Definition of Laplace Transform

The Laplace transform is an integral transform that converts a time-domain function $f(t)$ into a complex frequency-domain function $F(s)$, defined as:

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

where $s$ is a complex variable, $s = \sigma + i\omega$.

Fundamental Properties of Laplace Transform

The Laplace transform has several important properties that help convert time-domain integration and differentiation operations into algebraic operations in the frequency domain. The following are some common properties:

Linearity

The Laplace transform is linear with respect to addition and scalar multiplication. In other words, the transform of a composite function can be separated into the sum of transforms of individual functions.

$$ \mathcal{L}\{a f(t) + b g(t)\} = a \mathcal{L}\{f(t)\} + b \mathcal{L}\{g(t)\} $$

Time Shifting

A delay of $a$ units in the function leads to its Laplace transform being multiplied by an exponential decay term $e^{-as}$. This property is commonly used in analyzing time-delay systems.

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

where $u(t - a)$ is the unit step function, indicating the function starts acting from $t = a$.

Frequency Shifting

Multiplying the original function by $e^{at}$ results in a shift in $s$ in the Laplace domain, reflecting a transformation in the frequency domain.

$$ \mathcal{L}\{e^{a t} f(t)\} = \mathcal{L}\{f(t)\} \big|_{s \to s - a} $$

Convolution

The convolution of two functions in the time domain corresponds to the product of their Laplace transforms in the frequency domain. This is very useful for analyzing system input and impulse response.

$$ \mathcal{L}\{f(t) * g(t)\} = \mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\} $$

where convolution is defined as:

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

Initial Value Theorem

Used to quickly find the value of the original function at $t = 0^+$, provided the function and its transform exist and are non-divergent.

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

Final Value Theorem

Used to predict the steady-state value of the function ($t \to \infty$), applicable to stable systems where all poles lie in the left half-plane.

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

Laplace Transforms of Common Functions

Derivative Functions

First Derivative

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

Derivation:

$$ \begin{aligned} \mathcal{L}\{f’(t)\} &= \int_0^{\infty} e^{-st} f’(t) dt \\ &= \left[ e^{-st} f(t) \right]_0^{\infty} + s \int_0^{\infty} e^{-st} f(t) dt \\ &= 0 - f(0) + sF(s) \\ &= sF(s) - f(0) \end{aligned} $$

Second Derivative

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

Derivation:

$$ \begin{aligned} \mathcal{L}\{f’’(t)\} &= \mathcal{L}{f’(t)}’ \\ &= s \mathcal{L}{f’(t)} - f’(0) \\ &= s(sF(s) - f(0)) - f’(0) \\ &= s^2F(s) - sf(0) - f’(0) \end{aligned} $$

General n-th Derivative

$$ \mathcal{L}\{f^{(n)}(t)\} = s^n \mathcal{L}\{f(t)\} - \sum_{k=1}^{n} s^{n-k} f^{(k-1)}(0) $$

Integral Function

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

Derivation:

Let $g(t) = \int_0^t f(\tau) d\tau$, then $g’(t) = f(t)$, and $g(0) = 0$.

According to the derivative property of Laplace transforms:

$$ \mathcal{L}\{g’(t)\} = s \mathcal{L}\{g(t)\} - g(0) = s \mathcal{L}\{g(t)\} $$

But $g’(t) = f(t)$, thus:

$$ \mathcal{L}\{f(t)\} = s \mathcal{L}\{g(t)\} \Rightarrow \mathcal{L}\{g(t)\} = \frac{F(s)}{s} $$

Exponential Function

$$ \mathcal{L}\{e^{at}\} = \frac{1}{s - a}, \quad \text{for } \text{Re}(s) > a $$

Derivation:

$$ \begin{aligned} \mathcal{L}\{e^{at}\} &= \int_0^{\infty} e^{-st} e^{at} dt \\ &= \int_0^{\infty} e^{-(s - a)t} dt \\ &= \left[ \frac{e^{-(s - a)t}}{-(s - a)} \right]_0^{\infty} \\ &= \frac{1}{s - a} \end{aligned} $$

Example Application: Solving a First-Order Linear Differential Equation

Problem:

$$ y’(t) + 3y(t) = e^{2t}, \quad y(0) = 1 $$

Solution Steps:

1. Take the Laplace transform on both sides:

   $$ \mathcal{L}\{y’(t)\} + 3\mathcal{L}\{y(t)\} = \mathcal{L}\{e^{2t}\} $$

   Using the property:

   $$ (sY(s) - y(0)) + 3Y(s) = \frac{1}{s - 2} $$

   Plug in the initial condition $y(0) = 1$:

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

   Rearranging:

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

2. Solve the algebraic equation for $Y(s)$:

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

3. Partial fraction decomposition:

   $$ \frac{1}{(s - 2)(s + 3)} = \frac{1}{5} \cdot \frac{1}{s - 2} - \frac{1}{5} \cdot \frac{1}{s + 3} $$

   Combine:

   $$ Y(s) = \frac{1}{5} \cdot \frac{1}{s - 2} + \frac{4}{5} \cdot \frac{1}{s + 3} $$

4. Take the inverse Laplace transform to find $y(t)$:

   $$ y(t) = \frac{1}{5} e^{2t} + \frac{4}{5} e^{-3t} $$

Final Answer:

$$ y(t) = \frac{1}{5} e^{2t} + \frac{4}{5} e^{-3t} $$

Conclusion

The Laplace transform is a powerful mathematical tool that converts differential equations into algebraic equations, simplifying the solving process. Through the above derivations and examples, we can clearly see its application in solving differential equations. Mastering the properties and techniques of Laplace transforms is helpful for effectively analyzing and solving problems in engineering, physics, and mathematics.