3.1 HW 1

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3.1.1 Problem B 2-1

From Modern Control Engineering, 4th edition by Ogata

Question

1. Find Laplace transform for

   $\begin{array}{ccc}f\left(t\right)=0& & t<0\\ f\left(t\right)=e^{-0.4t}\cos 12t& & t\ge 0\end{array}$

2. Find Laplace transform for

   $\begin{array}{ccc}f\left(t\right)=0& & t<0\\ f\left(t\right)=\sin (4t+\frac{\pi }{3})& & t\ge 0\end{array}$

Solution

3.1.1.1 Part a

This is of the form $e^{-at}f\left(t\right)$, hence use the property of Laplace transform $$\begin{array}{}\text{(1)}& \mathcal{L}\left(e^{-at}f\left(t\right)\right)=F(s+a)\end{array}$$ Where $F(s)$ is Laplace transform of $f\left(t\right).$ But $\mathcal{L}(\cos \omega t)=\frac{s}{s^2+{\omega }^2}$, therefore $$F(s)=\mathcal{L}(\cos 12t)=\frac{s}{s^2+144}$$ Hence (1) becomes$$\begin{array}{rl}\mathcal{L}\left(e^{-at}f\left(t\right)\right)& =\mathcal{L}(e^{-0.4t}\cos 12t)\\ & =F(s+a)\\ & =\frac{(s+0.4)}{{(s+0.4)}^2+144}\end{array}$$

3.1.1.2 Part b

I can not solve $\mathcal{L}(\sin (4t+\frac{\pi }{3}))$ by using the property that $$\mathcal{L}\left(f(t-a)\right)=e^{-as}F\left(s\right)$$ Because here delay $\frac{\pi }{3}>0$ where the above property is valid for $a<0$. Instead, writing$$\begin{array}{rl}\sin (\omega t+\theta )& =\sin \left(\omega t\right)\cos \theta +\cos \left(\omega t\right)\sin \theta \\ \sin (4t+\frac{\pi }{3})& =\sin \left(4t\right)\cos \frac{\pi }{3}+\cos \left(4t\right)\sin \frac{\pi }{3}\\ \text{(2)}& \mathcal{L}(\sin (4t+\frac{\pi }{3}))& =\cos \frac{\pi }{3}\mathcal{L}(\sin 4t)+\sin \frac{\pi }{3}\mathcal{L}(\cos 4t)\end{array}$$

But $\cos \frac{\pi }{3}=\frac{1}{2}$ and $\sin \frac{\pi }{3}=\sqrt{3}$ and $\mathcal{L}(\sin \omega t)=\frac{\omega }{s^2+{\omega }^2}⟹\mathcal{L}(\sin 4t)=\frac{4}{s^2+16}$ and $\mathcal{L}(\cos \omega t)=\frac{s}{s^2+{\omega }^2}⟹\mathcal{L}(\cos 4t)=\frac{s}{s^2+16}$. Hence substituting into eq (2) gives$$\begin{array}{rl}\mathcal{L}(\sin (4t+\frac{\pi }{3}))& =\frac{1}{2}\frac{4}{s^2+16}+\sqrt{3}\frac{s}{s^2+16}\\ & =\frac{1}{2}\left(\frac{4+\sqrt{3}s}{s^2+16}\right)\end{array}$$

3.1.2 Problem B 2-6

Question

(a) find Laplace transform for

Figure 3.1:function

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 $g\left(t\right)=f(t-a)$ 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

$$e^{-as}\frac{1}{s^2}$$

3.1.3 Problem B 2-7

From Modern Control Engineering, 4th edition by Ogata

Question

(a) find Laplace transform for

Figure 3.2:function

Solution

The above function can be constructed as follows

Let $f\left(t\right)=t$ (the ramp function)

$g\left(t\right)=f\left(t\right)u\left(t\right)-f\left(t\right)u(t-T)+Tu(t-T)$

Where $u\left(t\right)$ is the unit step function

This is ilustrated in this diagram

Figure 3.3:function

Now $$\mathcal{L}\left(f\left(t\right)u\left(t\right)\right)=\frac{1}{s^2}$$ And$$\begin{array}{rl}\mathcal{L}\left(f\left(t\right)u(t-T)\right)& ={\int }_T^{\mathrm{\infty }}t{e}^{-st}dt={\frac{-t}{s}{e}^{-st}|}_T^{\mathrm{\infty }}-{\int }_T^{\mathrm{\infty }}\frac{e^{-st}}{s}dt\\ & =\frac{-1}{s}(0-T{e}^{-sT})-\frac{1}{s}\left(\frac{-1}{s}\right){e^{-st}|}_T^{\mathrm{\infty }}\\ & =\frac{T}{s}{e}^{-sT}+\frac{1}{s^2}(0-e^{-sT})=\frac{T}{s}{e}^{-sT}-\frac{e^{-sT}}{s^2}\end{array}$$

And$$\mathcal{L}\left(Tu(t-T)\right)=\frac{T}{s}{e}^{-sT}$$ Hence $$\begin{array}{rl}\mathcal{L}\left(g\left(t\right)\right)& =\frac{1}{s^2}-(\frac{T}{s}{e}^{-sT}-\frac{e^{-sT}}{s^2})+\frac{T}{s}{e}^{-Ts}\\ & =\frac{1}{s^2}-\frac{T}{s}{e}^{-sT}+\frac{e^{-sT}}{s^2}+\frac{T}{s}{e}^{-Ts}\\ & =\frac{1}{s^2}+\frac{e^{-sT}}{s^2}\\ & =\frac{1-e^{-sT}}{s^2}\end{array}$$

3.1.4 Problem B 2-15

From Modern Control Engineering, 4th edition by Ogata

Question Obtain partial-fraction using MATLAB for

$F\left(s\right)=\frac{10(s+2)(s+4)}{(s+1)(s+3){(s+5)}^2}$ and then find inverse laplace transform

Solution I used Mathematica to find Partial-fraction

Figure 3.4:Code

Hence the inverse laplace tranform is

$$f\left(t\right)=\frac{15}{16}{e}^{-t}+\frac{5}{4}{e}^{-3t}+\frac{15}{4}t{e}^{-5t}-\frac{35}{16}{e}^{-5t}$$

Here is a plot of the solution

Figure 3.5:Plot

The Matlab code to find Partial-fraction for this problem is below.

Figure 3.6:Matlab code

3.1.5 Problem B 2-16

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

Figure 3.7:Code

Now find the Inverse Laplace transform for each term in the above result as follows.$$\begin{array}{rl}\mathcal{L}\delta \left(t\right)& =1\\ \mathcal{L}\frac{23}{18}{e}^{-t}& =\frac{23}{18}\frac{1}{1+s}\\ \mathcal{L}\frac{3}{26}{e}^{-3t}& =\frac{3}{26}\frac{1}{1+s}\end{array}$$

The inverse laplace transform of the last term $\frac{-1450-253s}{117(10+2s+s^2)}$ is found by writing $10+2s+s^2={(s+1)}^2+3^2$. Therefore this terms becomes$$\begin{array}{rl}\frac{-1450-253s}{117(10+2s+s^2)}& =\frac{\frac{-1450}{117}-\frac{253}{117}s}{(s+1{)}^2+3^2}\\ & =\frac{\frac{-1450}{117}-\frac{253}{117}(s+1)+\frac{253}{117}}{(s+1{)}^2+3^2}\\ & =\frac{-\frac{1197}{117}-\frac{253}{117}(s+1)}{(s+1{)}^2+3^2}\\ & =-\frac{\frac{1197}{117}}{(s+1{)}^2+3^2}-\frac{\frac{253}{117}(s+1)}{(s+1{)}^2+3^2}\\ & =-\frac{1197}{\left(117\right)\left(3\right)}\frac{3}{(s+1{)}^2+3^2}-\frac{253}{117}\frac{(s+1)}{(s+1{)}^2+3^2}\end{array}$$

Hence $$\begin{array}{rl}{\mathcal{L}}^{-1}(-\frac{1197}{117\times 3}\frac{3}{(s+1{)}^2+3^2}-\frac{253}{117}\frac{(s+1)}{(s+1{)}^2+3^2})& =-\frac{1197}{\left(117\right)\left(3\right)}{\mathcal{L}}^{-1}\left(\frac{3}{(s+1{)}^2+3^2}\right)\\ & =-\frac{253}{117}{\mathcal{L}}^{-1}\left(\frac{(s+1)}{(s+1{)}^2+3^2}\right)\\ & =-\frac{1197}{\left(117\right)\left(3\right)}{e}^{-t}\sin 3t-\frac{253}{117}{e}^{-t}\cos 3t\\ & =-\frac{e^{-t}}{117}(399\sin 3t+253\cos 3t)\end{array}$$

Adding all of the above, gives the Inverse Laplace transform as$$f\left(t\right)=\delta \left(t\right)+\frac{23}{18}{e}^{-t}-\frac{3}{26}{e}^{-3t}-\frac{e^{-t}}{117}(399\sin 3t+253\cos 3t)$$ Here is a plot of the solution

Figure 3.8:Plot

The Matlab code to find Partial-fraction for this problem is below.

Figure 3.9:Matlab code

3.1.6 Problem B 2-17

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

Figure 3.10:Code

In Mathematica, the solution is as follows

Figure 3.11:Code

3.1.7 Problem B 2-23

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$$\begin{array}{rl}s(sX\left(s\right)-x\left(0\right))-x^{\prime }\left(0\right)+2(sX\left(s\right)-x\left(0\right))+10X\left(s\right)& =\mathcal{L}\left(e^{-t}\right)\\ s^{2}X\left(s\right)-sx\left(0\right)-x^{\prime }\left(0\right)+2sX\left(s\right)-2x\left(0\right)+10X\left(s\right)& =\frac{1}{s+1}\end{array}$$

Applying the initial conditions results in$$\begin{array}{rl}{s}^{2}X\left(s\right)+2sX\left(s\right)+10X\left(s\right)& =\frac{1}{s+1}\\ X\left(s\right)(s^2+2s+10)& =\frac{1}{s+1}\\ X\left(s\right)& =\frac{1}{(s+1)}\frac{1}{(s^2+2s+10)}\end{array}$$

Taking the inverse laplace transform of $X(s)$ and using partial fraction gives$$\begin{array}{}\text{(1)}& \frac{1}{(s+1)}\frac{1}{(s^2+2s+10)}=\frac{A}{s+1}+\frac{B}{s^2+2s+10}\end{array}$$ Multiplying (1) by $s+1$ gives$$\frac{1}{s^2+2s+10}=A+\frac{B(s+1)}{s^2+2s+10}$$ Evaluating at $s=-1$ gives$$\begin{array}{rl}\frac{1}{1-2+10}& =A\\ \frac{1}{9}& =A\end{array}$$

Multiplying eq (1) by $s^2+2s+10$ gives$$\frac{1}{s+1}=\frac{A(s^2+2s+10)}{s+1}+B$$ Evaluating at $s=3i-1$ gives$$\frac{1}{3i}=B$$ Therefore$$\begin{array}{rl}\frac{1}{s+1}\frac{1}{s^2+2s+10}& =\frac{1}{9}\frac{1}{s+1}+\frac{1}{3i}\frac{1}{{(s+1)}^2+3^2}\\ & =\frac{1}{9}\frac{1}{s+1}-\frac{i}{9}\frac{3}{{(s+1)}^2+3^2}\end{array}$$

Using tables the inverse Laplace transform is$$f\left(t\right)=\frac{1}{9}{e}^{-t}-\frac{i}{9}{e}^{-t}\sin 3t=\frac{1}{9}{e}^{-t}(1-i\sin 3t)$$