Laplace Transform

calendar_today July 14, 2025 article 634 words access_time 5 min replay Modified July 21, 2025

Table of Contents

Transforms differential equations into algebraic systems
- Solve the system and convert back to time domain
LT of a signal $x(t)$:
$$ X(s) = \int_{-\infty}^{\infty} x(t) e^{-st} \, dt $$
iLT of $X(s)$:
$$ x(t) = \frac{1}{j 2 \pi} \int_{c - j\infty}^{c + j\infty} X(s) e^{st} \, ds $$
Use the table of pairs and properties [Laplace transform table (course handout)] to evaluate LT and iLT

Example 08.1 (Find iLT of $\frac{7s - 6}{s^2 - s - 6}$).

Example 08.2 (Find iLT of $\frac{2s^2 + 5}{s^2 + 3s + 2}$).

Example 08.3 (Find iLT of $\frac{s+1}{s^2 + 5s + 6}$).

Solving Circuits Using Laplace Transform

Take the Laplace transform of the system’s equations including any nonzero initial conditions and solve for $Y(s)$
The $s$-domain output is:
$$ Y(s) = H(s) X(s) + Y_{\text{IC}}(s) $$
where:
- $H(s)$ is the transfer function
- $X(s)$ is the Laplace transform of the input signal
- $Y_{\text{IC}}(s)$ is the Laplace terms resulting from the initial conditions
  - This is obtained by setting $X(s) = 0$
Take the inverse Laplace transform to get the time-domain output:
$$ y(t) = \mathcal{L}^{-1} \left\{ H(s) X(s) + Y_{\text{IC}}(s) \right\} $$
This gives the total response as the sum of:
- Zero-state response: $$ y_{\text{ZS}}(t) = \mathcal{L}^{-1} \left\{ H(s) X(s) \right\} $$
- Zero-input response: $$ y_{\text{ZI}}(t) = \mathcal{L}^{-1} \left\{ Y_{\text{IC}}(s) \right\} $$
So the complete output is: $$ y(t) = y_{\text{ZS}}(t) + y_{\text{ZI}}(t) $$

Transform the circuit into the $s$-domain equivalent circuit:
- Resistors remain unchanged
- Inductors and capacitors may be modeled as $s$-impedances in series with an ideal independent voltage source
  - Capacitor:

$$V_{C} = \frac{1}{sC}i + \frac{V_{0}}{s}$$

	  - Inductor:

$$V_{L} = sL i - L i_{0}$$

Assume all sources are sinusoidal and operating at the same frequency
Transform the circuit into the phasor (frequency) domain
Replace $s$ with $j\omega$ to convert from Laplace to steady-state analysis
Replace elements with their steady-state impedances
- Resistor: $R$
- Inductor: $j\omega L$
- Capacitor: $\frac{1}{j\omega C}$
Use phasor circuit analysis to solve for voltages and currents
Convert phasor results back to time domain using
$$ x(t) = \Re \left\{ \tilde{X} e^{j\omega t} \right\} $$ where $\tilde{X}$ is the phasor representation of the signal

Useful so that you don’t have to solve for initial and final conditions of current/voltage using transient analysis
Using IVT/FVT, we may rewrite the complete response of a first-order circuit as:

$$ x(t) = \lim_{s \to 0} s X(s) + \left( \lim_{s \to \infty} s X(s) - \lim_{s \to 0} s X(s) \right) e^{-\frac{t}{\tau}} $$

$$ \lim_{t \to 0^+} f(t) = \lim_{s \to \infty} s F(s) $$

Requirements:

$$ \lim_{t \to \infty} f(t) = \lim_{s \to 0} s F(s) $$

Requirements:

$$ \forall p \in \mathcal{P} \setminus \{0\},\ \Re(p) < 0 $$