim(s→0)sY(s) may give a misleading finite number. Always verify stability before trusting the FVT.

📝 Worked example: Solve dy/dt + 3y(t) = u(t) with y(0) = 0, where u(t) is the unit step input.
  1. Step 1 — Transform both sides using the differentiation property and L{u(t)} = 1/s: sY(s) − y(0) + 3Y(s) = 1/s. Since y(0) = 0: sY(s) + 3Y(s) = 1/s.
  2. Step 2 — Factor and solve: Y(s)(s + 3) = 1/s, so Y(s) = 1/[s(s + 3)].
  3. Step 3 — Partial fractions: 1/[s(s+3)] = A/s + B/(s+3).
  4. Cover-up for A: multiply by s and evaluate at s = 0 → A = 1/(0+3) = 1/3.
  5. Cover-up for B: multiply by (s+3) and evaluate at s = −3 → B = 1/(−3) = −1/3.
  6. So Y(s) = (1/3)/s − (1/3)/(s+3).
  7. Step 4 — Invert using u(t) ↔ 1/s and e^(−3t)u(t) ↔ 1/(s+3): y(t) = (1/3)u(t) − (1/3)e^(−3t)u(t) = (1/3)(1 − e^(−3t))u(t).
  8. Sanity check (FVT): lim(s→0) sY(s) = lim(s→0) 1/(s+3) = 1/3, which matches y(∞) = (1/3)(1 − 0) = 1/3.
✓ y(t) = (1/3)(1 − e^(−3t))u(t)
✏️ Practice: Solve dy/dt + 4y(t) = u(t) with y(0) = 0. Find the steady-state value y(∞) using the Final Value Theorem (you do not need the full time-domain solution).
Solution
  1. Step 1 — Transform: sY(s) + 4Y(s) = 1/s, so Y(s) = 1/[s(s+4)].
  2. Apply the Final Value Theorem: y(∞) = lim(s→0) s·Y(s) = lim(s→0) s · 1/[s(s+4)].
  3. Cancel s: y(∞) = lim(s→0) 1/(s+4).
  4. Evaluate: y(∞) = 1/(0+4) = 1/4 = 0.25.

Check your understanding

1. In the three-step Laplace method, what does Step 2 (solving for Y(s)) involve?
After transforming, the ODE becomes an algebraic equation in s. Step 2 is simply rearranging to solve for Y(s) — polynomial algebra, no calculus.
2. Using the cover-up method on Y(s) = 1/[s(s+3)] = A/s + B/(s+3), what is the value of B?
Cover up (s+3) in the denominator and evaluate the remaining expression 1/s at s = −3: B = 1/(−3) = −1/3.
🔎 Related lessons
Transfer Functions, Poles & Zeros, Stability via the ROC · soon
✅ Key takeaways
  • The Laplace method has three steps: transform the ODE into an algebraic equation in s, solve for Y(s), then invert via partial fractions.
  • Partial-fraction decomposition expresses Y(s) = N(s)/D(s) as a sum A/(s−p₁) + B/(s−p₂) + ⋯, where p₁, p₂, … are the poles (roots of D(s)).
  • The cover-up method finds each coefficient by multiplying by (s−pᵢ) and evaluating at s = pᵢ.
  • Each term 1/(s+a) inverts to e^(−at)u(t), so the full solution is a sum of exponential modes.
  • The Final Value Theorem provides a quick steady-state check: y(∞) = lim(s→0) sY(s), when valid.
➡️ With Laplace inversion mastered, the next lesson zooms out from a single ODE to the transfer function H(s) itself — its poles, zeros, and the stability criterion they encode.
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