Laplace & Fourier Transform

LessonAbout 1.5 h3 of 3 in Advanced Engineering Mathematics

Laplace Transform

Key Ideas

  • Integral transform converting time-domain functions into s-domain algebraic expressions
  • Elementary transform pairs simplify differential equation solving techniques
  • Inverse transforms map s-domain expressions back using partial fractions and completing square

Laplace Transform an integral transform that maps a time-domain function \(f(t)\) into a complex frequency-domain function \(F(s)\).

Definition and Elementary Transforms

  • Direct Transform: converts \(f(t)\) into algebraic function \(F(s)\)
  • Exponential Shift: multiplies time function by \(e^{at}\) factor
  • Linearity Property: distributes transform across linear combinations

Definition of Laplace Transform

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

Transform of Exponential and Power Functions

\[ \mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}}, \qquad\qquad \mathcal{L}\{e^{at}\} = \dfrac{1}{s - a} \]

Example. Find the Laplace transform of \(f(t) = 3e^{2t} - 5t^2\).

Solution. Apply linearity to transform each term using basic transform formulas:

\[ \mathcal{L}\{3e^{2t} - 5t^2\} = 3 \mathcal{L}\{e^{2t}\} - 5 \mathcal{L}\{t^2\} \]
\[ \mathcal{L}\{e^{2t}\} = \dfrac{1}{s - 2}, \qquad\qquad \mathcal{L}\{t^2\} = \dfrac{2!}{s^3} = \dfrac{2}{s^3} \]
\[ F(s) = \dfrac{3}{s - 2} - \dfrac{10}{s^3} \]

Answer: \(\dfrac{3}{s - 2} - \dfrac{10}{s^3}\)

Trigonometric and Hyperbolic Transforms

Trigonometric Laplace Transforms

\[ \mathcal{L}\{\sin \omega t\} = \dfrac{\omega}{s^2 + \omega^2}, \qquad\qquad \mathcal{L}\{\cos \omega t\} = \dfrac{s}{s^2 + \omega^2} \]

Example. Determine the Laplace transform of \(f(t) = \cos 4t\).

Solution. Apply the cosine Laplace transform formula with parameter \(\omega = 4\):

\[ \mathcal{L}\{\cos 4t\} = \dfrac{s}{s^2 + 4^2} = \dfrac{s}{s^2 + 16} \]

Answer: \(\dfrac{s}{s^2 + 16}\)

Inverse Laplace Transforms

  • Direct Table Lookup: matches simple rational forms directly
  • Partial Fractions: decomposes complex denominators into sum
  • Completing the Square: handles irreducible quadratic denominator terms

Inverse Transform of Sine and Cosine Forms

\[ \mathcal{L}^{-1}\left\{\dfrac{k}{s^2 + k^2}\right\} = \sin kt, \qquad\qquad \mathcal{L}^{-1}\left\{\dfrac{s}{s^2 + k^2}\right\} = \cos kt \]

Example. Find the inverse Laplace transform of \(F(s) = \dfrac{4s - 15}{s^2 + 25}\).

Solution. Separate the expression into cosine and sine transform components:

\[ F(s) = 4 \left(\dfrac{s}{s^2 + 25}\right) - 3 \left(\dfrac{5}{s^2 + 25}\right) \]

Apply inverse transforms using \(k = 5\):

\[ f(t) = 4 \cos 5t - 3 \sin 5t \]

Answer: \(4 \cos 5t - 3 \sin 5t\)

Example. Find the inverse Laplace transform of \(F(s) = \dfrac{3}{s + 2} - \dfrac{5}{s + 4}\).

Solution. Apply the exponential inverse transform formula \(\mathcal{L}^{-1}\left\{\dfrac{1}{s - a}\right\} = e^{at}\) to each term individually:

\[ \mathcal{L}^{-1}\left\{\dfrac{3}{s + 2}\right\} = 3 e^{-2t}, \qquad\qquad \mathcal{L}^{-1}\left\{\dfrac{5}{s + 4}\right\} = 5 e^{-4t} \]
\[ f(t) = 3 e^{-2t} - 5 e^{-4t} \]

Answer: \(3 e^{-2t} - 5 e^{-4t}\)

First Shifting Theorem and Quadratic Forms

First Shifting Theorem

\[ \mathcal{L}\{e^{at} f(t)\} = F(s - a), \qquad\qquad \mathcal{L}^{-1}\{F(s - a)\} = e^{at} f(t) \]

Example. Determine the inverse Laplace transform of \(F(s) = \dfrac{150}{s^2 - 60s + 9900}\).

Solution. Complete the square in the denominator: \(s^2 - 60s + 9900 = (s - 30)^2 + 900 = (s - 30)^2 + 30^2\):

\[ F(s) = \dfrac{150}{(s - 30)^2 + 30^2} = 5 \left[ \dfrac{30}{(s - 30)^2 + 30^2} \right] \]

Apply the first shifting theorem with \(a = 30\) and \(k = 30\):

\[ f(t) = 5 e^{30t} \sin 30t \]

Answer: \(5 e^{30t} \sin 30t\)

Example. Determine the inverse Laplace transform of \(F(s) = \dfrac{1}{2s^2 - 8s}\).

Solution. Factor out constants and complete the square in denominator: \(2s^2 - 8s = 2(s^2 - 4s) = 2[(s - 2)^2 - 4]\):

\[ F(s) = \dfrac{1}{2[(s - 2)^2 - 2^2]} = \dfrac{1}{4} \left[ \dfrac{2}{(s - 2)^2 - 2^2} \right] \]

Recall that \(\mathcal{L}^{-1}\left\{\dfrac{k}{(s - a)^2 - k^2}\right\} = e^{at} \sinh kt\). Here \(a = 2\) and \(k = 2\):

\[ f(t) = \dfrac{1}{4} e^{2t} \sinh 2t \]

Answer: \(\dfrac{1}{4} e^{2t} \sinh 2t\)

Laplace Transform of Elementary Functions

Laplace Transform Table standard mathematical transform pairs for elementary time-domain functions.

Function \(f(t)\) Laplace Transform \(F(s)\)
\(1\) \(\dfrac{1}{s}\)
\(t\) \(\dfrac{1}{s^2}\)
\(t^n\) \(\dfrac{n!}{s^{n+1}}\)
\(e^{\pm at}\) \(\dfrac{1}{s \mp a}\)
\(t^n e^{\pm at}\) \(\dfrac{n!}{(s \mp a)^{n+1}}\)
\(e^{\pm at} \sin kt\) \(\dfrac{k}{(s \mp a)^2 + k^2}\)
\(e^{\pm at} \cos kt\) \(\dfrac{s \mp a}{(s \mp a)^2 + k^2}\)
\(\sin at\) \(\dfrac{a}{s^2 + a^2}\)
\(\cos at\) \(\dfrac{s}{s^2 + a^2}\)
\(\sinh at\) \(\dfrac{a}{s^2 - a^2}\)
\(\cosh at\) \(\dfrac{s}{s^2 - a^2}\)
\(t \sin at\) \(\dfrac{2as}{(s^2 + a^2)^2}\)
\(t \cos at\) \(\dfrac{s^2 - a^2}{(s^2 + a^2)^2}\)
\(\sin(\omega t + \theta)\) \(\dfrac{s \sin\theta + \omega \cos\theta}{s^2 + \omega^2}\)
\(\cos(\omega t + \theta)\) \(\dfrac{s \cos\theta - \omega \sin\theta}{s^2 + \omega^2}\)

Worked Examples using Elementary Transforms

Example. Find the Laplace transform of \(f(t) = 4t^3 - 2e^{3t} + 5\sin 2t\).

Solution. Apply linearity and lookup elementary transforms from the table:

\[ \mathcal{L}\{t^3\} = \dfrac{3!}{s^4} = \dfrac{6}{s^4}, \qquad\qquad \mathcal{L}\{e^{3t}\} = \dfrac{1}{s - 3}, \qquad\qquad \mathcal{L}\{\sin 2t\} = \dfrac{2}{s^2 + 4} \]
\[ F(s) = 4\left(\dfrac{6}{s^4}\right) - 2\left(\dfrac{1}{s - 3}\right) + 5\left(\dfrac{2}{s^2 + 4}\right) = \dfrac{24}{s^4} - \dfrac{2}{s - 3} + \dfrac{10}{s^2 + 4} \]

Answer: \(\dfrac{24}{s^4} - \dfrac{2}{s - 3} + \dfrac{10}{s^2 + 4}\)

Example. Determine the Laplace transform of \(f(t) = t \cos 3t\).

Solution. Use the elementary transform formula for \(t \cos at\) with \(a = 3\):

\[ \mathcal{L}\{t \cos 3t\} = \dfrac{s^2 - 3^2}{(s^2 + 3^2)^2} = \dfrac{s^2 - 9}{(s^2 + 9)^2} \]

Answer: \(\dfrac{s^2 - 9}{(s^2 + 9)^2}\)

Fourier Series and Complex Exponential Form

Key Ideas

  • Represents periodic functions as infinite sum of sines and cosines
  • Converts trigonometric terms into complex exponentials using Euler relation
  • Evaluates time-domain terms by substituting time and complex identities

Fourier Series an expansion of a periodic function into a sum of trigonometric terms or complex exponential components.

Trigonometric to Exponential Conversion

  • Euler Identity: relates complex exponentials to trigonometric functions
  • Cosine Term Conversion: decomposes cosine into symmetric exponential terms
  • Evaluation at Time t: evaluates complex exponential sum at specified time

Exponential Form of Cosine Term

\[ \cos \theta = \dfrac{e^{j\theta} + e^{-j\theta}}{2} \implies A \cos(\omega t) = \dfrac{A}{2} e^{j\omega t} + \dfrac{A}{2} e^{-j\omega t} \]

Example. Express the Fourier series cosine term \(12 \cos(30\pi t)\) in complex exponential form.

Solution. Apply Euler’s identity \(\cos\theta = \dfrac{e^{j\theta} + e^{-j\theta}}{2}\) with \(\theta = 30\pi t\):

\[ 12 \cos(30\pi t) = 12 \left[ \dfrac{e^{j30\pi t} + e^{-j30\pi t}}{2} \right] \]
\[ 12 \cos(30\pi t) = 6 e^{j30\pi t} + 6 e^{-j30\pi t} \]

Answer: \(6 e^{j30\pi t} + 6 e^{-j30\pi t}\)

Evaluation of Exponential Fourier Terms

Sum of Complex Exponentials

\[ e^{j\theta} + e^{-j\theta} = 2 \cos\theta \]

Example. Evaluate the exponential Fourier series expression \(3 e^{j 6\pi t} + 3 e^{-j 6\pi t}\) at \(t = 1\).

Solution. Factor out the constant and substitute \(t = 1\) into the expression:

\[ x = 3 \left( e^{j 6\pi (1)} + e^{-j 6\pi (1)} \right) = 3 \left( e^{j 6\pi} + e^{-j 6\pi} \right) \]

Apply Euler’s identity \(e^{j\theta} + e^{-j\theta} = 2 \cos\theta\) with \(\theta = 6\pi\) radians (\(6\pi\text{ rad} = 1080^\circ\)):

\[ x = 3 \left( 2 \cos(6\pi) \right) = 6 \cos(1080^\circ) = 6(1) = 6 \]

Answer: \(6\)