Functions are defined in some type of domain. Often in engineering mathematics we deal with functions defined in Time-Domain. Sometimes we face challenges trying to solve problems in time domain especially in controls engineering. Laplace transform is used to transform a set of functions from time domain to s-domain.
\[\Big[\text{Time Domain}\Big] \hspace{10mm} \underrightarrow{\text{Laplace Transform}} \hspace{10mm} \Big[\text{S Domain}\Big]\]
A dynamic system modeled by differential equations in time domain becomes a polynomial in S domain. Solving polynomial is lot easier than the differential equations. Once you solve the polynomial in S domain, transform the function from s domain to time domain using inverse Laplace Transform.
\[\Big[\text{S Domain}\Big] \hspace{10mm} \underrightarrow{\text{Inverse Laplace Transform}} \hspace{10mm} \Big[\text{Time Domain}\Big]\]
Here's the Laplace Transform: \[L[f(t)] = \int_{0}^{\infty}{f(t) e^{-st}dt}\tag{1}\] You have to put the function inside this integral and solve it to get the Laplace equation of the function. Here in the integral 's' is a constant.
Examples
| \[\mathscr{L}[A] = \frac{A}{S}\] | \[\mathscr{L}[At] = \frac{A}{S^2} \] |
| \[\mathscr{L}\Big[Acos(\omega t)\Big] = A\Bigg[\frac{S}{S^2+\omega^2}\Bigg] \] | \[\mathscr{L}\Big[Asin(\omega t)\Big] = A\Bigg[\frac{\omega}{S^2+\omega^2}\Bigg] \] |
Solving Differential Equation Using Laplace Transform
\[2\ddot{x} + 7\dot{x} + 3x = 0 \hspace{1cm} x(0) = 0 \hspace{1cm} \dot{x}(0)=0 \tag2 \]
\[\mathscr{L}\Big[2\ddot{x} + 7\dot{x} + 3x = 0\Big]\tag{2.1}\]Let's take the Laplace transforms of the individual terms first and then we will combine them in the end.
\[\mathscr{L}\Big[2\ddot{x}\Big] = 2\Big[s^2X(s) - sx(0) - \dot{x}(0)\Big] \hspace{1cm}\rightarrow \hspace{1cm} 2\Big[s^2X(s)-s(3)-0\Big] \\ =2s^2X(s)-6s \tag{2.2}\]
\[\mathscr{L}\Big[7\dot{x}\Big] = 7\Big[s X(s)-s x(0) \Big] \hspace{.5cm}\rightarrow \hspace{.5cm} 7\Big[s X(s) - 3\Big] \hspace{.5cm}\rightarrow \hspace{.5cm} 7sX(s)-21 \]
\[\mathscr{L}\Big[2x] = 3X(s)\tag{2.3}\] Combine the all the three terms
\[2s^2X(s) - 6s + 7sX(s) - 21 + 3X(s) = 0\tag{2.4} \]
\[X(s)\Big[2s^2 + 7s + 3\Big] = 6s + 21 \tag{2.5} \]
\[X(s) = \Bigg[\frac{6s + 21}{2s^2 + 7s + 3}\Bigg]\tag{2.6}\]
Now you have to use the Partial Fraction Decomposition to break this complicated equation in order to take the inverse Laplace transform of it to take the equation back to the original domain.
\[\Bigg[\frac{6s + 21}{2s^2 + 7s + 3}\Bigg] = \Bigg[\frac{6s + 21}{(s+3)(2s+1)}\Bigg]=\frac{A}{s+3} + \frac{B}{2s+1}\tag{2.7}\]
\[X(s) = \frac{A}{s+3} + \frac{B}{2s+1}\tag{2.8}\]
\[A(2s+1) + B(s+3) = 6s + 21\tag{2.9}\]
\[2As + A + Bs + 3B = 6s + 21 \tag{2.10}\]
\[(2A + B)s = 6s \hspace{1cm} A + 3B = 21 \tag{2.11}\]
Solve this systems of equation to find the values of A and B. The values of A and B are:
\[A = -0.6 \hspace{1cm} B = 7.2\tag{2.12}\]
Recall:
\[\mathscr{L}\Big[e^{at}\Big] = \frac{1}{s-a} \tag{2.13}\]
Let's substitute the values of A and B into Equation 2.8. Our goal here is to put these terms in the form of equation 2.13 so that we can directly write their inverse Laplace transform in original domain.
\[X(s) = \frac{-0.6}{s+3} + \frac{7.2}{2s+1} \tag{2.14}\]
\[X(s) = -0.6\Bigg[\frac{1}{s-(-3)}\Big] + \frac{7.2}{2} \Bigg[\frac{1}{s-(-1/2)} \Bigg]\tag{2.15}\]
Let's take the inverse Laplace transform of the Equation 2.15.
\[x(s) = 3.6e^{-0.5t}-0.6e^{-3t} \tag{2.16}\]
Equation 2.16 is the solution of the differential equation listed in Equation 2 above and satisfies both the boundary conditions.
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