Laplace Transform: Detailed Explanation, Proofs, and Derivations}+
The Laplace transform is an integral transform used to transform a function of time \( f(t) \) into a function of a complex variable \( s \). It is widely used in engineering and physics to solve differential equations and analyze systems.
Definition of the Laplace Transform
The Laplace transform of a function \( f(t) \), defined for \( t \geq 0 \), is given by:
$$ F(s) = \mathcal{L}\{f(t)\} = \int_{0}^{\infty} f(t) e^{-st} \, dt $$
where \( s \) is a complex number \( s = \sigma + i\omega \).
The inverse Laplace transform is given by:
$$ f(t) = \mathcal{L}^{-1}\{F(s)\} $$
Properties of the Laplace Transform
1. Linearity: If \( a \) and \( b \) are constants, then
$$ \mathcal{L}\{af(t) + bg(t)\} = a\mathcal{L}\{f(t)\} + b\mathcal{L}\{g(t)\} $$
2. First Shifting Theorem (Time Shifting): If \( t_0 \) is a constant, then
$$ \mathcal{L}\{f(t – t_0)u(t – t_0)\} = e^{-st_0}F(s) $$
where \( u(t) \) is the Heaviside step function.
3. Second Shifting Theorem (Frequency Shifting): If \( a \) is a constant, then
$$ \mathcal{L}\{e^{at}f(t)\} = F(s – a) $$
4. Scaling: If \( a \neq 0 \) is a constant, then
$$ \mathcal{L}\{f(at)\} = \frac{1}{|a|}F\left(\frac{s}{a}\right) $$
5. Differentiation in the Time Domain:
$$ \mathcal{L}\{f'(t)\} = sF(s) – f(0) $$
$$ \mathcal{L}\{f”(t)\} = s^2F(s) – sf(0) – f'(0) $$
6. Integration in the Time Domain:
$$ \mathcal{L}\{\int_{0}^{t} f(\tau) \, d\tau\} = \frac{F(s)}{s} $$
7. Convolution: If \( f(t) \) and \( g(t) \) are functions, then
$$ \mathcal{L}\{(f * g)(t)\} = F(s)G(s) $$
where \( (f * g)(t) = \int_{0}^{t} f(\tau)g(t – \tau) \, d\tau \) is the convolution of \( f \) and \( g \).
Derivations and Proofs
Derivation of the Laplace Transform of the Derivative
Starting with the definition of the Laplace transform:
$$ \mathcal{L}\{f'(t)\} = \int_{0}^{\infty} f'(t) e^{-st} \, dt $$
Integrating by parts, let \( u = e^{-st} \) and \( dv = f'(t) \, dt \), thus \( du = -se^{-st} \, dt \) and \( v = f(t) \):
$$ \int_{0}^{\infty} f'(t) e^{-st} \, dt = \left[ f(t)e^{-st} \right]_{0}^{\infty} + s\int_{0}^{\infty} f(t) e^{-st} \, dt $$
Assuming \( f(t) \) is of exponential order, the boundary term evaluates to zero as \( t \to \infty \):
$$ = -f(0) + sF(s) $$
Thus,
$$ \mathcal{L}\{f'(t)\} = sF(s) – f(0) $$
Derivation of the Convolution Theorem
The convolution of two functions \( f(t) \) and \( g(t) \) is defined as:
$$ (f * g)(t) = \int_{0}^{t} f(\tau)g(t – \tau) \, d\tau $$
Taking the Laplace transform of both sides:
$$ \mathcal{L}\{(f * g)(t)\} = \int_{0}^{\infty} \left( \int_{0}^{t} f(\tau)g(t – \tau) \, d\tau \right) e^{-st} \, dt $$
Interchanging the order of integration:
$$ = \int_{0}^{\infty} f(\tau) \left( \int_{\tau}^{\infty} g(t – \tau) e^{-st} \, dt \right) d\tau $$
Letting \( u = t – \tau \):
$$ = \int_{0}^{\infty} f(\tau) \left( \int_{0}^{\infty} g(u) e^{-s(u + \tau)} \, du \right) d\tau $$
$$ = \int_{0}^{\infty} f(\tau) e^{-s\tau} \left( \int_{0}^{\infty} g(u) e^{-su} \, du \right) d\tau $$
$$ = F(s)G(s) $$
Thus,
$$ \mathcal{L}\{(f * g)(t)\} = F(s)G(s) $$
