### Laplace Transform

• Find the Laplace transform of a function using the definition.

• Find the inverse Laplace function of a function.

• Use the Translation Theorems to find Laplace transforms.

• Find the Laplace transform of derivatives, integrals and periodic functions.

• Understand the Unilateral Laplace Transform of some commonly used signals.

### Solved Example:

#### 3-6-01

If f(t) is a function defined for all t $\geqq$ 0, its Laplace transform $\phi$(s) is defined as:

Solution:
The definition of Laplace transform does not involve complex numbers. So C and D are incorrect. Also, for first option, $e^{st}$ will become undefined for t = $\infty$. So A is also incorrect.

### Solved Example:

#### 3-6-02

The inverse Laplace transform of the function $F(s) =\dfrac{1}{s(s+1)}$ is given by:

Solution:
$F\left( s\right) =\dfrac {1}{s\left( s+1\right)}$ First using the partial fraction $\dfrac {1}{s\left( s+1\right)} =\dfrac {A}{s}+\dfrac {B}{s+1}$ $\dfrac {1}{s\left( s+1\right)} =\dfrac {A (s+1) +B s}{s (s+1)}$ Comparing numerators, $A\left( s+1\right) +Bs=1$
Comparing s coefficients, A + B = 0
Comparing constants, A = 1 which gives B = -1 $f\left( t\right) =\mathscr{L}^{-1}\left[ F\left( s\right) \right] =\mathscr{L}^{-1}\left[ \dfrac {1}{s}-\dfrac {1}{s+1}\right] =1-e^{-t}$

### Solved Example:

#### 3-6-03

A delayed unit step function is: \begin{align*} U(t -a)&=0; \mathrm{when\ } t < a,\\ &= 1; \mathrm{when\ } t \ge a. \end{align*} Its Laplace transform is:

Solution:
$\mathscr{L}\left[ f\left( t\right) \right] =\int ^{\infty }_{0}e^{-st}f\left( t\right) dt$ \begin{align*} \mathscr{L}\left( U\left( t-a\right) \right) &=\int ^{\infty}_{0} e^{-st} U\left( t-a\right) dt\\ &={\int ^{a}_{0}e^{-st}\left( 0\right) dt}+\int ^{\infty }_{a}e^{-st}\left( 1\right) dt\\ &=\left[ \dfrac {e^{-st}}{-s}\right] ^{\infty }_{a} =0-\left[ \dfrac {e^{-as}}{-s}\right]\\ &=\dfrac {e^{-as}}{s} \end{align*}

### Solved Example:

#### 3-6-04

Laplace transform of the function $\sin \omega t$ is:

Solution:
$\mathscr{L}[\cos \omega t] = \dfrac{s}{s^2 +\omega^2}$ $\mathscr{L}[\sin \omega t] = \dfrac{\omega}{s^2 +\omega^2}$ $\mathscr{L}[\cosh \omega t] = \dfrac{s}{s^2 -\omega^2}$ $\mathscr{L}[\sinh \omega t] = \dfrac{\omega}{s^2 -\omega^2}$

### Solved Example:

#### 3-6-05

If F(s) is the Laplace transform of function f (t), then Laplace transform of $\int ^{t}_{0}f\left( \tau \right) d\tau$ is:

Solution:
If F(s) is the Laplace transform of function f (t),then Laplace transform of derivative f'(t) is given by : $\mathscr{L}[f'(t)] = sF(s) - f (0)$ Laplace transform of integration $\int ^{t}_{0}f\left( \tau \right) d\tau$ is given by : $\dfrac{1}{s}F(s)$

### Solved Example:

#### 3-6-06

The Laplace transform of a function f(t)= $\dfrac{1}{s^2(s+1)}$. The function f(t) is:

Solution:
$f\left( t\right) =\mathscr{L}^{-1}\left[ \dfrac {1}{s^{2}\left( s+1\right) }\right]$ \begin{align*} \dfrac {1}{s^{2}\left( s+1\right) } &=\dfrac {A}{s}+\dfrac {B}{s^{2}}+\dfrac {C}{s+1}\\ &=\dfrac {As\left( 1+s\right) +B\left( s+1\right) +cs^{2}}{s^{2}\left( s+1\right) }\\ &=\dfrac{s^2(A+C)+ s(A+B)+ B}{s^{2}\left( s+1\right) } \end{align*} Compare the coefficients of s$^2$, s and constant terms and we get
$A + C = 0; A + B = 0 \mathrm{\ and\ } B = 1$ Solving above equation, we get A =- 1, B = 1 and C = 1 \begin{align*} f\left( t\right) &= \mathscr{L}^{-1}\left[ \dfrac {-1}{s}+\dfrac {1}{s^{2}}+\dfrac {1}{s+1}\right]\\ &= -1+t+e^{-t}\\ &= t-1+e^{-t} \end{align*}

### Solved Example:

#### 3-6-07

The inverse Laplace transform of $\dfrac{1}{(s^2+ s)}$ is:

Solution:
First let us convert using partial fractions. $\left[\dfrac{1}{(s^2+ s)}\right] = \left[\dfrac{1}{s(s+1)}\right] = \left[\dfrac{A}{s}+ \dfrac{B}{(s+1)}\right]$ $A(s) + B (s+1) = 1$ Put s = 0, $A= 1$ Put s = -1, $B= -1$ \begin{align*} \mathscr{L}^{-1}\left[\dfrac{1}{(s^2+ s)}\right] &= \mathscr{L}^{-1}\left[\dfrac{1}{s(s+1)}\right]\\ &= \mathscr{L}^{-1}\left[\dfrac{A}{s}+ \dfrac{B}{(s+1)}\right]\\ &= \mathscr{L}^{-1}\left[\dfrac{1}{s}+ \dfrac{-1}{(s+1)}\right]\\ &= \mathscr{L}^{-1}\left[\dfrac{1}{s}\right]- \mathscr{L}^{-1}\left[ \dfrac{1}{(s+1)}\right]\\ &= 1 - e^{-t} \end{align*}

### Solved Example:

#### 3-6-08

Find the Laplace transform of $f(t) = e^{3t}t^2$.

Solution:
$\mathscr{L}\left[t^{n}\right] =\dfrac{n!}{s^{n+1}}$ $\mathscr{L}\left[t^{2}\right] =\dfrac{2!}{s^{2+1}}=\dfrac{2}{s^{3}}$ Using first shifting theorem, when you multiply by $e^{3t}$, replace s by (s-3) $\mathscr{L}\left[e^{3t}t^{2}\right] =\dfrac{2}{\left(s-3\right)^{3}}$

### Solved Example:

#### 3-6-09

Find the inverse Laplace transform of $\phi(s) = \left[\dfrac{s}{s^2-2s+5}\right]$.

Solution:
\begin{align*} & \mathscr{L}^{-1} \left[\dfrac{s}{s^2-2s+5}\right]\\ &= \mathscr{L}^{-1} \left[\dfrac{s}{s^2-2s+1+4}\right]\\ &= \mathscr{L}^{-1} \left[\dfrac{s}{(s-1)^2+4}\right]\\ &=\mathscr{L}^{-1}\left[ \dfrac{\left(s-1\right) +1}{\left( s-1\right) ^{2}+4}\right]\\ &=\mathscr{L}^{-1}\left[ \dfrac{s-1}{\left( s-1\right) ^{2}+4}\right] &&+\mathscr{L}^{-1}\left[ \dfrac{1}{\left(s-1\right) ^{2}+4}\right]\\ &=e^{t}\mathscr{L}^{-1}\left[\dfrac{s}{s^{2}+4}\right] &&+e^{t} \mathscr{L}^{-1}\left[ \dfrac{1}{s^{2}+4}\right]\\ &= e^t \cos 2t &&+ \dfrac{1}{2}e^t \sin 2t \end{align*}

The solution of the differential equation: $y''- y =e^{2t}$ with the initial conditions f(0) = 0 and f'(0) = 1 is:
$y''-y =e^{2t}$ $\mathrm{Taking\ Laplace\ Transform\ of\ both\ sides,}$ \begin{align*} \mathscr{L}\left[ y''-y\right] &=\mathscr{L}\left[ e^{2t}\right] \\ s^{2}F\left( s\right) -sf\left( 0\right) -f'\left( 0\right) -F\left( s\right) &=\dfrac{1}{s-2} \end{align*} $\mathrm{Substituting\ the\ given\ initial\ conditions,}$ \begin{align*} \left(s^{2}-1\right) F\left( s\right) -0-1 &=\dfrac{1}{s-2}\\ \left(s^{2}-1\right) F\left( s\right) &=1+\dfrac{1}{s-2}\\ \left(s^{2}-1\right) F\left( s\right) &=\dfrac{s-2+1}{s-2}\\ \left(s^{2}-1\right) F\left( s\right) &=\dfrac{s-1}{s-2}\\ \left(s-1\right) \left( s+1\right) F\left( s\right) &=\dfrac{s-1}{s-2}\\ F\left(s\right) &=\dfrac{1}{\left( s+1\right) \left( s-2\right) }\\ F\left(s\right) &=\dfrac{A}{s+1}+\dfrac{B}{s-2} \end{align*} $\mathrm{Using\ partial\ fractions,}$ \begin{align*} F\left(s\right) &=\dfrac{\left(-\dfrac{1}{3}\right)}{s+1}+\dfrac{\left(\dfrac{1}{3}\right)}{s-2} \end{align*} $\mathrm{Taking\ inverse\ Laplace\ Transform\ of\ both\ sides,}$ \begin{align*} f\left(t\right) &=-\dfrac{1}{3}e^{-t}+\dfrac{1}{3}e^{2t}\\ f\left(t\right) &=\dfrac{1}{3}\left[ e^{2t}-e^{-t}\right] \end{align*}