Q21GATE 2010MCQ

The signal flow graph of a system is shown below: The transfer function of the system is

📖 Explanation

Forward path gain,

\begin{aligned} P_{1} &=2\left(\frac{1}{s}\right)\left(\frac{1}{s}\right)(0.5)=\frac{1}{s^{2}} \\ P_{2} &=2\left(\frac{1}{s}\right)(1)(0.5)=\frac{1}{s} \\ \Delta_{1} &=1 \\ \Delta_{2} &=1 \\ \Delta &=1-\left\{-\frac{1}{s}-\frac{1}{s^{2}}\right\} \\ &=1+\frac{1}{s}+\frac{1}{s^{2}} \end{aligned}

Transfer funct ion of the system,

\begin{aligned} \frac{Y(s)}{U(s)} &=\frac{P_{1} \Delta_{1}+P_{2} \Delta_{2}}{\Delta} \\ &=-\frac{1}{1+\frac{s^{2}}{s}+\frac{1}{s}}=\frac{s+1}{s^{2}+s+1} \end{aligned}

Q23GATE 2008MCQ

A signal flow graph of a system is given below. The set of equalities that corresponds to this signal flow graph is

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📖 Explanation

By solving this we get

\frac{d}{d t}\left(\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right)=\left[\begin{array}{ccc}-\gamma & 0 & \beta \\ \gamma & 0 & \alpha \\ -\beta & 0 & -\alpha\end{array}\right]\left(\begin{array}{c}x_{1} \\ x_{2} \\ x_{3}\end{array}\right)+\left[\begin{array}{cc}0 & 1 \\ 0 & 0 \\ 1 & 0\end{array}\right]\left(\begin{array}{c}u_{1} \\ u_{2}\end{array}\right)

Q24GATE 2008MCQ

Consider the matrix

P=\begin{bmatrix} 0 & 1\\ -2&-3 \end{bmatrix}

. Thr value of $e^{P}$ is

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Q25GATE 2007MCQ

Consider a linear system whose state space representation is $\dot{x}(t)=Ax(t)$ . If the initial state vector of the system is

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

, then the system response is

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

If the initial state vector of the system changes to

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

, then the system response becomes

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

The system matrix A is

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📖 Explanation

Multiplication of the eigen value= determinant of the matrix $\therefore$ from options it seems determinant should be $\pm 2.$ Only option (d) satisfies as det = 2.

Q26GATE 2007MCQ

The state space representation of a separately excited DC servo motor dynamics is given as

\begin{bmatrix} \frac{d\omega }{dt}\\ \frac{di_{a}}{dt} \end{bmatrix}=\begin{bmatrix} -1 & 1\\ -1&-10 \end{bmatrix}\begin{bmatrix} \omega \\ i_{a} \end{bmatrix}+\begin{bmatrix} 0\\ 10 \end{bmatrix}u

where $\omega$ is the speed of the motor, $i_{a}$ is the armature current and u is the armature voltage. The transfer function $\frac{\omega (s)}{U(s)}$ of the motor is

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📖 Explanation

\begin{aligned} \left[\begin{array}{c}\frac{d \omega}{d t} \\\frac{d i_{a}}{d t}\end{array}\right]&=\left[\begin{array}{cc} -1 & 1 \\ -1 & -10 \end{array}\right]\left[\begin{array}{l} \omega \\ i_{a} \end{array}\right]+\left[\begin{array}{c} 0 \\ 10 \end{array}\right] u\\ \Rightarrow \frac{d \omega}{d t}&=-\omega+i_{\mathrm{a}} &\ldots(i)\\ \Rightarrow \frac{d i_{\mathrm{a}}}{d t}&=-\omega-10 i_{\mathrm{a}}+10 u &\ldots(ii) \end{aligned}

Taking Laplace transform (i) & (ii)

\begin{array}{l} \Rightarrow s \omega(s)=-\omega(s)+I_{a}(s) \\ \Rightarrow(s+1) \omega(s)=I_{a}(s) \quad\ldots(iii)\\ \Rightarrow s I_{e}(s)=-\omega(s)-10 I_{a}(s)+10 U(s) \\ \Rightarrow \omega(s)=(-10-s) I_{a}(s)+10 U(s) \\ =(-10-s)(s+1) \omega(s)+10 U(s) \\ =-\left[s^{2}+11 s+10\right]\$ \omega(s)+10 U(s) \\ \Rightarrow\left[s^{2}+11 s+11\right]\$ \omega(s)=10 U(s) \\ \Rightarrow \frac{\omega(s)}{U(s)}=\frac{10}{\left(s^{2}+11 s+11\right)} \end{array}

Q27GATE 2007MCQ

Consider a linear system whose state space representation is $x(t)=Ax(t)$ . If the initial state vector of the system is

\dot{x}(0)=\begin{bmatrix} 1\\ -2 \end{bmatrix}

, then the system response is

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

If the initial state vector of the system changes to

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

, then the system response becomes

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

. The eigenvalue and eigenvector pairs ( $\lambda _{i},v_{i}$ ) for the system are

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Q28GATE 2006MCQ

A linear system is described by the following state equation

\dot{X}(t)=AX(t)+BU(t),A=\begin{bmatrix} 0 & 1\\ -1&0 \end{bmatrix}

The state transition matrix of the system is

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📖 Explanation

\begin{array}{l} \phi(t)=L^{-1}[S I-A]^{-1} \\ =L^{-1}\left[\left[\begin{array}{ll} s & 0 \\ 0 & s \end{array}\right]-\left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right]\right]^{-1}=L^{-1}\left[\begin{array}{cc} s & -1 \\ 1 & s \end{array}\right]^{-1} \\ =L^{-1}\left[\begin{array}{cc} \frac{s}{s^{2}+1} & \frac{1}{s^{2}+1} \\ \frac{-1}{s^{2}+1} & \frac{s}{s^{2}+1} \end{array}\right]\$\left[\begin{array}{c} D=s^{2}+1 \\ \end{array}\right]\$ \\ =\left[\begin{array}{cc} \cos t & \sin t \\ -\sin t & \cos t \end{array}\right] \end{array}

Q29GATE 2005MCQ

A linear system is equivalently represented by two sets of state equations : $\dot{X}=AX+BU \; and \; W=CW+DU$ The eigenvalues of the representations are also computed as $[\lambda]$ and $[\mu]$ . Which one of the following statements is true ?

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📖 Explanation

Eigen vlues of $A=[\lambda]$ Eigen values of $W=[\mu]$ The eigen values of a system are always unique. $\mathrm{So},[\lambda]=[\mu]$ But a system can be represented by different state models having different set of state variables.

\begin{array}{l} X=W \\ X \neq W \end{array}

Both are possible conditions.

Q30GATE 2004MCQ

The state variable equations of a system are : 1. $\dot{x_{1}}=-3x_{1}-x_{2}+u$ 2. $\dot{x_{2}}=2x_{1}$ $y=x_{1}+u$ The system is

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📖 Explanation

\begin{aligned} \left[\begin{array}{c}\dot{x}_{1} \\\dot{x}_{2}\end{array}\right]&=\left[\begin{array}{cc}-3 & -1 \\2 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]+\left[\begin{array}{l}1 \\0\end{array}\right] u \\ y&=\left[\begin{array}{ll}1 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]+\left[\begin{array}{l}1 \\0\end{array}\right] u \\ Q_{C}&=\left[\begin{array}{ll}B & A B\end{array}\right]\\ \end{aligned}

\begin{aligned} A B&=\left[\begin{array}{c}-3 \\ 2\end{array}\right] \\ \therefore \quad a_{c}&=\left[\begin{array}{cc}1 & -3 \\ 0 & 2\end{array}\right] \neq 0 \\ \therefore \text{Controllable } \\ Q_{0}&=\left[C^{\top} \quad A^{\top} C^{\prime}\right]\$ \\ C^{T}&=\left[\begin{array}{l}1 \\ 0\end{array}\right], \quad A^{T}&=\left[\begin{array}{ll}-3 & 2 \\ -1 & 0\end{array}\right] \\ A^{\top} C^{T}&=\left[\begin{array}{l}-3 \\ -1\end{array}\right] \\ \therefore \quad Q_{0}&=\left[\begin{array}{ll}1 & -3 \\ 0 & -1\end{array}\right] \neq 0 \\ \therefore \text{ Observable} \end{aligned}

Q31GATE 2004MCQ

Given

A=\begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix}

, the state transition matrix $e^{At}$ is given by

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📖 Explanation

\begin{aligned} [s I-A] &=\left[\begin{array}{ll}s & 0 \\0 & s\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right] \\ s I-A] & =\left[\begin{array}{cc}s-1 & 0 \\0 & s-1\end{array}\right] \\ e^{A t} & =[s I-A]^{-1}\\ &=\left[\begin{array}{cc}-\frac{1}{s-1} & 0 \\ 0 & \frac{1}{s-1}\end{array}\right]=\left[\begin{array}{ll}e^{t} & 0 \\ 0 & e^{t}\end{array}\right]\$ \end{aligned}

Q32GATE 2003MCQ

The zero-input response of a system given by the state-space equation

\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}

and

\begin{bmatrix} X_1(0)\\X_2(0) \end{bmatrix}=\begin{bmatrix} 1\\0 \end{bmatrix}

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📖 Explanation

\begin{aligned} (s I-A) &=\left[\begin{array}{ll} s & 0 \\ 0 & s \end{array}\right]-\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right] \\ &=\left[\begin{array}{cc} s-1 & 0 \\ -1 & s-1 \end{array}\right] \\ (s I-A)^{-1} &=\frac{\left[\begin{array}{cc} (s-1) & 0 \\ +1 & (s-1) \end{array}\right]}{(s-1)^{2}}\\ &=\left[\begin{array}{cc}\frac{1}{s-1} & 0 \\ \frac{+1}{(s-1)^{2}} & \frac{1}{s-1}\end{array}\right]\$ \\ L^{-1}[s I-A]^{-1}&=e^{A t}=\left[\begin{array}{cc}e^{t} & 0 \\ t e^{t} & e^{t}\end{array}\right]\$ \\ &=\left[\begin{array}{cc}e^{t} & 0 \\ t e^{t} & e^{t}\end{array}\right]\$\left[\begin{array}{l}1 \\ 0\end{array}\right]=\left[\begin{array}{c}e^{t} \\ t e^{t}\end{array}\right]\$ \end{aligned}

Q33GATE 2002MCQ

The transfer function Y(s)U(s) of a system described by the state equations x(t) = -2x(t) + 2u(t) and y(t) = 0.5x(t) is

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📖 Explanation

\begin{aligned} \dot{x}(t)=-2 x(t)+2 u(t) &\ldots(i)\\ y(t)=0.5 x(t)&\ldots(ii) \end{aligned}

From (i), Taking Laplace transform of (i)

\begin{aligned} s X(s)&=-2 X(s)+2 U(s)\\ X(s)[s+2]&=2 U(s)\\ \Rightarrow \quad X(s)&=\frac{2 U(s)}{(s+2)} \end{aligned}

Taking Laplace transform of (ii)

\begin{aligned} Y(s)&=0.5 X(s) \\ Y(s)&=\frac{0.5 \times 2 U(s)}{s+2} \\ \therefore \quad \frac{Y(s)}{U(s)}&=\frac{1}{(s+2)} \end{aligned}