Laplace Transforms and System Response

Introduction and Context

Differential equations describe the dynamic behaviour of mechanical, electrical, thermal, hydraulic, and electromechanical systems. In a time-domain formulation, solving these equations often requires repeated differentiation, integration, and careful treatment of initial conditions. The Laplace transform provides an alternative description in the complex-frequency domain. It converts differentiation into algebraic multiplication by the complex variable s, making many linear differential-equation problems easier to solve.

For mechatronics engineers, this method is especially important because sensors, actuators, motors, robots, and control systems are commonly represented by transfer functions. Once a model is expressed in the Laplace domain, engineers can determine transient response, steady-state response, stability, bandwidth, and the effect of poles and zeros.

This chapter develops the unilateral Laplace transform, inverse transforms, convolution, transfer functions, impulse and step responses, and the principal theorems used in continuous-time system analysis.

Learning Objectives

By the end of this chapter, you should be able to:

1. Define the Laplace transform and identify its region of convergence in basic cases.
2. Use transform properties and tables to transform derivatives and common functions.
3. Solve linear ordinary differential equations with initial conditions.
4. Compute inverse Laplace transforms using partial fractions and standard pairs.
5. Explain convolution and its relationship to system response.
6. Form transfer functions for linear time-invariant systems.
7. Interpret impulse response, step response, poles, and zeros.
8. Apply the initial-value and final-value theorems correctly.
9. Assess assumptions, limitations, and practical implications of Laplace-domain models.

1. The Laplace Transform

For a function f(t) defined for t ≥ 0, its unilateral Laplace transform is

where t is time in seconds, s = σ + jω is the complex-frequency variable, σ is the real part in s⁻¹, ω is angular frequency in rad·s⁻¹, j² = −1, and F(s) is the transformed function. The integral exists when f(t)e^(−σt) is integrable over [0,∞). A sufficient condition is that f(t) is piecewise continuous and of exponential order.

The unilateral transform is particularly convenient for initial-value problems because initial conditions appear explicitly when derivatives are transformed. In control engineering, signals are usually assumed to be zero for t < 0, and the unilateral and bilateral transforms then agree for the relevant causal signals.

Common transform pairs

Here δ(t) is the unit impulse. The frequency ω in the sine and cosine pairs is in rad·s⁻¹. The transform variable s has units of s⁻¹, so expressions such as s + a are dimensionally valid only when a also has units of s⁻¹.

Linearity

If a and b are constants and the transforms exist, then

This property permits complicated signals to be decomposed into simpler components.

Differentiation in time

For the unilateral transform,

and

The notation 0⁺ means the value immediately after t = 0. These formulas assume that the required derivatives are piecewise continuous and of exponential order.

Integration and multiplication by t

If the transform exists,

and

The variable τ is a dummy integration variable. The first relation shows why integration in time corresponds to division by s.

Time delay and exponential shifting

For a causal function delayed by T ≥ 0,

where u(t) is the unit-step function. Multiplication by e^(at) shifts the transform:

These properties are useful for modelling delayed commands and exponentially weighted signals.

2. Inverse Laplace Transforms

The inverse transform recovers f(t) from F(s). Although a complex contour integral defines the inverse formally, engineering calculations generally use transform tables, algebraic manipulation, partial fractions, and standard pairs.

Partial-fraction method

Suppose F(s) is a proper rational function: the degree of its numerator is less than the degree of its denominator. Factor the denominator, decompose F(s), and invert each term.

For distinct linear factors,

For repeated factors, terms such as A/(s+a) + B/(s+a)² are required. If the numerator degree is not smaller than the denominator degree, polynomial division must first be performed.

Worked Example 1: Inverse transform

Find the inverse transform of

Write

Multiplication by the denominator gives

Setting s = −1 gives A = 3/2. Setting s = −3 gives B = 1/2. Therefore,

and

The exponents are dimensionless because the numerical coefficients of t represent rates of s⁻¹.

Complex and repeated poles

A quadratic factor may be completed to a square. For example,

so its inverse is (1/3)e^(−2t)sin(3t). Repeated poles produce polynomial factors in time. For example,

A pole at s = p contributes a mode proportional to e^(pt), with additional powers of t for repeated poles.

3. Solving Differential Equations

The Laplace method follows a reliable sequence:

1. Define the time-domain input, output, and initial conditions.
2. Transform every term using the derivative rule.
3. Substitute initial values.
4. Solve algebraically for the transformed unknown.
5. Perform partial fractions or other algebraic simplification.
6. Invert the result and check the initial and long-time behaviour.

Worked Example 2: Mechanical-type equation

Consider

with x(0⁺) = 1 and x′(0⁺) = 0. For t ≥ 0, the transform of 4u(t) is 4/s. Thus,

Collecting terms,

Therefore,

Let

Substitution gives A = 2, B = −2, and C = 1. Hence

At t = 0, x(0) = 2 − 2 + 1 = 1, satisfying the specified initial displacement. As t tends to infinity, x(t) tends to 2, the equilibrium value associated with the constant input.

4. Convolution and Linear Time-Invariant Systems

For a causal linear time-invariant (LTI) system with input x(t), impulse response h(t), and output y(t), the zero-state response is the convolution

The two forms h(τ)x(t−τ) and x(τ)h(t−τ) are equivalent. Convolution is commutative, associative, and distributive under suitable existence conditions.

The convolution theorem states

Thus, multiplication in the Laplace domain corresponds to convolution in time. This is a central reason transfer functions are useful.

The unit impulse δ(t) satisfies

in the causal engineering convention. Since ℒ{δ(t)} = 1, the response of an initially relaxed system to an impulse is H(s) in the Laplace domain; consequently, h(t) is the inverse transform of H(s).

5. Transfer Functions and System Models

For a causal LTI system initially at rest, the transfer function is

where U(s) is the input transform and Y(s) is the output transform. “Initially at rest” means all independent initial conditions are zero. A transfer function therefore describes the zero-state input-output relationship, not the complete response when stored energy is present.

For a rational model,

where K is a gain, zᵢ are zeros, and pᵢ are poles. Pole and zero locations are expressed in the complex s-plane.

Deriving a motor-related model

A simplified armature-controlled DC motor may be represented by

and

where L is armature inductance in H, R is resistance in Ω, i is current in A, Kₑ is the back-emf constant, ω is angular speed in rad·s⁻¹, v is applied voltage in V, J is rotational inertia in kg·m², b is viscous-friction coefficient in N·m·s·rad⁻¹, Kₜ is torque constant in N·m·A⁻¹, and T_L is load torque in N·m. With zero initial conditions and T_L = 0,

Eliminating I(s),

This model demonstrates how physical laws become a transfer function. Parameter values, neglected nonlinearities, saturation, friction variation, and sampling effects determine how accurately the model predicts a real motor.

6. Impulse, Step, Ramp, and Sinusoidal Responses

For a system with transfer function G(s):

• Impulse response: input δ(t), so Y(s) = G(s) and y(t) = g(t).
• Unit-step response: input u(t), so U(s) = 1/s and Y(s) = G(s)/s.
• Unit-ramp response: input tu(t), so U(s) = 1/s² and Y(s) = G(s)/s².
• Sinusoidal response: input sin(ωt) or cos(ωt), analysed using the corresponding transforms or frequency response.

The impulse response reveals the complete zero-state dynamics of an LTI system. The step response is often easier to measure experimentally and is widely used to estimate rise time, settling time, overshoot, and steady-state gain.

For a stable first-order system,

where K is the static gain and τ > 0 is the time constant in seconds. Its unit-step response is

At t = τ, the response reaches 1 − e⁻¹ ≈ 63.2% of its final value. A practical five-time-constant rule states that the response is within approximately 0.67% of its final value at t = 5τ, because e⁻⁵ ≈ 0.0067.

For a standard second-order system,

where ωₙ > 0 is the undamped natural frequency in rad·s⁻¹ and ζ is the dimensionless damping ratio. The poles are

for 0 < ζ < 1. Such a system is underdamped and exhibits decaying oscillation. ζ = 1 gives critical damping; ζ > 1 gives overdamping; ζ = 0 gives undamped oscillation in the ideal model.

7. Poles, Zeros, and Stability

A pole is a value of s that makes the denominator of a transfer function zero. A zero makes its numerator zero. Poles determine natural modes and strongly govern stability and transient behaviour; zeros shape the response and may suppress or amplify particular dynamic features.

For a continuous-time rational LTI system, asymptotic stability requires every pole to lie strictly in the open left half-plane:

A pole on the imaginary axis generally produces sustained modes and fails asymptotic stability; repeated imaginary-axis poles produce unbounded terms. Any right-half-plane pole causes an exponentially growing mode and instability. Pole-zero cancellation requires caution: an algebraic cancellation may hide an internal unstable mode, especially in non-minimal models.

The distance of dominant poles from the imaginary axis indicates decay rate. A pole at −a contributes e^(−at), whose time constant is 1/a. Complex poles introduce oscillation at approximately the magnitude of their imaginary part.

8. Initial-Value and Final-Value Theorems

The initial-value theorem is

provided the required initial value exists and the transform does not contain inappropriate impulsive components at t = 0.

The final-value theorem is

only when all poles of sF(s) lie strictly in the open left half-plane, except possibly a simple pole at the origin associated with a finite constant limit. It must not be applied blindly to unstable or sustained oscillatory systems.

Worked Example 3: Final value

For the stable first-order step response,

The final-value theorem gives

This agrees with the time-domain result. For an undamped oscillator, however, sY(s) may have imaginary-axis poles, so the theorem is not applicable even if a formal substitution appears possible.

9. Frequency Response and the s-Plane

For a stable LTI system, the frequency response is obtained by evaluating the transfer function on the imaginary axis:

Its magnitude |G(jω)| indicates amplitude scaling and its phase ∠G(jω) indicates phase shift for sinusoidal steady-state input. This substitution is meaningful as a steady-state frequency response when the relevant response exists; unstable systems do not possess a bounded sinusoidal steady state in the usual sense.

Bode plots display 20log₁₀|G(jω)| in decibels and phase versus log frequency. These plots connect Laplace-domain models with bandwidth, filtering, sensor noise, and feedback design. A low-pass first-order system, for example, attenuates high-frequency components and introduces increasing phase lag as frequency rises.

10. Applied Case: Position Control of a Small Servo Axis

The following is an illustrative engineering case, not a report of a particular commercial product.

Background and problem

A laboratory robot arm uses a DC servo motor to position a lightweight link. The designer needs a simple model to predict the motor’s response to a commanded voltage and to determine whether a proportional controller will produce unacceptable oscillation.

Analysis

After linearisation about a nominal operating point and neglecting load torque, the motor model is approximated by

The pole at the origin reflects the integration from angular speed to position. The poles at −2 and −10 represent decaying electrical-mechanical modes. Since all poles are in the left half-plane except the integrator, the position plant is not asymptotically stable as an isolated input-to-position system: a constant voltage can produce continuing position change. This is physically consistent with a motor that maintains nonzero speed under a constant effective drive.

Suppose a proportional controller with gain K is used in unity negative feedback. The closed-loop characteristic equation is obtained from 1 + KG_m(s) = 0:

or

The coefficients show that increasing K changes the closed-loop poles. A design cannot be judged from gain alone; pole locations must be checked. For a third-order polynomial s³ + a₁s² + a₂s + a₃, the Routh-Hurwitz conditions for strict stability are a₁ > 0, a₂ > 0, a₃ > 0, and a₁a₂ > a₃. Here these conditions require K > 0 and 12×20 > 10K, giving K < 24. Thus, within this simplified model, positive proportional gains below 24 satisfy the strict stability test.

Approach and solution

The designer selects a preliminary gain K = 5, then computes or numerically estimates the closed-loop poles. The resulting response is tested against settling-time and overshoot requirements. If position error under constant disturbance remains excessive, integral action may be considered, but its effect on stability and wind-up must also be analysed. A derivative term may improve damping but can amplify measurement noise, so filtering and sensor bandwidth become important.

Outcome

The Laplace model provides a transparent path from physical parameters to poles, from poles to transient response, and from controller gain to closed-loop stability. Before implementation, the model should be validated experimentally using a small safe command, and saturation, friction, encoder quantisation, sampling, and current limits should be included.

Learning points

1. A transfer function is a model under stated assumptions, not a complete physical reality.
2. An integrator can be useful for tracking but changes stability and transient behaviour.
3. Closed-loop poles, rather than controller gain alone, determine dynamic performance.
4. Safety limits and experimental validation are essential when deploying a controller on moving machinery.

11. Limitations, Good Practice, and Contemporary Use

Laplace analysis assumes linearity and time invariance, or at least a valid local linear approximation. Real mechatronic systems may contain Coulomb friction, backlash, hysteresis, saturation, dead zones, switching, quantisation, and parameter variation. A transfer function also omits internal details that may matter for robustness or safety.

Best practice includes:

• state the operating point and modelling assumptions;
• distinguish zero-state and zero-input responses;
• retain units and check dimensional consistency;
• verify initial and final values independently;
• inspect pole locations before using the final-value theorem;
• avoid interpreting unstable pole-zero cancellations as physically safe;
• validate models with measured step or frequency-response data;
• include actuator saturation, sensor noise, delays, and sampling when relevant;
• document parameter uncertainty and safety constraints.

Computer algebra and simulation tools can perform transforms, partial fractions, root calculations, and response plots. They do not replace engineering judgement. An apparently elegant symbolic result may be invalid if a theorem’s pole conditions are violated or if the model ignores a dominant nonlinear effect. Modern control software also supports automated identification and optimisation, but engineers remain responsible for selecting experiments, checking identifiability, protecting equipment, and explaining design decisions.

Chapter Summary

The unilateral Laplace transform maps a causal time-domain function into a complex-frequency representation. Its derivative property converts linear differential equations into algebraic equations while explicitly incorporating initial conditions. Inverse transforms are commonly obtained through standard pairs, partial fractions, completing the square, and treatment of repeated factors.

Convolution gives the zero-state response of an LTI system and becomes multiplication in the Laplace domain. For an initially relaxed system, the transfer function is the ratio of output to input transforms. Its impulse response is the inverse transform of the transfer function, while step and ramp responses are obtained by multiplying the transfer function by 1/s and 1/s², respectively.

Poles determine natural modes, stability, and dominant transient behaviour; zeros shape the response. Continuous-time asymptotic stability requires all poles to lie strictly in the left half-plane. The initial-value and final-value theorems are efficient but conditional tools. Finally, Laplace models support mechatronic analysis and control design, provided assumptions, nonlinearity, saturation, uncertainty, and implementation constraints are addressed.

Key Terms

Laplace transform: Integral transformation from time domain to complex-frequency domain.

Unilateral transform: Laplace transform evaluated from t = 0 to infinity, useful for initial-value problems.

Inverse Laplace transform: Operation recovering a time-domain function from its transform.

Complex-frequency variable: The variable s = σ + jω used in Laplace analysis.

Region of convergence: Values of s for which the Laplace integral exists.

Unit impulse: Idealised signal δ(t) with unit area and zero duration.

Unit step: Signal u(t) that changes from zero to one at t = 0.

Convolution: Integral combining an input with an LTI system’s impulse response.

Linear time-invariant system: System satisfying superposition and time-shift invariance.

Transfer function: Zero-state ratio G(s) = Y(s)/U(s) for an initially relaxed LTI system.

Impulse response: Output of an initially relaxed system to a unit impulse.

Step response: Output resulting from a unit-step input.

Pole: Root of a transfer-function denominator.

Zero: Root of a transfer-function numerator.

Time constant: Parameter describing exponential response speed in a first-order system.

Natural frequency: Frequency parameter ωₙ of a standard second-order model.

Damping ratio: Dimensionless parameter ζ describing second-order damping.

Frequency response: Steady-state response represented by G(jω).

Asymptotic stability: Property that the natural response tends to zero as t tends to infinity.

Zero-state response: Response caused by the input when initial conditions are zero.

Review Questions

Short Answer Questions

1. State the unilateral Laplace transform and define the variable s.
2. What is the difference between a pole and a zero?
3. State the convolution theorem.
4. What does the time constant represent in a first-order system?
5. State the pole condition for continuous-time asymptotic stability.

Long Answer Questions

1. Derive the unilateral Laplace transform of the first derivative and explain the role of the initial condition.
2. Explain a systematic partial-fraction procedure for finding inverse Laplace transforms, including repeated poles.
3. Explain how impulse, step, and ramp responses are obtained from a transfer function.
4. Derive the transfer function of the simplified DC motor model given in this chapter.
5. Discuss the assumptions and limitations of using transfer functions for real mechatronic systems.

Analytical / Application Questions

1. Solve x″ + 4x′ + 3x = 6u(t), with x(0⁺) = 0 and x′(0⁺) = 1, using the Laplace transform. Check the initial and final values.
2. For G(s) = 25/(s² + 6s + 25), identify ωₙ and ζ, classify the response, and determine whether the system is asymptotically stable.
3. A measured servo step response has a final value of 2.5 units and reaches approximately 63.2% of that value after 0.08 s. Propose a first-order model and explain how you would validate it experimentally.

Multiple Choice Questions with Answers

Question 1

For a causal function with transform F(s), what is the unilateral Laplace transform of f′(t)?

A. sF(s)

B. sF(s) − f(0⁺)

C. F(s)/s

D. −dF(s)/ds

Correct Answer: B. sF(s) − f(0⁺)

Question 2

What is the inverse Laplace transform of 1/(s+a), assuming a is a positive constant?

A. e^(at)

B. te^(−at)

C. e^(−at)

D. sin(at)

Correct Answer: C. e^(−at)

Question 3

For an initially relaxed LTI system, which expression defines its transfer function?

A. U(s)/Y(s)

B. Y(s) + U(s)

C. Y(s)/U(s)

D. Y(s)U(s)

Correct Answer: C. Y(s)/U(s)

Question 4

If G(s) is a transfer function, what is the transform of its unit-step response?

A. sG(s)

B. G(s)/s

C. G(s)+s

D. 1/G(s)

Correct Answer: B. G(s)/s

Question 5

Which condition ensures continuous-time asymptotic stability of a rational LTI system?

A. All zeros lie in the left half-plane

B. All poles lie strictly in the left half-plane

C. All poles lie on the imaginary axis

D. The numerator degree exceeds the denominator degree

Correct Answer: B. All poles lie strictly in the left half-plane

Question 6

For G(s) = K/(τs+1), with τ > 0, the unit-step response at t = τ is what fraction of its final value?

A. Approximately 36.8%

B. Exactly 50%

C. Approximately 63.2%

D. Approximately 99.3%

Correct Answer: C. Approximately 63.2%

Question 7

The convolution theorem states that convolution in time corresponds to which operation in the Laplace domain?

A. Addition

B. Differentiation

C. Multiplication

D. Time reversal

Correct Answer: C. Multiplication

Question 8

For G(s) = 10/(s+2), what is the system pole?

A. 10

B. 2

C. −2

D. −10

Correct Answer: C. −2

Question 9

When is the final-value theorem valid in the usual control-system application?

A. When all poles of sF(s) are strictly in the open left half-plane, apart from an allowable simple pole at the origin

B. For every rational function F(s)

C. Only when all zeros are in the right half-plane

D. Only when F(s) has no denominator

Correct Answer: A. When all poles of sF(s) are strictly in the open left half-plane, apart from an allowable simple pole at the origin

Question 10

For the standard second-order denominator s² + 2ζωₙs + ωₙ², what does 0 < ζ < 1 indicate?

A. Overdamped response

B. Critically damped response

C. Underdamped response

D. An unstable response in every case

Correct Answer: C. Underdamped response