Laplace transforms for process dynamics

Why it matters

Process dynamics are described by linear ordinary differential equations, and the Laplace transform turns them into algebra. Every transfer function, block diagram, stability test and frequency-response plot in this subject is built on it, so speed with transforms, partial fractions and the limit theorems directly decides how fast you can solve control problems.

Key ideas

Definition. For a function f(t) defined for t ≥ 0, the Laplace transform is F(s) = ∫₀^∞ f(t)·e^(−st) dt, where s is a complex variable with units of 1/time. It is linear: L{a·f + b·g} = a·F + b·G.

Deviation variables. Control problems use deviation variables, y(t) = Y(t) − Y_ss, measured from the initial steady state. Then all initial conditions are zero, and differentiation becomes simple multiplication by s.

Derivatives. L{df/dt} = s·F(s) − f(0) and L{d²f/dt²} = s²·F(s) − s·f(0) − f′(0). With deviation variables starting at steady state, these reduce to s·F and s²·F.

Transfer function. For a linear system initially at steady state, G(s) = Y(s)/U(s), the ratio of output to input transforms. It depends only on the system, not on the input.

Poles and zeros. Poles are roots of the denominator of G(s); zeros are roots of the numerator. Each real pole p gives a term e^(pt) in the response; complex poles p = a ± jb give e^(at)·sin(bt) and e^(at)·cos(bt). Poles with negative real part decay (stable), poles with positive real part grow (unstable), and poles on the imaginary axis give sustained oscillation or a ramp (marginal).

Inverse transform by partial fractions. Factor the denominator, split into simple fractions, and invert each term from the table. For distinct real roots use the cover-up (Heaviside) rule; for repeated roots, include terms up to the root's multiplicity; for complex roots, complete the square.

Limit theorems. The final-value theorem gives the steady state without inversion, but only if s·Y(s) has all poles in the left half-plane (the limit must exist). The initial-value theorem gives f(0⁺).

Time delay. A pure dead time θ multiplies the transform by e^(−θs) (translation theorem). Shifting in s (multiplying f by e^(−at)) replaces s by s + a.

Formulas

F(s) = ∫₀^∞ f(t)·e^(−s·t) dt

Table of standard transforms (t ≥ 0):

• L{1} = 1/s (unit step)
• L{t} = 1/s² and L{tⁿ} = n!/sⁿ⁺¹
• L{δ(t)} = 1 (unit impulse)
• L{e^(−a·t)} = 1/(s + a)
• L{t·e^(−a·t)} = 1/(s + a)²
• L{sin(ω·t)} = ω/(s² + ω²) and L{cos(ω·t)} = s/(s² + ω²)
• L{e^(−a·t)·sin(ω·t)} = ω/((s + a)² + ω²)
• L{f(t − θ)·u(t − θ)} = e^(−θ·s)·F(s) (dead time θ, s)
• L{df/dt} = s·F(s) − f(0)
• L{∫₀ᵗ f dt} = F(s)/s

Limit theorems:

• lim(t→∞) f(t) = lim(s→0) s·F(s) (valid only if s·F(s) has no poles on or right of the imaginary axis)
• lim(t→0⁺) f(t) = lim(s→∞) s·F(s)

Symbols: t time (s); s Laplace variable (1/s); a decay rate (1/s); ω angular frequency (rad/s); θ dead time (s).

Worked examples

Example 1 (standard): solve an ODE. Solve d²y/dt² + 3·dy/dt + 2·y = 2 for t ≥ 0, with y(0) = 0 and dy/dt(0) = 0.

1. Transform: s²Y + 3sY + 2Y = 2/s, so Y(s) = 2 / (s·(s + 1)·(s + 2)).
2. Partial fractions: Y = A/s + B/(s + 1) + C/(s + 2).
3. Cover-up: A = 2/((1)(2)) = 1; B = 2/((−1)(1)) = −2; C = 2/((−2)(−1)) = 1.
4. Invert: y(t) = 1 − 2e^(−t) + e^(−2t).
5. Check: y(0) = 1 − 2 + 1 = 0, and the final value 1 agrees with lim s·Y = 2/(1·2) = 1.

y(t) = 1 − 2e^(−t) + e^(−2t)

Example 2 (GATE level): repeated pole. The response of a process is Y(s) = (s + 3) / (s·(s + 1)²), with t in minutes. Find y(t), its value at t = 1 min and its final value.

1. Write Y = A/s + B/(s + 1) + C/(s + 1)².
2. A = lim(s→0) s·Y = 3/1 = 3.
3. C = lim(s→−1) (s + 1)²·Y = (−1 + 3)/(−1) = −2.
4. Match the s² coefficients of (s + 3) = A(s + 1)² + B·s(s + 1) + C·s: 0 = A + B, so B = −3.
5. Invert: y(t) = 3 − 3e^(−t) − 2t·e^(−t).
6. At t = 1: y(1) = 3 − 3e^(−1) − 2e^(−1) = 3 − 5/e = 3 − 1.8394 = 1.161.
7. Final value: lim(s→0) s·Y = 3, valid since the remaining poles (s = −1) are in the left half-plane.

y(1 min) ≈ 1.161; y(∞) = 3

Example 3 (short): complex poles. F(s) = 2/(s² + 4s + 5) = 2/((s + 2)² + 1), so f(t) = 2e^(−2t)·sin t. At t = 0.5: f = 2·e^(−1)·sin(0.5) = 2·0.3679·0.4794 = 0.353.

Common mistakes

• Applying the final-value theorem to an unstable or oscillating system (for example F(s) = 1/(s − 1) or ω/(s² + ω²)); the theorem gives a number but the limit does not exist.
• Forgetting the 1/s of the step input when finding a step response.
• Missing the (s + a)² term for a repeated pole, which makes the partial fractions impossible to match.
• Writing the dead-time factor as e^(+θs) or applying it to the input instead of shifting the output.
• Using absolute variables with non-zero initial conditions and then dropping f(0) from the derivative rule.
• For complex poles, forgetting to complete the square before using the e^(−at)·sin(ωt) pair.

For GATE CH

Expect short numericals: inverse transforms by partial fractions, the final or initial value of a response, evaluating a response at a given time, and identifying poles to judge stability or oscillation. These underpin every later question on first- and second-order systems, closed-loop offset and stability, so practise partial fractions until they are routine and always check validity before using the final-value theorem.

Quick check

1. What is L{5e^(−2t)}?
2. Find the final value of y(t) if Y(s) = 6/(s·(2s + 3)).
3. Can the final-value theorem be applied to Y(s) = 1/(s·(s − 2))?
4. What time function corresponds to e^(−3s)/s?

Answers: 1. 5/(s + 2). 2. 2. 3. No, the pole at s = +2 makes y(t) grow without bound. 4. A unit step delayed by 3 time units, u(t − 3).