This is a part of the lecture notes in process control.
Laplace domain
The use of Laplace transformation is very useful in the analysis of dynamics and in the control system theory. It is an integral transformation defined by the equation:
![\[L[f(t)]=\int\limits_{0}^{\infty }{f(t){{e}^{-st}}}dt=F(s) tag{1}\]](https://softinery.com/wp-content/ql-cache/quicklatex.com-a3200375d350b11f03d82bbb9f2b6290_l3.png "Rendered by QuickLaTeX.com")
where t is time, L is Laplace transform operator and s is a complex variable. A function of complex variable F(s) is obtained as a result of the transformation. For example, if f(t) = 1, then, the Laplace transform of this function is:
(2) ![\[L[1]=\int\limits_{0}^{\infty }{1\cdot {{e}^{-st}}}dt=\left[ -\frac{1}{s}{{e}^{-st}} \right]_{t=0}^{t\to \infty }=\left[ 0+\frac{1}{s} \right]=\frac{1}{s} \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-6255a68a3da617f95f5fc0a01c06fb60_l3.png "Rendered by QuickLaTeX.com")
For the function f(t) = Kt, we have:
(3) ![\[L[Kt]=\int\limits_{0}^{\infty }{K\cdot t\cdot {{e}^{-st}}}dt=K\int\limits_{0}^{\infty }{t\cdot {{e}^{-st}}}dt \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-0b5e6054ee5da581f212cc2efe7c1b6a_l3.png "Rendered by QuickLaTeX.com")
Integrating by parts:
![\[K\int\limits_{0}^{\infty }{t\cdot {{e}^{-st}}}dt=\]](https://softinery.com/wp-content/ql-cache/quicklatex.com-6af23b7c2b73efab542a6dac369736f2_l3.png "Rendered by QuickLaTeX.com")
![\[=\left[ -\frac{Kt}{s}{{e}^{-st}} \right]_{t=0}^{t\to \infty }-K\int\limits_{0}^{\infty }{-\frac{1}{s}{{e}^{-st}}}dt=\]](https://softinery.com/wp-content/ql-cache/quicklatex.com-5a2a2337c284b1ccf36bf7e54f45aee5_l3.png "Rendered by QuickLaTeX.com")
![\[=\left[ -\frac{Kt}{s}{{e}^{-st}} \right]_{t=0}^{t\to \infty }+\frac{K}{s}\int\limits_{0}^{\infty }{{{e}^{-st}}}dt=\left[ -\frac{Kt}{s}{{e}^{-st}} \right]_{t=0}^{t\to \infty }-\left[ \frac{K}{{{s}^{2}}}{{e}^{-st}} \right]_{t=0}^{t\to \infty }=\]](https://softinery.com/wp-content/ql-cache/quicklatex.com-d688f17f3abd37a34b6a56b47549c944_l3.png "Rendered by QuickLaTeX.com")
(4) ![\[[0-0]+\frac{K}{{{s}^{2}}}=\frac{K}{{{s}^{2}}} \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-df7b46921f81e2f7e1f26a4bab6ba63a_l3.png "Rendered by QuickLaTeX.com")
It is necessary to know the properties of Laplace transform. One of them is linearity:
(5) ![\[L[{{f}_{1}}(t)+{{f}_{2}}(t)]=L[{{f}_{1}}(t)]+L[{{f}_{2}}(t)]={{F}_{1}}(s)+{{F}_{2}}(s) \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-16623b0393edca00305f3a05f5841dcc_l3.png "Rendered by QuickLaTeX.com")
It is also easy to prove that:
(6) ![\[L[a\cdot f(t)]=aL[f(t)] \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-75f6400aa348cdd9f146940d05656435_l3.png "Rendered by QuickLaTeX.com")
Another important property is the Laplace transform of function derivative:
(7) ![\[L\left[ \frac{df(t)}{dt} \right]=sL\left[ f(t) \right]-f(0) \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-602bbaf713901de11aa15ca7609c978f_l3.png "Rendered by QuickLaTeX.com")
and second-order derivative:
(8) ![\[L\left[ \frac{{{d}^{2}}f(t)}{d{{t}^{2}}} \right]={{s}^{2}}L\left[ f(t) \right]-sf(0)-\frac{df(0)}{dt} \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-6e966099c3f703f8be3642e8b234827f_l3.png "Rendered by QuickLaTeX.com")
Inverse Laplace transform
In order to transform the function of a complex variable s, an invert Laplace transform is used. It is calculated as follows:
(9) ![\[f(t)=\frac{1}{2\pi i}\int\limits_{\alpha -i\omega }^{\alpha +i\omega }{{{e}^{st}}F(s)ds} \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-f69b1627f2efea18ea68dedce951eb51_l3.png "Rendered by QuickLaTeX.com")
Commonly in practice, inverse Laplace transform is found using partial fraction expansion.
![\[f(t)={{L}^{-1}}[F(s)]={{L}^{-1}}[{{F}_{1}}(s)+{{F}_{2}}(s)+...+{{F}_{n}}(s)]= \\\]](https://softinery.com/wp-content/ql-cache/quicklatex.com-563a1e76d83d528bc8c126348c6123e4_l3.png "Rendered by QuickLaTeX.com")
(10) ![\[{{L}^{-1}}[{{F}_{1}}(s)]+{{L}^{-1}}[{{F}_{2}}(s)]+...+{{L}^{-1}}[{{F}_{n}}(s)] \\ \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-dbc98f9ee163bcdbdf540765f9f69e97_l3.png "Rendered by QuickLaTeX.com")
Above way makes it simpler to find the inverse Laplace transform, because the problem is changed to several simpler ones. Because calculating inverse Laplace transform using the definition (9) is difficult it is often easier to look up the function in mathematic tables.
Example: Find the inverse Laplace transform of the function:
(11) ![\[F(s)=\frac{K}{s(s+1/\tau )} \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-5f61ff60fe8df647a4fd7c9eaf7461ee_l3.png "Rendered by QuickLaTeX.com")
Using partial fraction expansion:
(12) ![\[F(s)=\frac{K}{s(s+1/\tau )}=\frac{A}{s}+\frac{B}{s+1/\tau } \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-8431a604c9292d4233455212b99f5ae5_l3.png "Rendered by QuickLaTeX.com")
The roots of the denominator are 0 and –1/τ.
![\[A=\underset{s\to 0}{\mathop{\lim }}\,\left[ sF(s) \right]=\underset{s\to 0}{\mathop{\lim }}\,\left[ s\frac{K}{s(s+1/\tau )} \right]=K\tau\]](https://softinery.com/wp-content/ql-cache/quicklatex.com-ab74aec564918a59460f0b5963bb070e_l3.png "Rendered by QuickLaTeX.com")
![\[B=\underset{s\to -1/\tau }{\mathop{\lim }}\,\left[ (s+1/\tau )F(s) \right]=\underset{s\to -1/\tau }{\mathop{\lim }}\,\left[ (s+1/\tau )\frac{K}{s(s+1/\tau )} \right]=-K\tau\]](https://softinery.com/wp-content/ql-cache/quicklatex.com-4044d8c88035c16a3956765b37dd1064_l3.png "Rendered by QuickLaTeX.com")
Therefore:
(13) ![\[F(s)=\frac{K\tau }{s}-\frac{K\tau }{(s+1/\tau )} \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-700e37d17ff798d4de4f986e2d16587f_l3.png "Rendered by QuickLaTeX.com")
More examples on partial fraction expansion can be found here:
https://lpsa.swarthmore.edu/BackGround/PartialFraction/PartialFraction.html
The inverse Laplace transform is:
(14) ![\[L[F(s)]=L\left[ \frac{K\tau }{s} \right]-L\left[ \frac{K\tau }{(s+1/\tau )} \right]=K\tau L\left[ \frac{1}{s} \right]-K\tau L\left[ \frac{1}{(s+1/\tau )} \right] \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-b772639dd4ba3ab9e38257cc6c881257_l3.png "Rendered by QuickLaTeX.com")
We will use mathematic tables (attached to the lecture), part of which is shown below.
According to row 2, we know that ![L\left[ \frac{1}{s} \right]=1](https://softinery.com/wp-content/ql-cache/quicklatex.com-1264acf0e2888848e8caa42c32675139_l3.png "Rendered by QuickLaTeX.com")
(a is equal 1) and according to row 5, that (a is equal 1/τ). Finally, we have:
(15) ![\[f(t)={{L}^{-1}}\left[ F(s) \right]=K\tau (1+{{e}^{-t/\tau }}) \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-4abbdaa892840c788ec61f2ea23413d9_l3.png "Rendered by QuickLaTeX.com")
Applications of Laplace transform
Laplace transform can be used for solving differential equations. The procedure is following:
- Transform the differential equation.
- Solve the algebraic equation with regard to Y(s).
- Apply inverse Laplace transform to obtain y(t).
Example: Solve the differential equation using the Laplace transform.
(16) ![\[\frac{dy}{dt}-1={{e}^{-t}}\ \ \ \ y(0)=1 \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-5d4a64e44a16d0ab9888d1f25f274c8a_l3.png "Rendered by QuickLaTeX.com")
After applying the Laplace transform:
(17) ![\[L\left[ \frac{dy}{dt}-1 \right]=L\left[ {{e}^{-t}} \right]\ \ \ \ y(0)=1 \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-1babaeb4bb4a95effdd7e3cb41d15808_l3.png "Rendered by QuickLaTeX.com")
After using the properties of Laplace transform:
(18) ![\[L\left[ \frac{dy}{dt} \right]-L\left[ 1 \right]=L\left[ {{e}^{-t}} \right] \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-fff8c35a19ae2403063ea7ce29bcd70a_l3.png "Rendered by QuickLaTeX.com")
Using the property described by Eq. (7):
![\[sL\left[ y \right]-y(0)-L\left[ 1 \right]=L\left[ {{e}^{-t}} \right]\Rightarrow \\\]](https://softinery.com/wp-content/ql-cache/quicklatex.com-ea1010200ded94c6e3c5edd80deefb13_l3.png "Rendered by QuickLaTeX.com")
(19) ![\[sL\left[ y \right]-1-L\left[ 1 \right]=L\left[ {{e}^{-t}} \right] \\ \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-349a52c5a1f1c246ce089e0f045cebc9_l3.png "Rendered by QuickLaTeX.com")
Laplace transform of function y(t) is not know, so we will denote it by Y(s). After applying formulas from the table (rows 2 and 5):
(20) ![\[sY(s)-1-\frac{1}{s}=\frac{1}{s+1} \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-9e483322ec248599a60ea565c9545f0f_l3.png "Rendered by QuickLaTeX.com")
Then, we solve the equation with regard to Y(s):
(21) ![\[Y(s)=\frac{1}{s(s+1)}+\frac{1}{s}+\frac{1}{{{s}^{2}}} \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-b105357f224bd64813da5c4da07064cd_l3.png "Rendered by QuickLaTeX.com")
The function y(t) is equal to inverse Laplace transform:
![\[y(t)={{L}^{-1}}[y(t)]={{L}^{-1}}\left[ \frac{1}{s(s+1)}+\frac{1}{s}+\frac{1}{{{s}^{2}}} \right]= \\\]](https://softinery.com/wp-content/ql-cache/quicklatex.com-e61d41dbd7dd2facf29daec1573f6b1d_l3.png "Rendered by QuickLaTeX.com")
(22) ![\[{{L}^{-1}}\left[ \frac{1}{s(s+1)} \right]+{{L}^{-1}}\left[ \frac{1}{s} \right]+{{L}^{-1}}\left[ \frac{1}{{{s}^{2}}} \right] \\ \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-fd69ad6ec1e4ca8462b7507d11b299ac_l3.png "Rendered by QuickLaTeX.com")
The first term can be split by partial fraction expansion, but this is not necessary, as the adequate formula is available in the table (row no. 7, substitute a =1). Finally, after using formulas in the table, we get:
(23) ![\[y(t)=1-{{e}^{-t}}+1+t=2-{{e}^{-t}}+t \]](https://softinery.com/wp-content/ql-cache/quicklatex.com-3825ba43a4a12032c10d78dab44aa50e_l3.png "Rendered by QuickLaTeX.com")
