Lecture 2 MATH 223 (Lt Col Tahmina Sultana)

 

 

 MATH 223

Lecture 02 

Laplace Transformation

   Let $F(t)$ be a function of $t$ specified for $t>0$ , then the laplace transform of $F(t)$, 

denoted by  $\mathcal{L} \{f(t)\}$ , is defined by 

  $$\mathcal{L} \left\{ F(t) \right\}= \int^\infty_0 e^{- st} ~ F(t) ~dt = f(s )$$  

 

---

    

     Example: If $F(t)= 1$, then find out  $\mathcal{L}\{F(t)\}$ .

Solution:  By the definition of Laplace Transformation

$$\begin{align*} \mathcal{L} \{F(t)\} & = \int^\infty_0 e^{- st} F(t)~dt \\ \mathcal{L} \{1\} & = \int^\infty_0 e^{- st} ~1~dt \\ & = \left[\frac{e^{- st}}{ - s}\right]_0^\infty\\ \textrm{Putting the value of t} & ~ \\ & = \left[ \frac{e^{- s. \infty}}{- s} - \frac{e^{- s.0}}{- s}\right] \\ & = \left[\frac{e^{- \infty}}{- s}- \frac{1}{- s }\right] \\ & = 0 - \frac{1}{- s} ~~~~[ e^{- \infty = \frac{1}{e^\infty}= \frac{1}{\infty}= 0 }]\\ & = \frac{1}{s} \\ & = f(s);~~~ s>0 \end{align*}$$  

---

    

    Example: If  $F(t) = t$, then find out $\mathcal{L} \{F(t)\}$.

Solution:  By Definition 

$$\begin{align*} \mathcal{L}\{F(t)\} & = \int^\infty_0 e^{- st} ~ F(t) ~dt \\ \mathcal{L} \{t\} & = \int^\infty_0 e^{- st } ~t~dt \\ & = \left[t~\int e^{- st}~dt - \int \left[\frac{d}{dt}t ~\int e^{- st} ~dt\right]dt\right] \\ & = \left[\frac{t~e^{- st}}{- s} - \int \frac{e^{- st}}{- s}~dt\right]_0^\infty \\ & = \left[\frac{t~e^{- st}}{- s}- \frac{e^{- st}}{s^2}\right]_0^\infty \\ & = \left[\frac{\infty~ e^{- \infty}}{- s} - \frac{e^{- \infty}}{s^2} - \frac{0.e^{0}}{- s}+ \frac{e^0}{s^2} \right]\\ & = 0 - 0 - 0 - \frac{1}{s^2} \\ \therefore ~ \mathcal{L}\{t\} & = \frac{1}{s^2} \end{align*}$$  

---

    Home Work: Find $\mathcal{L} \{F(t)\}$, where $F(t) = t^2$.

Solution:  $$\begin{align*} \textrm{by the definition of laplace Transformation,}\\ \mathcal{L}\{F(t)\} & = \int^\infty_0 e^{- st} ~F(t) ~dt \\ \mathcal{L}\{\sin~at\} & = \int^\infty_0 e^{- st} ~ \sin at ~ dt\\ & ~\\ & ~~\\ \textrm{Using the property of Integration,} & \\ ~~~~~~~~~~~ \int e^{at} \sin \beta t ~dt = & \frac{e^{\alpha t}}{\alpha ^2 + \beta^2} ( \alpha \sin \beta t - \beta~ \cos \beta t)\\ & ~\\ \textrm{Comparing them,}&~ ~~ \alpha = -s \textrm{ and } \beta = a \\ \textrm{so, we'll get,}& ~\\ & ~ \left[\frac{e^{- st}}{s^2 + a^2} ( - s \sin at - a \cos at )\right]_0^\infty \\ = & ~ \left[\frac{e^{- st}}{s^2 + a^2} ( - s \sin \infty - a \cos \infty) \right] - \left[ \frac{e^{- 0}}{s^2 + a^2}( - s \sin 0 - a \cos 0 )\right] \\ = & ~ 0 - \frac{1}{s^2 + a^2} ( 0 - a) \\ = & ~ \frac{a}{s^2 + a^2} \end{align*}$$  

---

   

    Example: Find $\mathcal{L} \{F(t)\}$, where $F(t) = e^{at}$

Solution: By the definition of Laplace Transformation, 

  $$\begin{align*} \mathcal{L}\{F(t)\} & = \int^\infty_0 e^{- st} ~F(t) ~dt \\ \mathcal{L}\{e^{at}\} & = \int^\infty_0 e^{- st} . e^{at} ~dt ~~~~~~~~~~[ e^a \times e^b = e^{a + b}] \\ & = \int^\infty_0 e^{- (s - a)t } ~dt \\ & = \left[\frac{e^{- (s - a)t}}{- (s - a)}\right]_0^\infty \\ & = \left[ \frac{e^{- (s - a)\infty}}{- (s - a)} - \frac{e^{- (s - a).0}}{- (s - a)}\right]\\ & = \frac{e^{- \infty}}{- (s - a)} - \frac{1}{- (s - a)} \\ & = 0 + \frac{1}{s - a} \\ \therefore \mathcal{L}\{e^{at}\} & = \frac{1}{s - a} \end{align*}$$  

   

---

    Example: If $F(t) = \sin at$, then find $\mathcal{L} \{F(t)\}$.

Solution: By the definition of Laplace Transformation,

  $$\begin{align*} \mathcal{L}\{F(t)\} & = \int^\infty_0 e^{- st} ~F(t) ~dt \\ \mathcal{L}\{\sin~at\} & = \int^\infty_0 e^{- st} ~ \sin at ~ dt\\ & ~\\ & ~~\\ \textrm{Property of Integration} & \\ ~~~~~~~~~~~ \int e^{at} \sin \beta t ~dt = & \frac{e^{\alpha t}}{\alpha ^2 + \beta^2} ( \alpha \sin \beta t - \beta~ \cos \beta t)\\ & ~\\ \textrm{Comparing them,}&~ ~~ \alpha = -s \textrm{ and } \beta = a \\ \textrm{so, we'll get,}& ~\\ & ~ \left[\frac{e^{- st}}{s^2 + a^2} ( - s \sin at - a \cos at )\right]_0^\infty \\ = & ~ \left[\frac{e^{- \infty}}{s^2 + a^2} ( - s \sin \infty - a \cos \infty) \right]\\ & ~~~~~~~~~~- \left[ \frac{e^{- 0}}{s^2 + a^2}( - s \sin 0 - a \cos 0 )\right] \\ = & ~ 0 - \frac{1}{s^2 + a^2} ( 0 - a) \\ = & ~ \frac{a}{s^2 + a^2} \end{align*}$$