LinSys Assignment 2

Problem 1

Transclude of Exact-Discretization

To find $A_d$ we can do the following transform

$$\begin{array}{rl}A& =Q\Lambda Q^{-1}\end{array}$$

Where we can find $\Lambda$ and $Q$ by finding the Eigenvectors and Eigenvalues

$$\det (A-\lambda I)=\left|\begin{array}{cc}-\lambda & 1\\ -2& -2-\lambda \end{array}\right|={\lambda }^2+2\lambda +2=(\lambda +(1+j))(\lambda +(1-j))$$

giving us ${\lambda }_1=1-j$ and ${\lambda }_2=1+j$

$$ker({\lambda }_{1}I-A)=\cdots =ker(\left[\begin{array}{cc}2& 1+j\\ 0& 0\end{array}\right])\to q_1=\left[\begin{array}{c}-(1+j)\\ 2\end{array}\right]$$ $$ker({\lambda }_2-A)=\cdots =ker(\left[\begin{array}{cc}2& 1-j\\ 0& 0\end{array}\right])\to q_2=\left[\begin{array}{c}-1+j\\ 2\end{array}\right]$$

Which gives us

$$A=Q\Lambda Q^{-1}=\frac{1}{4}\left[\begin{array}{cc}-1-j& -1+j\\ 2& 2\end{array}\right]\left[\begin{array}{cc}1-j& 0\\ 0& 1+j\end{array}\right]\left[\begin{array}{cc}2j& 1+j\\ -2j& 1-j\end{array}\right]$$

And most importantly

$$\begin{array}{rl}{A}_d& =e^{AT}\\ A_d& =Q{e}^{\Lambda T}{Q}^{-1}\\ A_d& =\frac{1}{4}\left[\begin{array}{cc}-1-j& -1+j\\ 2& 2\end{array}\right]\left[\begin{array}{cc}{e}^{(1-j)T}& 0\\ 0& e^{(-1-j)T}\end{array}\right]\left[\begin{array}{cc}2j& 1+j\\ -2j& 1-j\end{array}\right]\end{array}$$

And after inserting $T=\frac{\pi }{2}$ and solving the equation with $e^{jT}=\cos (T)+j\sin (T)$ we get:

$$A_d=\left[\begin{array}{cc}{e}^{-\pi /2}& e^{-\pi /2}\\ -2e^{-\pi /2}& -e^{-\pi /2}\end{array}\right]$$

Which is what we were supposed to prove. For $B_d$ we now how to calculate the integral where A is nonsingular

$$\begin{array}{rl}{B}_d& =\left({\int }_0^T{e}^{A\tau }d\tau \right)B\\ B_d& =A^{-1}(e^{AT}-{\mathbb{I}}_2)B\\ B_d& =\left[\begin{array}{cc}0& \frac{1}{2}-\frac{1}{2}{e}^{-\pi /2}\\ 0& e^{-\pi /2}\end{array}\right]\end{array}$$

$C_d$ and $D_d$ are quite simple to find and are

$$C_d=C=\left[\begin{array}{cc}4& 0\end{array}\right],D_d=D=\left[\begin{array}{cc}0& 0\end{array}\right]$$

Problem 2

Given a system on the form

Link to original

Where

$$A=\left[\begin{array}{cc}-2& 4\\ -1& 3\end{array}\right],B=\left[\begin{array}{c}8\\ 2\end{array}\right],C=\left[\begin{array}{cc}1& -1\end{array}\right],D=2$$

We do the following transformation

$$\bar{x}=Tx,T=\left[\begin{array}{cc}0& -2\\ 1& -1\end{array}\right]$$

a) Which gives us $\dot{\bar{x}}=T\dot{x}$ and $x=T^{-1}\bar{x}$

\dot{\bar{x}} &= TAT^{-1}x + TBu \\ y &= CT^{-1}x + Du \end{align}

Meaning that our new matrices will be defined as

\bar{A} &= TAT^{-1} =\begin{bmatrix} 0 & -2 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} -2 & 4 \\ -1 & 3 \end{bmatrix} \frac{1}{0*(-1)-(-2)*1} \begin{bmatrix} -1 & 2 \\ -1 & 0 \end{bmatrix} \\ \bar{A} &= \begin{bmatrix} 2 & 2 \\ 0 & -1 \end{bmatrix} \end{align} \bar{B} &= TB \\ \bar{B} &= \begin{bmatrix} 0 & -2 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 8 \\ 2 \end{bmatrix} \\ \bar{B} &= \begin{bmatrix} -4 \\ 6 \end{bmatrix} \end{align}

and

$$\begin{array}{rl}\bar{C}& =C{T}^{-1}\\ \bar{C}& =\left[\begin{array}{cc}1& -1\end{array}\right]\frac{1}{0\ast (-1)-(-2)\ast 1}\left[\begin{array}{cc}-1& 2\\ -1& 0\end{array}\right]\\ \bar{C}& =\left[\begin{array}{cc}0& 1\end{array}\right]\end{array}$$

And $\bar{D}=D=2$ $■$

b) The similarity transform means that the systems are algebraically equivalent aka the systems are the same.

c) Since the matrices aren’t even the same dimensions, they cannot be algebraically equivalent as there exists no $T$ that can do a transformation $\tilde{x}=Tx$. However, they can still be zero-state equivalent and we can check this by finding the transfer function: General equation for the transfer function of a system on State Space Representation

Link to original

G(s) = C(sI-A)^{-1}B+D= \dots = \frac{2s+8}{s+1} \end{align}

and

\tilde{G}(s) &= \tilde{C}(sI-\tilde{A})^{-1}\tilde{B}+\tilde{D} \\ \tilde{G}(s) &= 3(s-(-1))^{-1}*2+2 \\ \tilde{G}(s) &= \frac{2s+8}{s+1} \end{align}

Since $\tilde{G}(s)=G(s)$ the systems are zero state equivalent

Problem 3

Given

$$\dot{x}(t)=Ax(t)+Bu(t)$$

Where

$$A=\left[\begin{array}{cc}2& -3\\ 4& -5\end{array}\right]andB=\left[\begin{array}{cc}0& 0\\ 2& 2\end{array}\right]$$

a) The Controllability Matrix is given by

Controllability Matrix

The general equation for a controllability matrix given a system on State Space Representation form is: $\mathcal{C}=\left[\begin{array}{c}B|AB|A^{2}B|...|A^{n-1}B\end{array}\right]$ And the system is controllable if $rank(\mathcal{C})=n$, aka we can controll each state.

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Multi-input Controllability

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$$\mathcal{C}=[B,AB]=\left[\begin{array}{cccc}0& 0& -6& -6\\ 2& 2& -10& -10\end{array}\right]$$

We can see that the $rank(\mathcal{C})=n=2$ which means that the system is controllable

b) We will now find the eigenvalues of A

To find the eigenvalues of a matrix, $A$, we need to find $det(A-\lambda I_n)=0$ To find the eigenvectors of a matrix, $A$, we need to find $A{q}_i={\lambda }_i{q}_i$ → $ker({\lambda }_i{I}_n-A)$

Link to original

$$\det (A-\lambda {\mathbb{I}}_2)=\det (\left[\begin{array}{cc}2-\lambda & -3\\ 4& -5-\lambda \end{array}\right])={\lambda }^2+3\lambda +2$$

Which gives us ${\lambda }_1=-1$ and ${\lambda }_2=-2$

c) We first find

&[A-\lambda \mathbb{I}_{2} | B] =\left[ \begin{array}{cc|cc} 2-\lambda & -3 & 0 & 0 \\ 4 & -5-\lambda & 2 & 2 \end{array} \right] \end{align}

For $\lambda ={\lambda }_1=-1$ we get:

$$rank(\left[\begin{array}{cccc}3& -3& 0& 0\\ 4& -4& 2& 2\end{array}\right])=2=n$$

For $\lambda ={\lambda }_2=-2$ we get:

$$rank(\left[\begin{array}{cccc}4& -3& 0& 0\\ 4& -3& 2& 2\end{array}\right])=2=n$$

Since the Popov-Belevitch-Hautus Test has full rank for all $\lambda$, the system is controllable.

d)

Lyapunov Test for Controllability

If the eigenvalues of $A$ have strictly negative real parts, then the system is controllable if and only if the matrix $W$ is a Positive-Definite Matrix

Positive-Definite Matrix

A symmetric matrix with all positive eigenvalues. NOTE: All symmetric matrices have only real eigenvalues

For a matrix, $M$, to be positive-semidefinite the following has to hold

Criterion

$x^{T}Mx>0,\forall x\in {\mathbb{R}}^n\backslash \{0\}$ Given that $M$ is a $n\times n$ symmetric real matrix

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Can be written generally as

$$M=\left[\begin{array}{cc}{m}_1& m_2\\ m_2& m_3\end{array}\right]$$Link to original

$$AW+W{A}^T=-B{B}^T$$

Where $W$ is a Symmetric Matrix

Link to original

We can then write

$$W=W^T=\left[\begin{array}{cc}{w}_{11}& w_{12}\\ w_{21}& w_{22}\end{array}\right]$$

And solve for the coefficients $w_{11},w_{12},w_{21},w_{22}$.

\begin{bmatrix} 2 & -3 \\ 4 & -5 \end{bmatrix} \begin{bmatrix} w_{11} & w_{12} \\ w_{21} & w_{22} \end{bmatrix} + \begin{bmatrix} w_{11} & w_{12} \\ w_{21} & w_{22} \end{bmatrix} \begin{bmatrix} 2 & 4 \\ -3 & -5 \end{bmatrix} &= -\begin{bmatrix} 0 & 0 \\ 2 & 2 \end{bmatrix}\begin{bmatrix} 0 & 2 \\ 0 & 2 \end{bmatrix} \\ \begin{bmatrix} 2w_{11}-3w_{21} & 2w_{12}-3w_{22} \\ 4w_{11}-5w_{21} & 4w_{12} - 5w_{22} \end{bmatrix} + \begin{bmatrix} 2w_{11}-3w_{12} & 4w_{11}-5w_{12} \\ 2w_{21}-3w_{22} & 4w_{21} -5w_{22} \end{bmatrix} &=\begin{bmatrix} 0 & 0 \\ 0 & 8 \end{bmatrix} \end{align}

Solving the four equations for the four variables gives us $w_{11}=6,w_{12}=4,w_{21}=4,w_{22}=4$

$$W=\left[\begin{array}{cc}6& 4\\ 4& 4\end{array}\right]$$

We can see that transposing $W$ gives us

$$W^T=\left[\begin{array}{cc}6& 4\\ 4& 4\end{array}\right]=W$$

And since the eigenvalues of $A$ are strictly negative, Lyapunovs test for controllability deems the system as controllable

Problem 4

Given

$$\dot{x}(t)=Ax(t)+Bu(t)$$

Where

$$A=\left[\begin{array}{ccc}1& 2& 0\\ 0& 3& 4\\ 0& 0& 5\end{array}\right],andB=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$$

a)

Controllability Matrix

The general equation for a controllability matrix given a system on State Space Representation form is: $\mathcal{C}=\left[\begin{array}{c}B|AB|A^{2}B|...|A^{n-1}B\end{array}\right]$ And the system is controllable if $rank(\mathcal{C})=n$, aka we can controll each state.

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Multi-input Controllability

Link to original

We calculate $\mathcal{C}$

$$\mathcal{C}=\left[\begin{array}{c}B|AB|A^{2}B\end{array}\right]\left[\begin{array}{ccc}0& 0& 8\\ 0& 4& 32\\ 1& 5& 25\end{array}\right]$$

Which has full rank → Controllable

b)

&\det(\bar{A}-\lambda \mathbb{I_{3}}) \\ &=\det(A-BK-\lambda \mathbb{I}_{3}) \\ &=-\lambda^{3}+(9-k_{3})\lambda²+(4k_{3}-4k_{2}-23)\lambda-8k_{1}+4k_{2}-3k_{3}+15 \end{align}

c) We want to place our poles such that $\overline{{\lambda }_1}=-1,\overline{{\lambda }_2}=2,\overline{{\lambda }_3}=-3$.

$$\det (\bar{A}-\lambda {\mathbb{I}}_3)=(-1-\lambda )(-2-\lambda )(-3-\lambda )=-{\lambda }^3-6{\lambda }^2-11\lambda -6$$

Solving this with our equation from b) gives us

$$\begin{array}{rl}9-k_3& =-6\\ 4k_3-4k_2-23& =-11\\ -8k_1+4k_2-3k_3+15& =-6\end{array}$$

Where $k_1=3,k_2=12,k_3=15$

and

$$K=\left[\begin{array}{ccc}3& 12& 15\end{array}\right]$$

Q.E.D