Laplace Transforms in Control Systems

Control engineers design cruise controllers, drone autopilots, temperature regulators and robotic arms, and every one of these systems is described by differential equations. Solving those equations by hand in the time domain is slow and error-prone. The Laplace transform converts them into algebraic equations in a new variable \( s \), where the hard work becomes simple fractions, and then brings the answer back into time.

The Definition

For a function \( f(t) \) defined for \( t \ge 0 \), the Laplace transform is

$$ F(s) = \mathcal{L}\{f(t)\} = \int_0^{\infty} f(t)\, e^{-st}\, dt $$

where \( s = \sigma + j\omega \) is a complex variable. You can think of \( e^{-st} \) as a probe that tests how much of each growing, decaying or oscillating exponential is present in \( f(t) \).

A Short Table of Transforms

These few pairs cover most control problems:

• Unit step \( u(t) \to \dfrac{1}{s} \)
• Unit impulse \( \delta(t) \to 1 \)
• Ramp \( t \to \dfrac{1}{s^2} \)
• Exponential \( e^{-at} \to \dfrac{1}{s + a} \)
• Sine \( \sin\omega t \to \dfrac{\omega}{s^2 + \omega^2} \)
• Cosine \( \cos\omega t \to \dfrac{s}{s^2 + \omega^2} \)

For example, the exponential pair follows directly from the definition:

$$ \int_0^{\infty} e^{-at}e^{-st}\,dt = \left[\frac{-e^{-(s+a)t}}{s+a}\right]_0^{\infty} = \frac{1}{s+a} $$

valid when the real part of \( s + a \) is positive.

The Properties That Make It Powerful

Differentiation Becomes Multiplication

$$ \mathcal{L}\left\{\frac{df}{dt}\right\} = sF(s) - f(0), \qquad \mathcal{L}\left\{\frac{d^2f}{dt^2}\right\} = s^2F(s) - sf(0) - f'(0) $$

Initial conditions are built in automatically, which is a major advantage over the classical method.

Linearity

\( \mathcal{L}\{af + bg\} = aF(s) + bG(s) \), so each term of an equation can be transformed separately.

Final and Initial Value Theorems

$$ \lim_{t\to\infty} f(t) = \lim_{s\to 0} sF(s), \qquad f(0^+) = \lim_{s\to\infty} sF(s) $$

The final value theorem tells you the steady-state output without solving the whole problem, provided the system is stable.

The Standard Procedure

1. Write the differential equation of the system.
2. Take the Laplace transform of every term, inserting initial conditions.
3. Solve algebraically for the output \( Y(s) \).
4. Split \( Y(s) \) into partial fractions.
5. Use the table to invert each fraction back to the time domain.

Worked Example 1: A First-Order System

A small DC motor's speed \( y(t) \) (in thousands of rpm) responds to a step input voltage according to

$$ 0.5\frac{dy}{dt} + y = 2u(t), \qquad y(0) = 0 $$

Find \( y(t) \) and the speed after 1.5 s.

Step 1: Transform.

$$ 0.5\left[sY(s) - 0\right] + Y(s) = \frac{2}{s} $$

Step 2: Solve for \( Y(s) \).

$$ Y(s)(0.5s + 1) = \frac{2}{s} \quad \Rightarrow \quad Y(s) = \frac{2}{s(0.5s + 1)} = \frac{4}{s(s + 2)} $$

Step 3: Partial fractions. Write \( \dfrac{4}{s(s+2)} = \dfrac{A}{s} + \dfrac{B}{s+2} \). Cover-up gives \( A = 4/2 = 2 \) and \( B = 4/(-2) = -2 \):

$$ Y(s) = \frac{2}{s} - \frac{2}{s + 2} $$

Step 4: Invert.

$$ y(t) = 2\left(1 - e^{-2t}\right) \ \text{thousand rpm} $$

Step 5: Evaluate at \( t = 1.5 \) s.

$$ y(1.5) = 2\left(1 - e^{-3}\right) = 2(1 - 0.0498) = 1.90 \ \text{thousand rpm} $$

Sanity check. The time constant is \( \tau = 0.5 \) s, and after \( 3\tau = 1.5 \) s a first-order system should reach about 95% of its final value. Here \( 1.90/2 = 95\% \). The final value theorem also gives \( \lim_{s\to0} s \cdot \dfrac{4}{s(s+2)} = 2 \), matching \( y(\infty) = 2 \).

Worked Example 2: A Mass–Spring–Damper Platform

A 1000 kg machine platform rests on springs and dampers with total stiffness \( k = 2000 \ \text{N/m} \) and damping \( c = 3000 \ \text{N·s/m} \). A steady force of 40 N is suddenly applied at \( t = 0 \), with the platform initially at rest. Find the displacement \( x(t) \).

Step 1: Equation of motion.

$$ 1000\ddot{x} + 3000\dot{x} + 2000x = 40\,u(t) $$

Dividing by 1000:

$$ \ddot{x} + 3\dot{x} + 2x = 0.04\,u(t) $$

Step 2: Transform with \( x(0) = 0 \), \( \dot{x}(0) = 0 \):

$$ (s^2 + 3s + 2)X(s) = \frac{0.04}{s} \quad \Rightarrow \quad X(s) = \frac{0.04}{s(s + 1)(s + 2)} $$

Step 3: Partial fractions \( \dfrac{A}{s} + \dfrac{B}{s+1} + \dfrac{C}{s+2} \) by cover-up:

$$ A = \frac{0.04}{(1)(2)} = 0.02, \quad B = \frac{0.04}{(-1)(1)} = -0.04, \quad C = \frac{0.04}{(-2)(-1)} = 0.02 $$

Step 4: Invert.

$$ x(t) = 0.02 - 0.04e^{-t} + 0.02e^{-2t} \ \text{m} $$

Sanity checks.

• At \( t = 0 \): \( x = 0.02 - 0.04 + 0.02 = 0 \) m, correct.
• Velocity at \( t = 0 \): \( \dot{x} = 0.04 - 0.04 = 0 \) m/s, correct.
• Final value: \( x(\infty) = 0.02 \) m = 20 mm, equal to the static deflection \( F/k = 40/2000 = 0.02 \) m.

The poles at \( s = -1 \) and \( s = -2 \) are real and negative, so the platform settles smoothly without oscillating (it is overdamped).

Why Control Engineers Love the s-Domain

The real payoff is the transfer function, the ratio \( Y(s)/U(s) \) with zero initial conditions. In Example 2 it is \( 1/(1000s^2 + 3000s + 2000) \). Blocks in series multiply, feedback loops combine with a simple formula, and the pole locations reveal stability and speed at a glance. Poles in the left half of the s-plane mean decaying responses; poles in the right half mean the system runs away.

Did You Know?

• Pierre-Simon Laplace used integrals of this type in his work on probability around the 1780s, long before anyone applied them to circuits or machines.
• Oliver Heaviside solved circuit equations in the 1880s and 1890s with an "operational calculus" that worked brilliantly but lacked proof; mathematicians such as Bromwich and Doetsch later justified it using the Laplace transform.
• MATLAB, Python and calculators all use the same partial-fraction method shown here to compute step responses.

Key Takeaways

• The Laplace transform turns linear differential equations into algebraic equations in \( s \).
• Derivatives become multiplication by \( s \), with initial conditions included automatically.
• Partial fractions plus a short table invert most control-system answers.
• The final value theorem gives steady-state results quickly for stable systems.
• Pole locations in the s-plane determine whether and how fast a system settles.

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