What is Laplace transform?

A method that transforms a function of time f(t) (t ≥ 0) into a function of the complex variable s, F(s) = ∫₀^∞ f(t)e^(−st) dt. With zero initial values, differentiation corresponds to multiplying by s and integration to dividing by s, so linear differential equations can be handled as algebraic expressions in s. It is the basic tool for finding transfer functions in control engineering, and for discrete-time systems the z-transform plays the same role. A unit step transforms to 1/s and e^(−at) to 1/(s + a).

Japanese name: ラプラス変換

Laplace transform appears in 10 practice questions across 3 topics, including Circuit Analysis: AC, Linear Circuits and Transients, Control Engineering Basics, Digital Signal Processing Basics.

Field: Control Engineering Basics

Commonly used transform pairs

A unit impulse δ(t) becomes 1, a unit step 1/s, a ramp t 1/s², e^(−at) 1/(s + a) and sin ωt ω/(s² + ω²). Delaying in time by L multiplies by e^(−Ls), and multiplying by e^(−at) replaces s with s + a. Combining these pairs with partial-fraction expansion lets many responses be inverse-transformed by looking them up in a table.

The final value and initial value theorems

The final value of an output can be found as lim[t→∞] f(t) = lim[s→0] sF(s), but only when all poles of sF(s) lie in the left half of the s-plane. For example, the final value of the step response of a system with transfer function G(s) is G(0), called the steady-state gain (DC gain). The initial value is found as lim[t→+0] f(t) = lim[s→∞] sF(s).

Relation to the Fourier transform and the z-transform

Evaluating the Laplace transform on the imaginary axis s = jω gives the Fourier transform (the frequency response). For sampled signals the z-transform with z = e^(sT) plays the same role, and the left half of the s-plane corresponds to the inside of the unit circle in the z-plane. The continuous-time rule "stable if all poles have negative real parts" therefore becomes "stable if all poles lie inside the unit circle" in discrete time.

Try it: questions about Laplace transform

The Laplace transform of a voltage in a circuit is V(s) = 10(s + 3)/(s(s + 2)(s + 5)). Which of the following is the voltage v(∞) after a long time?

A0 V
B3 V
C10 V

Show answer and explanation

Answer: B. 3 V

Method 1 (final-value theorem): v(∞) = lim(s→0) sV(s) = 10 × 3/(2 × 5) = 3 V. The theorem can be used when all the poles of sV(s) are in the left half-plane (the response converges). Method 2 (check with partial fractions): writing V(s) = A/s + B/(s + 2) + C/(s + 5) gives A = 3, B = −5/3 and C = −4/3, so v(t) = 3 − (5/3)e^(−2t) − (4/3)e^(−5t). As t → ∞ the exponential terms vanish, leaving 3 V, and at t = 0 the value is 3 − 5/3 − 4/3 = 0, which agrees with the initial-value theorem lim(s→∞) sV(s) = 0.

With all initial values zero, which of the following is the correct pair of operations that represent differentiation and integration of a time function in the s domain after the Laplace transform?

ADifferentiation is multiplying by s, and integration is dividing by s
BDifferentiation is dividing by s, and integration is multiplying by s
CDifferentiation is adding s, and integration is subtracting s

Show answer and explanation

Answer: A. Differentiation is multiplying by s, and integration is dividing by s

Under the Laplace transform, the derivative of f(t) corresponds to sF(s) − f(0) and the integral to F(s)/s. With zero initial values, differentiation becomes multiplication by s and integration becomes division by s, so a differential equation turns into calculations with polynomials in s. For example, the RC-circuit equation RC·dv/dt + v = vin becomes (RCs + 1)V(s) = Vin(s), which gives the transfer function 1/(1 + RCs). The unit step function transforms to 1/s, and e^(−at) to 1/(s + a).

FAQ

Why is the Laplace transform used in control engineering?

Because it turns differential equations into algebraic expressions, so the transfer function of a system of connected elements can be built up by multiplication and division. The pole positions also show the stability and speed of response.

What happens when the initial values are not zero?

The transform of a derivative becomes sF(s) − f(0), and an initial-value term remains. Transfer functions are defined with zero initial values, so the response due to initial values is considered and added separately.

Laplace transform in the explanations

Under the Laplace transform, the derivative of f(t) corresponds to sF(s) − f(0) and the integral to F(s)/s.
With all initial values zero, which of the following is the…
Method 2 (Laplace transform): in the s-domain, VC(s) = (E/s) × (1/(sC))/(R + 1/(sC)) = E/(s(1 + sRC)), which splits into partial fractions as E(1/s − 1/(s + 1/RC)) and transforms back to the same formula.
In the RC circuit in the figure, E = 10 V, R = 10 kΩ and C …
Method 2 (Laplace transform): representing the capacitor as 1/(sC) in series with a voltage source V0/s gives I(s) = ((E − V0)/s)/(R + 1/(sC)) = ((E − V0)/R)/(s + 1/(RC)), whose inverse transform is the same i(t).
In the RC circuit in the figure, E = 10 V, R = 2 kΩ and C =…
Method 2 (Laplace transform): I(s) = E/(s(R + sL)), so VL(s) = sL I(s) = E/(s + R/L), whose inverse transform is vL(t) = E e^(−Rt/L).
In the RL circuit in the figure, E = 12 V, R = 6 Ω and L = …
Method 2 (Laplace transform with the initial- and final-value theorems): with the impedance seen from the source Z(s) = R1 + (sL + R2 in parallel with 1/(sC)), I(s) = E/(s Z(s)).
In the circuit in the figure, E = 12 V, R1 = 2 Ω and R2 = 4…

Questions that mention Laplace transform (10)

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