Laplace Transforms – Dr. Mohamed Mabrok

1. Definition of Laplace Transform

The Laplace transform converts a function of time \( f(t) \) into a function of complex frequency \( F(s) \):

\[ \mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) \, dt \]

The transform exists if the integral converges for some values of \( s \).

2. Time Domain to Frequency Domain Examples

Example 1: Exponential Function

Find \( \mathcal{L}\{e^{at}\} \):

\[ \mathcal{L}\{e^{at}\} = \int_{0}^{\infty} e^{-st} e^{at} \, dt = \int_{0}^{\infty} e^{-(s-a)t} \, dt = \left. \frac{-1}{s-a} e^{-(s-a)t} \right|_{0}^{\infty} = \frac{1}{s-a} \]

For \( \text{Re}(s) > \text{Re}(a) \).

Example 2: Sine Function

Find \( \mathcal{L}\{\sin(\omega t)\} \):

\[ \mathcal{L}\{\sin(\omega t)\} = \int_{0}^{\infty} e^{-st} \sin(\omega t) \, dt = \frac{\omega}{s^2 + \omega^2} \]

Common Laplace Transforms

\( \mathcal{L}\{1\} = \frac{1}{s} \)

\( \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \)

\( \mathcal{L}\{\cos(\omega t)\} = \frac{s}{s^2 + \omega^2} \)

\( \mathcal{L}\{t^n e^{at}\} = \frac{n!}{(s-a)^{n+1}} \)

3. Inverse Laplace Transform

The inverse Laplace transform converts back from frequency domain to time domain:

\[ \mathcal{L}^{-1}\{F(s)\} = f(t) \]

Example 1: Rational Function

Find \( \mathcal{L}^{-1}\left\{\frac{1}{s(s+1)}\right\} \):

\[ \frac{1}{s(s+1)} = \frac{A}{s} + \frac{B}{s+1} = \frac{1}{s} - \frac{1}{s+1} \] \[ \mathcal{L}^{-1}\left\{\frac{1}{s} - \frac{1}{s+1}\right\} = 1 - e^{-t} \]

Example 2: Quadratic Denominator

Find \( \mathcal{L}^{-1}\left\{\frac{s}{s^2 + 4s + 13}\right\} \):

\[ \frac{s}{s^2 + 4s + 13} = \frac{s}{(s+2)^2 + 9} = \frac{s+2}{(s+2)^2 + 9} - \frac{2}{(s+2)^2 + 9} \] \[ \mathcal{L}^{-1}\left\{\frac{s+2}{(s+2)^2 + 9} - \frac{2}{(s+2)^2 + 9}\right\} = e^{-2t}\cos(3t) - \frac{2}{3}e^{-2t}\sin(3t) \]

4. Solving Initial Value Problems (IVP)

Steps to solve an IVP using Laplace transforms:

Take the Laplace transform of both sides of the ODE
Solve for \( Y(s) \) (the transform of the solution)
Apply the inverse transform to find \( y(t) \)

Example: Second Order ODE

Solve \( y'' + 3y' + 2y = 0 \) with \( y(0) = 1 \), \( y'(0) = 0 \):

1. Transform the equation:

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

2. Solve for \( Y(s) \):

\[ Y(s) = \frac{s + 3}{s^2 + 3s + 2} = \frac{s + 3}{(s + 1)(s + 2)} = \frac{2}{s + 1} - \frac{1}{s + 2} \]

3. Inverse transform:

\[ y(t) = 2e^{-t} - e^{-2t} \]

5. Unit Step Function and Piecewise Functions

The unit step function (Heaviside function) is defined as:

\[ u(t - a) = \begin{cases} 0 & \text{if } t < a \\ 1 & \text{if } t \geq a \end{cases} \]

We can express piecewise functions using step functions:

Example: Piecewise to Step Function

Express \( f(t) = \begin{cases} t & \text{if } 0 \leq t < 2 \\ 2 & \text{if } t \geq 2 \end{cases} \) using step functions:

\[ f(t) = t - (t - 2)u(t - 2) \]

Laplace transform of shifted functions:

\[ \mathcal{L}\{u(t - a)f(t - a)\} = e^{-as}F(s) \]

Example: Transform of Step Function

Find \( \mathcal{L}\{u(t - 3)(t - 3)^2\} \):

\[ \mathcal{L}\{t^2\} = \frac{2}{s^3} \] \[ \mathcal{L}\{u(t - 3)(t - 3)^2\} = e^{-3s}\frac{2}{s^3} \]

6. Solving IVP with Piecewise Forcing Function

Example: ODE with Piecewise Forcing

Solve \( y'' + y = f(t) \), \( y(0) = 0 \), \( y'(0) = 0 \) where:

\( f(t) = \begin{cases} 1 & \text{if } 0 \leq t < \pi \\ 0 & \text{if } t \geq \pi \end{cases} \)

1. Express \( f(t) \) using step functions:

\[ f(t) = 1 - u(t - \pi) \]

2. Take Laplace transform:

\[ \mathcal{L}\{f(t)\} = \frac{1}{s} - \frac{e^{-\pi s}}{s} \] \[ (s^2 + 1)Y(s) = \frac{1}{s} - \frac{e^{-\pi s}}{s} \] \[ Y(s) = \frac{1}{s(s^2 + 1)} - \frac{e^{-\pi s}}{s(s^2 + 1)} \]

3. Partial fractions for first term:

\[ \frac{1}{s(s^2 + 1)} = \frac{A}{s} + \frac{Bs + C}{s^2 + 1} = \frac{1}{s} - \frac{s}{s^2 + 1} \]

4. Inverse transform:

\[ y(t) = (1 - \cos t) - u(t - \pi)(1 - \cos(t - \pi)) \] \[ y(t) = \begin{cases} 1 - \cos t & \text{if } 0 \leq t < \pi \\ -\cos t - (1 - (-\cos t)) = -1 & \text{if } t \geq \pi \end{cases} \]