Training Electrical Engineering The Laplace Transform in Circuits
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The Laplace Transform in Circuits

24 min Electrical Engineering

The Laplace Transform in Circuits

The Laplace transform converts differential equations into algebraic equations, making circuit analysis of transient and steady-state responses straightforward.

Laplace Transform

$$\mathcal{L}\{f(t)\} = F(s) = \int_0^{\infty} f(t) e^{-st}\,dt$$

Key Transform Pairs
$f(t)$$F(s)$
$1$ (step)$1/s$
$e^{-at}$$1/(s+a)$
$t$$1/s^2$
$\sin(\omega t)$$\omega/(s^2+\omega^2)$
$\cos(\omega t)$$s/(s^2+\omega^2)$
$t e^{-at}$$1/(s+a)^2$
Example 1

An RC circuit with $R = 1$ kΩ, $C = 1\,\mu$F is driven by a 5 V step. Find $v_C(t)$.

Transfer function: $H(s) = \frac{1}{\tau s + 1}$ where $\tau = RC = 10^{-3}$ s.

$V_C(s) = H(s) \cdot \frac{5}{s} = \frac{5}{s(\tau s + 1)}$

Partial fractions: $\frac{5}{s} - \frac{5}{s + 1/\tau}$

$$v_C(t) = 5(1 - e^{-t/\tau}) = 5(1 - e^{-1000t}) \text{ V}$$

Example 2

Find the inverse Laplace transform of $F(s) = \dfrac{3}{(s+1)(s+4)}$.

Partial fractions: $\frac{3}{(s+1)(s+4)} = \frac{A}{s+1} + \frac{B}{s+4}$

$A = 3/(4-1) = 1$, $B = 3/(1-4) = -1$

$$f(t) = e^{-t} - e^{-4t}$$

Example 3

An RL circuit: $L = 0.5$ H, $R = 10\,\Omega$, step input $V = 20$ V. Find $i(t)$.

$I(s) = \frac{V/L}{s(s + R/L)} = \frac{40}{s(s + 20)}$

Partial fractions: $\frac{2}{s} - \frac{2}{s+20}$

$$i(t) = 2(1 - e^{-20t}) \text{ A}$$

Final value: $i(\infty) = V/R = 2$ A ✓

Practice Problems

1. Find $\mathcal{L}\{3e^{-2t}\}$.
2. Find $\mathcal{L}^{-1}\left\{\dfrac{5}{s+3}\right\}$.
3. Find $\mathcal{L}^{-1}\left\{\dfrac{s}{s^2+9}\right\}$.
4. Solve $\frac{di}{dt} + 5i = 10$ with $i(0) = 0$ using Laplace transforms.
5. RC circuit: $R = 500\,\Omega$, $C = 2\,\mu$F, step 10 V. Find $v_C(t)$.
6. Find the final value of $F(s) = \dfrac{10}{s(s+2)}$ without inverting.
7. Find $\mathcal{L}\{t^2\}$.
8. An RLC series circuit with $R=10$, $L=1$ H, $C=0.01$ F. Step input 12 V. Find the characteristic equation and classify the response.
9. Find $\mathcal{L}^{-1}\left\{\dfrac{2s+1}{s^2+4s+5}\right\}$.
10. If $F(s) = \frac{1}{(s+1)^2}$, find $f(t)$.
11. Use the initial value theorem to find $f(0^+)$ for $F(s) = \frac{3s+2}{s^2+5s+6}$.
12. A second-order system has step response $y(t) = 1 - e^{-t}(\cos 2t + 0.5\sin 2t)$. Find the transfer function.
Show Answer Key

1. $3/(s+2)$

2. $5e^{-3t}$

3. $\cos 3t$

4. $sI - 0 + 5I = 10/s$; $I = 10/(s(s+5))$; $i(t) = 2(1-e^{-5t})$

5. $\tau = 10^{-3}$ s; $v_C(t) = 10(1 - e^{-1000t})$ V

6. FVT: $\lim_{s \to 0} sF(s) = 10/2 = 5$

7. $2/s^3$

8. $s^2 + 10s + 100 = 0$; $\Delta = 100 - 400 < 0$ → underdamped; $\zeta = 0.5$, $\omega_n = 10$.

9. Complete the square: $s^2+4s+5 = (s+2)^2+1$; $F = \frac{2(s+2)-3}{(s+2)^2+1}$; $f(t) = e^{-2t}(2\cos t - 3\sin t)$

10. $f(t) = te^{-t}$

11. IVT: $f(0^+) = \lim_{s\to\infty} sF(s) = \lim \frac{3s^2+2s}{s^2+5s+6} = 3$

12. Poles at $s = -1 \pm 2j$; $H(s) = \frac{5}{s^2+2s+5}$ (since $\omega_n^2 = 5$)

Transfer Functions, Poles & Zeros Placement Test Practice — Electrical Engineering