From Modern Control Engineering, 4th edition by Ogata
Question
\(\begin{array} [c]{ccc}f\left ( t\right ) =0 & & t<0\\ f\left ( t\right ) =e^{-0.4t}\cos 12t & & t\geq 0 \end{array} \)
\(\begin{array} [c]{ccc}f\left ( t\right ) =0 & & t<0\\ f\left ( t\right ) =\sin \left ( 4t+\frac{\pi }{3}\right ) & & t\geq 0 \end{array} \)
Solution
This is of the form \(e^{-at}f\left ( t\right ) \), hence use the property of Laplace transform \begin{equation} \mathcal{\mathcal{L}}\left ( e^{-at}f\left ( t\right ) \right ) =F\left ( s+a\right ) \tag{1} \end{equation} Where \(F(s)\) is Laplace transform of \(f\left ( t\right ) .\) But \(\mathcal{\mathcal{L}}\left ( \cos \omega t\right ) =\frac{s}{s^{2}+\omega ^{2}}\), therefore \[ F(s)=\mathcal{\mathcal{L}}\left ( \cos 12t\right ) =\frac{s}{s^{2}+144}\] Hence (1) becomes\begin{align*} \mathcal{\mathcal{L}}\left ( e^{-at}f\left ( t\right ) \right ) & =\mathcal{\mathcal{L}}\left ( e^{-0.4t}\cos 12t\right ) \\ & =F(s+a)\\ & =\frac{\left ( s+0.4\right ) }{\left ( s+0.4\right ) ^{2}+144} \end{align*}
I can not solve \(\mathcal{\mathcal{L}}\left ( \sin \left ( 4t+\frac{\pi }{3}\right ) \right ) \) by using the property that \[ \mathcal{\mathcal{L}}\left ( f\left ( t-a\right ) \right ) =e^{-as}F\left ( s\right ) \] Because here delay \(\frac{\pi }{3}>0\) where the above property is valid for \(a<0\). Instead, writing\begin{align} \sin \left ( \omega t+\theta \right ) & =\sin \left ( \omega t\right ) \cos \theta +\cos \left ( \omega t\right ) \sin \theta \nonumber \\ \sin \left ( 4t+\frac{\pi }{3}\right ) & =\sin \left ( 4t\right ) \cos \frac{\pi }{3}+\cos \left ( 4t\right ) \sin \frac{\pi }{3}\nonumber \\ \mathcal{\mathcal{L}}\left ( \sin \left ( 4t+\frac{\pi }{3}\right ) \right ) & =\cos \frac{\pi }{3}\mathcal{\mathcal{L}}\left ( \sin 4t\right ) +\sin \frac{\pi }{3}\mathcal{\mathcal{L}}\left ( \cos 4t\right ) \tag{2} \end{align}
But \(\cos \frac{\pi }{3}=\frac{1}{2}\) and \(\sin \frac{\pi }{3}=\sqrt{3}\) and \(\mathcal{\mathcal{L}}\left ( \sin \omega t\right ) =\frac{\omega }{s^{2}+\omega ^{2}}\Longrightarrow \mathcal{\mathcal{L}}\left ( \sin 4t\right ) =\frac{4}{s^{2}+16}\) and \(\mathcal{\mathcal{L}}\left ( \cos \omega t\right ) =\frac{s}{s^{2}+\omega ^{2}}\Longrightarrow \mathcal{\mathcal{L}}\left ( \cos 4t\right ) =\frac{s}{s^{2}+16}\). Hence substituting into eq (2) gives\begin{align*} \mathcal{\mathcal{L}}\left ( \sin \left ( 4t+\frac{\pi }{3}\right ) \right ) & =\frac{1}{2}\frac{4}{s^{2}+16}+\sqrt{3}\frac{s}{s^{2}+16}\\ & =\frac{1}{2}\left ( \frac{4+\sqrt{3}s}{s^{2}+16}\right ) \end{align*}
Question
solution (a)This is a delayed ramp with slope=1. Hence ramp equation is \(f\left ( t\right ) =t\). The amount of delay is \(a\)
Hence we want to find Laplace transform for \(g\left ( t\right ) =f\left ( t-a\right ) \) which is, from Laplace properties, is \(e^{-as}F\left ( s\right ) \)
But \(F\left ( s\right ) =\mathcal{L}\left ( t\right ) =\frac{1}{s^{2}}\) therefore the answer is
\[ e^{-as}\frac{1}{s^{2}} \]
From Modern Control Engineering, 4th edition by Ogata
Question
Solution
The above function can be constructed as follows
Let \(f\left ( t\right ) =t\) (the ramp function)
\(g\left ( t\right ) =f\left ( t\right ) u\left ( t\right ) -f\left ( t\right ) u\left ( t-T\right ) +Tu\left ( t-T\right ) \)
Where \(u\left ( t\right ) \) is the unit step function
This is ilustrated in this diagram
Now \[\mathcal{L}\left ( f\left ( t\right ) u\left ( t\right ) \right ) =\frac{1}{s^{2}}\] And\begin{align*} \mathcal{L}\left ( f\left ( t\right ) u\left ( t-T\right ) \right ) & =\int _{T}^{\infty }te^{-st}dt=\left . \frac{-t}{s}e^{-st}\right \vert _{T}^{\infty }-\int _{T}^{\infty }\frac{e^{-st}}{s}dt\\ & =\frac{-1}{s}\left ( 0-Te^{-sT}\right ) -\frac{1}{s}\left ( \frac{-1}{s}\right ) \ \left . e^{-st}\right \vert _{T}^{\infty }\\ & =\frac{T}{s}e^{-sT}+\frac{1}{s^{2}}\left ( 0-e^{-sT}\right ) =\frac{T}{s}e^{-sT}-\frac{e^{-sT}}{s^{2}} \end{align*}
And\[\mathcal{L}\left ( Tu\left ( t-T\right ) \right ) =\frac{T}{s}e^{-sT}\] Hence \begin{align*} \mathcal{L}\left ( g\left ( t\right ) \right ) & =\frac{1}{s^{2}}-\left ( \frac{T}{s}e^{-sT}-\frac{e^{-sT}}{s^{2}}\right ) +\frac{T}{s}e^{-Ts}\\ & =\frac{1}{s^{2}}-\frac{T}{s}e^{-sT}+\frac{e^{-sT}}{s^{2}}+\ \frac{T}{s}e^{-Ts}\\ & =\frac{1}{s^{2}}+\frac{e^{-sT}}{s^{2}}\ \\ & =\frac{1-e^{-sT}}{s^{2}} \end{align*}
From Modern Control Engineering, 4th edition by Ogata
Question Obtain partial-fraction using MATLAB for
\(F\left ( s\right ) =\frac{10\left ( s+2\right ) \left ( s+4\right ) }{\left ( s+1\right ) \left ( s+3\right ) \left ( s+5\right ) ^{2}}\) and then find inverse laplace transform
Solution I used Mathematica to find Partial-fraction
Hence the inverse laplace tranform is
\[ f\left ( t\right ) =\frac{15}{16}e^{-t}+\frac{5}{4}e^{-3t}+\frac{15}{4}te^{-5t}-\frac{35}{16}e^{-5t} \]
Here is a plot of the solution
The Matlab code to find Partial-fraction for this problem is below.
From Modern Control Engineering, 4th edition by Ogata
Question Obtain partial-fraction using MATLAB for
\[ F\left ( s\right ) =\frac{s^{4}+5s^{3}+6s^{2}+9s+30}{s^{4}+6s^{3}+21s^{2}+46s+30}\] And then find inverse laplace transform Solution
Using Mathematica to find Partial-fraction
Now find the Inverse Laplace transform for each term in the above result as follows.\begin{align*} \mathcal{L}\delta \left ( t\right ) & =1\\\mathcal{L}\frac{23}{18}e^{-t} & =\frac{23}{18}\frac{1}{1+s}\\\mathcal{L}\frac{3}{26}e^{-3t} & =\frac{3}{26}\frac{1}{1+s} \end{align*}
The inverse laplace transform of the last term \(\frac{-1450-253s}{117\left ( 10+2s+s^{2}\right ) }\) is found by writing \(10+2s+s^{2}=\left ( s+1\right ) ^{2}+3^{2}\). Therefore this terms becomes\begin{align*} \frac{-1450-253s}{117\left ( 10+2s+s^{2}\right ) } & =\frac{\frac{-1450}{117}-\frac{253}{117}s}{(s+1)^{2}+3^{2}}\\ & =\frac{\frac{-1450}{117}-\frac{253}{117}\left ( s+1\right ) +\frac{253}{117}}{(s+1)^{2}+3^{2}}\\ & =\frac{-\frac{1197}{117}-\frac{253}{117}\left ( s+1\right ) }{(s+1)^{2}+3^{2}}\\ & =-\frac{\frac{1197}{117}}{(s+1)^{2}+3^{2}}-\frac{\frac{253}{117}\left ( s+1\right ) }{(s+1)^{2}+3^{2}}\\ & =-\frac{1197}{\left ( 117\right ) \left ( 3\right ) }\frac{3}{(s+1)^{2}+3^{2}}-\frac{253}{117}\frac{\left ( s+1\right ) }{(s+1)^{2}+3^{2}} \end{align*}
Hence \begin{align*} \mathcal{L}^{-1}\left ( -\frac{1197}{117\times 3}\frac{3}{(s+1)^{2}+3^{2}}-\frac{253}{117}\frac{\left ( s+1\right ) }{(s+1)^{2}+3^{2}}\right ) & =-\frac{1197}{\left ( 117\right ) \left ( 3\right ) }\mathcal{L}^{-1}\left ( \frac{3}{(s+1)^{2}+3^{2}}\right ) \\ & =-\frac{253}{117}\mathcal{L}^{-1}\left ( \frac{\left ( s+1\right ) }{(s+1)^{2}+3^{2}}\right ) \\ & =-\frac{1197}{\left ( 117\right ) \left ( 3\right ) }e^{-t}\sin 3t-\frac{253}{117}e^{-t}\cos 3t\\ & =-\frac{e^{-t}}{117}\left ( 399\sin 3t+253\cos 3t\right ) \end{align*}
Adding all of the above, gives the Inverse Laplace transform as\[ f\left ( t\right ) =\delta \left ( t\right ) +\frac{23}{18}e^{-t}-\frac{3}{26}e^{-3t}-\frac{e^{-t}}{117}\left ( 399\sin 3t+253\cos 3t\right ) \] Here is a plot of the solution
The Matlab code to find Partial-fraction for this problem is below.
From Modern Control Engineering, 4th edition by Ogata
Question
A function \(B\left ( s\right ) /A\left ( s\right ) \) consists of the following zeros, poles, and gain K. Zeros at \(s=-1,s=-2\), poles at \(s=0,s=-4,s=-6\), gain \(k=5\).
Obtain an expression for \(B\left ( s\right ) /A\left ( s\right )\) using Matlab.
Solution
In Matlab
In Mathematica, the solution is as follows
From Modern Control Engineering, 4th edition by Ogata
Question
Solve the following ODE
\(x^{\prime \prime }+2x^{\prime }+10x=e^{-t}\)
\(x\left ( 0\right ) =0\)
\(x^{\prime }\left ( 0\right ) =0\)
The forcing function \(e^{-t}\) is given at \(t=0\) when the system is at rest.
Solution
Taking laplace transform of the differential equation gives\begin{align*} s\left ( sX\left ( s\right ) -x\left ( 0\right ) \right ) -x^{\prime }\left ( 0\right ) +2\left ( sX\left ( s\right ) -x\left ( 0\right ) \right ) +10X\left ( s\right ) & =\mathcal{L}\left ( e^{-t}\right ) \\ s^{2}X\left ( s\right ) -sx\left ( 0\right ) -x^{\prime }\left ( 0\right ) +2sX\left ( s\right ) -2x\left ( 0\right ) +10X\left ( s\right ) & =\frac{1}{s+1} \end{align*}
Applying the initial conditions results in\begin{align*} s^{2}X\left ( s\right ) +2sX\left ( s\right ) +10X\left ( s\right ) & =\frac{1}{s+1}\\ X\left ( s\right ) \left ( s^{2}+2s+10\right ) & =\frac{1}{s+1}\\ X\left ( s\right ) & =\frac{1}{\left ( s+1\right ) }\frac{1}{\left ( s^{2}+2s+10\right ) } \end{align*}
Taking the inverse laplace transform of \(X(s)\) and using partial fraction gives\begin{equation} \frac{1}{\left ( s+1\right ) }\frac{1}{\left ( s^{2}+2s+10\right ) }=\frac{A}{s+1}+\frac{B}{s^{2}+2s+10}\tag{1} \end{equation} Multiplying (1) by \(s+1\) gives\[ \frac{1}{s^{2}+2s+10}=A+\frac{B\left ( s+1\right ) }{s^{2}+2s+10}\] Evaluating at \(s=-1\) gives\begin{align*} \frac{1}{1-2+10} & =A\ \\ \frac{1}{9} & =A \end{align*}
Multiplying eq (1) by \(s^{2}+2s+10\) gives\[ \frac{1}{s+1}=\frac{A\left ( s^{2}+2s+10\right ) }{s+1}+B \] Evaluating at \(s=3i-1\) gives\[ \frac{1}{3i}=B \] Therefore\begin{align*} \frac{1}{s+1}\frac{1}{s^{2}+2s+10} & =\frac{1}{9}\frac{1}{s+1}+\frac{1}{3i}\frac{1}{\left ( s+1\right ) ^{2}+3^{2}}\\ & =\frac{1}{9}\frac{1}{s+1}-\frac{i}{9}\frac{3}{\left ( s+1\right ) ^{2}+3^{2}} \end{align*}
Using tables the inverse Laplace transform is\[ f\left ( t\right ) =\frac{1}{9}e^{-t}-\frac{i}{9}e^{-t}\sin 3t=\frac{1}{9}e^{-t}\left ( 1-i\sin 3t\right ) \]