Laplace Transform Notes

Laplace Transform

• The Laplace transform converts a function of a real variable (often time) to a function of a complex variable (frequency).

• It is an integral transform named after Pierre-Simon Laplace.

• Laplace transforms are useful for solving linear ODEs and analyzing frequency response and stability.

Definition of Laplace Transform

• The integral transform of a function $f(t)$ in the interval $a \leq x \leq b$ is given by:
  $L[f(t)] = \int_{a}^{b} K(t, s) f(t) dt = F(s)$
  where \infty \leq a < b \leq \infty

• Given a function $f(t)$ defined for t > 0 ($0 \leq t \leq \infty$), and $K(t, s) = e^{-st}$, the Laplace transform is defined as: $L[f(t)] = \int_{0}^{\infty} e^{-st} f(t) dt = F(s)$, where Re(s) > 0

  • $s$ is the transform variable, a complex number.

• The Laplace transform converts time domain functions and operations into the frequency domain: $f(t) \rightarrow F(s)$, where $t \in \mathbb{R}, s \in \mathbb{C}$

Laplace Transform of Elementary Functions

• Example 1: Find the Laplace transform of $f(t) = 1$ for t > 0

  • Solution:
    $L[f(t)] = \int<em>{0}^{\infty} e^{-st} f(t) dt$ $L[1] = \int</em>{0}^{\infty} e^{-st} (1) dt = \left[ \frac{e^{-st}}{-s} \right]_{0}^{\infty} = 0 - \frac{1}{-s} = \frac{1}{s}$
    $L[1] = \frac{1}{s}$

• Example 2: Find the Laplace transform of $f(t) = e^{at}$ for t > 0, where $a$ is a constant.

  • Solution:
    $L[e^{at}] = \int<em>{0}^{\infty} e^{-st} e^{at} dt = \int</em>{0}^{\infty} e^{-(s-a)t} dt$
    $= \left[ \frac{e^{-(s-a)t}}{-(s-a)} \right]_{0}^{\infty} = 0 - \frac{1}{-(s-a)} = \frac{1}{s-a}$, for s > a
    $L[e^{at}] = \frac{1}{s-a}$, for s > a

• Example 3: Find the Laplace transform of $f(t) = \sin(at)$, where $a$ is a real constant.

  • Solution:
    $L[\sin(at)] = \int<em>{0}^{\infty} e^{-st} \sin(at) dt$ Since $\sin(at) = \frac{e^{iat} - e^{-iat}}{2i}$ $L[\sin(at)] = \frac{1}{2i} \int</em>{0}^{\infty} e^{-st} (e^{iat} - e^{-iat}) dt = \frac{1}{2i} \int<em>{0}^{\infty} (e^{-(s-ia)t} - e^{-(s+ia)t}) dt$ $= \frac{1}{2i} \left[ \frac{e^{-(s-ia)t}}{-(s-ia)} - \frac{e^{-(s+ia)t}}{-(s+ia)} \right]</em>{0}^{\infty} = \frac{1}{2i} \left[ \frac{1}{s-ia} - \frac{1}{s+ia} \right]$
    $= \frac{1}{2i} \left[ \frac{s+ia - (s-ia)}{(s-ia)(s+ia)} \right] = \frac{1}{2i} \left[ \frac{2ia}{s^2 + a^2} \right] = \frac{a}{s^2 + a^2}$
    $L[\sin(at)] = \frac{a}{s^2 + a^2}$

• Example 4: Find the Laplace transform of $f(t) = \cos(at)$, where $a$ is a real constant.

  • Solution:
    $L[\cos(at)] = \frac{s}{s^2 + a^2}$

• Example 5: Find the Laplace transform of $f(t) = \cosh(at)$, where $a$ is a real constant.

  • Solution:
    $L[\cosh(at)] = \int<em>{0}^{\infty} e^{-st} \cosh(at) dt$ Since $\cosh(at) = \frac{e^{at} + e^{-at}}{2}$ $L[\cosh(at)] = \frac{1}{2} \int</em>{0}^{\infty} e^{-st} (e^{at} + e^{-at}) dt = \frac{1}{2} \int<em>{0}^{\infty} (e^{-(s-a)t} + e^{-(s+a)t}) dt$ $= \frac{1}{2} \left[ \frac{e^{-(s-a)t}}{-(s-a)} + \frac{e^{-(s+a)t}}{-(s+a)} \right]</em>{0}^{\infty} = \frac{1}{2} \left[ \frac{1}{s-a} + \frac{1}{s+a} \right]$
    $= \frac{1}{2} \left[ \frac{s+a + s-a}{(s-a)(s+a)} \right] = \frac{1}{2} \left[ \frac{2s}{s^2 - a^2} \right] = \frac{s}{s^2 - a^2}$
    $L[\cosh(at)] = \frac{s}{s^2 - a^2}$

• Example 6: Find the Laplace transform of $f(t) = \sinh(at)$, where $a$ is a real constant.

  • Solution:
    $L[\sinh(at)] = \frac{a}{s^2 - a^2}$

• Note:

  1. The gamma function is defined by the improper integral: \Gamma(n) = \int_{0}^{\infty} x^{n-1} e^{-x} dx, n > 0

  2. If $n$ is a positive integer, then $\Gamma(n+1) = n!$

  3. $\Gamma(n+1) = n \Gamma(n)$

  4. $\Gamma(\frac{1}{2}) = \sqrt{\pi}$

• Find the Laplace transform of $f(t) = t^n$, where n > -1

  • Solution:
    $L[f(t)] = \int<em>{0}^{\infty} e^{-st} f(t) dt = F(s)$ Let $f(t) = t^n$, then $L[t^n] = \int</em>{0}^{\infty} e^{-st} t^n dt$
    Let $x = st$, so $dt = \frac{dx}{s}$
    $L[t^n] = \int<em>{0}^{\infty} e^{-x} \left( \frac{x}{s} \right)^n \frac{dx}{s} = \frac{1}{s^{n+1}} \int</em>{0}^{\infty} e^{-x} x^n dx = \frac{\Gamma(n+1)}{s^{n+1}}$
    If $n$ is a positive integer, then $\Gamma(n+1) = n!$
    $L[t^n] = \frac{n!}{s^{n+1}}$

Linearity of Laplace Transforms

• Let $f<em>1(t)$ and $f</em>2(t)$ be two functions with Laplace Transforms $F<em>1(s)$ and $F</em>2(s)$. If $c<em>1$ and $c</em>2$ are any constants, then
  $L[c<em>1 f</em>1(t) + c<em>2 f</em>2(t)] = c<em>1 L[f</em>1(t)] + c<em>2 L[f</em>2(t)] = c<em>1 F</em>1(s) + c<em>2 F</em>2(s)$

• Example 8: Find the Laplace Transform of $f(t) = 1 + 2e^{3t}$.

  • Solution:
    $L[1 + 2e^{3t}] = L[1] + 2L[e^{3t}] = \frac{1}{s} + 2 \cdot \frac{1}{s-3}$

• Example 9: Find the Laplace Transform of $f(t) = t^3 + e^{-3t} + \sqrt{t}$.

  • Solution:
    $L[t^3 + e^{-3t} + \sqrt{t}] = L[t^3] + L[e^{-3t}] + L[t^{1/2}]$
    $= \frac{3!}{s^4} + \frac{1}{s+3} + \frac{\Gamma(3/2)}{s^{3/2}} = \frac{6}{s^4} + \frac{1}{s+3} + \frac{\sqrt{\pi}}{2s^{3/2}}$

Heaviside First Shifting Theorem

• If $L[f(t)] = F(s)$, then $L[e^{at} f(t)] = F(s-a)$, where $a$ is a real constant.

  • Proof:
    $L[f(t)] = \int<em>{0}^{\infty} e^{-st} f(t) dt = F(s)$ $L[e^{at} f(t)] = \int</em>{0}^{\infty} e^{-st} e^{at} f(t) dt = \int_{0}^{\infty} e^{-(s-a)t} f(t) dt = F(s-a)$

• Example: Find $L[t^3 e^{4t}]$.

  • Solution:
    Let $f(t) = t^3$, then $L[f(t)] = L[t^3] = \frac{3!}{s^4} = F(s)$
    $L[t^3 e^{4t}] = \frac{3!}{(s-4)^4} = \frac{6}{(s-4)^4}$

Change of Scale Property

• If $L[f(t)] = F(s)$, then $L[f(at)] = \frac{1}{a} F(\frac{s}{a})$, a > 0

  • Proof:
    $L[f(t)] = \int<em>{0}^{\infty} e^{-st} f(t) dt = F(s)$ $L[f(at)] = \int</em>{0}^{\infty} e^{-st} f(at) dt$
    Let $x = at$, then $t = \frac{x}{a}$, and $dt = \frac{dx}{a}$. If $t \rightarrow 0$, then $x \rightarrow 0$, and if $t \rightarrow \infty$, then $x \rightarrow \infty$
    $L[f(at)] = \int<em>{0}^{\infty} e^{-s(x/a)} f(x) \frac{dx}{a} = \frac{1}{a} \int</em>{0}^{\infty} e^{-(s/a)x} f(x) dx = \frac{1}{a} F(\frac{s}{a})$

• Example: If $L[f(t)] = \log(\frac{s+3}{s})$, find $L[f(2t)]$.

  • Solution:
    $L[f(2t)] = \frac{1}{2} \log\left( \frac{\frac{s}{2} + 3}{\frac{s}{2}} \right) = \frac{1}{2} \log\left( \frac{s+6}{s} \right) = \log \sqrt{\frac{s+6}{s}}$

Second Shifting Theorem

• If $L[f(t)] = F(s)$ and
  $g(t) = \begin{cases} f(t-a), &amp; t &gt; a \ 0, &amp; t &lt; a \end{cases}$
  then $L[g(t)] = e^{-as} F(s)$

• Example: Find the Laplace transform of
  $g(t) = \begin{cases} \sin(t-\pi), &amp; t &gt; \pi \ 0, &amp; t &lt; \pi \end{cases}$

  • Solution:
    Here, $f(t) = \sin(t)$, $a = \pi$
    $L[g(t)] = e^{-as} F(s) = e^{-\pi s} \cdot \frac{1}{s^2 + 1}$

• Example: Find the Laplace transform of
  $g(t) = \begin{cases} t^3, &amp; t &gt; 1 \ 0, &amp; t &lt; 1 \end{cases}$

  • Solution:
    Here, $f(t) = t^3$, $a = 1$
    $L[g(t)] = e^{-as} F(s) = e^{-s} \cdot \frac{3!}{s^4} = \frac{6e^{-s}}{s^4}$

Differentiation of Laplace Transform (Multiplication by t)

• If $L[f(t)] = F(s)$, then $L[t^n f(t)] = (-1)^n \frac{d^n}{ds^n} [F(s)], n = 1, 2, 3, …$

• Example: Find $L[t \sin^3(t)]$.

  • Solution:
    Here, $f(t) = \sin^3(t) = \frac{3\sin(t) - \sin(3t)}{4}$
    $L[t \sin^3(t)] = L\left[ t \left( \frac{3\sin(t) - \sin(3t)}{4} \right) \right] = \frac{3}{4}L[t\sin(t)] - \frac{1}{4} L[t\sin(3t)]$
    $= \frac{3}{4} \left( -\frac{d}{ds} \left( \frac{1}{s^2+1} \right) \right) - \frac{1}{4} \left( -\frac{d}{ds} \left( \frac{3}{s^2+9} \right) \right) = \frac{3s}{2(s^2+1)^2} - \frac{3s}{2(s^2+9)^2}$

*Example: Find $L[te^{2t} \cos(3t)]$

• Solution:
  Let $f(t) = e^{2t}\cos(3t)$
  $L[e^{2t} \cos(3t)] = \frac{s-2}{(s-2)^2 + 9} = \frac{s-2}{s^2-4s+13} \quad \text{[Using 1st shifting theorem]}$
  Now, $L[te^{2t}\cos(3t)] = (-1) \frac{d}{ds}[L[e^{2t}\cos(3t)]]$

  $= -\frac{d}{ds} \left[ \frac{s-2}{s^2-4s+13} \right]$
  $= -\frac{(1)(s^2-4s+13)-(s-2)(2s-4)}{(s^2-4s+13)^2}$
  $= -\frac{s^2-4s+13-(2s^2-8s+8)}{(s^2-4s+13)^2}$
  $= \frac{s^2-4s-5}{(s^2-4s+13)^2}$

Integration of Laplace Transform (Division by t)

• If $L[f(t)] = F(s)$, then $L\left[ \frac{f(t)}{t} \right] = \int_{s}^{\infty} F(x) dx$

• Example: Find $L\left[ \frac{\sin(at)}{t} \right]$.

  • Solution:
    Let $f(t) = \sin(at)$
    $L[\sin(at)] = \frac{a}{s^2+a^2} = F(s)$
    $L\left[ \frac{\sin(at)}{t} \right] = \int<em>{s}^{\infty} \frac{a}{x^2+a^2} dx = \left[ \arctan\left( \frac{x}{a} \right) \right]</em>{s}^{\infty} = \frac{\pi}{2} - \arctan\left( \frac{s}{a} \right)$

*Example: Find $L\left[ \frac{\cos(2t) - \cos(3t)}{t} \right]$

Solution:

$L\left[ \frac{\cos(2t) - \cos(3t)}{t} \right] = \int<em>{s}^{\infty} L[\cos(2t)] ds - \int</em>{s}^{\infty} L[\cos(3t)] ds$
$= \int<em>{s}^{\infty} \frac{x}{x^2+4} dx - \int</em>{s}^{\infty} \frac{x}{x^2+9} dx$
$= \frac{1}{2} \left[ \log(x^2+4) \right]<em>{s}^{\infty} - \frac{1}{2} \left[ \log(x^2+9) \right]</em>{s}^{\infty}$
$= \frac{1}{2} \left[ \lim_{b \to \infty} \log\left(\frac{b^2+4}{b^2+9} \right) - \log\left(\frac{s^2+4}{s^2+9} \right) \right]$
$= \frac{1}{2} \left[ \log(1) - \log\left(\frac{s^2+4}{s^2+9} \right) \right]$
$= -\frac{1}{2} \log\left(\frac{s^2+4}{s^2+9} \right) = \frac{1}{2} \log\left(\frac{s^2+9}{s^2+4} \right)$

Laplace Transform of Derivatives

• When solving differential equations using Laplace transforms, we need to evaluate quantities like $L[\frac{dy}{dt}]$ and $L[\frac{d^2y}{dt^2}]$.

• If $f'(t)$ is continuous for $t \geq 0$, then integration by parts gives
  $L[f'(t)] = \int<em>{0}^{\infty} e^{-st} f'(t) dt = [e^{-st} f(t)]</em>{0}^{\infty} - \int<em>{0}^{\infty} (-s) e^{-st} f(t) dt$ $= -f(0) + s \int</em>{0}^{\infty} e^{-st} f(t) dt = -f(0) + s L[f(t)]$
  Or, $L[f'(t)] = sF(s) - f(0)$

• Similarly,
  $L[f''(t)] = \int<em>{0}^{\infty} e^{-st} f''(t) dt = [e^{-st} f'(t)]</em>{0}^{\infty} - \int_{0}^{\infty} (-s) e^{-st} f'(t) dt$
  $= -f'(0) + s L[f'(t)] = -f'(0) + s [sF(s) - f(0)]$
  Or, $L[f''(t)] = s^2 F(s) - sf(0) - f'(0)$

• In a similar manner,
  $L[f'''(t)] = s^3 F(s) - s^2 f(0) - sf'(0) - f''(0)$

• Theorem: If $f, f', f'', …, f^{(n-1)}$ are continuous on $(0, \infty)$ and are of exponential order, and if $f^{(n)}(t)$ is piecewise continuous on $(0, \infty)$, then
  $L[f^{(n)}(t)] = s^n F(s) - s^{n-1} f(0) - s^{n-2} f'(0) - … - f^{(n-1)}(0)$

*Example: If $L[\frac{1}{\sqrt{t}}] = \sqrt{\frac{\pi}{s}}$, then show that $L[\sqrt{t}] = \frac{\sqrt{\pi}}{2s^{3/2}}$

Solution:

Let $f(t) = \frac{1}{\sqrt{t}}$, then $f(0) = 1$
We know that,
$f'(t) = -\frac{1}{2t^{3/2}} = \frac{1}{2\sqrt{t^3}}$
$L[f'(t)] = sL[f(t)] - f(0)$
$L[\frac{1}{-2\sqrt{t^3}}] = s \sqrt{\frac{\pi}{s}} - 1$
$-\frac{1}{2} L[\frac{1}{\sqrt{t^3}}]= \sqrt{s\pi} - 1$

Laplace transform of integrals $L[\int_{0}^{t} f(u)du] = \frac{F(s)}{s}$

If $L[f(t)] = F(s)$, then

Proof:

Let $G(t) = \int<em>{0}^{t} f(u) du$ Now, $G(0) = 0 \implies G'(t) = f(t)$ $L[G'(t)] = sL[G(t)] - G(0)$ $= sL[G(t)]$ But, $L[G'(t)] = L[f(t)]$ Now,$sL[G(t)] = L[f(t)]$ $L[G(t)] = \frac{L[f(t)]}{s}$ $L[ \int</em>{0}^{t} f(u) du] = \frac{F(s)}{s}$

Example: Evaluate $L[ \int<em>{0}^{t} \frac{Sint}{t} dt]$ Solution: $f(t) = Sint$ Now $L[Sint]= \frac{1}{s^2 + 1}$ So, L[ \int{0}^{t} \frac{Sint}{t} dt]=\int{s}^{\infty}L \frac{Sint}{t} ds$$=\int{s}^{\infty}\frac{1}{s^2 + 1}ds$<br>$=[tan^{-1} s]_{s}^{\infty}$<br>$=-tan^{-1} s = Cot^{-1}s$</p><ul><li><p>Evaluate$L[\int_{0}^{t} e^{-u} \cos(u) du]$</p></li><li><p>Solution: <br>Let$f(t) = e^{-t} \cos(t)$<br>$L[e^{-t} \cos(t)] = \frac{s+1}{(s+1)^2 + 1} = F(s)$<br>Then$L[ \int_{0}^{t} e^{-u} \cos(u) du] = \frac{F(s)}{s}= \frac{s+1}{s((s+1)^2 + 1)}$</p></li></ul><h3 id="25cd87e6-7e19-49fd-a434-e7eda510d72e" data-toc-id="25cd87e6-7e19-49fd-a434-e7eda510d72e" collapsed="false" seolevelmigrated="true">Unit Step Function or Heaviside's Unit Function</h3><ul><li><p>The unit step or Heaviside's unit step function$u(t-a)$is defined as follows:<br>$u(t-a) = \begin{cases} 0, & 0 \leq t < a \ 1, & t \geq a \end{cases}$</p></li><li><p>In particular, if$a = 0$, then<br>$u(t) = \begin{cases} 0, & t < 0 \ 1, & t > 0 \end{cases}$</p></li></ul><h3 id="83da52eb-d466-4c3f-a17c-83be8c639842" data-toc-id="83da52eb-d466-4c3f-a17c-83be8c639842" collapsed="false" seolevelmigrated="true">Laplace Transform of Unit Step Function</h3><ul><li><p>$L[u(t-a)] = \int{0}^{\infty} e^{-st} u(t-a) dt = \int{0}^{a} e^{-st} u(t-a) dt + \int{a}^{\infty} e^{-st} u(t-a) dt$$= \int{0}^{a} e^{-st} (0) dt + \int{a}^{\infty} e^{-st} (1) dt = 0 + \left[ \frac{e^{-st}}{-s} \right]{a}^{\infty} = 0 - \frac{e^{-sa}}{-s} = \frac{e^{-sa}}{s}$<br>$L[u(t-a)] = \frac{e^{-sa}}{s}$<br>In particular, if$a = 0$,$L[u(t)] = \frac{1}{s}$</p></li></ul><h3 id="657fb90a-1d89-4595-8693-7a9ef3b7101c" data-toc-id="657fb90a-1d89-4595-8693-7a9ef3b7101c" collapsed="false" seolevelmigrated="true">Inverse Laplace Transform</h3><ul><li><p>If$F(s)$is the Laplace transform of a function$f(t)$, that is, if$L[f(t)] = F(s)$, then$f(t)$is called the inverse Laplace transform of$F(s)$and is written as$f(t) = L^{-1}[F(s)]$, and$L^{-1}$is called the inverse Laplace transform operator.</p></li><li><p>For example,$L[e^{at}] = \frac{1}{s-a} \implies e^{at} = L^{-1}\left[ \frac{1}{s-a} \right]$</p></li></ul><h3 id="a3f2ec49-2c83-43a5-b272-cf6f675fb6a0" data-toc-id="a3f2ec49-2c83-43a5-b272-cf6f675fb6a0" collapsed="false" seolevelmigrated="true">Linearity Property of Inverse Laplace Transform</h3><ul><li><p>If$F1(s)$and$F2(s)$are the Laplace transforms of the functions$f1(t)$and$f2(t)$, respectively, then we have<br>$L^{-1}[c1 F1(s) + c2 F2(s)] = c1 L^{-1}[F1(s)] + c2 L^{-1}[F2(s)]$<br>where$c1$and$c2$are constants.</p></li></ul><h3 id="c7c4e85e-2957-437e-a83a-f8cbe7e6f08a" data-toc-id="c7c4e85e-2957-437e-a83a-f8cbe7e6f08a" collapsed="false" seolevelmigrated="true">First Shifting Property for Inverse Laplace Transform</h3><ul><li><p>If$L^{-1}[F(s)] = f(t)$, then$L^{-1}[F(s-a)] = e^{at} f(t)$.</p><ul><li><p>Proof:<br>$F(s) = \int{0}^{\infty} e^{-st} f(t) dt$$F(s-a) = \int{0}^{\infty} e^{-(s-a)t} f(t) dt = \int{0}^{\infty} e^{-st+at} f(t) dt = \int{0}^{\infty} e^{-st} (e^{at} f(t)) dt = L[e^{at} f(t)]$<br>$L^{-1}[F(s-a)] = e^{at} f(t)$</p></li></ul></li></ul><h3 id="fb2e4495-376a-4cc3-af5c-bb247746a415" data-toc-id="fb2e4495-376a-4cc3-af5c-bb247746a415" collapsed="false" seolevelmigrated="true">Summary Table of Laplace Transforms</h3><table style="min-width: 75px"><colgroup><col style="min-width: 25px"><col style="min-width: 25px"><col style="min-width: 25px"></colgroup><tbody><tr><th colspan="1" rowspan="1"><p>S.No.</p></th><th colspan="1" rowspan="1"><p>F(s)</p></th><th colspan="1" rowspan="1"><p>$f(t)$</p></th></tr><tr><td colspan="1" rowspan="1"><p>1</p></td><td colspan="1" rowspan="1"><p>1/s</p></td><td colspan="1" rowspan="1"><p>1</p></td></tr><tr><td colspan="1" rowspan="1"><p>2</p></td><td colspan="1" rowspan="1"><p>$1/(s-a)$</p></td><td colspan="1" rowspan="1"><p>$e^{at}$</p></td></tr><tr><td colspan="1" rowspan="1"><p>3</p></td><td colspan="1" rowspan="1"><p>1/(s+a)</p></td><td colspan="1" rowspan="1"><p>$e^{-at}$</p></td></tr><tr><td colspan="1" rowspan="1"><p>4</p></td><td colspan="1" rowspan="1"><p>a/(s^2+a^2)</p></td><td colspan="1" rowspan="1"><p>sin at</p></td></tr><tr><td colspan="1" rowspan="1"><p>5</p></td><td colspan="1" rowspan="1"><p>s/(s^2+a^2)</p></td><td colspan="1" rowspan="1"><p>cos at</p></td></tr><tr><td colspan="1" rowspan="1"><p>6</p></td><td colspan="1" rowspan="1"><p>a/(s^2-a^2)</p></td><td colspan="1" rowspan="1"><p>sinh at</p></td></tr><tr><td colspan="1" rowspan="1"><p>7</p></td><td colspan="1" rowspan="1"><p>s/(s^2-a^2)</p></td><td colspan="1" rowspan="1"><p>cosh at</p></td></tr><tr><td colspan="1" rowspan="1"><p>8</p></td><td colspan="1" rowspan="1"><p>$n!/s^{n+1}$</p></td><td colspan="1" rowspan="1"><p>$t^n$</p></td></tr><tr><td colspan="1" rowspan="1"><p>9</p></td><td colspan="1" rowspan="1"><p>$\frac{b}{(s-a)^2+b^2}$</p></td><td colspan="1" rowspan="1"><p>$e^{at}sin(bt)$</p></td></tr><tr><td colspan="1" rowspan="1"><p>10</p></td><td colspan="1" rowspan="1"><p>$\frac{s-a}{(s-a)^2+b^2}$</p></td><td colspan="1" rowspan="1"><p>$e^{at}cos(bt)$</p></td></tr><tr><td colspan="1" rowspan="1"><p>11</p></td><td colspan="1" rowspan="1"><p>$\frac{b}{(s-a)^2-b^2}$</p></td><td colspan="1" rowspan="1"><p>$e^{at}sinh(bt)$</p></td></tr><tr><td colspan="1" rowspan="1"><p>12</p></td><td colspan="1" rowspan="1"><p>$\frac{s-a}{(s-a)^2-b^2}$</p></td><td colspan="1" rowspan="1"><p>$e^{at}cosh(bt)$</p></td></tr></tbody></table><ul><li><p>Find the inverse Laplace transform of the following functions:</p><ul><li><p>$L^{-1}\left[ \frac{3}{s} - \frac{3}{s^2} + \frac{4}{s^3} \right] = 3L^{-1}\left[ \frac{1}{s} \right] - 3L^{-1}\left[ \frac{1}{s^2} \right] + 2L^{-1}\left[ \frac{2}{s^3} \right] = 3 - 3t + 2t^2$</p></li><li><p>$L^{-1}\left[ \frac{2s-5}{9s^2-25} \right] = L^{-1}\left[ \frac{2s}{9s^2-25} - \frac{5}{9s^2-25} \right] = \frac{2}{9} L^{-1}\left[ \frac{s}{s^2-(5/3)^2} \right] - \frac{5}{9} L^{-1}\left[ \frac{1}{s^2-(5/3)^2} \right]$<br>$= \frac{2}{9} \cosh\left( \frac{5}{3}t \right) - \frac{5}{9} \cdot \frac{3}{5} \sinh\left( \frac{5}{3}t \right) = \frac{2}{9} \cosh\left( \frac{5}{3}t \right) - \frac{1}{3} \sinh\left( \frac{5}{3}t \right)$</p></li></ul></li><li><p>$L^{-1}[ \frac{1}{(s+1/2)^2 - 1/4} ] = L^{-1}[ \frac{1}{(s+1/2)^2 - (1/2)^2} ] = e^{-t/2}Sinh(t/2)$</p></li><li><p>$L^{-1}[ \frac{3s+1}{(s+1)^4} ]$<br>=$L^{-1}[ \frac{3(s+1) - 2}{(s+1)^4} ]$<br>=$L^{-1}[ \frac{3(s+1)}{(s+1)^4}] + L^{-1}[ \frac{-2}{(s+1)^4} ]$<br>=$3L^{-1}[ \frac{1}{(s+1)^3}] + L^{-1}[ \frac{-2}{(s+1)^4} ]$<br>=$3e^{-t}\frac{t^2}{2!} + (-2)e^{-t}\frac{t^3}{3!}$<br>=$e^{-t}\frac{3}{2}t^2 + e^{-t}\frac{-1}{3}t^3$</p></li></ul><p><em>Evaluate$2^{-1}[\frac{8s+29}{s^2-12s+32}]$Solution:$2^{-1}[\frac{8s+29}{s^2-12s+32}]$$=2^{-1}[\frac{8s+29}{(s-4)(s-8)}] $Using partial fraction decomposition.$\frac{8s+29}{(s-4)(s-8)} = \frac{A}{s-4} + \frac{B}{s-8}$Cross multiplying and collecting like terms.$8s+29=A(s-8) +B(s-4)$Let s=4:$84+29 = A(4-8)
32+29=-4A
61=-4A
A=-61/4
Let s=8:
$8*8+29=B(8-4)$
64+29=4B$<br>$B=93/4$<br>Therefore,<br>$2^{-1}[\frac{8s+29}{s^2-12s+32}]$<br>$=2^{-1}[\frac{-61}{4(s-4)} + \frac{93}{4(s-8)}] $<br>$=( \frac{-61}{4}e^{4t} + \frac{93}{4}e^{8t})$</p><p>*Evaluate$2^{-1}[\frac{5s^2-15s-11}{(s+1)(s-2)^3}]$<br>Solution:</p><p>$\frac{5s^2-15s-11}{(s+1)(s-2)^3} = \frac{A}{s+1} + \frac{B}{s-2} + \frac{C}{(s-2)^2} + \frac{D}{(s-2)^3}$</p><p>Multiplying both sides by$(s+1)(s-2)^3$gives<br>$5s^2-15s-11 = A(s-2