Maths Formulae Transforms Laplace Transform Pairs

Laplace Transform Pairs – Time-Domain and S-Domain

Access a list of common Laplace transform pairs to convert time-domain functions into their s-domain equivalents.
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Definition of Laplace Transform Pairs

Standard Laplace Transform Pairs are the fundamental building blocks of s-domain analysis, providing essential relationships between common time-domain functions, f(t), and their s-domain counterparts, F(s). These pairs serve as a reference library, enabling rapid conversion between time and frequency representations without performing complex integrations. They form the foundation for solving differential equations, analyzing system responses, and designing control systems.

\[ F(s) = \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \,dt \]
Laplace Transform Definition
SymbolDescription
\[ f(t) \]A function of time (t) in the time-domain.
\[ F(s) \]The Laplace transform of f(t), a function in the complex frequency domain (s-domain).
\[ s \]The complex frequency variable, s = σ + jω, where σ is the neper frequency and ω is the angular frequency.
\[ \delta(t) \]Unit Impulse (Dirac delta function), an infinitely sharp spike at t=0 with an area of 1.
\[ u(t) \]Unit Step (Heaviside function), a function that is 0 for t < 0 and 1 for t ≥ 0.
\[ e^{-at} \]An exponential function representing decay (if a > 0) or growth (if a < 0).
\[ \omega \]Angular Frequency, representing oscillation in radians per second.
\[ \leftrightarrow \]Symbol indicating a transform pair, denoting the bidirectional relationship between f(t) and F(s).
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Key Transform Pairs

\[ \delta(t) \leftrightarrow 1 \]
Unit Impulse
\[ u(t) \leftrightarrow \frac{1}{s} \]
Unit Step
\[ t^n u(t) \leftrightarrow \frac{n!}{s^{n+1}} \]
Power Functions (for integer n ≥ 0)
\[ e^{-at}u(t) \leftrightarrow \frac{1}{s+a} \]
Exponential Decay
\[ \sin(\omega t)u(t) \leftrightarrow \frac{\omega}{s^2 + \omega^2} \]
Sine Function
\[ \cos(\omega t)u(t) \leftrightarrow \frac{s}{s^2 + \omega^2} \]
Cosine Function
\[ e^{-at}\sin(\omega t)u(t) \leftrightarrow \frac{\omega}{(s+a)^2 + \omega^2} \]
Damped Sine Function
\[ e^{-at}\cos(\omega t)u(t) \leftrightarrow \frac{s+a}{(s+a)^2 + \omega^2} \]
Damped Cosine Function
\[ \sinh(at)u(t) \leftrightarrow \frac{a}{s^2 - a^2} \]
Hyperbolic Sine
\[ \cosh(at)u(t) \leftrightarrow \frac{s}{s^2 - a^2} \]
Hyperbolic Cosine
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Visualization

f(t) (t ≥ 0) F(s) = ℒ{f(t)} 1 (unit step) 1/s t 1/s² e^(−at) 1/(s+a) sin(ωt) ω / (s²+ω²) cos(ωt) s / (s²+ω²) δ(t) impulse 1
Common Laplace Transform pairs — the core toolkit for solving ODEs and analysing control systems

Unlike a geometric shape, a Laplace Transform pair does not have a physical diagram. Instead, the relationship is visualized on the complex s-plane. The s-domain function, F(s), is analyzed by plotting its poles (values of 's' where F(s) → ∞) and zeros (values of 's' where F(s) = 0). The location of these poles and zeros on the s-plane reveals critical information about the time-domain system's stability and behavior (e.g., oscillation, decay, growth).

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Properties

Linearity: The transform of a sum of functions is the sum of their individual transforms. This allows complex functions to be broken down into simpler, standard pairs. \[ \mathcal{L}\{af_1(t) + bf_2(t)\} = aF_1(s) + bF_2(s) \]

Time Shift: Shifting a function in the time domain corresponds to multiplication by an exponential in the s-domain. \[ \mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s) \]

Frequency Shift: Multiplication by an exponential in the time domain corresponds to a shift in the s-domain. This is key for damped functions. \[ \mathcal{L}\{e^{-at}f(t)\} = F(s+a) \]

Time Differentiation: Differentiating in the time domain corresponds to multiplication by 's' in the s-domain (minus initial conditions). This property transforms differential equations into algebraic equations. \[ \mathcal{L}\{f'(t)\} = sF(s) - f(0) \]

Time Integration: Integrating in the time domain corresponds to division by 's' in the s-domain. \[ \mathcal{L}\left\{ \int_0^t f(\tau)d\tau \right\} = \frac{F(s)}{s} \]

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Proof of the Unit Step Transform

We can derive the Laplace Transform for the unit step function, f(t) = u(t), using the fundamental definition of the transform.

\[ F(s) = \int_0^\infty e^{-st} f(t) \,dt \]
Step 1: Start with the definition of the Laplace Transform.

The unit step function u(t) is equal to 1 for all t ≥ 0. Since the integral starts at 0, we can substitute f(t) with 1.

\[ \mathcal{L}\{u(t)\} = \int_0^\infty e^{-st} (1) \,dt = \int_0^\infty e^{-st} \,dt \]
Step 2: Substitute f(t) = u(t) = 1.

Now, we evaluate the standard exponential integral.

\[ \int_0^\infty e^{-st} \,dt = \left[ -\frac{1}{s}e^{-st} \right]_0^\infty \]
Step 3: Perform the integration.

Finally, we apply the limits of integration. For the integral to converge, we require the real part of s to be greater than 0, so that e-st approaches 0 as t approaches ∞.

\[ \left[ -\frac{1}{s}e^{-st} \right]_0^\infty = \lim_{t \to \infty} \left(-\frac{1}{s}e^{-st}\right) - \left(-\frac{1}{s}e^{-s(0)}\right) = 0 - \left(-\frac{1}{s}\right) = \frac{1}{s} \]
Step 4: Evaluate the limits.

Thus, the Laplace transform of the unit step function is 1/s.

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Worked Example

Find the Laplace Transform of the function \[ f(t) = 5u(t) - 2e^{-3t} + 4\sin(2t) \]
  1. Apply the linearity property: \[ \mathcal{L}\{f(t)\} = \mathcal{L}\{5u(t)\} - \mathcal{L}\{2e^{-3t}\} + \mathcal{L}\{4\sin(2t)\} = 5\mathcal{L}\{u(t)\} - 2\mathcal{L}\{e^{-3t}\} + 4\mathcal{L}\{\sin(2t)\} \]
  2. Use the standard pair for the unit step: \[ \mathcal{L}\{u(t)\} = \frac{1}{s} \]
  3. Use the standard pair for the exponential function, with a = 3: \[ \mathcal{L}\{e^{-3t}\} = \frac{1}{s+3} \]
  4. Use the standard pair for the sine function, with ω = 2: \[ \mathcal{L}\{\sin(2t)\} = \frac{2}{s^2 + 2^2} = \frac{2}{s^2+4} \]
  5. Combine the results with their coefficients: \[ F(s) = 5\left(\frac{1}{s}\right) - 2\left(\frac{1}{s+3}\right) + 4\left(\frac{2}{s^2+4}\right) \]
\[ F(s) = \frac{5}{s} - \frac{2}{s+3} + \frac{8}{s^2+4} \]
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Applications

Circuit Analysis & Design: Laplace transforms are used to analyze the transient and steady-state response of RLC (Resistor-Inductor-Capacitor) circuits. They simplify the analysis of circuits with capacitors and inductors by transforming integro-differential equations into algebraic equations in the s-domain.

Control Systems Engineering: The concept of a transfer function, which is the ratio of the output's Laplace transform to the input's Laplace transform, is central to control theory. It is used to analyze system stability, design controllers (like PID controllers), and predict system response to various inputs.

Mechanical System Analysis: The transform is used to solve the differential equations governing mechanical vibrations, such as in spring-mass-damper systems. It helps in analyzing the dynamic response of structures to forces and initial conditions.

Signal Processing: In signal processing, Laplace transforms are used to characterize linear time-invariant (LTI) systems and design analog filters. The s-plane representation of a system's transfer function provides deep insight into its filtering characteristics.

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Real-World Examples

An RLC series circuit has R = 10 Ω, L = 0.5 H, and C = 0.02 F. The circuit is initially at rest. At t=0, a 12V DC voltage source is connected. Find the s-domain expression for the current, I(s).
  1. Write the KVL equation in the time domain: \[ L\frac{di}{dt} + Ri + \frac{1}{C}\int i \,dt = v(t) \]
  2. Take the Laplace Transform of the entire equation. The input voltage is a step function: \[ v(t) = 12u(t) \implies V(s) = \frac{12}{s} \]
  3. The transformed equation (with zero initial conditions) is: \[ LsI(s) + RI(s) + \frac{1}{Cs}I(s) = V(s) \]
  4. Substitute the component values: \[ 0.5sI(s) + 10I(s) + \frac{1}{0.02s}I(s) = \frac{12}{s} \]
  5. Factor out I(s): \[ I(s) \left( 0.5s + 10 + \frac{50}{s} \right) = \frac{12}{s} \]
  6. Solve for I(s): \[ I(s) = \frac{12/s}{0.5s + 10 + 50/s} = \frac{12}{0.5s^2 + 10s + 50} = \frac{24}{s^2 + 20s + 100} \]
\[ I(s) = \frac{24}{(s+10)^2} \] (This corresponds to a critically damped response of the form \[ i(t) = 24te^{-10t} \])
A 2 kg mass is attached to a spring with a spring constant k = 8 N/m and a damper with a damping coefficient c = 4 Ns/m. The system is pushed from equilibrium with an initial velocity of 5 m/s. Find the s-domain expression for its position, X(s).
  1. The equation of motion is: \[ m\ddot{x} + c\dot{x} + kx = 0 \]
  2. Take the Laplace Transform, with initial conditions x(0)=0 and x'(0)=5: \[ m(s^2X(s) - sx(0) - x'(0)) + c(sX(s) - x(0)) + kX(s) = 0 \]
  3. Substitute the values and initial conditions: \[ 2(s^2X(s) - 5) + 4(sX(s)) + 8X(s) = 0 \]
  4. Simplify the equation: \[ 2s^2X(s) - 10 + 4sX(s) + 8X(s) = 0 \]
  5. Factor out X(s): \[ X(s)(2s^2 + 4s + 8) = 10 \]
  6. Solve for X(s): \[ X(s) = \frac{10}{2s^2 + 4s + 8} = \frac{5}{s^2 + 2s + 4} \]
\[ X(s) = \frac{5}{s^2 + 2s + 4} = \frac{5}{(s+1)^2 + (\sqrt{3})^2} \] (This corresponds to a damped sinusoidal motion)
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Real-World Scenarios

Σ PID Plant r G(s)=Kp+Ki/s+Kd·s step→1/s, ramp→1/s²
PID Controller Design
PID controllers are built from Laplace pairs: step input = 1/s, integrator = 1/s², derivative = s. The controller G(s) = Kp + Ki/s + Kd·s directly uses these standard pairs to achieve desired closed-loop performance.
σ × × −σ+jω₀ −σ−jω₀ zero cos(ωt)↔s/(s²+ω²)
Audio Equaliser (s-plane)
Each audio EQ band is a bandpass filter with conjugate poles at s = −σ ± jω₀. The pair cos(ω₀t) ↔ s/(s²+ω₀²) defines the frequency response — poles near the jω axis give sharp, high-Q resonance for precise frequency sculpting.
e^(−at) N/2 e^(−at) ↔ 1/(s+a) nuclear & pharmacology
Radioactive Decay
N(t) = N₀e^(−λt) is the most universal Laplace pair: e^(−at) ↔ 1/(s+a). The same pair models drug elimination, RC discharge, and population decline — the single pole at s = −a fully characterises the decay rate.

Automotive Suspension Systems: Engineers model a car's suspension as a spring-mass-damper system. They use Laplace transforms to analyze how the car will respond to bumps and potholes (impulse inputs) or gradual hills (step inputs). This analysis helps them choose the right spring stiffness and shock absorber damping to provide a ride that is both comfortable and safe.

Audio Equalizers: The filters in an audio equalizer (which adjust bass, midrange, and treble) are designed using s-domain analysis. Each frequency band corresponds to a specific range of poles and zeros on the s-plane. The Laplace transform pairs for RLC circuits are used directly to design the electronic filters that boost or cut these specific audio frequencies.

Power Grid Stability: When a large power plant or load connects to or disconnects from the electrical grid, it creates a transient disturbance. Power systems engineers use Laplace transforms to model these events and ensure the grid remains stable. They analyze the system's response to prevent oscillations that could lead to blackouts.

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Classification of Transform Pairs

Function TypeTime-Domain f(t)s-Domain F(s)
Singularity Functions\[ \delta(t), u(t) \]\[ 1, \frac{1}{s} \]
Power Functions\[ t^n u(t) \]\[ \frac{n!}{s^{n+1}} \]
Exponential Functions\[ e^{-at}u(t) \]\[ \frac{1}{s+a} \]
Trigonometric Functions\[ \sin(\omega t), \cos(\omega t) \]\[ \frac{\omega}{s^2+\omega^2}, \frac{s}{s^2+\omega^2} \]
Hyperbolic Functions\[ \sinh(at), \cosh(at) \]\[ \frac{a}{s^2-a^2}, \frac{s}{s^2-a^2} \]
Damped Functions\[ e^{-at}\sin(\omega t), e^{-at}\cos(\omega t) \]\[ \frac{\omega}{(s+a)^2+\omega^2}, \frac{s+a}{(s+a)^2+\omega^2} \]
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Common Mistakes

⚠️ Forgetting Causality (u(t)): The one-sided Laplace transform assumes functions are zero for t < 0. Many tables omit the explicit u(t) multiplier, but it's crucial. Forgetting this can lead to incorrect results when applying the time-shift property.
💡 Confusing Sine and Cosine Transforms: A common error is mixing up the numerators. Remember: sin(ωt) has the constant ω on top, while cos(ωt) has the variable s on top. A mnemonic is 'S for s-variable, S for coSine'.
💡 Incorrect Factorial for Power Rule: The transform for t^n involves n! (n factorial), which is only defined for non-negative integers. For non-integer powers, the more general Gamma function (Γ(n+1)) must be used instead of n!.
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Study Strategy

1 🧠 Build the Foundation
  • Thoroughly review the 'Definition of Laplace Transform Pairs' to understand the integral relationship between the time domain f(t) and the frequency domain F(s).
  • Study the 'Classification of Transform Pairs' to distinguish between algebraic, exponential, and trigonometric function types.
  • Analyze the 'Visualization' section to build an intuitive connection between the graph of a function and its corresponding transform.
  • Follow the 'Proof of the Unit Step Transform' to grasp the logical derivation process behind a fundamental pair.
2 ✍️ Memorize the Essentials
  • Focus on the 'Key Transform Pairs' table, creating flashcards for the top 10 most common functions (e.g., step, ramp, exponential, sine, cosine).
  • Practice writing the key pairs from memory, verifying against the formula sheet to identify and correct any errors.
  • Leverage the 'Properties' of the transform (linearity, time shift) to see how new pairs can be derived from the ones you've memorized.
  • Group related pairs together during memorization, such as sine and cosine, to reinforce their structural similarities.
3 ✏️ Practice with Examples
  • Work through the provided 'Worked Example' step-by-step, replicating the solution process on your own.
  • Find practice problems and apply the key pairs and properties to find the forward (f(t) to F(s)) and inverse (F(s) to f(t)) transforms.
  • Consult the 'Common Mistakes' section before starting practice problems to actively avoid frequent pitfalls like sign errors or incorrect constants.
  • Challenge yourself by solving problems that require combining multiple pairs using the linearity property.
4 ⚙️ Apply to Real Scenarios
  • Read the 'Applications' section to understand how transform pairs are crucial in fields like electrical engineering and control systems.
  • Connect the 'Real-World Examples' (e.g., RLC circuits) to specific transform pairs, identifying which pair models each physical behavior.
  • Analyze the 'Real-World Scenarios' and predict which transform pair would best represent the system's input signal or response.
  • Explore 'Related Formulas' like the Fourier Transform to understand how Laplace pairs fit into the broader context of signal analysis.
Mastering Laplace Transform Pairs becomes achievable by systematically building from core definitions to practical, real-world applications.

Frequently Asked Questions

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