2016 AC-2 P1. (a) $X=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=\begin{bmatrix} y \\ \dot{y} \end{bmatrix}$

$\begin{cases} \dot{x}=\begin{bmatrix} \dot{x_1} \\ \dot{x_2} \end{bmatrix}=\begin{bmatrix} x_2 \\ -2x_1 x_2-u x_1+2u \end{bmatrix}\\ y=x_1 \end{cases}$

(b) $u \equiv 2$

$\dot{x} =\begin{bmatrix} x_2 \\ -2x_1 x_2-2x_1+4 \end{bmatrix}=\begin{bmatrix} x_2 \\ -2x_1 (x_2+1)+4 \end{bmatrix}$

let $\begin{cases} -2x_1 (x_2+1)+4=0 \\ x_2=0 \end{cases} \Rightarrow \begin{cases} -2x_1 +4=0 \\ x_2=0 \end{cases} \Rightarrow \begin{cases} x_1=2 \\ x_2=0 \end{cases}$

$\therefore The \; equilibrum\; point\; is \;x_e=\begin{bmatrix} 2 \\ 0 \end{bmatrix}$

(c) $u \equiv 2 \quad x_e=\begin{bmatrix} 2 \\ 0 \end{bmatrix}, \quad let \;x=f(x)$

The Jacobin of $\dot{x}$ is: \begin{align} Df(x)= \begin{bmatrix} 0 & 1 \\ -2x_1-2 & -2x_1 \end{bmatrix} \end{align}

The linear dynamics around $x_e$ is $\frac{d}{dt}f(x)=\begin{bmatrix} 0 & 1 \\ -2 & -4 \end{bmatrix} f(x)$

which is stable, locally stable at $x_e$.

P2. i) $x[k+1]=A x[k]$

$y[k]=\begin{bmatrix} -1 & 1 \end{bmatrix}x[k]$

let $x_[k]=\begin{bmatrix} a\\ b \end{bmatrix} \quad y[0]=\begin{bmatrix} -1 & 1 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix}=1$

$-a+b=1$

$y[1]=\begin{bmatrix} -1 & 1 \end{bmatrix}x[1]=\begin{bmatrix} -1 & 1 \end{bmatrix}\begin{bmatrix} -2 & 4 \\ -1 & 2 \end{bmatrix}x[0]=0$

$x[1]=Ax[0]$

$3a+2b=0$

$\therefore a=-\frac{2}{5} \quad b=\frac{3}{5}$

$x[0]=\begin{bmatrix} -\frac{2}{5}\\ \frac{3}{5} \end{bmatrix}$

ii)$x[0]=\begin{bmatrix} a\\ b \end{bmatrix} \quad x[2]=Ax[1]=A^2x[0]$

$A^2=0 \quad x[2]=0$

$y[2]=[-1 \quad 1]\quad x[2]=0 \quad y[1]=[-1 \quad 1] \quad x[1]=1$

$[-1\quad 1]\quad \begin{bmatrix} 2 &4\\ -1&2 \end{bmatrix}\quad\begin{bmatrix} a\\ b \end{bmatrix}=1$

we only have -3a-2b=1,so we can't uniquely determine a,b.

P3 (a)$\lambda I-A=\begin{bmatrix} \lambda+2&-4\\ 1&\lambda-2 \end{bmatrix}$

$\lambda_1=\lambda_2=0$

$\begin{bmatrix} -2-\lambda_1 & 4\\ -1 & 2-\lambda_1 \end{bmatrix}\begin{bmatrix} u_1 \\ u_2 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$

$\begin{cases} -2u_1-\lambda_1u_1+4u_2=0\\ -u_1+2u_2-\lambda_1u_2=0 \end{cases}$

$u_1=2u_2$

$\therefore eigenvector \begin{bmatrix} u_1\\ u_2 \end{bmatrix}=\begin{bmatrix} 2\\ 1 \end{bmatrix}$

$J=MAM^{-1}$

(b)$e^{At}=L^{-1}\begin{bmatrix} (SI-A)^{-1} \end{bmatrix}=L^{-1}\begin{bmatrix} \frac{s-2}{s^2} & \frac{4}{s^2} \\ \frac{-1}{s^2} & \frac{s+2}{s^2} \end{bmatrix}$

$L^{-1}\begin{bmatrix} \frac{s-2}{s^2} \end{bmatrix}= L^{-1}\begin{bmatrix} \frac{1}{s}-\frac{2}{s^2} \end{bmatrix}=1-2t$

$L^{-1}\begin{bmatrix} \frac{4}{s^2} \end{bmatrix}=4t$

$L^{-1}\begin{bmatrix} \frac{-1}{s^2} \end{bmatrix}=-t$

$L^{-1}\begin{bmatrix} \frac{s+2}{s^2} \end{bmatrix}=L^{-1}\begin{bmatrix} \frac{1}{s}+\frac{2}{s^2} \end{bmatrix}=1+2t$

$e^{At}=\begin{bmatrix} 1-2t & 4t\\ -t & 1+2t \end{bmatrix}$

(c)T(s)=$C(SI-A)^{-1}B$

=$[-1 \quad 1] \quad \begin{bmatrix} \frac{s-2}{s^2} & \frac{4}{s^2} \\ \frac{-1}{s^2} & \frac{s+2}{s^2} \end{bmatrix} \quad \begin{bmatrix} 2\\ 1 \end{bmatrix}$

$=\frac{\begin{bmatrix} -1 & 1 \end{bmatrix}}{s^2}\begin{bmatrix} 2s-4+4 \\ -2+s+2 \end{bmatrix}$

$T(s)=\begin{bmatrix} -1 & 1 \end{bmatrix}\begin{bmatrix} \frac{2}{s} \\ \frac{1}{s} \end{bmatrix}=\frac{-2}{s}+\frac{1}{s}$

$T(s)=\frac{-1}{s}$

pole is at s=0.

$\therefore marginally \; stable.$

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

$Ax=\begin{bmatrix} -2 & u \\ -1 & 2 \end{bmatrix} x,\quad Bu=\begin{bmatrix} 2 \\ 1 \end{bmatrix} u$

$C=\begin{bmatrix} B & AB & A^2B \end{bmatrix}=\begin{bmatrix} 2 & 0 \\ 1 & 0 \end{bmatrix}$

$\therefore not \; controllable$

To make $x(1)=\begin{bmatrix} -u & -2 \end{bmatrix}$

corres ponding characteristic equation.

$(s+u)(s+2)=0$

$s^2+6s+8=0$

$u=-kx$

$\dot{x}=(A-Bk) x$

$\begin{vmatrix} SI-(A-Bk) \end{vmatrix}=0$

$A-BK=\begin{bmatrix} -2 & 4\\ -1 & 2 \end{bmatrix}-\begin{bmatrix} 2K_1 \\ 1K_2 \end{bmatrix}$

$\begin{vmatrix} S+2+2k_1 & -4 \\ 1+k_2 & S-2 \end{vmatrix}=0 \qquad \begin{bmatrix} A-Bk \end{bmatrix}=\begin{bmatrix} -2-2k_1 & 4 \\ -1-k_2 & 2 \end{bmatrix}$

$(S+2+2k_1)(S-2)+4(1+k_2)=0$

$2k_1=6$

$k_1=3$

$uk_2=20$

$k_2=5$

$\therefore$ control $u_{(t)}=-\begin{bmatrix} k_1 & k_2 \end{bmatrix} x$

or $u_{(t)}=-\begin{bmatrix} 3 & 5 \end{bmatrix} x$

our goal is to have $x_{(1)}= -\begin{bmatrix} 1 & 2 \end{bmatrix}$

$(s-1)(s+2)=0$

$2k_1=-3$

$k_1=-\frac{3}{2}$

$uk_2=-4$

$2k_1=-3$

$k_2=-1$

$\therefore u_{(t)}=-\begin{bmatrix} k_1 & k_2 \end{bmatrix} x =\begin{bmatrix} -\frac{3}{2} & -1 \end{bmatrix} x$

e) consider $\dot{x}=A_{C1}X$

$A_{C1}=A-Bk$

$s^2+2k_1s-uk_1+uk_2=0$

$1+ \frac{C_t(s)}{t(s)}=0$

$\frac{C_t(s)}{t(s)}=s^2+2k_1s-uk_1+uk_2-1$

$\lim_{s \to 0} s\frac{C_t(s)}{t(s)}=\lim_{s \to 0} s(s^2+2k_1s-uk_1+uk_2-1)$

Final value is zero and no such controller is possible to design.

But for Bounded time period,it is possible to have solution of x(t). As we have already seen in previous problem.

$\alpha_1+\alpha_2=2k_1, \quad \alpha_1\alpha_2=-uk_1+uk_2.$

$(s+\alpha_1)(s+\alpha_2)=0$

$\begin{cases} \alpha^2+(\alpha_1+\alpha_2)s+\alpha_1\alpha_2=0 \\ s^2+2k_1s-uk_1+uk_2=0 \end{cases}$

(f) $\dot{x}=\begin{bmatrix} -2 & 4 \\ -1 & 2 \end{bmatrix} x+\begin{bmatrix} 2 \\ 1 \end{bmatrix} u$

$y=\begin{bmatrix} -1 & 1 \end{bmatrix} x \quad D=[0]$

$N=\begin{bmatrix} C \\ CA \end{bmatrix}=\begin{bmatrix} -1 & 1 \\ 1 & -2 \end{bmatrix}$

The system is observable.

(g) $\frac{Y(S)}{U(S)}=C[SI-A]^{-1}B+D$

$SI-A=\begin{bmatrix} S & 0 \\ 0 & S \end{bmatrix}-\begin{bmatrix} -2 & 4 \\ -1 & 2 \end{bmatrix}$

$[SI-A]^{-1}=\frac{1}{S^2-4+4}\begin{bmatrix} S-2 & 4 \\ -1 & S+2 \end{bmatrix}$

$\frac{Y(s)}{u(s)}==\begin{bmatrix} -1 & 1 \end{bmatrix}\begin{bmatrix} \frac{s-2}{s^2} & \frac{4}{s^2} \\ \frac{-1}{s^2} & \frac{s+2}{s^2} \end{bmatrix}\begin{bmatrix} 2 \\ 1 \end{bmatrix}$

$H (s)= \frac{-1}{s}$ marginally stable ,not BIBO.

(h) Cant not resolve.

## Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett