4.1 Laplace Transforms, Inverse Transforms & Circuit Applications

4.1 Laplace Transforms, Inverse Transforms & Circuit Applications

The Laplace transform is one of the most powerful analytical tools in electrical engineering and applied mathematics. It converts complex continuous-time differential equations governing dynamic physical systems into simple linear algebraic equations in the complex frequency domain ($s$-domain, where $s = \sigma + j\omega$). For the PRC Registered Electrical Engineer (REE) Licensure Examination, mastering Laplace transforms is essential for solving circuit transients, transfer functions, stability criteria, and control system responses.

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1. Definition and Region of Convergence

Direct Unilateral Laplace Transform

For a time-domain function $f(t)$ defined for $t \ge 0$, the unilateral direct Laplace transform is defined by integral: $\mathcal{L}\{f(t)\} = F(s) = \int_{0^-}^\infty f(t) e^{-st} \, dt$ where $s = \sigma + j\omega$ is the complex frequency parameter in $\text{rad/s}$, and lower limit $0^-$ captures potential instantaneous discontinuities (such as impulse functions $\delta(t)$) at $t = 0$.

Region of Convergence (ROC)

The Region of Convergence (ROC) consists of all points in the complex $s$-plane for which the defining integral converges absolutely. For causal signals, the ROC is a half-plane $\text{Re}{s} > \sigma_0$, where $\sigma_0$ is the axis of convergence.

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2. Standard Laplace Transform Pairs

The table below summarizes elementary transform pairs routinely encountered in engineering mathematics and circuit transient problems:

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3. Operational Theorems & Properties

Linearity

$\mathcal{L}\{a f_1(t) + b f_2(t)\} = a F_1(s) + b F_2(s)$

First Shifting Theorem (Frequency Shift)

$\mathcal{L}\{e^{-at} f(t)\} = F(s + a)$

Second Shifting Theorem (Time Shift)

$\mathcal{L}\{f(t - a) u(t - a)\} = e^{-as} F(s) \quad \text{for } a \ge 0$

Differentiation in Time Domain

• First Derivative: $\mathcal{L}\{f'(t)\} = s F(s) - f(0^-)$
• Second Derivative: $\mathcal{L}\{f''(t)\} = s^2 F(s) - s f(0^-) - f'(0^-)$
• $n$-th Derivative: $\mathcal{L}\{f^{(n)}(t)\} = s^n F(s) - s^{n-1} f(0^-) - s^{n-2} f'(0^-) - \dots - f^{(n-1)}(0^-)$

Integration in Time Domain

$\mathcal{L}\left\{ \int_0^t f(\tau) \, d\tau \right\} = \frac{F(s)}{s}$

Differentiation in $s$-Domain (Frequency Differentiation)

$\mathcal{L}\{t \cdot f(t)\} = -\frac{d}{ds} F(s)$ $\mathcal{L}\{t^n \cdot f(t)\} = (-1)^n \frac{d^n}{ds^n} F(s)$

Convolution Theorem

The Laplace transform of the time-domain convolution of two signals equals the algebraic product of their individual transforms: $\mathcal{L}\{f_1(t) * f_2(t)\} = \mathcal{L}\left\{ \int_0^t f_1(\tau) f_2(t - \tau) \, d\tau \right\} = F_1(s) \cdot F_2(s)$

Boundary Value Theorems

• Initial Value Theorem (IVT): If $f(t)$ and $f'(t)$ are Laplace transformable, then: $f(0^+) = \lim_{t \to 0^+} f(t) = \lim_{s \to \infty} s F(s)$
• Final Value Theorem (FVT): If all poles of $s F(s)$ lie strictly in the left-half of the $s$-plane (system is stable), then: $f(\infty) = \lim_{t \to \infty} f(t) = \lim_{s \to 0} s F(s)$

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4. Inverse Laplace Transform & Partial Fractions

To convert an algebraic $s$-domain expression $F(s) = \frac{P(s)}{Q(s)}$ back into the time domain $f(t)$, Partial Fraction Expansion decomposes proper rational functions into sum terms matching standard table forms.

Case I: Distinct Real Poles

If $Q(s) = (s - p_1)(s - p_2) \dots (s - p_n)$ with distinct real roots: $F(s) = \frac{A_1}{s - p_1} + \frac{A_2}{s - p_2} + \dots + \frac{A_n}{s - p_n}$ Residue formula for coefficient $A_k$: $A_k = \lim_{s \to p_k} (s - p_k) F(s)$

Case II: Repeated Real Poles

If $Q(s)$ contains a repeated factor $(s - p)^m$ of multiplicity $m$: $F(s) = \frac{A_1}{s - p} + \frac{A_2}{(s - p)^2} + \dots + \frac{A_m}{(s - p)^m} + \text{other terms}$ Residue formula for $A_{m-k}$: $A_{m-k} = \frac{1}{k!} \lim_{s \to p} \frac{d^k}{ds^k} \left[ (s - p)^m F(s) \right]$

Case III: Complex Conjugate Poles

If $Q(s)$ contains quadratic term $(s + a)^2 + \omega^2$ with roots $s = -a \pm j\omega$: $\frac{B s + C}{(s + a)^2 + \omega^2} = \frac{B(s + a) + (C - a B)}{(s + a)^2 + \omega^2} = B \frac{s + a}{(s + a)^2 + \omega^2} + \frac{C - a B}{\omega} \frac{\omega}{(s + a)^2 + \omega^2}$ Taking the inverse transform yields: $f(t) = B e^{-at} \cos(\omega t) + \frac{C - a B}{\omega} e^{-at} \sin(\omega t)$

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5. $s$-Domain Electrical Circuit Models

By transforming circuit elements into $s$-domain impedances along with initial condition energy sources, Kirchhoff's Laws (KVL and KCL) can be applied directly using standard linear algebra.

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Step-by-Step Worked Numerical Examples

Example 4.1.1: Inverse Laplace Transform via Partial Fraction Expansion

Problem: Determine the inverse Laplace transform $f(t) = \mathcal{L}^{-1}{F(s)}$ for: $F(s) = \frac{10 s + 40}{s^2 + 6 s + 25}$

Solution:

1. Analyze the denominator quadratic: Discriminant $\Delta = 6^2 - 4(1)(25) = 36 - 100 = -64 < 0$. The poles are complex conjugate pairs.
2. Complete the square for the denominator: $s^2 + 6 s + 25 = (s^2 + 6 s + 9) + 16 = (s + 3)^2 + 4^2$
3. Rewrite the numerator in terms of $(s + 3)$: $10 s + 40 = 10(s + 3) + 10$
4. Split into standard damped cosine and sine forms: $F(s) = \frac{10(s + 3) + 10}{(s + 3)^2 + 4^2} = 10 \cdot \frac{s + 3}{(s + 3)^2 + 4^2} + \frac{10}{4} \cdot \frac{4}{(s + 3)^2 + 4^2}$ $F(s) = 10 \cdot \frac{s + 3}{(s + 3)^2 + 4^2} + 2.5 \cdot \frac{4}{(s + 3)^2 + 4^2}$
5. Apply inverse transform matching table pairs: $\mathcal{L}^{-1}\left\{ \frac{s + 3}{(s + 3)^2 + 4^2} \right\} = e^{-3t} \cos(4t)$ $\mathcal{L}^{-1}\left\{ \frac{4}{(s + 3)^2 + 4^2} \right\} = e^{-3t} \sin(4t)$ $f(t) = 10 e^{-3t} \cos(4t) + 2.5 e^{-3t} \sin(4t) \quad \text{for } t \ge 0$

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Example 4.1.2: $s$-Domain Transient Response of Series RLC Circuit

Problem: A series RLC circuit consists of $R = 6\ \Omega$, $L = 1\text{ H}$, and $C = 0.04\text{ F} = \frac{1}{25}\text{ F}$. A step voltage source $v(t) = 100 u(t)\text{ V}$ is connected at $t = 0$. Assuming zero initial energy ($i(0^-) = 0\text{ A}, v_C(0^-) = 0\text{ V}$), derive the current response $i(t)$.

Solution:

1. Formulate the $s$-domain KVL equation: $V(s) = I(s) \left[ R + s L + \frac{1}{s C} \right]$ Substitute $V(s) = \frac{100}{s}$, $R = 6$, $L = 1$, and $\frac{1}{C} = 25$: $\frac{100}{s} = I(s) \left[ 6 + s + \frac{25}{s} \right] = I(s) \left[ \frac{s^2 + 6s + 25}{s} \right]$
2. Solve algebraically for $I(s)$: $I(s) = \frac{100/s}{(s^2 + 6s + 25)/s} = \frac{100}{s^2 + 6s + 25}$
3. Complete the square in the denominator: $s^2 + 6s + 25 = (s + 3)^2 + 4^2$
4. Format $I(s)$ for inverse transformation: $I(s) = \frac{100}{(s + 3)^2 + 4^2} = \frac{100}{4} \cdot \frac{4}{(s + 3)^2 + 4^2} = 25 \cdot \frac{4}{(s + 3)^2 + 4^2}$
5. Take inverse Laplace transform: $i(t) = 25 e^{-3t} \sin(4t) \text{ A} \quad \text{for } t \ge 0$

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Example 4.1.3: Boundary Analysis via Initial and Final Value Theorems

Problem: A control system closed-loop output in the frequency domain is given by: $Y(s) = \frac{5 s + 20}{s (s^2 + 4 s + 13)}$ Find the initial value $y(0^+)$ and the steady-state final value $y(\infty)$.

Solution:

1. Apply Initial Value Theorem (IVT): $y(0^+) = \lim_{s \to \infty} s Y(s) = \lim_{s \to \infty} s \left[ \frac{5 s + 20}{s (s^2 + 4 s + 13)} \right] = \lim_{s \to \infty} \frac{5 s + 20}{s^2 + 4 s + 13}$ Divide numerator and denominator by $s^2$: $y(0^+) = \lim_{s \to \infty} \frac{5/s + 20/s^2}{1 + 4/s + 13/s^2} = \frac{0 + 0}{1 + 0 + 0} = 0$
2. Apply Final Value Theorem (FVT): Check pole stability: Poles of $s Y(s)$ are roots of $s^2 + 4s + 13 = 0 \implies s = -2 \pm j3$, which both reside in the left-half $s$-plane (stable system). $y(\infty) = \lim_{s \to 0} s Y(s) = \lim_{s \to 0} s \left[ \frac{5 s + 20}{s (s^2 + 4 s + 13)} \right] = \lim_{s \to 0} \frac{5 s + 20}{s^2 + 4 s + 13}$ Substitute $s = 0$: $y(\infty) = \frac{5(0) + 20}{0^2 + 4(0) + 13} = \frac{20}{13} \approx 1.5385$
3. Conclusion: $y(0^+) = 0$ and $y(\infty) = 1.5385$.