Q2GATE 2024MSQ

Consider a system $S$ represented in state space as

\frac{d x}{d t}=\left[\begin{array}{ll} 0 & -2 \\ 1 & -3 \end{array}\right] x+\left[\begin{array}{l} 1 \\ 0 \end{array}\right] r, y=\left[\begin{array}{ll} 2 & -5 \end{array}\right] x

Which of the state space representations given below has/have the same transfer function as that of $S$ ?

🚀 Solve in Practice Mode

📖 Explanation

From given state model, (observable form) $T F=C\left[(s I-A)^{-1} B\right]\$ +D=\frac{(2 s+1)}{s^{2}+3 s+2}$ From this controllable form can be written as

\begin{aligned} X^{0} & =\left[\begin{array}{cc} 0 & 1 \\ -2 & -3 \end{array}\right] X+\left[\begin{array}{l} 0 \\ 1 \end{array}\right] u \\ Y & =\left[\begin{array}{ll} 1 & 2 \end{array}\right] X+[0] u \end{aligned}

Another possibility, $\quad \mathrm{TF}=\frac{2 s+1}{s^{2}+3 s+2}=\frac{-1}{s+1}+\frac{3}{s+2}$

\begin{aligned} X^{0} & =\left[\begin{array}{cc} -1 & 0 \\ 0 & -2 \end{array}\right] X+\left[\begin{array}{c} -1 \\ 3 \end{array}\right] v \\ Y & =\left[\begin{array}{ll} 1 & 1 \end{array}\right] X+[0] u \end{aligned}

Diagonal form of state mode.

Q5GATE 2019MCQ

Let the state-space representation of an LTI system be $\dot{x}(t)=Ax(t)+Bu(t)$ , $\dot{y}(t)=Cx(t)+du(t)$ where A, B, C are matrices, d is a scalar, u(t) is the input to the system, and y(t) is its output. Let $B=[0\;0\;1]^T$ and d=0. Which one of the following options for A and C will ensure that the transfer function of this LTI system is $H(s)=\frac{1}{s^{3+}3s^{2+}2s+1}$ ?

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} X(t) &=A x(t)+B u(t) \\ y(t) &=C x(t) \\ B &=\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right] \\ \frac{\gamma(s)}{U(s)} &=\\ \frac{\gamma(s)}{X_{1}(s)} \times \frac{X_{1}(s)}{U(s)}=& 1 \times \frac{1}{s^{3}+3 s^{2}+2 s+1} \\ x_{1}(s)\left(s^{3}+3 s^{2}+2 s+1\right]&=U(s) \\ x_{2} &=\dot{x}_{1}(t) \\ x_{2}(s) &=s x_{1}(s) \\ x_{3} &=\dot{x}_{2}(t) \\ \Rightarrow \quad x_{3}(s) &=s x_{2}(s)=s^{2} X_{1}(s) \\ \text { So, } \quad s x_{3}(s) &=-x_{1}(s)-2 x_{2}(s)-3 x_{3}(s)+u(s) \\ \dot{x}_{3}(t) &=-x_{1}(t)-2 r_{2}(t)-3 x_{3}(t)+u(t) \\ y(t) &=x_{1}(t) \\ \dot{x}(t) &=\left[\begin{array}{ccc}0 & 1 & 0 \\ 0 & 0 & 1 \\ -1 & -2 & -3\end{array}\right] x(t)+\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right] u(t) \\ y(t) &=[1 \quad 0 \quad 0] x(t) \end{aligned}

Q7GATE 2017NAT

Consider the state space realization

\begin{bmatrix} x_{1}(t)\\ x_{2}(t) \end{bmatrix}=\begin{bmatrix} 0 & 0\\ 0&-9 \end{bmatrix} \begin{bmatrix} x_{1}(t)\\ x_{2}(t) \end{bmatrix}+ \begin{bmatrix} 0\\ 45 \end{bmatrix}u(t)

, with the initial condition

\begin{bmatrix} x_{1}(0)\\ x_{2}(0) \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}

where u(t) denotes the unit step function. The value of $\lim_{t\rightarrow \infty }|\sqrt{x_{1}^{2}(t)+x_{2}^{2}(t)}|$ is _______.

🚀 Solve in Practice Mode

📖 Explanation

\left[\begin{array}{l} \dot{x}_{1}(t) \\ \dot{x}_{2}(t) \end{array}\right]=\left[\begin{array}{cc} 0 & 0 \\ 0 & -9 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right]+\left[\begin{array}{c} 0 \\ 45 \end{array}\right] u(t)

By applying Laplace transform on both sides. get

\begin{aligned} s x_{1}(s)-x_{1}(0)&=0\\ X_{1}(s)&=\frac{x_{1}(0)}{s}=0 \quad \because x_{1}(0)=0 \\ \text{So,} x_{1}(t)&=0 \\ \text{and }\quad s X_{\lambda}(s)-x_{2}(0)&=-9 X_{1}(s)+\frac{45}{s} \\ X_{2}(s) &=\frac{45}{s(s+9)} \quad \because x_{2}(0)=0 \\ \text { Required value } &=\lim _{t \rightarrow \infty }|\sqrt{x_{1}^{2}(t)+x_{2}^{2}(t)}| \\ &=\left|\lim _{t \rightarrow \infty } x_{2}(t)\right| \quad \because x_{1}(t)=0 \\ \lim _{t \rightarrow \infty } x_{2}(t) &=\lim _{s \rightarrow 0} s X_{2}(s)=\frac{45}{9}=5 \end{aligned}

So, Required value $= |5| =5$

Q8GATE 2017MCQ

A second - order LTI system is described by the following state equations, $\frac{d}{dt}x_{1}(t)- x_{2}(t)=0$ $\frac{d}{dt}x_{2}(t)+2x_{1}(t)+3x_{2}(t)=r(t)$ Where $x_{1}(t)$ and $x_{2}(t)$ are the two state variables and r(t) denotes the input. The output $c(t) =x_{1}(t)$ . The system is.

🚀 Solve in Practice Mode

📖 Explanation

Given state variable equations are as follows:

\begin{aligned} &\frac{d}{d t} x_{1}(t)-x_{2}(t)=0 &\ldots(i)\\ &\frac{d}{d t} x_{2}(t)+2 x_{2}(t)+3 x_{2}(t)=r(t) &\ldots(ii) \end{aligned}

Also given that, input $=r(t)$ and output $=x_{1}(t)$ By applying Laplace transform to equation (i), we get, $s X_{1}(s)=X_{2}(s) \qquad \ldots(iii)$ By applying Laplace transform to equation (ii). we get,

\begin{aligned} s X_{2}(s)+2 X_{1}(s)+3 X_{2}(s)&=R(s) \\ x_{2}(s)&=\frac{R(s)-2 X_{1}(s)}{s+3} &\ldots(iv) \end{aligned}

By substituting equation (iv) in equation (iii), we get,

\begin{aligned} s X_{\mathrm{i}}(s)&=\frac{R(s)-2 X_{1}(s)}{s+3}\\ \left(s^{2}+3 s+2\right) X_{1}(s)&=R(s) \end{aligned}

So, the transfer function of the given system is,

\begin{aligned} \frac{X_{1}(s)}{R(s)} &=\frac{1}{s^{2}+3 s+2} \\ v r^{2} &=2 \\ 2 \zeta_{n} &=3 \\ \xi &=\frac{3}{2 \omega_{n}}=\frac{3}{2 \sqrt{2}}=1.06 \end{aligned}

As $\xi \gt 1,$ the given system is overdamped.

Q10GATE 2015MCQ

A network is described by the state model as $\dot{x_1}=2x_{1}-x_{2}+3u$ $\dot{x_2}=-4x_{2}-u$ $y=3x_{1}-2x_{2}$ The transfer function $H(s)(=\frac{Y(s)}{U(s)})$ is

🚀 Solve in Practice Mode

📖 Explanation

State matrix form:

\begin{array}{l} \dot{x}=\left[\begin{array}{cc} 2 & -1 \\ 0 & -4 \end{array}\right] x+\left[\begin{array}{c} 3 \\ -1 \end{array}\right] u \\ y=\left[\begin{array}{ll} 3 & -2 \end{array}\right] x \end{array}

The transfer function from tate matrix can be calculated as: $T(S)=C(s l-A]^{-1} B+D$ Here,

[s l-A]^{-1}=\left[\begin{array}{cc} s-2 & 1 \\ 0 & s+4 \end{array}\right]^{-1}

\begin{aligned} &=\frac{1}{(s-2)(s+4)}\left[\begin{array}{cc} (s+4) & -1 \\ 0 & (s-2) \end{array}\right] \\ \Rightarrow &C[s /-A]^{-1} B \\ =& \frac{1}{(s-2)(s+4)}[3-2]\left[\begin{array}{cc} (s+4) & -1 \\ 0 & (s-2) \end{array}\right]\left[\begin{array}{c} 3 \\ -1 \end{array}\right] \\ =& \frac{1}{(s-2)(s+4)}[3-2]\left[\begin{array}{c} 3 s+12+1 \\ -s+2 \end{array}\right] \\ =& \frac{1}{(s-2)(s+4)}[9 s+39+2 s-4] \\ =& \frac{11 s+35}{(s-2)(s+4)} \end{aligned}

Q11GATE 2015MCQ

The state variable representation of a system is given as

x=\begin{bmatrix} 0 & 1\\ 0&-1 \end{bmatrix}x; \; x(0)=\begin{bmatrix} 1\\ 0 \end{bmatrix} \; y= \begin{bmatrix} 0 &1 \end{bmatrix}x

The response y(t) is

🚀 Solve in Practice Mode

📖 Explanation

The response of the system is given by $x(t)=L^{-1}\left[(s l-A)^{-1} x(0)+(s /-A)^{-1} B \cdot U(s)\right]$ and the complete response $y(t)=[C] x(t)$ From the given state model,

\begin{array}{l} A=\left[\begin{array}{ll}0 & 1 \\ 0 & -1\end{array}\right] ; B=0: C=[0 \quad 1] \; and \; x(0)=\left[\begin{array}{c}1 \\ 0\end{array}\right] \\ \therefore \quad(s l-A)^{-1}=\left[\begin{array}{cc}s & -1 \\ 0 & s+1\end{array}\right]^{-1} \\ =\frac{1}{s(s+1)}\left[\begin{array}{cc}s+1 & 1 \\ 0 & s\end{array}\right] \\ (s l-A)^{-1} x(0)=\frac{1}{s(s+1)}\left[\begin{array}{cc} s+1 & 1 \\ 0 & s \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \end{array}\right] \\ =\frac{1}{s(s+1)}\left[\begin{array}{c} s+1 \\ 0 \end{array}\right]=\left[\begin{array}{c} 1 / s \\ 0 \end{array}\right] \\ \Rightarrow x(t)=L^{-1}\left[(s I-A)^{-1} x(0)\right]\$ \\ =\left[\begin{array}{c} 1 \\ 0 \end{array}\right] u(t) \end{array}

and the complete response

y(t)=\left[\begin{array}{lll}0 & 1\end{array}\right] x(t)=\left[\begin{array}{ll}0 & 1\end{array}\right]\left[\begin{array}{l}1 \\ 0\end{array}\right]=0

Q12GATE 2014MCQ

Consider the state space model of a system, as given below

\begin{bmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3} \end{bmatrix}=\begin{vmatrix} -1 & 1&0 \\ 0 & -1 & 0\\ 0&0 & -2 \end{vmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}+\begin{bmatrix} 0\\ 4\\ 0 \end{bmatrix}u;y\begin{bmatrix} 1 & 1 &1 \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}

The system is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \left[\begin{array}{c}\dot{x}_{1} \\ \dot{x}_{2} \\ \dot{x}_{3}\end{array}\right]&=\left[\begin{array}{rrr}-1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]+\left[\begin{array}{l}0 \\ 4 \\ 0\end{array}\right] u \\ y&=[1 \;1 \;1 ]\left[\begin{array}{r}x_{3} \\ x_{3} \\ x_{3}\end{array}\right] \\ A&=\left[\begin{array}{rrr}-1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2\end{array}\right], B=\left[\begin{array}{l}x_{1} \\ x_{2} \\ 0\end{array}\right], C=\left[1\;1\;1\right] \\ \end{aligned}

Check for controllability:

\begin{aligned} Q_{C} &=\left[B: A B: A^{2} B\right]\$ \\ &=\left[\begin{array}{rrr} 0 & 4 & -8 \\ 4 & -4 & 4 \\ 0 & 0 & 0 \end{array}\right] \end{aligned}

For controllable. $\left|Q_{c}\right| \neq 0$ Here , $\quad\left|Q_{c}\right|=4(0)=0 \quad \therefore$ Uncontrollable Check for observability:

\begin{aligned} Q_{0} &=\left[C^{T}: A^{T} C^{T}: A^{2 T} C^{T}\right]\$ \\ &=\left[\begin{array}{rrr} 1 & -1 & 1 \\ 1 & 0 & -1 \\ 1 & -2 & 4 \end{array}\right] \end{aligned}

For observable, $\left|Q_{0}\right| \neq 0$ Here, $\left|Q_{o}\right|=1 \quad \therefore$ Observable

Q13GATE 2014MCQ

Consider the state space system expressed by the signal flow diagram shown in the figure. The corresponding system is

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📖 Explanation

The state equation from the given state diagram is

\begin{aligned} \dot{x}_{1} &=x_{2} \\ \dot{x}_{2} &=x_{3} \\ \dot{x}_{3} &=a_{3} x_{3}+a_{2} \cdot x_{2}+a_{1} x_{1}+u \\ \text { Also, } y &=c_{1} x_{1}+c_{2} x_{2}+c_{3} x_{3} \end{aligned}

Thus, state matrix

\left[\begin{array}{c}\dot{x}_{1} \\ \dot{x}_{2} \\ \dot{x}_{3}\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & 0 \\ 0 & 0 & 1 \\ a_{1} & a_{2} & a_{3}\end{array}\right]\left[\begin{array}{c}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]+\left[\begin{array}{c}0 \\ 0 \\ 1\end{array}\right] u

and

y=\left[\begin{array}{lll}c_{1} & c_{2} & c_{3}\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]

Check for controllability: The system is said to be controllable if, the rank of controllability matrix $Q_{c}$ is equal to the rank of the state matrix A . However, if the controllability matrix $Q_{c}$ is a square matrix then the condition for controllability is $\left|o_{c}\right| \neq 0$ where

\begin{aligned} Q_{c}&=\left[B: A B: A^{2} B\right]\$\\ \therefore Q_{c}&=\left[\begin{array}{ccc}0 & 0 & 1 \\ 0 & 1 & a_{3} \\ 1 & a_{3} & a_{2}+a_{3}^{2}\end{array}\right]\$ \\ \because\left|Q_{C}\right| \neq & 0 \text{ and Rank of } Q_{c}=\text{ Rank of }A=1 \end{aligned}

$\therefore$ The system is controllable. Check for observability: The system is said to be observable if, the rank of observability matrix $Q_{o}$ is equal to the rank of the state matrix A . However, if the observability matrix $Q_{o}$ is a square matrix then the condition for observability is $\left|0_{0}\right| \neq 0$ where

\begin{aligned} Q_{0}&=\left[C^{T}: A^{T} C^{T}:\left(A^{T}\right)^{2} C^{T}\right]\$ \\ O_{0}&=\left[\begin{array}{ccc}c_{1} & c_{2} & c_{3} \\ s_{3} a_{1} & c_{1}+c_{2} a_{2} & c_{2}+c_{2} a_{3} \\ c_{1}\left(c_{2}+c_{3} a_{3}\right) & c_{3} a_{1}+\left(c_{2}+c_{3} a_{3}\right) a_{2} & c_{1}+c_{2} a_{2}+\left(c_{2}+c_{3} a_{3}\right) a_{3}\end{array}\right] \end{aligned}

$\because\left|Q_{0}\right|$ depends on value of unknown. Hence, not observable always.

Q14GATE 2014MCQ

The state equation of a second-order linear system is given by $\dot{x}(t) = Ax(t), \;\; x(0) = x_{0}$ For

x_{0}=\begin{bmatrix} 1\\ -1 \end{bmatrix}, \; x(t)=\begin{bmatrix} e^{-t}\\ -e^{-t} \end{bmatrix}

and for

x_{0}=\begin{bmatrix} 0\\ 1 \end{bmatrix},

x(t)=\begin{bmatrix} e^{-t}-e^{-2t}\\ -e^{-t}+2e^{-2t} \end{bmatrix}

when

x_{0}=\begin{bmatrix} 3\\ 5 \end{bmatrix}

, x(t) is

🚀 Solve in Practice Mode

📖 Explanation

Given $\dot{x}(t)=A x(t), \quad x(0)=x_{0}$ Taking the Laplace transform

\begin{aligned} s X(s)-x(0) &=A X(s) \\ [s I-A] X(s) &=x(0) \\ X(s) &=[s I-A]^{-1} x(0) \\ x(t) &=\mathcal{L}^{-1}[s I-A]^{-1} x(0) &\ldots(i) \end{aligned}

Condition given are

\begin{aligned} For x_{0}&=\left[\begin{array}{c}1 \\ -1\end{array}\right], x(t)=\left[\begin{array}{c}e^{-t} \\ -e^{-t}\end{array}\right]\$ \\ For x_{0}&=\left[\begin{array}{l}0 \\ 1\end{array}\right], x(t)=\left[\begin{array}{l}e^{-t}-e^{-2 t} \\ -e^{-t}+2 e^{-2 t}\end{array}\right]\$ \end{aligned}

Using the linearity property in equation (i) $K_{1} x_{1}(t)=\mathcal{L}^{-1}[s I-A]^{-1} x_{1}(0) K_{1}$ $K_{2} x_{2}(t)=\mathcal{L}^{-1}[s I-A]^{-1} x_{2}(0) K_{2}$ Using the linearity property as

\begin{aligned} K_{1} x_{1}(t)+K_{2} x_{2}(t)&=\mathcal{L}^{-1}[s I-A]^{-1} \\ \left[K_{1} x_{1}(0)+K_{2} x_{2}(0)\right] & \quad \ldots(ii)\\ \text { Also } \quad X_{3}(s)&=[s I-A]^{-1} x_{3}(0) \\ \text{So, } \quad K_{1}\left[\begin{array}{r}1 \\ -1\end{array}\right]+&K_{2}\left[\begin{array}{l}0 \\ 1\end{array}\right]=\left[\begin{array}{l}3 \\ 5\end{array}\right] \\ \left[\begin{array}{l}K_{1}+O K_{2} \\ -K_{1}+K_{2}\end{array}\right]&=\left[\begin{array}{l}3 \\ 5\end{array}\right] \\ \Rightarrow \quad K_{1}&=3 \\ K_{2}&=8 \end{aligned}

So, from equation (ii), we get x(t)

\begin{aligned} x(t) &=K_{1} x_{1}(t)+K_{2} x_{2}(t) \\ &=3\left[\begin{array}{c} e^{-t} \\ -e^{-t} \end{array}\right]\$+8\left[\begin{array}{c} e^{-t}-e^{-2 t} \\ -e^{-t}+2 e^{-2 t} \end{array}\right]\$ \\ &=\left[\begin{array}{c} 11 e^{-t}-8 e^{-2 t} \\ -1 t e^{-t}+16 e^{-2 t} \end{array}\right]\$ \end{aligned}

Q15GATE 2014MCQ

An unforced linear time invariant (LTI) system is represented by

\begin{vmatrix} \dot{x}_{1}\\ \dot{x}_{2} \end{vmatrix}=\begin{vmatrix} -1 & 0\\ 0&-2 \end{vmatrix}\begin{vmatrix} x_{1}\\ x_{2} \end{vmatrix}

If the initial conditions are $x_{1}(0)=1 \; and \; x_{2}(0)=-1$ , the solution of the state equation is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned}\left[\begin{array}{l}\dot{x}_{1} \\ \dot{x}_{2}\end{array}\right] &=\left[\begin{array}{cc}-1 & 0 \\ 0 & -2\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2}\end{array}\right] \\ \dot{x}_{1} &=-x_{1}+0 x_{2} \\ \dot{x}_{2} &=0 x_{1}-2 x_{2} \end{aligned}

Applying Laplace transform on both equations

\begin{aligned} \Rightarrow s X_{1}(s)-x_{1}(0)&=-X_{1}(s) \\ X_{1}(s)[s+1]&=X_{1}(0) \quad &\text{given }x_{1}(0)=1 \\ X_{1}(s)&=\frac{1}{s+1} \\ \Rightarrow \quad e^{-t}&=x_{1}(t)\\ \Rightarrow \quad s X_{2}(s)-X_{2}(0)&=-2 X_{2}(s) \\ (s+2) X_{2}(s)&=X_{2}(0) &X_{2}(0)=-1 \\ X_{2}(s)&=-\frac{1}{s+2} \\ x_{2}(t)&=-e^{-2 t} \end{aligned}

Q18GATE 2012MCQ

The state variable description of an LTI system is given by

\begin{pmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3} \end{pmatrix}=\begin{pmatrix} 0 & a_{1}& 0\\ 0& 0 & a_{2}\\ a_{3} & 0 & 0 \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3} \end{pmatrix}+\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}u

y=\begin{pmatrix} 1 &0 &0 \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3} \end{pmatrix}

where y is the output and u is the input. The system is controllable for

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \left[\begin{array}{l}\dot{x}_{1} \\\dot{x}_{2} \\\dot{x}_{3}\end{array}\right]&=\left[\begin{array}{lll} 0 & a_{1} & 0 \\0 & 0 & a_{2} \\a_{3} & 0 & 0\end{array}\right]\left[\begin{array}{l} x_{1} \\x_{2} \\x_{3}\end{array}\right]+\left[\begin{array}{l}0 \\0 \\1 \end{array}\right] u\\ y&=\left[\begin{array}{lll}1 & 0 & 0\end{array}\right]\left[\begin{array}{l} x_{1} \\x_{2} \\x_{3}\end{array}\right]\\ Q_{c}&=\left[B A B A^{2} B\right]\$\\ A B&=\left[\begin{array}{lll} 0 & a_{1} & 0 \\ 0 & 0 & a_{2} \\ a_{3} & 0 & 0 \end{array}\right]\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right]=\left[\begin{array}{l} 0 \\ a_{2} \\ 0 \end{array}\right]\\ A^{2} B&=A A B=\left[\begin{array}{lll} 0 & a_{1} & 0 \\ 0 & 0 & a_{2} \\ a_{3} & 0 & 0 \end{array}\right]\left[\begin{array}{l} 0 \\ a_{2} \\ 0 \end{array}\right]\\ A^{2} B&=\left[\begin{array}{c} a_{1} a_{2} \\ 0 \\ 0 \end{array}\right]\\ Q_{c}&=\left[\begin{array}{ccc}0 & 0 & a_{1} a_{2} \\ 0 & a_{2} & 0 \\ 1 & 0 & 0\end{array}\right] \end{aligned}

For system to be controllable,

\begin{array}{l} |Q_{i}| \neq 0 \\ \Rightarrow\left(0-a, a_{i}\right) \neq 0 \\ \Rightarrow a_{1} \neq 0 \\ a_{2} \neq 0 \end{array}

Q19GATE 2011MCQ

The block diagram of a system with one input it and two outputs $y_1 \; and \; y_2$ is given below. A state space model of the above system in terms of the state vector $\underline{x}$ and the output vector $\underline{y}= [y_{1} \; y_{2}]^{T}$ is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \frac{Y_{1}(s)}{U(s)} &=\frac{1}{s+2} \\ \frac{Y_{2}(s)}{U(s)} &=\frac{2}{s+2} \\ \frac{Y_{1}(s)}{x_{1}(s)} \frac{x_{1}(s)}{U(s)} &=\frac{1}{s+2} \\ \frac{r_{2}(s)}{x_{2}(s)} \frac{x_{2}(s)}{U(s)} &=\frac{2}{s+2} \end{aligned}

\begin{aligned} \frac{X_{1}(s)}{U(s)} &=\frac{1}{s+2} \text { and } \frac{Y_{1}(s)}{X_{1}(s)}=1 \\ \frac{x_{2}(s)}{U(s)} &=\frac{1}{s+2} \text { and } \frac{Y_{2}(s)}{U(s)}=2 \\ s X_{1}(s)+2 X_{1}(s) &=U(s) \\ \text{and} \quad Y_{1}(s) &=x_{1}(s) \\ s X_{2}(s)+2 X_{2}(s) &=U(s) \\ \text{and }\quad Y_{2}(s)&=2 X_{2}(s) \\ \dot{x}_{1}(t)+2 x_{1}(t)&=U(t) \\ \text{and }y_{1}(t)&=x_{1}(t) \\ \dot{x}_{2}(t)+2 x_{2}(t)&=U(t) \\ y_{2}(t)&=2 x_{2}(t) \end{aligned}

From Question,

\begin{aligned} y &=\left[y_{1} y_{2}\right]^{\top}=[12]^{\top}=\left[\begin{array}{l}1 \\2\end{array}\right] \\ \dot{x}_{1}(t) &=-2 x_{1}(t)+u(t) \\ \dot{x}_{2}(t) &=-2 x_{2}(t)+u(t) \\ \left[\begin{array}{c}\dot{x}_{1} \\\dot{x}_{2}\end{array}\right]&=\left[\begin{array}{cc}-2 & 0 \\-2 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2} \end{array}\right]+\left[\begin{array}{l}1 \\1\end{array}\right] u(t) \\ \dot{x} &=[-2] x+[1] u \end{aligned}

only option (B) is satisfied.

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