Q21GATE 2003MCQ

The following equation defines a separately excited dc motor in the form of a differential equation $\frac{d^{2}\omega }{dt}+\frac{B}{J}\frac{d\omega }{dt}+\frac{K^{2}}{LJ}\omega =\frac{K}{LJ}V_{a}$ The above equation may be organized in the state-space form as follows

\begin{bmatrix} \frac{d^2\omega}{dt^2}\\ \frac{d\omega}{dt} \end{bmatrix}=P\begin{bmatrix} \\frac{d\omega}{dt}\\ \omega \end{bmatrix}+QV_a

where the P matrix is given by

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📖 Explanation

Let, $x_1=\frac{d\omega }{dt}$ and $x_2=\omega$ $\;\;\; \dot{x_2}=x_1$ $\frac{d^2\omega }{dt^2}+\frac{B}{J}\frac{d\omega }{dt}+\frac{k^2}{LJ}\omega =\frac{k}{LJ}V_a$ $\frac{d^2\omega }{dt^2}=-\frac{B}{J}\frac{d\omega }{dt}-\frac{k^2}{LJ}\omega + \frac{k}{LJ}V_a$ $\Rightarrow \;\; \dot{x_1}=-\frac{B}{J}x_1-\frac{k^2}{LJ}x_2+\frac{k}{LJ}V_a$ $\;\;\dot{x_2}=x_1$

\begin{bmatrix} \dot{x_1}\\ \dot{x_2} \end{bmatrix}=P\begin{bmatrix} x_1\\ x_2 \end{bmatrix}+QV_a

where,

P=\begin{bmatrix} -\frac{B}{J} & -\frac{k^2}{LJ}\\ 1 & 0 \end{bmatrix}

and

Q=\begin{bmatrix} \frac{k}{LJ}\\ 0 \end{bmatrix}

Q22GATE 2003MCQ

A second order system starts with an initial condition of

\begin{bmatrix} 2\\ 3 \end{bmatrix}

without any external input. The state transition matrix for the system is given by

\begin{bmatrix} e^{-2t} & 0\\ 0& e^{-t} \end{bmatrix}

. The state of the system at the end of 1 second is given by

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📖 Explanation

State transition matrix,

\phi (t)=\begin{bmatrix} e^{-2t} & 0\\ 0& e^{-t} \end{bmatrix}

Initial conditions,

x(0)=\begin{bmatrix} 2\\ 3 \end{bmatrix}

Zero input response is given by $x(t)=\phi (t)x(0)$

\;\;=\begin{bmatrix} e^{-2t} & 0\\ 0& e^{-t} \end{bmatrix}\begin{bmatrix} 2\\ 3 \end{bmatrix} \;\;=\begin{bmatrix} 2e^{-2t}\\ 3e^{-t} \end{bmatrix}

State of the system at t=1s

X(t)|_{t=1}=\begin{bmatrix} 2e^{-2}\\ 3e^{-1} \end{bmatrix}=\begin{bmatrix} 0.271\\ 1.100 \end{bmatrix}

Q23GATE 2002MCQ

For the system

\dot{x}=\begin{bmatrix} 2 & 3\\ 0&5 \end{bmatrix} X +\begin{bmatrix} 1\\ 0 \end{bmatrix}u

, which of the following statements is true?

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📖 Explanation

|Q_C|=\begin{bmatrix} 1& 2\\ 0 & 0 \end{bmatrix}=0

$\Rightarrow$ Uncontrollable. Characteristic equation: $|sI-A|=0$

\begin{bmatrix} s-2& -3\\ 0 & s-5 \end{bmatrix}=0

$s^{2-}7s+13=0$ $\Rightarrow$ eigen values, $s =3.5 \pm j0.866$ i.e. roots lies on right side of s-plane . $\Rightarrow$ unstable.

Q24GATE 2002MCQ

For the system

\dot{x}=\begin{bmatrix} 2 & 0\\ 0&4 \end{bmatrix}+\begin{bmatrix} 1\\ 1 \end{bmatrix} u; y=\begin{bmatrix} 4 & 0 \end{bmatrix}X

, with u as unit impulse and with zero initial state, the output, y, becomes

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📖 Explanation

$x(t)=e^{At}X(0)+\int_{0}^{t}e^{A(t-\tau) }BU(\tau )d\tau$ $\because \;\; X(0)=0$ (as given)

\therefore \;\; x(t)=0+\int_{0}^{t}\begin{bmatrix} e^{2(t-\tau )} &0 \\ 0 & e^{4(t-\tau )} \end{bmatrix}\begin{bmatrix} 1\\ 1 \end{bmatrix}\delta (t)dt

\;\;=\int_{0}^{t}\begin{bmatrix} e^{2(t-\tau )}\\ e^{2(t-\tau )} \end{bmatrix}\delta (t)dt

\;\;=\begin{bmatrix} e^{2t}\\ e^{4t} \end{bmatrix}

Output

y(t)=\begin{bmatrix} 4 &0 \end{bmatrix}x(t)

\;\;=\begin{bmatrix} 4 &0 \end{bmatrix}\begin{bmatrix} e^{2t}\\ e^{4t} \end{bmatrix}

Hence, $y(t)=4e^{2t}$

Q25GATE 2002MCQ

The state transition matrix for the system $\dot{X}=AX$ with initial state X(0) is

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📖 Explanation

$e^{At}=L^{-1}[sI-A]^{-1}$

Q26GATE 2001MCQ

Given the homogeneous state-space equaion

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

the steady state value $x_{ss}=\lim_{t\rightarrow \infty }x(t)$ , given the initial state value of $x\left ( t \right )=\left[ 10-10 \right]^{T}$ , is

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📖 Explanation

[sI-A]^{-1}=\begin{bmatrix} s+3 &-1 \\ 0 & s+2 \end{bmatrix}^{-1} =\begin{bmatrix} \frac{1}{s+3} &\frac{1}{(s+2)(s+3)} \\ 0& \frac{1}{s+2} \end{bmatrix}

e^{At}=L^{-1}[sI-A]^{-1}=\begin{bmatrix} e^{-3t} & e^{-2t}-e^{-3t}\\ 0 & e^{-2t} \end{bmatrix}

x(t)=e^{At}X(0) =\begin{bmatrix} e^{-3t} & e^{-2t}-e^{-3t}\\ 0 & e^{-2t} \end{bmatrix}\begin{bmatrix} 10\\ -10 \end{bmatrix}

\therefore \;\; x(t)=\begin{bmatrix} 20e^{-3t}-10e^{-2t}\\ -10e^{-2t} \end{bmatrix}

\therefore \;\; X_{ss}=\lim_{t \rightarrow \infty }\begin{bmatrix} 0\\ 0 \end{bmatrix}