Laplace Transform Examples | Proveiff IT Solutions

Laplace Transform Examples : Laplace Transforms is an operator that is performed on a function. Laplace Transforms of a function is a real number.

Definition:

Laplace transform of a function or signal f(t) is denoted as \(\mathcal{L}(f(t))\) and defined as follows

\[\mathcal{L}(f(t))= \mathcal{F}(s)=\int_{0}^{\infty}{e^{-st}f\left(t\right)dt}\hspace{5 cm}(1.1)\]

where \(s\ne 0\) is a complex number.

Exponential type function:

A function satisfying \( \left|f(t)\right| < A e^{\gamma t} \forall t \ge 0\) for some constant \(A, \gamma\), is known as an exponential function type \(\gamma \).

Existence of Laplace transform:

If \(f(t)\) is an exponential type function then \(\int_{0}^{\infty}{e^{-st}f\left(t\right)dt} \) converges absolutely for \(Re(s)> \gamma\)that is Laplace transform of \(f(t)\) exists.

Example (1):

Show that Laplace transform of \(f(t) = 1\) is \(\mathcal{F}(s) =\frac{1}{s}\)

Solution:

Laplace transform of \(f(t)\) is determined as

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

Substituting \(f(t) = 1\) into above formula

\[\Rightarrow \mathcal{F}(s)= \mathcal{L}(1)= \int_{0}^{\infty}{e^{-st}1dt}\]

\[=\int_{0}^{\infty}{e^{-st}dt}\]

\[=\left[-\frac{e^{-st}}{s}\right]_{t=0}^{\infty}\]

\[=-\frac{1}{s}\left[e^{-\infty}-e^{0}\right]\]

Since \(e^{-\infty}=0,e^{0}=0\)

\[\Rightarrow \mathcal{F}(s)=-\frac{1}{s}\left[0-1\right]\]

\[\mathcal{F}(s)=\frac{1}{s}\]

Example (2):

Prove that Laplace transform of \(f(t) =t^n\) is \(\mathcal{F}(s) =\frac{n!}{s^{n+1}}, n\ge 0\)

Solution:

We know Laplace transform of \(f(t)\) is given as

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

given \(f(t) =t^n\)

\[\Rightarrow \mathcal{F}(s)= \mathcal{L}(t^n)= \int_{0}^{\infty}{e^{-st}t^ndt}\]

\[=\int_{0}^{\infty}{e^{-st}t^ndt}\]

\[= \left[t^n\int e^{-st}dt\right]_{t=0}^{\infty} – \int_{0}^{\infty}\left(\frac{d}{dt}(t^n)\int e^{-st} dt\right)dt\]

\[= \left[-t^n\frac{ e^{-st}}{s}\right]_{t=0}^{\infty} +\frac{n}{s} \int_{0}^{\infty}t^{n-1}e^{-st}dt\]

since \( limit_{t \rightarrow \infty} t^n e^{-st} =0, limit_{t \rightarrow 0} t^n e^{-st} =0 \)

\[\Rightarrow \mathcal{F}(s) =\frac{n}{s} \int_{0}^{\infty}t^{n-1}e^{-st}dt\]

again integrating as above

\[\Rightarrow \mathcal{F}(s)=\frac{n(n-1)}{s^2} \int_{0}^{\infty}t^{n-2}e^{-st}dt\]

and continuing same way

\[\Rightarrow \mathcal{F}(s)=\frac{n(n-1)(n-2)…..3.2.1}{s^{n}} \int_{0}^{\infty}e^{-st}t^0dt\]

since \(\int_{0}^{\infty}e^{-st}t^0dt=\frac{1}{s},n(n-1)(n-2)…..3.2.1 = n! \) therefore

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

Example (3):

Find Laplace transform of \(f(t)=e^{at}\) where \(a\) is a constant.

Solution:

Using \(f(t)=e^{at}\) in to Laplace transform formula \((1.1)\)

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

\[= \int_{0}^{\infty}{e^{-(s-a)t}dt}\]

{“textColor”:”very-dark-gray”} –>

let \(s-a = s_1\)

\[= \int_{0}^{\infty}{e^{- s_1 t}dt}\]

\[=\left[ -\frac {e^{-s_1 t}}{s_1}\right]_{t=0}^{\infty}\]

\[=-\frac{1}{s_1}\left[e^{-\infty}-e^{0}\right]\]

\[=-\frac{1}{s_1}\left[0-1\right]\]

\[=\frac{1}{s_1}\]

{“textColor”:”very-dark-gray”} –>

substitute back \(s_1 =s-a\)

\[\Rightarrow \mathcal{L}(e^{at})= \mathcal{F}(s)=\frac{1}{s-a}\]

Example (4):

Prove that \(\mathcal{L}(e^{-at}) =\frac{1}{s+a}\) for any constant \(a\).

Solution:

\[\mathcal{L}(e^{-at})= \int_{0}^{\infty}{e^{-st}e^{-at}dt}\]

\[= \int_{0}^{\infty}{e^{-(s+a)t}dt}\]

substitue \(s-a = s_2\)

\[= \int_{0}^{\infty}{e^{- s_2 t}dt}\]

\[=\left[ -\frac {e^{-s_2 t}}{s_2}\right]_{t=0}^{\infty}\]

\[=-\frac{1}{s_2}\left[e^{-\infty}-e^{0}\right]\]

\[=-\frac{1}{s_2}\left[0-1\right]\]

\[=\frac{1}{s_2}\]

Since \(s_2=s+a\)

\[\Rightarrow \mathcal{L}(e^{-at})=\frac{1}{s+a}\]

Properties of Laplace transforms

\(\mathcal{L}(0)=0\)Laplace transform is Linear that is \[\mathcal{L}(cf) =c \mathcal{F}(s) and \mathcal{L}(c_1 f\pm c_2 g) =c_1\mathcal{F}(s) \pm Scaling property \[\mathcal{L}\left(f(ct)\right) =\frac{1}{c}\mathcal{F(\_2\mathcal{G}(s), c, c_1, c_2\] are constants.frac{s}{c})}\] c is a constant.First shifting property \[\mathcal{L}\left( e^{ct}f(t) \right)=\mathcal{F(s-c)}\] c is a constant.Second shifting properties \[\mathcal{L}\left(f(t) H(t-c)\right)=e^{-cs}\mathcal{F(s+c)}\] and \[\mathcal{L}\left(f(t-c) H(t-c)\right)=e^{-cs}\mathcal{F(s)}\] where \(H(t)\) denotes Heaviside functionLaplace transform of derivatives \[\mathcal{L}\left( \frac{df}{dt}\right)=s\mathcal{F(s)}-f(0)\] and in general \[ \mathcal{L}\left(\frac{d^k f}{dt^k}\right)=s^k\mathcal{F(s)}-s^{k-1}f'(0)-s^{k-2}f”(0)-…..sf^{k-2}-f^{k-1}(0), \forall k\ge 0\]Differentiation of Laplace transform \[ \frac{d\mathcal{F(s)}}{ds}=-\mathcal{L}(tf(t))\] and in general \[ \frac{d^k\mathcal{F(s)}}{ds^k}=(-1)^k\mathcal{L}(t^kf(t)), \forall k\ge 0\]Laplace transform of an integral \[\mathcal{L}\left( \int_{0}^{t}f(t)dt \right)=\frac{1}{s}\mathcal{F(s)}\]Laplace transform of periodic function with period P \[\mathcal{L}\left(f(t)\right)=\frac{\int_{0}^{P}e^{-st}f(t)dt}{1-e^{-sP}}\]Laplace transform of convolution \[\mathcal{L}\left( (f \star g) (t)\right)=\mathcal{F(s)}\mathcal{G(s)}\] where \((f \star g) (t) =\int_{0}^{t}f(v)g(t-v)dv\) is known as convolution of \(f , g\).

Inverse Laplace transform

Definition : If \( \mathcal{F}(s)\) be Laplace transform of the function \(f(t)\) then inverse Laplace transform \( \mathcal{L^{-1}}\) is an operator that operates on \( \mathcal{F}(s)\) and returns function \(f(t)\).

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

Examples:

\[ \mathcal{L^{-1}} \left(\frac{1}{s}\right)=1, \mathcal{L^{-1}} \left(\frac{3}{s}\right)=3\]\[ \mathcal{L^{-1}} \left(\frac{1}{s-2}\right)=e^{2t}, \mathcal{L^{-1}} \left(\frac{1}{s+3}\right)=e^{-3t}\]\[ \mathcal{L^{-1}} \left(\frac{1}{s^2}\right)=t, \mathcal{L^{-1}} \left(\frac{6}{s^4}\right)=t^3\]\[ \mathcal{L^{-1}} \left(\frac{4}{s^2+4^2}\right)=\sin(4t)\]\[ \mathcal{L^{-1}} \left(\frac{s}{s^2+5^2}\right)=\cos(5t)\]\[ \mathcal{L^{-1}} \left(t^3e^{-3t}\right)=\frac{6}{(s+3)^4}\]

Problem (1):

Find inverse Lapalce transform of \(\mathcal{F}(s)=\frac{1}{s^2+2s-3}\)

Solution:

\[ \mathcal{L}^{-1}\left( \mathcal{F}(s) \right)= \mathcal{L}^{-1}\left(\frac{1}{s^2+2s-3} \right)\]

Since \(s^2+2s-3=\left(s-1\right)\left(s+3\right)\)

Hence \(\frac{1}{s^2+2s-3}=\frac{1}{\left(s-1\right)\left(s+3\right)}=\frac{1}{4}\left[\frac{1}{s-1}-\frac{1}{s+3}\right]\)

\[\Rightarrow \mathcal{L}^{-1}\left(\frac{1}{s^2+2s-3} \right)= \frac{1}{4}\left[\mathcal{L}^{-1}\left(\frac{1}{s-1}\right)-\mathcal{L}^{-1}\left(\frac{1}{s+3}\right)\right]\]

substituting \(a=1, -3\) into formulas \(e^{at} =\mathcal{L}^{-1}\left(\frac{1}{s-a}\right) \Rightarrow e^{t} =\mathcal{L}^{-1}\left(\frac{1}{s-1}\right), e^{-3t}=\mathcal{L}\left(\frac{1}{s+3}\right)\)

{“textColor”:”very-dark-gray”} –>

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

is the required answer.

Problem (2):

Determine inverse Lapalce transform of \(\mathcal{F}(s)=\frac{s-2}{s^2+9}\)

Solution:

Since inverse Laplace transform is given as

\[ \mathcal{L}^{-1}\left( \mathcal{F}(s) \right)= \mathcal{L}^{-1}\left(\frac{s-2}{s^2+9} \right)\]

first determining partial fraction of \(\frac{s-2}{s^2+9}=\frac{s}{s^2+3^2}-\frac{2}{3}\frac{3}{s^2+3^2}\)

now substituing \(a =3\) into formulas of inverse laplace transforms \(\mathcal{L}^{-1}\left[\frac{s}{s^2+a^2}\right]=\cos(at) ,\mathcal{L}^{-1}\left[\frac{a}{s^2+a^2}\right]=\sin(at)\)

we have \(\mathcal{L}^{-1}\left[\frac{s}{s^2+3^2}\right]=\cos(3t) ,\mathcal{L}^{-1}\left[\frac{3}{s^2+3^2}\right]=\sin(3t)\)

therefore

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

Solving ODEs by Laplace transform method:

Example (1):

Solve \(y^{\prime\prime}-2y^\prime-8y=0, y\left(0\right)=0, y^\prime(0)=1 \)

Solution: to find solution of

\[y^{\prime\prime}-2y^\prime-8y=0\hspace{10 cm}(1)\]

that satisfies \( y\left(0\right)=0, y^\prime(0)=1\)

applying Lapalce transform both sides of eqn. (1)

{“textColor”:”very-dark-gray”} –>

\[\Rightarrow \mathcal{L}\left[y^{\prime\prime}-2y^\prime-8y\right]=\mathcal{L}\left[0\right]\]

by linear property of Laplace transform

\[\Rightarrow \mathcal{L}\left(y^{\prime\prime}\right)- 2\mathcal{L}\left(y^\prime\right)-8 \mathcal{L}\left(y\right)=\mathcal{L}\left(0\right)\]

Since Laplace transform of derivatives rule \(\mathcal{L}\left(y^{\prime}\right)=sY\left(s\right)-y\left(0\right), \mathcal{L}\left(y^{\prime\prime}\right)= s^2Y\left(s\right)-sy\left(0\right)-y^\prime\left(0\right), \mathcal{L}\left(0\right)=0\)

where \(\mathcal{L}\left(y(t)\right)=Y(s)\) denotes Lapalce transform of \(y(t)\).

\[\Rightarrow s^2Y\left(s\right)-sy\left(0\right)-y^\prime\left(0\right)- 2\left[sY\left(s\right)-y\left(0\right)\right]-8Y\left(s\right)=0\]

substitute \(y\left(0\right)=0, y^\prime(0)=1 \)

\[\Rightarrow s^2Y\left(s\right)-0-1-2\left[sY\left(s\right)-0\right]-8Y\left(s\right)=0\]

\[\Rightarrow (s^2-2s-8)Y(s)-1=0\]

\[\Rightarrow Y(s)=\frac{1}{s^2-2s-8}\]

\[\Rightarrow Y(s)=\frac{1}{6}\left[ \frac{1}{s-4}-\frac{1}{s+2} \right]\]

applying inverse Laplace trnsform in above eqn.

\[y(t)=\mathcal{L}^{-1}\left(Y(s)\right)=\frac{1}{6}\left[\mathcal{L}^{-1}\left( \frac{1}{s-4}\right)-\mathcal{L}^{-1}\left( \frac{1}{s+2}\right)\right]\]

now inverse Laplace transform formulas \(\mathcal{L}^{-1}\left(\frac{1}{s-4}\right) =e^{4t}, \mathcal{L}^{-1}\left(\frac{1}{s+2}\right) =e^{-2t} \)

\[\Rightarrow y(t)=\frac{1}{6}\left(e^{4t}- e^{-2t} \right)\]

is solution of the differential equation that satisfies given initial conditions.

Example (2):

Solve \(y^{\prime\prime}+7y^\prime+12y=t, y\left(0\right)=0, y^\prime(0)=0\)

Solution: given differential equation

{“textColor”:”very-dark-gray”} –>

\[y^{\prime\prime}+7y^\prime+12y=t\hspace{10 cm}(1)\]

\( y\left(0\right)=0, y^\prime(0)=0\)

operating \( \mathcal{L}\) both sides in eqn. (1)

\[\Rightarrow \mathcal{L}\left[y^{\prime\prime}-y^\prime-6y\right]=\mathcal{L}\left[0\right]\]

{“textColor”:”very-dark-gray”} –>

by linear property

\[\Rightarrow \mathcal{L}\left(y^{\prime\prime}\right)+7y^\prime+12y=\mathcal{L}\left(t\right)\]

Since \( \mathcal{L}\left(y^{\prime}\right)=sY\left(s\right)-y\left(0\right), \mathcal{L}\left(y^{\prime\prime}\right)= s^2Y\left(s\right)-sy\left(0\right)-y^\prime\left(0\right), \mathcal{L}\left(t\right)=\frac{1}{s^2}\)

{“textColor”:”very-dark-gray”} –>

where \(Y(s)\) is Lapalce transform of \(y(t)\).

\[\Rightarrow s^2Y\left(s\right)-sy\left(0\right)-y^\prime\left(0\right)+7\left[sY\left(s\right)-y\left(0\right)\right]+8Y\left(s\right)=\frac{1}{s^2}\]

using conditions \(y\left(0\right)=0, y^\prime(0)=0 \)

\[\Rightarrow\ \ s^2Y\left(s\right)+7s+8Y\left(s\right)=\frac{1}{s^2}\]

\[\Rightarrow (s^2+7s+8)Y(s)=\frac{1}{s^2}\]

\[\Rightarrow Y\left(s\right)=\frac{1}{s^2 (s^2+7s+8)}=\frac{1}{s^2(s+3)(s+4)}\]

\[=\frac{1}{s^2(s+3)(s+4)}\]

\[\Rightarrow Y\left(s\right)=\frac{1}{12 s^2}+\frac{1}{9 (s+3)}-\frac{1}{16 (s+4)}-\frac{7}{144 s}\]

operating \(\mathcal{L}^{-1}\)

{“textColor”:”very-dark-gray”} –>

\[\Rightarrow y(t)=\mathcal{L}^{-1}\left(\frac{1}{12 s^2}\right)+\mathcal{L}^{-1}\left(\frac{1}{9 (s+3)}\right)-\mathcal{L}^{-1}\left(\frac{1}{16 (s+4)}\right)-\mathcal{L}^{-1}\left(\frac{7}{144 s}\right)\]

is the solution of the differential equation satisfying given initial conditions .

Example (3)

Solve \( \frac{dx}{dt}-3x = 6\cos(2t) \) satisfying condition \( x\left(0\right) = 0\) by Laplace transforms

Solution:

To solve \( \frac{dx}{dt}-3x = 6\cos(2t) \hspace{7.0 cm}(1)\)

with the initial condition \( x\left(0\right)=0\)

applying Laplace transform both sides

\[\Rightarrow L\left[\frac{dx}{dt} – 3x\right]=L\left[6\cos(2t)\right]\]

by linear property

\[\Rightarrow L\left[\frac{dx}{dt} \right]-3 L\left[x\right]=6L\left[cos(2t)\right]\]

Let \( L\left[x\right] =X(s) \) and formulas \( L\left[cos⁡(at)\right] =\frac{s}{s^2+a^2} , L\left[\frac{dx}{dt} \right]=sX(s)-x(0)\)

\[\Rightarrow sX(s)-x(0)-3 X(s)=6\frac{s}{s^2+2^2} =6\frac{s}{s^2+4} \]

substitute initial condition \( x\left(0\right) = 0\)

\[\Rightarrow sX(s)-3 X(s)=6\frac{s}{s^2+4} \]

\[\Rightarrow (s-3) X(s)=6\frac{s}{s^2+4} \]

\[\Rightarrow X(s)=6\frac{s}{(s-3)(s^2+4)} \]

Applying inverse Laplace transform both sides

\[\Rightarrow L^{-1}\left[X\left(s\right)\right]=6 L^{-1}\left[\frac{1}{\left(s-3\right)}\frac{s}{(s^2+4)}\right]\]

Let \( F\left(s\right)=\frac{1}{s-3} , G\left(s\right)=\frac{s}{s^2+4}\)

Since \(L^{-1}\left\{F\left(s\right)\right]=e^{3t}=f\left(t\right) , L^{-1}\left[G\left(s\right)\right]=\cos(2t)=g\left(t\right)\)

Now by convolution theorem \( L^{-1}\left[F\left(s\right)G(s)\right]=\int_{0}^{t}f\left(t-y\right)g\left(y\right)dy\)

\[\Rightarrow L^{-1}\left[X\left(s\right)\right]= x(t) = 6 L^{-1}\left[\frac{1}{\left(s-3\right)}\frac{s}{(s^2+4)}\right] =6 \int_{0}^{t}{e^{3(t-y)}\cos(2y)dy}\]

\[x(t) =6e^{3t}\left[\int_{0}^{t}{e^{-3y}\cos(2y)dy}\right]\]

\[\Rightarrow x(t) = \frac{6e^{3t}}{13} \left(e^{-3 t} (2 \sin (2 t)-3 \cos (2 t))+3\right)\]

\[\Rightarrow x(t) = \frac{6}{13} \left(3e^{3t}+ 2 \sin (2 t)-3 \cos (2 t)\right)\]

\(x(t)\) is the solution of the given differential equation.

Example (4):

Solve IVP \(y^\prime-y=te^t\sin(t), y\left(0\right)=0 \)

Solution:

Initial value problem

\[x^\prime-x=te^t\sin(t), x\left(0\right)=0\hspace{5.0 cm}(1) \]

Applying Laplace transform in eqn. (1)

\[L[x^\prime-x]=L[te^{t}sin⁡(t)]\]

\[\Rightarrow L[x^\prime]-L[x]=L[te^{t}sin⁡(t)] \hspace{5.0 cm}(2)\]

Let’s denote \( L\left[x\left(t\right)\right]=X(s)\) and the facts \(L\left[x^\prime\right]=sX\left(s\right)-x\left(0\right) , L\left[e^{ct}\sin{\left(dt\right)}\right]=\frac{d}{\left(s-c\right)^2+d^2} = F(s) , L\left[tf\left(t\right)\right]=-\frac{dF(s)}{ds}\)

Therefore \( L[te^{t}sin⁡(t)] =- \frac{d}{ds}\left(\frac{1}{\left(s-1\right)^2+1}\right) =\frac{1}{\left(\left(s-1\right)^2+1\right)^2}\left(2\left(s-1\right)\right) \)

Substitute in eqn. (2)

\( sX\left(s\right)-x\left(0\right)-X(s)=\frac{1}{\left(\left(s-1\right)^2+1\right)^2}\left(2\left(s-1\right)\right)\)

Since given that \( x\left(0\right)=0\)

\(\Rightarrow \left(s-1\right)X\left(s\right)=\frac{1}{\left(\left(s-1\right)^2+1\right)^2}\left(2\left(s-1\right)\right)\)

dividing by \((s-1)\)

\(\Rightarrow X\left(s\right)=2\frac{1}{\left(\left(s-1\right)^2+1\right)^2}\)

applying inverse Laplace transform both sides to this equation

\(\Rightarrow x\left(t\right)=L^{-1}\left[\frac{2\ast1^3}{\left(\left(s-1\right)^2+1\right)^2}\right]=e^t(-tcos(t)+sin(t))\)

\(\Rightarrow x(t)=e^t(-tcos\left(t\right)+sin(t))\)

This is the solution of the initial value problem.