15. Laplace Transform

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15. Laplace Transform

Learning Objectives

  • Understand the definition and existence conditions of the Laplace transform, and explain the concept of the region of convergence
  • Derive Laplace transforms of basic functions and use transform tables to calculate transforms of complex functions
  • Prove and apply major properties of the Laplace transform: shifting theorems, differentiation properties, convolution theorem, etc.
  • Systematically solve initial value problems for ordinary differential equations using partial fraction decomposition and inverse Laplace transforms
  • Understand the basic principles of linear system analysis and stability determination using transfer functions
  • Apply Laplace transforms to physics and engineering problems such as RLC circuits and damped oscillations

Importance in physics and engineering: The Laplace transform is a core tool that enables systematic solution of initial value problems by converting differential equations into algebraic equations. It is essential in nearly all engineering fields including circuit analysis, control engineering, signal processing, mechanical vibration, and heat conduction. As a generalization of the Fourier transform, it is particularly powerful for analyzing transient responses.

1. Definition and Existence Conditions

1.1 Definition

The Laplace transform of a function $f(t)$ ($t \geq 0$) is defined as:

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

where $s = \sigma + i\omega$ is a complex variable. $F(s)$ is defined in the region of $s$ where this integral converges.

Intuitively, the Laplace transform converts a function $f(t)$ in the time domain to a function $F(s)$ in the complex frequency domain. The real part $\sigma$ represents exponential decay/growth, and the imaginary part $\omega$ represents oscillation.

1.2 Existence Conditions

Sufficient conditions for the existence of the Laplace transform:

  1. Piecewise continuous: $f(t)$ is continuous except at a finite number of discontinuities on any finite interval $[0, T]$
  2. Exponential order: There exist constants $M > 0$, $a \in \mathbb{R}$, $T > 0$ such that

$$|f(t)| \leq M e^{at}, \quad t > T$$

If these conditions are satisfied, the integral converges for all $s$ with $\text{Re}(s) > a$.

1.3 Region of Convergence

The region of convergence (ROC) is the region in the complex plane where the Laplace integral converges:

$$\text{ROC} = \{ s \in \mathbb{C} \mid \text{Re}(s) > \sigma_c \}$$

where $\sigma_c$ is called the abscissa of convergence. For example: - $f(t) = 1$: $\sigma_c = 0$ (i.e., $\text{Re}(s) > 0$) - $f(t) = e^{at}$: $\sigma_c = a$ (i.e., $\text{Re}(s) > a$) - $f(t) = e^{-t}\sin t$: $\sigma_c = -1$

import sympy as sp
import numpy as np
import matplotlib.pyplot as plt

# SymPy๋ฅผ ์ด์šฉํ•œ ๋ผํ”Œ๋ผ์Šค ๋ณ€ํ™˜ ๊ณ„์‚ฐ
t, s = sp.symbols('t s', positive=True)

# ๊ธฐ๋ณธ ํ•จ์ˆ˜๋“ค์˜ ๋ผํ”Œ๋ผ์Šค ๋ณ€ํ™˜
funcs = {
    '1': 1,
    't': t,
    't^2': t**2,
    'exp(-2t)': sp.exp(-2*t),
    'sin(3t)': sp.sin(3*t),
    'cos(3t)': sp.cos(3*t),
}

print("=" * 50)
print("๊ธฐ๋ณธ ํ•จ์ˆ˜์˜ ๋ผํ”Œ๋ผ์Šค ๋ณ€ํ™˜")
print("=" * 50)
for name, f in funcs.items():
    F = sp.laplace_transform(f, t, s, noconds=True)
    print(f"L{{{name}}} = {F}")

1.4 Relationship with Fourier Transform

The Laplace transform is a generalization of the Fourier transform. Setting $s = i\omega$ (and if $f(t) = 0$ for $t < 0$):

$$F(i\omega) = \int_0^\infty f(t) e^{-i\omega t} \, dt$$

This is identical to the one-sided Fourier transform. The Laplace transform multiplies by an additional convergence factor $e^{-\sigma t}$, making it applicable to functions where the Fourier transform doesn't exist (e.g., $f(t) = e^{2t}$).

2. Laplace Transforms of Basic Functions

2.1 Transform Table

The table below summarizes the most important Laplace transform pairs:

$f(t)$ ($t \geq 0$) $F(s) = \mathcal{L}\{f(t)\}$ Convergence condition
$1$ $\dfrac{1}{s}$ $\text{Re}(s) > 0$
$t^n$ ($n = 0, 1, 2, \ldots$) $\dfrac{n!}{s^{n+1}}$ $\text{Re}(s) > 0$
$e^{at}$ $\dfrac{1}{s - a}$ $\text{Re}(s) > a$
$\sin(bt)$ $\dfrac{b}{s^2 + b^2}$ $\text{Re}(s) > 0$
$\cos(bt)$ $\dfrac{s}{s^2 + b^2}$ $\text{Re}(s) > 0$
$\sinh(bt)$ $\dfrac{b}{s^2 - b^2}$ $\text{Re}(s) > |b|$
$\cosh(bt)$ $\dfrac{s}{s^2 - b^2}$ $\text{Re}(s) > |b|$
$t^n e^{at}$ $\dfrac{n!}{(s-a)^{n+1}}$ $\text{Re}(s) > a$
$e^{at}\sin(bt)$ $\dfrac{b}{(s-a)^2 + b^2}$ $\text{Re}(s) > a$
$e^{at}\cos(bt)$ $\dfrac{s-a}{(s-a)^2 + b^2}$ $\text{Re}(s) > a$

Example derivation: Computing $\mathcal{L}\{e^{at}\}$ directly

$$\int_0^\infty e^{at} e^{-st} \, dt = \int_0^\infty e^{-(s-a)t} \, dt = \left[ \frac{e^{-(s-a)t}}{-(s-a)} \right]_0^\infty = \frac{1}{s-a}$$

provided that $\text{Re}(s) > a$ so the exponential converges to 0 as $t \to \infty$.

2.2 Unit Step Function (Heaviside Function)

Heaviside function definition:

$$u(t - a) = \begin{cases} 0, & t < a \\ 1, & t \geq a \end{cases}$$

Laplace transform:

$$\mathcal{L}\{u(t-a)\} = \int_a^\infty e^{-st} \, dt = \frac{e^{-as}}{s}, \quad a \geq 0$$

2.3 Dirac Delta Function

$\delta(t - a)$ represents a unit impulse at $t = a$:

$$\mathcal{L}\{\delta(t - a)\} = \int_0^\infty \delta(t - a) e^{-st} \, dt = e^{-as}, \quad a \geq 0$$

In particular, when $a = 0$, $\mathcal{L}\{\delta(t)\} = 1$. This means the Laplace transform of an impulse input is the constant 1.

import sympy as sp

t, s, a = sp.symbols('t s a', positive=True)

# ๋‹จ์œ„ ๊ณ„๋‹จ ํ•จ์ˆ˜์˜ ๋ผํ”Œ๋ผ์Šค ๋ณ€ํ™˜
heaviside_transform = sp.laplace_transform(sp.Heaviside(t - a), t, s, noconds=True)
print(f"L{{u(t-a)}} = {heaviside_transform}")

# ๋””๋ž™ ๋ธํƒ€ ํ•จ์ˆ˜์˜ ๋ผํ”Œ๋ผ์Šค ๋ณ€ํ™˜
delta_transform = sp.laplace_transform(sp.DiracDelta(t - a), t, s, noconds=True)
print(f"L{{delta(t-a)}} = {delta_transform}")

# ๊ธฐ๋ณธ ๋ณ€ํ™˜ ์Œ ์ˆ˜์น˜ ๊ฒ€์ฆ: L{sin(3t)} = 3/(s^2+9)
import numpy as np
from scipy import integrate

def numerical_laplace(f, s_val, upper=50):
    """๋ผํ”Œ๋ผ์Šค ๋ณ€ํ™˜์˜ ์ˆ˜์น˜ ๊ณ„์‚ฐ"""
    integrand = lambda tau: f(tau) * np.exp(-s_val * tau)
    result, _ = integrate.quad(integrand, 0, upper)
    return result

s_test = 2.0
# ์ˆ˜์น˜ ๊ฒฐ๊ณผ
numerical = numerical_laplace(lambda tau: np.sin(3*tau), s_test)
# ํ•ด์„์  ๊ฒฐ๊ณผ: 3/(s^2+9)
analytical = 3 / (s_test**2 + 9)

print(f"\nL{{sin(3t)}} ๊ฒ€์ฆ (s={s_test}):")
print(f"  ์ˆ˜์น˜ ์ ๋ถ„: {numerical:.6f}")
print(f"  ํ•ด์„์ :    {analytical:.6f}")
print(f"  ์˜ค์ฐจ:      {abs(numerical - analytical):.2e}")

3. Properties of the Laplace Transform

3.1 Linearity

The Laplace transform is a linear operator:

$$\mathcal{L}\{\alpha f(t) + \beta g(t)\} = \alpha F(s) + \beta G(s)$$

This follows directly from the linearity of integration.

3.2 First Shifting Theorem (s-shift)

If $\mathcal{L}\{f(t)\} = F(s)$, then:

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

Proof:

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

Application example: To find $\mathcal{L}\{e^{-2t}\cos(3t)\}$

$$\mathcal{L}\{\cos(3t)\} = \frac{s}{s^2 + 9}$$

Replace $s$ with $s + 2$:

$$\mathcal{L}\{e^{-2t}\cos(3t)\} = \frac{s+2}{(s+2)^2 + 9}$$

3.3 Second Shifting Theorem (t-shift)

If $\mathcal{L}\{f(t)\} = F(s)$, then:

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

Proof: Using the substitution $\tau = t - a$

$$\int_0^\infty f(t-a)\,u(t-a)\,e^{-st} \, dt = \int_a^\infty f(t-a)\,e^{-st} \, dt = \int_0^\infty f(\tau)\,e^{-s(\tau+a)} \, d\tau = e^{-as}F(s)$$

This theorem is fundamental for transforms of time-delayed signals.

3.4 Differentiation Property

The most powerful property of the Laplace transform is that it converts differentiation into algebraic operations:

$$\mathcal{L}\{f'(t)\} = sF(s) - f(0)$$

$$\mathcal{L}\{f''(t)\} = s^2 F(s) - sf(0) - f'(0)$$

Generally, the transform of the $n$-th derivative is:

$$\mathcal{L}\{f^{(n)}(t)\} = s^n F(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \cdots - f^{(n-1)}(0)$$

Proof (first derivative): Applying integration by parts

$$\int_0^\infty f'(t)e^{-st} \, dt = \left[f(t)e^{-st}\right]_0^\infty + s\int_0^\infty f(t)e^{-st} \, dt = -f(0) + sF(s)$$

3.5 Integration Property

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

Integration in the time domain corresponds to division by $s$ in the $s$-domain.

3.6 Convolution Theorem

The convolution of two functions is:

$$(f * g)(t) = \int_0^t f(\tau) \, g(t - \tau) \, d\tau$$

Convolution theorem:

$$\mathcal{L}\{f * g\} = F(s) \cdot G(s)$$

That is, convolution in the time domain corresponds to multiplication in the $s$-domain.

Proof: Using interchange of integration order

$$\mathcal{L}\{f * g\} = \int_0^\infty \left(\int_0^t f(\tau)g(t-\tau) \, d\tau \right) e^{-st} \, dt$$

Substituting $u = t - \tau$ and interchanging integration order yields $F(s) \cdot G(s)$.

3.7 Initial Value and Final Value Theorems

Initial value theorem: $f(0^+) = \lim_{s \to \infty} sF(s)$

Final value theorem: $\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s)$

However, the final value theorem is valid only when all poles of $sF(s)$ are in the left half of the complex plane (i.e., when the system is stable).

import sympy as sp

t, s = sp.symbols('t s')
a_sym = sp.Symbol('a', positive=True)

# ์ œ1์ด๋™ ์ •๋ฆฌ ๊ฒ€์ฆ: L{e^(-2t)*cos(3t)}
f1 = sp.exp(-2*t) * sp.cos(3*t)
F1_direct = sp.laplace_transform(f1, t, s, noconds=True)
F1_shift = (s + 2) / ((s + 2)**2 + 9)  # s-์ด๋™ ์ ์šฉ
print("=== ์ œ1์ด๋™ ์ •๋ฆฌ ๊ฒ€์ฆ ===")
print(f"์ง์ ‘ ๋ณ€ํ™˜:  {F1_direct}")
print(f"์ด๋™ ์ •๋ฆฌ:  {F1_shift}")
print(f"๋™์ผ ์—ฌ๋ถ€:  {sp.simplify(F1_direct - F1_shift) == 0}")

# ๋ฏธ๋ถ„ ์„ฑ์งˆ ๊ฒ€์ฆ: L{f'(t)} = sF(s) - f(0)
# f(t) = t*exp(-t), f(0) = 0
f2 = t * sp.exp(-t)
f2_prime = sp.diff(f2, t)  # (1-t)*exp(-t)

F2 = sp.laplace_transform(f2, t, s, noconds=True)  # 1/(s+1)^2
F2_prime_direct = sp.laplace_transform(f2_prime, t, s, noconds=True)
F2_prime_property = s * F2 - 0  # f(0) = 0

print("\n=== ๋ฏธ๋ถ„ ์„ฑ์งˆ ๊ฒ€์ฆ ===")
print(f"L{{f'(t)}} ์ง์ ‘:  {F2_prime_direct}")
print(f"sF(s) - f(0):    {sp.simplify(F2_prime_property)}")

# ํ•ฉ์„ฑ๊ณฑ ์ •๋ฆฌ ๊ฒ€์ฆ: L{sin(t) * sin(t)} = [1/(s^2+1)]^2
conv_result = sp.laplace_transform(
    sp.integrate(sp.sin(t - sp.Symbol('tau')) * sp.sin(sp.Symbol('tau')),
                 (sp.Symbol('tau'), 0, t)), t, s, noconds=True
)
product_result = 1 / (s**2 + 1)**2
print("\n=== ํ•ฉ์„ฑ๊ณฑ ์ •๋ฆฌ ===")
print(f"F(s)*G(s) = 1/(s^2+1)^2 = {product_result}")

4. Inverse Laplace Transform

4.1 Partial Fraction Decomposition

The key technique for inverse Laplace transforms is to decompose $F(s)$ into partial fractions and then apply the transform table in reverse.

Case 1: Distinct real roots

$$\frac{P(s)}{(s-a_1)(s-a_2)\cdots(s-a_n)} = \frac{A_1}{s-a_1} + \frac{A_2}{s-a_2} + \cdots + \frac{A_n}{s-a_n}$$

Case 2: Repeated roots

$$\frac{P(s)}{(s-a)^n} = \frac{A_1}{s-a} + \frac{A_2}{(s-a)^2} + \cdots + \frac{A_n}{(s-a)^n}$$

Case 3: Complex conjugate roots

$$\frac{P(s)}{s^2 + bs + c} = \frac{As + B}{s^2 + bs + c}$$

Then complete the square and apply $\sin$, $\cos$ transform pairs.

4.2 Heaviside Cover-up Method

For distinct first-order factors, coefficients can be quickly found. The coefficient of the term with denominator $(s - a_k)$ in $F(s)$:

$$A_k = \left[(s - a_k) F(s)\right]_{s = a_k}$$

Example: Find the inverse transform of $F(s) = \dfrac{3s + 2}{(s+1)(s-2)}$

$$A_1 = \left[\frac{3s+2}{s-2}\right]_{s=-1} = \frac{-1}{-3} = \frac{1}{3}$$

$$A_2 = \left[\frac{3s+2}{s+1}\right]_{s=2} = \frac{8}{3}$$

Therefore $f(t) = \frac{1}{3}e^{-t} + \frac{8}{3}e^{2t}$

4.3 Bromwich Integral

The rigorous formula for the inverse Laplace transform is given by a complex integral:

$$f(t) = \mathcal{L}^{-1}\{F(s)\} = \frac{1}{2\pi i} \int_{\gamma - i\infty}^{\gamma + i\infty} F(s) e^{st} \, ds$$

where $\gamma$ is a real number greater than the real part of all singularities of $F(s)$. This integral can be computed using the residue theorem, a direct application of complex analysis from Chapter 12.

import sympy as sp

s, t = sp.symbols('s t')

print("=== ๋ถ€๋ถ„๋ถ„์ˆ˜ ๋ถ„ํ•ด์™€ ์—ญ ๋ผํ”Œ๋ผ์Šค ๋ณ€ํ™˜ ===\n")

# ์˜ˆ์ œ 1: ์„œ๋กœ ๋‹ค๋ฅธ ์‹ค์ˆ˜ ๊ทผ
F1 = (3*s + 2) / ((s + 1) * (s - 2))
print(f"F(s) = {F1}")
pf1 = sp.apart(F1, s)
print(f"๋ถ€๋ถ„๋ถ„์ˆ˜: {pf1}")
f1 = sp.inverse_laplace_transform(F1, s, t)
print(f"f(t) = {f1}\n")

# ์˜ˆ์ œ 2: ์ค‘๊ทผ
F2 = (2*s + 3) / (s + 1)**2
print(f"F(s) = {F2}")
pf2 = sp.apart(F2, s)
print(f"๋ถ€๋ถ„๋ถ„์ˆ˜: {pf2}")
f2 = sp.inverse_laplace_transform(F2, s, t)
print(f"f(t) = {f2}\n")

# ์˜ˆ์ œ 3: ๋ณต์†Œ ์ผค๋ ˆ๊ทผ
F3 = (s + 3) / (s**2 + 2*s + 5)
print(f"F(s) = {F3}")
# ์™„์ „์ œ๊ณฑ์‹: (s+1)^2 + 4 -> e^(-t)cos(2t) + e^(-t)sin(2t)
f3 = sp.inverse_laplace_transform(F3, s, t)
print(f"f(t) = {f3}\n")

# ์˜ˆ์ œ 4: ๊ณ ์ฐจ ๋ถ„๋ชจ
F4 = 1 / (s * (s**2 + 4))
print(f"F(s) = {F4}")
pf4 = sp.apart(F4, s)
print(f"๋ถ€๋ถ„๋ถ„์ˆ˜: {pf4}")
f4 = sp.inverse_laplace_transform(F4, s, t)
print(f"f(t) = {f4}")

5. Solving Ordinary Differential Equations

5.1 Solution Procedure

General procedure for solving ODEs using the Laplace transform:

  1. Apply the Laplace transform to both sides of the ODE
  2. Use the differentiation property to substitute initial conditions
  3. Solve algebraically for $Y(s)$
  4. Find $y(t)$ by inverse Laplace transform

5.2 Second-Order Constant Coefficient ODE

Example: Solve the initial value problem

$$y'' + 3y' + 2y = 0, \quad y(0) = 1, \quad y'(0) = 0$$

Solution:

Apply Laplace transform to both sides:

$$[s^2 Y(s) - sy(0) - y'(0)] + 3[sY(s) - y(0)] + 2Y(s) = 0$$

Substitute initial conditions:

$$s^2 Y - s + 3sY - 3 + 2Y = 0$$

Solve for $Y(s)$:

$$Y(s)(s^2 + 3s + 2) = s + 3$$

$$Y(s) = \frac{s + 3}{s^2 + 3s + 2} = \frac{s + 3}{(s+1)(s+2)}$$

Partial fraction decomposition: $\frac{s+3}{(s+1)(s+2)} = \frac{2}{s+1} - \frac{1}{s+2}$

Inverse transform:

$$y(t) = 2e^{-t} - e^{-2t}$$

5.3 Nonhomogeneous ODE

Example: $y'' + y = \sin(2t)$, $y(0) = 0$, $y'(0) = 0$

Apply Laplace transform to both sides:

$$s^2 Y(s) + Y(s) = \frac{2}{s^2 + 4}$$

$$Y(s) = \frac{2}{(s^2 + 1)(s^2 + 4)}$$

Partial fraction decomposition:

$$\frac{2}{(s^2+1)(s^2+4)} = \frac{2}{3} \cdot \frac{1}{s^2+1} - \frac{2}{3} \cdot \frac{1}{s^2+4}$$

Inverse transform:

$$y(t) = \frac{2}{3}\sin t - \frac{1}{3}\sin 2t$$

5.4 Coupled Differential Equations

Example: Coupled system

$$\begin{cases} x' = 2x - y, \quad x(0) = 1 \\ y' = x, \quad\quad\;\;\; y(0) = 0 \end{cases}$$

Apply Laplace transform:

$$sX - 1 = 2X - Y \quad \Rightarrow \quad (s-2)X + Y = 1$$

$$sY = X \quad \Rightarrow \quad X = sY$$

Substituting:

$$s(s-2)Y + Y = 1 \quad \Rightarrow \quad Y(s) = \frac{1}{s^2 - 2s + 1} = \frac{1}{(s-1)^2}$$

$$X(s) = sY(s) = \frac{s}{(s-1)^2} = \frac{1}{s-1} + \frac{1}{(s-1)^2}$$

Inverse transform:

$$x(t) = e^t + te^t = (1 + t)e^t, \quad y(t) = te^t$$

import sympy as sp

t, s = sp.symbols('t s')
Y = sp.Function('Y')

print("=== ODE ํ’€์ด: y'' + 3y' + 2y = 0, y(0)=1, y'(0)=0 ===\n")

# ๋ฐฉ๋ฒ• 1: SymPy์˜ dsolve๋กœ ์ง์ ‘ ํ’€๊ธฐ
y = sp.Function('y')
ode = sp.Eq(y(t).diff(t, 2) + 3*y(t).diff(t) + 2*y(t), 0)
sol = sp.dsolve(ode, y(t), ics={y(0): 1, y(t).diff(t).subs(t, 0): 0})
print(f"dsolve ๊ฒฐ๊ณผ: {sol}\n")

# ๋ฐฉ๋ฒ• 2: ๋ผํ”Œ๋ผ์Šค ๋ณ€ํ™˜์„ ๋‹จ๊ณ„๋ณ„๋กœ ์ˆ˜ํ–‰
print("--- ๋‹จ๊ณ„๋ณ„ ๋ผํ”Œ๋ผ์Šค ๋ณ€ํ™˜ ํ’€์ด ---")

# Y(s) ๊ตฌํ•˜๊ธฐ
Ys = (s + 3) / (s**2 + 3*s + 2)
print(f"Y(s) = {Ys}")

# ๋ถ€๋ถ„๋ถ„์ˆ˜ ๋ถ„ํ•ด
pf = sp.apart(Ys, s)
print(f"๋ถ€๋ถ„๋ถ„์ˆ˜: {pf}")

# ์—ญ ๋ผํ”Œ๋ผ์Šค ๋ณ€ํ™˜
yt = sp.inverse_laplace_transform(Ys, s, t)
print(f"y(t) = {yt}\n")

# ๊ฒ€์ฆ: ์ดˆ๊ธฐ ์กฐ๊ฑด๊ณผ ODE ๋งŒ์กฑ ์—ฌ๋ถ€
print("--- ๊ฒ€์ฆ ---")
print(f"y(0) = {yt.subs(t, 0)}")
print(f"y'(0) = {sp.diff(yt, t).subs(t, 0)}")
residual = sp.simplify(sp.diff(yt, t, 2) + 3*sp.diff(yt, t) + 2*yt)
print(f"y'' + 3y' + 2y = {residual}")

print("\n=== ๋น„์ œ์ฐจ ODE: y'' + y = sin(2t) ===\n")
Ys2 = 2 / ((s**2 + 1) * (s**2 + 4))
pf2 = sp.apart(Ys2, s)
print(f"๋ถ€๋ถ„๋ถ„์ˆ˜: {pf2}")
yt2 = sp.inverse_laplace_transform(Ys2, s, t)
print(f"y(t) = {yt2}")

6. Transfer Functions and System Analysis

6.1 Transfer Function Definition

In linear time-invariant (LTI) systems, the transfer function is the ratio of the Laplace transforms of output to input:

$$H(s) = \frac{Y(s)}{X(s)}$$

where all initial conditions are assumed to be zero. The transfer function represents the intrinsic characteristics of the system, independent of the input.

6.2 Poles and Zeros

  • Zeros: Values of $s$ where $H(s) = 0$ (roots of the numerator)
  • Poles: Values of $s$ where $H(s) \to \infty$ (roots of the denominator)

The locations of the poles determine the dynamic characteristics of the system.

6.3 Impulse Response and Step Response

  • Impulse response $h(t) = \mathcal{L}^{-1}\{H(s)\}$: output when input is $\delta(t)$
  • Step response: output when input is $u(t)$, $Y(s) = H(s)/s$

6.4 Stability Analysis

For a system to be BIBO stable (Bounded-Input Bounded-Output stable):

All poles of the transfer function $H(s)$ must be located in the left half of the complex plane (i.e., $\text{Re}(s) < 0$).

Response characteristics based on pole locations: - Left half-plane ($\text{Re}(s) < 0$): Decay โ†’ stable - Imaginary axis ($\text{Re}(s) = 0$): Sustained oscillation โ†’ marginally stable - Right half-plane ($\text{Re}(s) > 0$): Divergence โ†’ unstable

import numpy as np
import matplotlib.pyplot as plt
from scipy import signal

# 2์ฐจ ์‹œ์Šคํ…œ: H(s) = omega_n^2 / (s^2 + 2*zeta*omega_n*s + omega_n^2)
omega_n = 2.0  # ๊ณ ์œ  ์ง„๋™์ˆ˜
zeta_values = [0.1, 0.3, 0.7, 1.0, 2.0]  # ๊ฐ์‡ ๋น„

fig, axes = plt.subplots(1, 2, figsize=(14, 5))

for zeta in zeta_values:
    # ์ „๋‹ฌํ•จ์ˆ˜ ์ •์˜
    num = [omega_n**2]
    den = [1, 2*zeta*omega_n, omega_n**2]
    sys = signal.TransferFunction(num, den)

    # ๊ณ„๋‹จ ์‘๋‹ต
    t_step, y_step = signal.step(sys)
    axes[0].plot(t_step, y_step, label=f'zeta={zeta}')

    # ๊ทน์  ํ‘œ์‹œ
    poles = np.roots(den)
    axes[1].plot(poles.real, poles.imag, 'x', markersize=10,
                 label=f'zeta={zeta}: {poles[0]:.2f}')

axes[0].set_xlabel('์‹œ๊ฐ„ t')
axes[0].set_ylabel('y(t)')
axes[0].set_title('๊ณ„๋‹จ ์‘๋‹ต (๊ฐ์‡ ๋น„์— ๋”ฐ๋ฅธ ๋ณ€ํ™”)')
axes[0].legend()
axes[0].grid(True)
axes[0].axhline(y=1, color='k', linestyle='--', alpha=0.3)

axes[1].axvline(x=0, color='k', linestyle='-', alpha=0.3)
axes[1].axhline(y=0, color='k', linestyle='-', alpha=0.3)
axes[1].set_xlabel('Re(s)')
axes[1].set_ylabel('Im(s)')
axes[1].set_title('๊ทน์  ์œ„์น˜ (s-ํ‰๋ฉด)')
axes[1].legend()
axes[1].grid(True)
axes[1].set_aspect('equal')

plt.tight_layout()
plt.savefig('transfer_function_analysis.png', dpi=150)
plt.show()
print("๊ฐ์‡ ๋น„๊ฐ€ ํด์ˆ˜๋ก ๊ทน์ ์ด ์‹ค์ˆ˜์ถ•์— ๊ฐ€๊นŒ์›Œ์ง€๊ณ , ์‘๋‹ต์ด ๋น ๋ฅด๊ฒŒ ์•ˆ์ •ํ™”๋จ")

7. Physical Applications

7.1 RLC Circuit Analysis

Applying Kirchhoff's law to a series RLC circuit:

$$L\frac{di}{dt} + Ri + \frac{1}{C}\int_0^t i(\tau) \, d\tau = V(t)$$

In terms of charge $q$ (where $i = dq/dt$):

$$L q'' + R q' + \frac{q}{C} = V(t)$$

Applying Laplace transform (with $q(0) = 0$, $q'(0) = 0$):

$$\left(Ls^2 + Rs + \frac{1}{C}\right) Q(s) = V(s)$$

Transfer function:

$$H(s) = \frac{Q(s)}{V(s)} = \frac{1}{Ls^2 + Rs + \frac{1}{C}} = \frac{1/L}{s^2 + \frac{R}{L}s + \frac{1}{LC}}$$

Setting natural frequency $\omega_0 = 1/\sqrt{LC}$ and damping ratio $\zeta = R/(2\sqrt{L/C})$ gives the same form as a standard second-order system.

import numpy as np
import matplotlib.pyplot as plt
from scipy import signal

# RLC ํšŒ๋กœ ํŒŒ๋ผ๋ฏธํ„ฐ
R = 10       # Ohm (์ €ํ•ญ)
L = 0.1      # H (์ธ๋•ํ„ด์Šค)
C = 1e-3     # F (์ปคํŒจ์‹œํ„ด์Šค)

omega_0 = 1 / np.sqrt(L * C)  # ๊ณ ์œ  ์ง„๋™์ˆ˜
zeta = R / (2 * np.sqrt(L / C))  # ๊ฐ์‡ ๋น„
print(f"RLC ํšŒ๋กœ ํŒŒ๋ผ๋ฏธํ„ฐ:")
print(f"  ๊ณ ์œ  ์ง„๋™์ˆ˜ omega_0 = {omega_0:.2f} rad/s")
print(f"  ๊ฐ์‡ ๋น„ zeta = {zeta:.4f}")
print(f"  ์ƒํƒœ: {'๊ณผ๊ฐ์‡ ' if zeta > 1 else '์ž„๊ณ„๊ฐ์‡ ' if zeta == 1 else '๋ถ€์กฑ๊ฐ์‡ '}")

# ์ „๋‹ฌํ•จ์ˆ˜: H(s) = (1/L) / (s^2 + (R/L)s + 1/(LC))
num = [1/L]
den = [1, R/L, 1/(L*C)]
sys_rlc = signal.TransferFunction(num, den)

# ๋‹จ์œ„ ๊ณ„๋‹จ ์ „์•• ์ž…๋ ฅ์— ๋Œ€ํ•œ ์‘๋‹ต (์Šค์œ„์น˜ ON)
t_sim = np.linspace(0, 0.1, 1000)
t_out, q_out = signal.step(sys_rlc, T=t_sim)

# ์ „๋ฅ˜ i(t) = dq/dt
i_out = np.gradient(q_out, t_out)

fig, axes = plt.subplots(2, 1, figsize=(10, 8))

axes[0].plot(t_out * 1000, q_out * 1e6, 'b-', linewidth=2)
axes[0].set_xlabel('์‹œ๊ฐ„ (ms)')
axes[0].set_ylabel('์ „ํ•˜ q (uC)')
axes[0].set_title('RLC ์ง๋ ฌ ํšŒ๋กœ - ๊ณ„๋‹จ ์‘๋‹ต')
axes[0].grid(True)

axes[1].plot(t_out * 1000, i_out * 1000, 'r-', linewidth=2)
axes[1].set_xlabel('์‹œ๊ฐ„ (ms)')
axes[1].set_ylabel('์ „๋ฅ˜ i (mA)')
axes[1].set_title('์ „๋ฅ˜ ์‘๋‹ต')
axes[1].grid(True)

plt.tight_layout()
plt.savefig('rlc_circuit_response.png', dpi=150)
plt.show()

7.2 Damped Oscillation (Mass-Spring-Damper System)

A system with mass $m$, spring constant $k$, and damping coefficient $c$:

$$m\ddot{x} + c\dot{x} + kx = F(t)$$

For free vibration ($F = 0$) with initial displacement $x_0$:

$$ms^2 X - msx_0 + csX - cx_0 + kX = 0$$

$$X(s) = \frac{(ms + c)x_0}{ms^2 + cs + k} = \frac{x_0(s + c/m)}{s^2 + (c/m)s + k/m}$$

With damping ratio $\zeta = c/(2\sqrt{mk})$, natural frequency $\omega_n = \sqrt{k/m}$, and damped frequency $\omega_d = \omega_n\sqrt{1-\zeta^2}$:

$$x(t) = x_0 e^{-\zeta\omega_n t}\left(\cos\omega_d t + \frac{\zeta}{\sqrt{1-\zeta^2}}\sin\omega_d t\right)$$

import numpy as np
import matplotlib.pyplot as plt

# ์งˆ๋Ÿ‰-์Šคํ”„๋ง-๋Œํผ ํŒŒ๋ผ๋ฏธํ„ฐ
m = 1.0   # kg
k = 100.0  # N/m
x0 = 0.05  # m (์ดˆ๊ธฐ ๋ณ€์œ„ 5cm)

# ๋‹ค์–‘ํ•œ ๊ฐ์‡  ์กฐ๊ฑด
c_values = [0.5, 5.0, 20.0, 25.0]  # N*s/m

t = np.linspace(0, 3, 1000)
plt.figure(figsize=(10, 6))

for c in c_values:
    omega_n = np.sqrt(k / m)
    zeta = c / (2 * np.sqrt(m * k))

    if zeta < 1:  # ๋ถ€์กฑ๊ฐ์‡ 
        omega_d = omega_n * np.sqrt(1 - zeta**2)
        x = x0 * np.exp(-zeta * omega_n * t) * (
            np.cos(omega_d * t) +
            (zeta / np.sqrt(1 - zeta**2)) * np.sin(omega_d * t)
        )
        label = f'c={c} (zeta={zeta:.2f}, ๋ถ€์กฑ๊ฐ์‡ )'
    elif zeta == 1:  # ์ž„๊ณ„๊ฐ์‡ 
        x = x0 * (1 + omega_n * t) * np.exp(-omega_n * t)
        label = f'c={c} (zeta={zeta:.2f}, ์ž„๊ณ„๊ฐ์‡ )'
    else:  # ๊ณผ๊ฐ์‡ 
        r1 = -zeta * omega_n + omega_n * np.sqrt(zeta**2 - 1)
        r2 = -zeta * omega_n - omega_n * np.sqrt(zeta**2 - 1)
        A = x0 * r2 / (r2 - r1)
        B = -x0 * r1 / (r2 - r1)
        x = A * np.exp(r1 * t) + B * np.exp(r2 * t)
        label = f'c={c} (zeta={zeta:.2f}, ๊ณผ๊ฐ์‡ )'

    plt.plot(t, x * 100, linewidth=2, label=label)

plt.xlabel('์‹œ๊ฐ„ (s)')
plt.ylabel('๋ณ€์œ„ (cm)')
plt.title('์งˆ๋Ÿ‰-์Šคํ”„๋ง-๋Œํผ ์‹œ์Šคํ…œ์˜ ์ž์œ  ์ง„๋™')
plt.legend()
plt.grid(True)
plt.axhline(y=0, color='k', linestyle='-', alpha=0.3)
plt.savefig('damped_oscillation.png', dpi=150)
plt.show()

7.3 Heat Conduction Problem

Heat conduction equation when temperature $T_0$ is suddenly applied to one end of a rod of length $L$:

$$\frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial x^2}$$

Applying Laplace transform with respect to time (with $u(x, 0) = 0$):

$$sU(x, s) = \alpha \frac{d^2 U}{dx^2}$$

This is an ODE in $x$:

$$U(x, s) = A e^{-x\sqrt{s/\alpha}} + B e^{x\sqrt{s/\alpha}}$$

Applying boundary conditions $U(0, s) = T_0/s$, $U(\infty, s) = 0$:

$$U(x, s) = \frac{T_0}{s} e^{-x\sqrt{s/\alpha}}$$

Inverse transform:

$$u(x, t) = T_0 \, \text{erfc}\left(\frac{x}{2\sqrt{\alpha t}}\right)$$

where $\text{erfc}$ is the complementary error function.

import numpy as np
import matplotlib.pyplot as plt
from scipy.special import erfc

# ์—ด์ „๋„ ํŒŒ๋ผ๋ฏธํ„ฐ
T0 = 100.0        # ๊ฒฝ๊ณ„ ์˜จ๋„ (deg C)
alpha = 1.14e-4   # ์—ดํ™•์‚ฐ๊ณ„์ˆ˜ (m^2/s, ์ฒ )

# ์‹œ๊ฐ„๋ณ„ ์˜จ๋„ ๋ถ„ํฌ
x = np.linspace(0, 0.1, 200)  # ์œ„์น˜ (m)
times = [1, 10, 60, 300, 1800]  # ์‹œ๊ฐ„ (s)

plt.figure(figsize=(10, 6))
for t_val in times:
    u = T0 * erfc(x / (2 * np.sqrt(alpha * t_val)))
    plt.plot(x * 100, u, linewidth=2, label=f't = {t_val}s')

plt.xlabel('์œ„์น˜ x (cm)')
plt.ylabel('์˜จ๋„ u (deg C)')
plt.title('๋ฐ˜๋ฌดํ•œ ๋ด‰์˜ ์—ด์ „๋„ (๋ผํ”Œ๋ผ์Šค ๋ณ€ํ™˜ ํ’€์ด)')
plt.legend()
plt.grid(True)
plt.savefig('heat_conduction_laplace.png', dpi=150)
plt.show()

8. Numerical Inverse Laplace Transform

8.1 Need

In many practical problems, the analytical inverse transform of $F(s)$ is impossible or very complex. In such cases, numerical inverse Laplace transform algorithms are needed.

8.2 Stehfest Algorithm

The Stehfest algorithm approximates $f(t)$ using only values of $F(s)$ on the real axis:

$$f(t) \approx \frac{\ln 2}{t} \sum_{k=1}^{N} V_k \, F\left(\frac{k \ln 2}{t}\right)$$

where the weights $V_k$ are computed using binomial coefficients. $N$ is chosen to be even (typically $N = 10 \sim 18$).

8.3 Talbot Method

The Talbot method is a deformation of the Bromwich integral path, numerically integrating along a parabolic path in the complex plane. It is generally more accurate than the Stehfest method.

import numpy as np
import matplotlib.pyplot as plt
from math import factorial

def stehfest_weights(N):
    """์Šคํ…ŒํŽ˜์ŠคํŠธ ์•Œ๊ณ ๋ฆฌ์ฆ˜์˜ ๊ฐ€์ค‘์น˜ ๊ณ„์‚ฐ"""
    V = np.zeros(N)
    for k in range(1, N + 1):
        s = 0
        for j in range(int((k + 1) / 2), min(k, N // 2) + 1):
            numer = j**(N // 2) * factorial(2 * j)
            denom = (factorial(N // 2 - j) *
                     factorial(j) *
                     factorial(j - 1) *
                     factorial(k - j) *
                     factorial(2 * j - k))
            s += numer / denom
        V[k - 1] = (-1)**(k + N // 2) * s
    return V

def numerical_inverse_laplace(F_func, t_values, N=12):
    """์Šคํ…ŒํŽ˜์ŠคํŠธ ์•Œ๊ณ ๋ฆฌ์ฆ˜์„ ์ด์šฉํ•œ ์ˆ˜์น˜์  ์—ญ ๋ผํ”Œ๋ผ์Šค ๋ณ€ํ™˜

    Parameters:
        F_func: F(s)๋ฅผ ๊ณ„์‚ฐํ•˜๋Š” ํ•จ์ˆ˜
        t_values: ์—ญ๋ณ€ํ™˜์„ ๊ตฌํ•  ์‹œ๊ฐ„๊ฐ’ ๋ฐฐ์—ด
        N: ์Šคํ…ŒํŽ˜์ŠคํŠธ ์ฐจ์ˆ˜ (์ง์ˆ˜, ๊ธฐ๋ณธ๊ฐ’ 12)
    """
    V = stehfest_weights(N)
    ln2 = np.log(2)
    result = np.zeros_like(t_values, dtype=float)

    for i, t in enumerate(t_values):
        if t <= 0:
            result[i] = 0
            continue
        s_vals = np.arange(1, N + 1) * ln2 / t
        F_vals = np.array([F_func(sv) for sv in s_vals])
        result[i] = (ln2 / t) * np.sum(V * F_vals)

    return result

# ๊ฒ€์ฆ: F(s) = 1/(s+1) -> f(t) = e^(-t)
F_exp = lambda sv: 1.0 / (sv + 1.0)
t_test = np.linspace(0.01, 5, 200)
f_numerical = numerical_inverse_laplace(F_exp, t_test)
f_exact = np.exp(-t_test)

plt.figure(figsize=(10, 6))
plt.plot(t_test, f_exact, 'b-', linewidth=2, label='ํ•ด์„์ : exp(-t)')
plt.plot(t_test, f_numerical, 'r--', linewidth=2, label='์Šคํ…ŒํŽ˜์ŠคํŠธ (N=12)')
plt.xlabel('์‹œ๊ฐ„ t')
plt.ylabel('f(t)')
plt.title('์ˆ˜์น˜์  ์—ญ ๋ผํ”Œ๋ผ์Šค ๋ณ€ํ™˜ ๊ฒ€์ฆ')
plt.legend()
plt.grid(True)
plt.savefig('numerical_inverse_laplace.png', dpi=150)
plt.show()

# ์˜ค์ฐจ ๋ถ„์„
max_error = np.max(np.abs(f_numerical - f_exact))
print(f"์ตœ๋Œ€ ์˜ค์ฐจ: {max_error:.2e}")

# ๋” ๋ณต์žกํ•œ ์˜ˆ: F(s) = 1/(s^2+1) -> f(t) = sin(t)
F_sin = lambda sv: 1.0 / (sv**2 + 1.0)
f_sin_numerical = numerical_inverse_laplace(F_sin, t_test)
f_sin_exact = np.sin(t_test)

print(f"\nsin(t) ์—ญ๋ณ€ํ™˜ ์ตœ๋Œ€ ์˜ค์ฐจ: {np.max(np.abs(f_sin_numerical - f_sin_exact)):.2e}")

Practice Problems

Basic

Problem 1. Find the Laplace transform of the following functions.

(a) $f(t) = 3t^2 - 2e^{-t} + 5\cos(4t)$

(b) $f(t) = t^3 e^{2t}$

(c) $f(t) = e^{-3t}\sin(5t)$

Problem 2. Find the inverse Laplace transform of the following $F(s)$.

(a) $F(s) = \dfrac{5}{s^3}$

(b) $F(s) = \dfrac{2s + 1}{s^2 + 4s + 13}$

(c) $F(s) = \dfrac{3}{(s-1)(s+2)(s-3)}$

Problem 3. Use the convolution theorem to find $\mathcal{L}^{-1}\left\{\dfrac{1}{s(s+1)}\right\}$.

Intermediate

Problem 4. Solve the following initial value problem using the Laplace transform.

$$y'' - 4y' + 4y = e^{2t}, \quad y(0) = 0, \quad y'(0) = 1$$

Problem 5. Use the second shifting theorem to find the Laplace transform of:

$$f(t) = \begin{cases} 0, & 0 \leq t < 2 \\ t - 2, & t \geq 2 \end{cases}$$

Problem 6. Solve the following coupled differential equations using the Laplace transform.

$$x' + y = e^t, \quad x + y' = 0, \quad x(0) = 1, \quad y(0) = 0$$

Problem 7. For the transfer function $H(s) = \dfrac{s + 3}{s^2 + 4s + 8}$, find the poles and zeros, determine system stability, and find the impulse response $h(t)$.

Advanced

Problem 8. Solve the integral equation using the Laplace transform.

$$y(t) = 1 + \int_0^t y(\tau) \sin(t - \tau) \, d\tau$$

Problem 9. For a series RLC circuit ($R = 4\,\Omega$, $L = 1\,$H, $C = 1/5\,$F) driven by $V(t) = 10u(t)$ (unit step voltage), find the current $i(t)$ using the Laplace transform.

Problem 10. Use the final value theorem to find the steady-state value of the unit step response for a system with transfer function:

$$H(s) = \frac{10(s + 2)}{s^2 + 5s + 6}$$

Advanced Topics

Bilateral Laplace Transform

The standard Laplace transform is unilateral, defined only for $t \geq 0$, but the bilateral Laplace transform is:

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

The bilateral transform has a strip-shaped region of convergence and is important in signal processing and probability theory.

Relationship with Z-Transform

The discrete-time counterpart of the Laplace transform is the Z-transform:

$$X(z) = \sum_{n=0}^{\infty} x[n] z^{-n}$$

Setting $z = e^{sT}$ (where $T$ is the sampling period) establishes the relationship between the two transforms. It plays a key role in digital signal processing and discrete control systems.

References

Textbooks: - Boas, M. L. Mathematical Methods in the Physical Sciences, 3rd ed., Ch. 8 (Sec. 10-12) - Arfken, Weber Mathematical Methods for Physicists, Ch. 15

Supplementary materials: - Schiff, J. L. The Laplace Transform: Theory and Applications - Balanced coverage of theory and applications - Dyke, P. P. G. An Introduction to Laplace Transforms and Fourier Series - Mathematical physics approach

Key Formula Summary

Property Time domain $s$-domain
Definition $f(t)$ $F(s) = \int_0^\infty f(t)e^{-st}\,dt$
Linearity $\alpha f + \beta g$ $\alpha F + \beta G$
First shift $e^{at}f(t)$ $F(s-a)$
Second shift $f(t-a)u(t-a)$ $e^{-as}F(s)$
Differentiation $f'(t)$ $sF(s) - f(0)$
Integration $\int_0^t f(\tau)\,d\tau$ $F(s)/s$
Convolution $(f*g)(t)$ $F(s) \cdot G(s)$
Initial value $f(0^+)$ $\lim_{s\to\infty} sF(s)$
Final value $\lim_{t\to\infty} f(t)$ $\lim_{s\to 0} sF(s)$

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