State Variable Analysis Practice Questions

We have, transfer function $\mathrm{T}(\mathrm{s})=\mathrm{C}[\mathrm{SI}-\mathrm{A}]^{-1} \mathrm{~B}+\mathrm{D} \quad (1)$ where,

$$[S I-A]=\left[\begin{array}{ll}S & 0 \\ 0 & S\end{array}\right]-\left[\begin{array}{cc}0 & 1 \\ -1 & -2\end{array}\right]$$ $$\begin{aligned} & =\left[\begin{array}{cc} \mathrm{s} & -1 \\ 1 & \mathrm{~s}+2 \end{array}\right] \\ \therefore \quad[\mathrm{sl}-\mathrm{A}]^{-1} & =\frac{1}{\mathrm{~s}^{2}+2 \mathrm{~s}+1}\left[\begin{array}{cc} \mathrm{s}+2 & 1 \\ -1 & \mathrm{~s} \end{array}\right] \end{aligned}$$

From eqn. (1), we get

$$\begin{aligned} \mathrm{T}(\mathrm{s})&=\left[\begin{array}{ll} 3 & -2 \end{array}\right]\left[\begin{array}{cc} \frac{\mathrm{s}+2}{(\mathrm{~s}+1)^{2}} & \frac{1}{(s+1)^{2}} \\ -\frac{1}{(s+1)^{2}} & \frac{s}{(s+1)^{2}} \end{array}\right]\$\left[\begin{array}{l} 0 \\ 1 \end{array}\right]+1 \\ &=\left[\begin{array}{ll} 3 & -2 \end{array}\right]\left[\begin{array}{c} \frac{1}{(s+1)^{2}} \\ \frac{s}{(s+1)^{2}} \end{array}\right]\$+1 \\ &=\frac{3}{(s+1)^{2}}-\frac{2 s}{(s+1)}+1 \\ &=\frac{3-2 s+s^{2}+2 s+1}{s^{2}+2 s+1} \\ &=\frac{s^{2}+4}{s^{2}+2 s+1} \end{aligned}$$

Put, $s=j \omega$ , $T(j \omega)=\frac{-\omega^{2}+4}{-\omega^{2}+\mathrm{j} 2 \omega+1}$ Now, condition for output is zero,

$$\begin{aligned} -\omega^{2}+4 & =0 \\ \Rightarrow \quad \omega & =2 \mathrm{rad} / \mathrm{sec} . \end{aligned}$$

$\dot{x}=-x-K x=x(-k I-I)$ Characteristic equation,

$$\begin{aligned} |S I+K I+I| &=0 \\ |(S+1+K) I| &=0 \\ \therefore \qquad\qquad S+1+K &=0 \\ S &=-1-K \\ -2 &=-1-K \\ K &=1 \end{aligned}$$

Characteristic equation is, $|sI-A|=0$

$$|sI-A|=\begin{vmatrix} s &-1 \\ \alpha &s+2\beta \end{vmatrix}$$

$\;\;=s^{2+}2s\beta +\alpha =0$ $\therefore \;\; \omega _n^2=\alpha$ $\;\;\; \omega _n=\sqrt{\alpha }$ $\;\;\;2\xi \omega _n=2\beta$ $\;\;\;\xi =\frac{\beta }{\sqrt{\alpha }}$

$$y(t)=\begin{bmatrix} 1 &0 \end{bmatrix}x(t)$$

x(t)=L^{-1}\[(sI-A)^{-1}]$x(0)+L^{-1}$[(sI-A)^{-1}]$BU(s)$

$$L^{-1}\$[(sI-A)^{-1}]\$x(0)=\begin{bmatrix} 1\\ 0 \end{bmatrix}$$ $$L^{-1}\$[(sI-A)^{-1}]\$BU(s)=\begin{bmatrix} -0.25+0.5t+0.25e^{-2t}\\ 0.5-0.5e^{-2t} \end{bmatrix}$$ $$\therefore \;\; x(t)=\begin{bmatrix} 0.75+0.5t+0.25e^{-2t}\\ 0.5-0.5e^{-2t} \end{bmatrix}$$

$\therefore \;\; y(t)=0.75+0.5t+0.25e^{-2t}$ $\therefore \;\;y(1)=1.284$

$\dot{x}=Ax$ Eigen values are $\lambda _1$ and $\lambda _2$ we can write,

$$\phi (t)=\begin{bmatrix} e^{\lambda _1 t} & 0\\ 0& e^{\lambda _2 t} \end{bmatrix}$$

Response due to initial conditions, $x(t)=\phi (t)x(0)$

$$x(t)=\begin{bmatrix} e^{\lambda _1 t} & 0\\ 0& e^{\lambda _2 t} \end{bmatrix}\begin{bmatrix} \alpha \\ 0 \end{bmatrix}$$

$=\alpha e^{\lambda _1 t}$

Equations from the flow diagram, $\dot{x}_1=5x_1-2x_2+u_1$ $y_1=x_1$ $\dot{x}_2=5x_1+x_2+u_1+u_2$ $y_2=x_1-x_2$

$$\dot{x}=\begin{bmatrix} 5 & -2\\ 2 & 1 \end{bmatrix}x+\begin{bmatrix} 1 &0 \\ 1& 1 \end{bmatrix}u$$ $$y=\begin{bmatrix} 1 & 0\\ 1 & -1 \end{bmatrix}x$$

Considering the SISO cases:[/latex] $1. \;\;u_1(input)$ and $y_1(output):$

$$A\triangleq\begin{bmatrix} 5 & -2\\ 2 & 1 \end{bmatrix}$$ $$B\triangleq\begin{bmatrix} 1\\ 1 \end{bmatrix}$$

and

$$C=\begin{bmatrix} 1 &0 \end{bmatrix}$$ $$O\triangleq\begin{bmatrix} C\\ CA \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 5 & -2 \end{bmatrix}$$

$\Rightarrow \; Rank (O)=2$ $\Rightarrow \; Observable$

$$C\triangleq\begin{bmatrix} B & AB \end{bmatrix}=\begin{bmatrix} 1 &3 \\ 1 & 3 \end{bmatrix}$$

$\Rightarrow \; Rank(C)\neq 2$ $\Rightarrow \; non-controllable$ $2.\;\;u_2(input)$ and $y_1(output):$

$$A\triangleq\begin{bmatrix} 5 & -2\\ 2 & 1 \end{bmatrix}$$ $$B\triangleq\begin{bmatrix} 0\\ 1 \end{bmatrix}$$

and

$$C\triangleq \begin{bmatrix} 1 &-1 \end{bmatrix}$$ $$O\triangleq\begin{bmatrix} 1 & 0\\ 5 & -2 \end{bmatrix}$$

$\Rightarrow \; Rank (O)=2$ $\Rightarrow \; Observable$

$$C\triangleq\begin{bmatrix} 0 &-2 \\ 1 & 1 \end{bmatrix}$$

$\Rightarrow \; Rank(C)=2$ $\Rightarrow \; controllable$ $3. \;\;u_2(input)$ and $y_2(output) :$ $Rank(O)\neq 2\Rightarrow$ non-observable $Rank(C) =2 \Rightarrow$ controllable $4. \; \; u_1 (input)$ and $y_2(output)$ : non-observale and non-controllable.

Given

$$A=\begin{bmatrix} 0 & 1\\ -1&-1 \end{bmatrix},$$ $$B=\begin{bmatrix} 0\\ 1 \end{bmatrix},$$ $$C=\begin{bmatrix} 1 & 0 \end{bmatrix}$$

Transfer function of the given system is given by $\frac{Y(s)}{U(s)}=C(sI-A)^{-1}B$ Now,

$$(sI-A)=\begin{bmatrix} s & 0\\ 0 & s \end{bmatrix}-\begin{bmatrix} 0 & 1\\ -1&-1 \end{bmatrix}$$ $$\;\;\;=\begin{bmatrix} s & -1\\ 1&s+1 \end{bmatrix}$$

$(sI-A)^{-1}=\frac{1}{|sI-A|}(Adj[A])$ $|sI-A|=s(s+1)+1=(s^{2+}s+1)$ and

$$Adj[A]=\begin{bmatrix} s+1 & 1\\ -1&s \end{bmatrix}$$ $$\therefore \;\;(sI-A)^{-1}=\frac{1}{(s^{2+}s+1)}\begin{bmatrix} s+1 & 1\\ -1&s \end{bmatrix}$$

$\therefore$ Transfer function,

$$\frac{Y(s)}{U(s)}=\frac{1}{(s^{2+}s+1)}\left \{ \begin{bmatrix} 1 &0 \end{bmatrix}\begin{bmatrix} s+1 & 1\\ -1&s \end{bmatrix}\begin{bmatrix} 0\\ 1 \end{bmatrix} \right \}$$ $$\;\;=\frac{1}{(s^{2+}s+1)}\left \{ \begin{bmatrix} (s+1) & 1 \end{bmatrix} \begin{bmatrix} 0\\ 1 \end{bmatrix}\right \}$$

$\;\;=\frac{1}{(s^{2+}s+1)} \times 1$ $\frac{Y(s)}{U(s)}=\frac{1}{(s^{2+}s+1)}$ Given, input=unit step $\therefore \;\; U(s)=\frac{1}{s}$ $\therefore \;\; Y(s)=\frac{1}{s}\frac{1}{(s^{2+}s+1)}=\frac{1}{s(s^{2+}s+1)}$ $\therefore$ Final Value $\lim_{s\rightarrow 0}[sY(s)]$ =\lim_{s\rightarrow 0}\left[ \frac{1}{(s^{2+}s+1)} \right]\=1$$\therefore ;;$Error=Final value -Initial value$e_{ss}=0$

Given:

$$\begin{bmatrix} \dot{x}_1\\ \dot{x}_2 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 1&1 \end{bmatrix}\begin{bmatrix} x_1\\ x_2 \end{bmatrix}+\begin{bmatrix} 1\\ 1 \end{bmatrix}u$$

or, $\;\; \dot{x}(t)=Ax(t)+Bu$ Here,

$$[A]=\begin{bmatrix} 1 & 0\\ 1&1 \end{bmatrix}, \;\; B=\begin{bmatrix} 1\\ 1 \end{bmatrix}$$

State transition matric is given by, $e^{At}=L^{-1}[sI-A]^{-1}$ Now,

$$[sI-A]=\begin{bmatrix} s-1 & \\ 0 -1 & s-1 \end{bmatrix}$$ $$Adj[sI-A]=\begin{bmatrix} s-1 & \\ 1 & s-1 \end{bmatrix}$$

and $|sI-A|=(s-1)^2$ $\therefore \;\; [sI-A]^{-1}=\frac{1}{|sI-A|}Adj[sI-A]$

$$[sI-A]^{-1}=\frac{1}{(s-1)^2}\begin{bmatrix} s-1 & \\ 1 & s-1 \end{bmatrix}$$ $$=\begin{bmatrix} \frac{1}{s-1} & \\ \frac{1}{(s-1)^2} & \frac{1}{s-1} \end{bmatrix}$$ $$\therefore \;\; L^{-1}[sI-A]^{-1}=\begin{bmatrix} e^t & 0\\ te^t& e^t \end{bmatrix}$$

$\therefore$ State transition matrix

$$=\begin{bmatrix} e^t & 0\\ te^t& e^t \end{bmatrix}$$

Given: $\dot{x}(t)=Px+Qu$ $y=Rx$

$$P=\begin{bmatrix} -1 &1 \\ 0& -3 \end{bmatrix}$$

,

$$Q=\begin{bmatrix} 0\\ 1 \end{bmatrix}$$

,

$$R=\begin{bmatrix} 0 & 1 \end{bmatrix}$$

For controllability,

$$Q_c=\begin{bmatrix} Q &PQ \end{bmatrix}$$ $$[PQ]=\begin{bmatrix} -1 &1 \\ 0& -3 \end{bmatrix}\begin{bmatrix} 0\\ 1 \end{bmatrix}=\begin{bmatrix} 1\\ -3 \end{bmatrix}$$ $$\therefore \;\; Q_c=\begin{bmatrix} 0 &1 \\ 1& -3 \end{bmatrix}$$

$\therefore \;\;|Q_c|=-1\neq 0\Rightarrow \;\; controllable$ Also, for observability,

$$Q_0=\begin{bmatrix} R^T & P^TR^T \end{bmatrix}$$ $$\begin{bmatrix} P^T & R^T \end{bmatrix}=\begin{bmatrix} -1 & 0\\ 1 & -3 \end{bmatrix}\begin{bmatrix} 0\\ 1 \end{bmatrix}=\begin{bmatrix} 0\\ -3 \end{bmatrix}$$ $$\therefore \;\;Q_0=\begin{bmatrix} 0 & 0\\ 1 & -3 \end{bmatrix}$$

$\therefore \;\; |Q_0|=(0 \times -3 -0 \times 1)=0\Rightarrow \;\; non-observable$ Since, $Q_c\neq 0$ and $Q_0=0$ $\therefore$ System is controllable but not observable.

$$A=\begin{bmatrix} 0 &a_1 &0 \\ 0 & 0& a_2\\ a_3& 0 & 0 \end{bmatrix},$$ $$B=\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix},$$ $$C=\begin{bmatrix} 1 & 0 & 0 \end{bmatrix}^T$$

For stable system to be controllable, the metric $Q_c$ must be non singular.

$$Q_c=\begin{bmatrix} B & AB & A^2B \end{bmatrix}$$ $$AB=\begin{bmatrix} 0 & a_2 &0 \end{bmatrix}^T$$ $$A^2=\begin{bmatrix} 0 &0 &a_1a_2 \\ a_2a_3& 0& 0\\ 0& a_1a_3 & 0 \end{bmatrix}$$ $$A^2B=\begin{bmatrix} a_1a_2 & 0 & 0 \end{bmatrix}^2$$ $$Q_c=\begin{bmatrix} 0 & 0 & a_1a_2\\ 0& a_2 &0 \\ 1 & 0 & 0 \end{bmatrix}$$

$|Q_c|=-a_1a_2^2$ For $Q_c$ to be nonsingular $|Q_c|\neq 0$ $\therefore \;\; a_1\neq 0$ and $a_2\neq 0$ and $a_3\in R$

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

and

$$B=\begin{bmatrix} 0\\ 1 \end{bmatrix}$$ $$[sI-A]=\begin{bmatrix} s & 0\\ 0 & s \end{bmatrix}-\begin{bmatrix} -1 &2 \\ 0 & 2 \end{bmatrix}$$ $$=\begin{bmatrix} s+1 &-2 \\ 0 & s-2 \end{bmatrix} [sI-A]^{-1}$$ $$=\frac{\begin{bmatrix} s-1 &2 \\ 0 & s+1 \end{bmatrix}}{(s+1)(s+2)}\;\;...(i)$$

Transfer function $= C[sI-A]^{-1}B$ So, denominator of equation (i) gives pole of the system. $(s+1)(s-2)=0$ $s=-1 \; and \; 2$ One pole lies in RHS of s-plane. Hence, system is unstable. For controllability, $Q_c$ is defined as

$$Q_c=\begin{bmatrix} B\; : & AB \end{bmatrix}$$ $$AB=\begin{bmatrix} -1 &2 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} 0\\ 1 \end{bmatrix}=\begin{bmatrix} 2\\ 2 \end{bmatrix}$$ $$Q_C=\begin{bmatrix} 0 &2 \\ 1 & 2 \end{bmatrix}$$

$|Q_c|\neq 0$ Hence the system is controllable.