Laplace Transform — Definition and Basics
Bilateral Laplace Transform
X(s) = \int_{-\infty}^{\infty} x(t)\, e^{-st}\, dt, \quad s = \sigma + j\omega
| Symbol | Description | Unit |
|---|---|---|
| X(s) | Laplace transform | depends on x(t) |
| s | Complex frequency variable | rad/s |
| \sigma | Real part of s (determines ROC) | Neper/s |
Worked example
Find the Laplace transform of x(t) = e^{-4t}u(t).
Given: x(t)=e^{-4t}u(t), a=4
- X(s) = ∫₀^∞ e^{-4t}e^{-st} dt = ∫₀^∞ e^{-(s+4)t} dt
- = [e^{-(s+4)t}/(-(s+4))]₀^∞
- Converges when Re(s+4) > 0, i.e., σ > −4
- X(s) = 1/(s+4)
Answer: X(s) = 1/(s+4), ROC: Re(s) > −4
Unilateral Laplace Transform
X(s) = \mathcal{L}\{x(t)\} = \int_{0^-}^{\infty} x(t)\, e^{-st}\, dt
| Symbol | Description | Unit |
|---|---|---|
| 0^- | Instant just before t=0 (captures initial conditions) | s |
Worked example
Find L{δ(t)} and L{u(t)}.
Given: δ(t): unit impulse; u(t): unit step
- L{δ(t)} = ∫₀^∞ δ(t)e^{-st}dt = e^{-s×0} = 1
- L{u(t)} = ∫₀^∞ e^{-st}dt = [e^{-st}/(−s)]₀^∞ = 0−(−1/s) = 1/s (for Re(s)>0)
Answer: L{δ(t)}=1; L{u(t)}=1/s, ROC: Re(s)>0
Inverse Laplace Transform (Bromwich Integral)
x(t) = \frac{1}{2\pi j} \int_{\sigma_0 - j\infty}^{\sigma_0 + j\infty} X(s)\, e^{st}\, ds
| Symbol | Description | Unit |
|---|---|---|
| \sigma_0 | Real number in the ROC of X(s) | Neper/s |
Worked example
Find x(t) by partial fractions for X(s) = (s+6)/[(s+2)(s+3)], ROC: Re(s) > −2.
Given: X(s)=(s+6)/[(s+2)(s+3)]
- X(s) = A/(s+2) + B/(s+3)
- A = (s+6)/(s+3)|_{s=-2} = (−2+6)/(−2+3) = 4/1 = 4
- B = (s+6)/(s+2)|_{s=-3} = (−3+6)/(−3+2) = 3/(−1) = −3
- x(t) = [4e^{-2t} − 3e^{-3t}]u(t)
Answer: x(t) = (4e^{-2t} − 3e^{-3t})u(t)
Laplace Transform Properties
Linearity
\mathcal{L}\{a\,x_1(t) + b\,x_2(t)\} = a\,X_1(s) + b\,X_2(s)
| Symbol | Description | Unit |
|---|---|---|
| a, b | Real or complex constants | dimensionless |
Worked example
Find L{5cos(3t) − 2sin(3t)} using linearity.
Given: L{cos(ωt)}=s/(s²+ω²), L{sin(ωt)}=ω/(s²+ω²), ω=3
- L{cos(3t)} = s/(s²+9)
- L{sin(3t)} = 3/(s²+9)
- L{5cos(3t)−2sin(3t)} = 5s/(s²+9) − 2×3/(s²+9)
- = (5s−6)/(s²+9)
Answer: X(s) = (5s−6)/(s²+9)
Time-Shifting (Delay) Property
x(t - a)\,u(t - a) \xrightarrow{\mathcal{L}} e^{-as} X(s), \quad a \geq 0
| Symbol | Description | Unit |
|---|---|---|
| a | Time delay | s |
Worked example
Find L{e^{-2(t-3)}u(t-3)}.
Given: X(s)=L{e^{-2t}u(t)}=1/(s+2), a=3
- Apply time-shift: L{e^{-2(t-3)}u(t-3)} = e^{-3s} × 1/(s+2)
- = e^{-3s}/(s+2)
Answer: L{e^{-2(t-3)}u(t-3)} = e^{-3s}/(s+2)
Differentiation in Time
\mathcal{L}\left\{\frac{dx}{dt}\right\} = s\,X(s) - x(0^-)
| Symbol | Description | Unit |
|---|---|---|
| x(0^-) | Initial value of x(t) just before t=0 | same as x(t) |
Worked example
Find L{dx/dt} if X(s)=5/(s+3) and x(0⁻)=2.
Given: X(s)=5/(s+3), x(0⁻)=2
- L{dx/dt} = s·X(s) − x(0⁻)
- = s·5/(s+3) − 2
- = 5s/(s+3) − 2(s+3)/(s+3)
- = (5s − 2s − 6)/(s+3) = (3s−6)/(s+3)
Answer: L{dx/dt} = (3s−6)/(s+3)
Integration in Time
\mathcal{L}\left\{\int_0^t x(\tau)\, d\tau\right\} = \frac{X(s)}{s}
| Symbol | Description | Unit |
|---|---|---|
| X(s)/s | Division by s in s-domain represents time integration | depends on x(t) |
Worked example
Find L{∫₀ᵗ e^{-3τ}dτ} without evaluating the integral.
Given: L{e^{-3t}} = 1/(s+3)
- L{∫₀ᵗ e^{-3τ}dτ} = L{e^{-3t}}/s
- = [1/(s+3)] / s
- = 1/[s(s+3)]
Answer: L = 1/[s(s+3)] (verify: partial fractions give (1/3)[1/s − 1/(s+3)] → (1/3)(1−e^{-3t}) ✓)
Initial Value Theorem
x(0^+) = \lim_{s \to \infty} s\, X(s)
| Symbol | Description | Unit |
|---|---|---|
| x(0^+) | Initial value of x(t) at t=0⁺ | same as x(t) |
Worked example
Find x(0⁺) for X(s) = (3s+5)/(s²+4s+3).
Given: X(s)=(3s+5)/(s²+4s+3)
- x(0⁺) = lim_{s→∞} s·(3s+5)/(s²+4s+3)
- = lim_{s→∞} (3s²+5s)/(s²+4s+3)
- Divide numerator and denominator by s²:
- = lim_{s→∞} (3+5/s)/(1+4/s+3/s²) = 3/1 = 3
Answer: x(0⁺) = 3
Final Value Theorem
x(\infty) = \lim_{s \to 0} s\, X(s) \quad \text{(valid only if poles of } sX(s) \text{ in LHP)}
| Symbol | Description | Unit |
|---|---|---|
| x(\infty) | Steady-state value of x(t) as t → ∞ | same as x(t) |
Worked example
A system has transfer function H(s) = 10/(s(s+5)). Find the final value of the step response.
Given: Y(s) = H(s)/s = 10/(s²(s+5))
- For unit step input: Y(s) = H(s)·(1/s) = 10/(s²(s+5))
- Wait — H(s)=10/(s(s+5)) already has a 1/s factor; step response Y(s)=H(s)/s=10/(s²(s+5)) — pole at s=0 is repeated. FVT requires poles of sY(s) in LHP.
- sY(s)=10/(s(s+5)) — still has pole at s=0; FVT not applicable directly. Instead use directly: sX(s)=s·10/[s(s+5)]=10/(s+5) → lim_{s→0}=10/5=2
- For H(s)=10/(s+5) with step: Y(s)=10/(s(s+5)), sY(s)=10/(s+5) → x(∞)=10/5=2
Answer: x(∞) = 2 (steady-state step response)
Common Laplace Transform Pairs
Damped Sinusoid
e^{-\alpha t}\cos(\omega_d t)\,u(t) \xrightarrow{\mathcal{L}} \frac{s + \alpha}{(s+\alpha)^2 + \omega_d^2}
| Symbol | Description | Unit |
|---|---|---|
| \alpha | Damping coefficient | Neper/s |
| \omega_d | Damped natural frequency | rad/s |
Worked example
Find x(t) for X(s)=(s+2)/[(s+2)²+16], ROC: Re(s)>−2.
Given: α=2, ω_d=4
- X(s) matches the pair with α=2, ω_d=4
- x(t) = e^{-2t}cos(4t)u(t)
Answer: x(t) = e^{-2t}cos(4t)u(t)
Polynomial in t (Ramp and Higher Powers)
t^n u(t) \xrightarrow{\mathcal{L}} \frac{n!}{s^{n+1}}, \quad n = 0, 1, 2, \ldots
| Symbol | Description | Unit |
|---|---|---|
| n! | Factorial of n | dimensionless |
Worked example
Find L{3t²u(t) + 5tu(t)} and identify the partial fraction expansion.
Given: n=2: 2!/s³=2/s³; n=1: 1!/s²=1/s²
- L{3t²u(t)} = 3×2!/s³ = 6/s³
- L{5tu(t)} = 5×1/s² = 5/s²
- X(s) = 6/s³ + 5/s² = (6+5s)/s³
Answer: X(s) = (5s+6)/s³
Shifted Exponential Times Cosine
e^{-\alpha t}\sin(\omega_d t)\,u(t) \xrightarrow{\mathcal{L}} \frac{\omega_d}{(s+\alpha)^2 + \omega_d^2}
| Symbol | Description | Unit |
|---|---|---|
| \omega_d | Damped frequency | rad/s |
| \alpha | Damping factor | Neper/s |
Worked example
Find x(t) for X(s) = 6/[(s+1)²+9].
Given: α=1, ω_d²=9 → ω_d=3, numerator=6=2×3
- Write 6 = 2×ω_d = 2×3
- X(s) = 2 × 3/[(s+1)²+9]
- x(t) = 2e^{-t}sin(3t)u(t)
Answer: x(t) = 2e^{-t}sin(3t)u(t)
Transfer Functions and System Analysis
Transfer Function Definition
H(s) = \frac{Y(s)}{X(s)} = \frac{\sum_{k=0}^{M} b_k s^k}{\sum_{k=0}^{N} a_k s^k}
| Symbol | Description | Unit |
|---|---|---|
| H(s) | Transfer function (ratio of output to input Laplace transforms with zero initial conditions) | dimensionless (or appropriate unit ratio) |
| b_k, a_k | Numerator and denominator polynomial coefficients | dimensionless |
Worked example
A system is described by: d²y/dt² + 5 dy/dt + 6y = 2 dx/dt + x. Find H(s).
Given: d²y/dt²+5dy/dt+6y = 2dx/dt+x, zero ICs
- Take Laplace: s²Y+5sY+6Y = 2sX+X
- Y(s²+5s+6) = X(2s+1)
- H(s) = Y/X = (2s+1)/(s²+5s+6) = (2s+1)/[(s+2)(s+3)]
Answer: H(s) = (2s+1)/[(s+2)(s+3)]
Poles, Zeros, and BIBO Stability
\text{BIBO stable} \Leftrightarrow \text{all poles have } \text{Re}(p_i) < 0
| Symbol | Description | Unit |
|---|---|---|
| p_i | Poles of H(s) (roots of denominator) | rad/s |
Worked example
Determine stability of H(s) = (s+1)/[(s+2)(s²+4s+5)].
Given: Denominator: (s+2)(s²+4s+5)
- Poles from (s+2)=0: s₁=−2 (Re=−2<0 ✓)
- Poles from (s²+4s+5)=0: s=−4/2 ± √(16−20)/2 = −2 ± j1
- Re(−2±j1)=−2<0 ✓
- All poles in LHP → BIBO stable
Answer: System is BIBO stable; poles at s=−2, −2±j1
Step Response from Transfer Function
Y_{step}(s) = \frac{H(s)}{s}
| Symbol | Description | Unit |
|---|---|---|
| Y_{step}(s) | Step response in s-domain | depends on H(s) |
Worked example
Find step response of H(s)=5/(s+5).
Given: H(s)=5/(s+5), step input → 1/s
- Y(s) = H(s)/s = 5/[s(s+5)]
- Partial fractions: A/s + B/(s+5)
- A = 5/(s+5)|_{s=0} = 5/5 = 1
- B = 5/s|_{s=-5} = 5/(−5) = −1
- Y(s) = 1/s − 1/(s+5)
- y(t) = (1 − e^{-5t})u(t)
Answer: y(t) = (1−e^{-5t})u(t) — first-order step response with τ=0.2 s
Circuit Analysis Using Laplace Transform
Impedance in s-Domain
Z_R = R, \quad Z_L = sL - Li(0^-), \quad Z_C = \frac{1}{sC} + \frac{v_C(0^-)}{s}
| Symbol | Description | Unit |
|---|---|---|
| i(0^-) | Initial inductor current | A |
| v_C(0^-) | Initial capacitor voltage | V |
Worked example
Series RLC: R=2Ω, L=1H, C=0.5F, zero initial conditions, step input V(s)=1/s. Find I(s).
Given: R=2, L=1, C=0.5, V(s)=1/s, ICs=0
- Z_total = R + sL + 1/(sC) = 2 + s + 2/s = (2s+s²+2)/s
- I(s) = V(s)/Z_total = (1/s)/[(s²+2s+2)/s] = 1/(s²+2s+2)
- s²+2s+2 = (s+1)²+1
- I(s) = 1/[(s+1)²+1]
- i(t) = e^{-t}sin(t)u(t)
Answer: i(t) = e^{-t}sin(t)u(t) — underdamped oscillation
Second-Order System Standard Form
H(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}
| Symbol | Description | Unit |
|---|---|---|
| \omega_n | Natural frequency | rad/s |
| \zeta | Damping ratio (ζ<1 underdamped, ζ=1 critically damped, ζ>1 overdamped) | dimensionless |
Worked example
Identify ω_n and ζ for H(s) = 100/(s²+10s+100). Classify the response.
Given: Denominator: s²+10s+100
- ω_n² = 100 → ω_n = 10 rad/s
- 2ζω_n = 10 → ζ = 10/(2×10) = 0.5
- ζ=0.5 < 1 → underdamped response
- ω_d = ω_n√(1-ζ²) = 10√(1-0.25) = 10×0.866 = 8.66 rad/s
Answer: ω_n=10 rad/s, ζ=0.5 (underdamped), ω_d=8.66 rad/s
Partial Fractions
Distinct Real Poles
X(s) = \frac{N(s)}{(s+a_1)(s+a_2)\cdots} = \frac{K_1}{s+a_1} + \frac{K_2}{s+a_2} + \cdots
| Symbol | Description | Unit |
|---|---|---|
| K_i | Residue at pole −a_i | depends on X(s) |
| K_i = (s+a_i)X(s)|_{s=-a_i} | Cover-up rule for residue | depends on X(s) |
Worked example
Find x(t) for X(s) = 10(s+3)/[(s+1)(s+2)(s+5)].
Given: Poles at s=−1,−2,−5
- K₁ = 10(s+3)/[(s+2)(s+5)]|_{s=-1} = 10(2)/[(1)(4)] = 20/4 = 5
- K₂ = 10(s+3)/[(s+1)(s+5)]|_{s=-2} = 10(1)/[(−1)(3)] = 10/(−3) = −10/3
- K₃ = 10(s+3)/[(s+1)(s+2)]|_{s=-5} = 10(−2)/[(−4)(−3)] = −20/12 = −5/3
- x(t) = [5e^{-t} − (10/3)e^{-2t} − (5/3)e^{-5t}]u(t)
Answer: x(t) = [5e^{-t} − (10/3)e^{-2t} − (5/3)e^{-5t}]u(t)
Repeated Poles
\frac{N(s)}{(s+a)^2} = \frac{K_1}{(s+a)^2} + \frac{K_2}{s+a}
| Symbol | Description | Unit |
|---|---|---|
| K_1 = (s+a)^2 X(s)|_{s=-a} | Residue at the repeated pole | depends on X(s) |
| K_2 = \frac{d}{ds}[(s+a)^2 X(s)]|_{s=-a} | Residue derivative rule | depends on X(s) |
Worked example
Find x(t) for X(s) = 4/[(s+2)²(s+4)].
Given: Poles: s=−2 (double), s=−4
- K₁ = 4/(s+4)|_{s=-2} = 4/2 = 2
- K₂ = d/ds[4/(s+4)]|_{s=-2} = −4/(s+4)²|_{s=-2} = −4/4 = −1
- K₃ = 4/(s+2)²|_{s=-4} = 4/4 = 1
- X(s) = 2/(s+2)² + (−1)/(s+2) + 1/(s+4)
- x(t) = [2te^{-2t} − e^{-2t} + e^{-4t}]u(t)
Answer: x(t) = [(2t−1)e^{-2t} + e^{-4t}]u(t)
Stability Analysis
Routh-Hurwitz Stability Criterion
\text{Stable if all elements of the first column of the Routh array are positive}
| Symbol | Description | Unit |
|---|---|---|
| a_n s^n + a_{n-1}s^{n-1} + \cdots + a_0 = 0 | Characteristic polynomial | dimensionless |
Worked example
Test stability of s³+2s²+3s+6=0 using Routh-Hurwitz.
Given: Coefficients: a3=1, a2=2, a1=3, a0=6
- Row 1: 1, 3
- Row 2: 2, 6
- Row 3: b1 = (2×3 − 1×6)/2 = (6−6)/2 = 0 ← zero row — system is marginally stable
- Auxiliary equation from row 2: 2s²+6=0 → s²=−3 → s=±j√3
- Poles on jω axis → marginally stable (not BIBO stable)
Answer: Marginally stable; poles at s = ±j√3 (jω-axis)
Characteristic Equation from Closed-Loop
1 + G(s)H(s) = 0 \Rightarrow \text{characteristic polynomial}
| Symbol | Description | Unit |
|---|---|---|
| G(s) | Forward path transfer function | dimensionless |
| H(s) | Feedback path transfer function | dimensionless |
Worked example
Unity feedback system: G(s)=K/(s(s+3)). Find range of K for stability.
Given: H(s)=1 (unity feedback)
- Characteristic equation: 1 + K/(s(s+3)) = 0 → s(s+3)+K=0 → s²+3s+K=0
- Routh array: Row 1: 1, K; Row 2: 3, 0
- b₁ = (3K−0)/3 = K
- For stability: all first-column terms positive: 1>0, 3>0, K>0
- Also K > 0 required
Answer: System stable for K > 0
Quick reference
| Formula | Expression |
|---|---|
| Unilateral Laplace Transform | X(s) = \int_{0^-}^{\infty}x(t)e^{-st}dt |
| Exponential Pair | e^{-at}u(t) \leftrightarrow 1/(s+a) |
| Unit Step Pair | u(t) \leftrightarrow 1/s |
| Ramp Pair | tu(t) \leftrightarrow 1/s^2 |
| Damped Cos Pair | e^{-\alpha t}\cos(\omega_d t)u(t) \leftrightarrow (s+\alpha)/[(s+\alpha)^2+\omega_d^2] |
| Differentiation Property | dx/dt \leftrightarrow sX(s)-x(0^-) |
| Integration Property | \int_0^t x\,d\tau \leftrightarrow X(s)/s |
| Time Shift | x(t-a)u(t-a) \leftrightarrow e^{-as}X(s) |
| Initial Value Theorem | x(0^+) = \lim_{s\to\infty}sX(s) |
| Final Value Theorem | x(\infty) = \lim_{s\to 0}sX(s) |
| Transfer Function | H(s) = Y(s)/X(s) |
| Standard 2nd-Order Form | H(s) = \omega_n^2/(s^2+2\zeta\omega_n s+\omega_n^2) |
| BIBO Stability | \text{All poles Re}(p_i)<0 |
| Closed-Loop CE | 1+G(s)H(s)=0 |
| Routh Stability | \text{All first-column elements positive} |
Exam tips
- GATE consistently awards marks for correctly applying initial and final value theorems — always verify that poles of sX(s) lie strictly in the LHP before stating a finite final value, since the theorem fails for marginal or unstable systems.
- In second-order system problems, examiners specifically check that you extract both ω_n and ζ before classifying the response; writing only ζ without ω_n loses partial credit in most marking schemes.
- For Routh-Hurwitz, a zero in the first column signals a root on the jω-axis (marginal stability) — treat it by substituting ε and taking the limit, or use the auxiliary equation approach.
- Partial fraction problems with repeated poles require the derivative rule for the second residue; the most common error is applying the cover-up rule to both coefficients of a double pole.
- When solving circuit problems using Laplace, model the capacitor as V_C(0⁻)/s in series and the inductor as LI(0⁻) in series with sL — ignoring initial conditions is the most frequent sign of an incomplete answer.
- Transfer function derivations from differential equations must assume zero initial conditions; examiners penalise responses that include x(0⁻) or y(0⁻) terms in H(s).