State Variables MCQ Quiz - Objective Question with Answer for State Variables - Download Free PDF

Last updated on Mar 11, 2025

Latest State Variables MCQ Objective Questions

State Variables Question 1:

The minimum number of states required to describe the network shown in the figure is-

control systems7 1

  1. 3
  2. 2
  3. 1
  4. 0
  5. 4

Answer (Detailed Solution Below)

Option 2 : 2

State Variables Question 1 Detailed Solution

Concept:

  • States can be considered as variables that carry sufficient information about the history of the system.
  • A purely resistive network has no states at all as it has no storage property.
  • Only a network with energy storage elements or reactive elements will have states as they store the information about the past states.

 

Number of the states required = Order of the network

Also, the Order of the network = Number of energy storage elements

Since the given circuit has two energy storage elements (L and C), the minimum number of required states will be 2.

State Variables Question 2:

The minimum number of sates required to describe the network shown in the figure is-

control systems7 1

  1. 3
  2. 2
  3. 1
  4. 0

Answer (Detailed Solution Below)

Option 2 : 2

State Variables Question 2 Detailed Solution

Concept:

  • States can be considered as variables that carry sufficient information about the history of the system.
  • A purely resistive network has no states at all as it has no storage property.
  • Only a network with energy storage elements or reactive elements will have states as they store the information about the past states.

 

Number of the states required = Order of the network

Also, the Order of the network = Number of energy storage elements

Since the given circuit has two energy storage elements (L and C), the minimum number of required states will be 2.

State Variables Question 3:

Number of state variables of discrete-time system, described by

\(y[n] - \frac{3}{4}y[n - 1] + \frac{1}{8}y[n - 2] = x[n]\) is

  1. 2
  2. 3
  3. 4
  4. 1

Answer (Detailed Solution Below)

Option 1 : 2

State Variables Question 3 Detailed Solution

Concept:

Number of state variables = order of the system

= number of independent energy storage elements

= number of poles of the system.

Analysis:

\(y[n] - \frac{3}{4}y[n - 1] + \frac{1}{8}y[n - 2] = x[n]\)

Taking Z transform we get:

Y(Z) - 0.75 Z-1 Y(Z) + 0.125 Z-2 Y(Z) = X(Z)

Y(Z) [ 1 - 0.75 Z-1 + 0.125 Z-2 ] = X(Z)

\({Y(Z) \over X(Z)} = \frac{1}{0.125 Z^{-2} - 0.75 Z^{-1} + 1}\)

The number of poles = 2

Hence, number of state variables = 2

State Variables Question 4:

Consider the network shown below

F1 Neha.B 05-01-21 Savita D1

The state variables are:

  1. i1, i2, iR
  2. i3, i2, V1
  3. Vi, V1, V2
  4. V1, i3

Answer (Detailed Solution Below)

Option 4 : V1, i3

State Variables Question 4 Detailed Solution

Concept:

Number of state variables = order of the system

= number of independent energy storage elements

= number of poles of the system.

The state variable for an inductor is the current through the inductor, while that for a capacitor is the voltage across the capacitor.

As the resistor is not a storage element, we cannot take the current passing through a resistor as a state variable.

Analysis:

The voltage across the capacitor (1 F) = V1

The current across the inductor (1 H) = i3

The state-variables are V1 and i3 only.

State Variables Question 5:

The transfer function Y(s) / U(s) of a system described by the state equations

ẋ(t) = - 2x(t) + u(t) and

y(t) = 0.5x(t), is 

  1. \(\frac{0.5}{(s-2) }\)
  2. \(\frac{1}{(s-2) }\)
  3. \(\frac{0.5}{(s+2) }\)
  4. \(\frac{1}{(s+2) }\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{0.5}{(s+2) }\)

State Variables Question 5 Detailed Solution

Concept:

A standard form of State-space matrix:

\(\dot X = AX + BU\)

It is called a State equation or Dynamic equation

Y = CX + DU 

It is known as the output equation

\(\dot X\): Differential State Vector

Y: Output vector

U: Input vector

A: State matrix

B: Input matrix

C: Output matrix

D: Transition matrix

Transfer Function:

For the state model defined by we have to write the Transfer function for the analysis.

It is given by

\(TF = C{\left[ {sI - A} \right]^{ - 1}}B + D\)

 

The characteristic equation is defined by

|sI – A| = 0

Application:

The state equations in the question can be easily solved as follows:

ẋ(t) = - 2x(t) + 2u(t)

Convert the above equation into Laplace domain, we get

sX(s) = - 2X(s) + U(s)

(s+2) X(s) = U(s)

\(X(s) = \frac{U(s)}{(s+2)}\)    ---------    (1)

y(t) = 0.5x(t)

Y(s) = 0.5 X(s)            ----------    (2)

From (1)  and (2), we get

\(Y(s) = 0.5 \frac{U(s)}{(s+2)}\)

\(\frac{Y(s)}{U(s)} = \frac{0.5}{(s+2)}\)

Top State Variables MCQ Objective Questions

Number of state variables of discrete-time system, described by

\(y[n] - \frac{3}{4}y[n - 1] + \frac{1}{8}y[n - 2] = x[n]\) is

  1. 2
  2. 3
  3. 4
  4. 1

Answer (Detailed Solution Below)

Option 1 : 2

State Variables Question 6 Detailed Solution

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Concept:

Number of state variables = order of the system

= number of independent energy storage elements

= number of poles of the system.

Analysis:

\(y[n] - \frac{3}{4}y[n - 1] + \frac{1}{8}y[n - 2] = x[n]\)

Taking Z transform we get:

Y(Z) - 0.75 Z-1 Y(Z) + 0.125 Z-2 Y(Z) = X(Z)

Y(Z) [ 1 - 0.75 Z-1 + 0.125 Z-2 ] = X(Z)

\({Y(Z) \over X(Z)} = \frac{1}{0.125 Z^{-2} - 0.75 Z^{-1} + 1}\)

The number of poles = 2

Hence, number of state variables = 2

Consider the following state-space representation of a linear time-invariant system.

\(\dot x\left( t \right) = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&2 \end{array}} \right]x\left( t \right),y\left( t \right) = {c^T}x\left( t \right),c = \left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right]and\;x\left( 0 \right) = \left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right]\)

The value  of \(y\left( t \right)\) for \(t = {\log _{\rm{e}}}2\) is __________.

Answer (Detailed Solution Below) 5.9 - 6.1

State Variables Question 7 Detailed Solution

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\(\dot x\left( t \right) = Ax\left( t \right)\)

Taking Laplace transform

\(sX\left( s \right)\;-\;{x_0}\; = \;Ax\left( t \right)\)

\(x\left( t \right)\; = \;\left\{ {{L^{ - 1}}\;{{\left( {sI\;-\;A} \right)}^{ - 1}}\;x\left( 0 \right)} \right\}\)

\(= \left[ {\begin{array}{*{20}{c}} {{e^t}}\\ {{e^{2t}}\;} \end{array}} \right]\)

\(y\left( t \right)\; = \;{c^T}\;x\left( t \right)\)

\(= \;{e^t}\; + \;{e^{2t}}\)

at \(t = {\log _{\rm{e}}}2,\;y\left( t \right) = 2 + 4 = 6\)

Read the following statements regarding state variable analysis

(i) A purely resistive network has no states at all.

(ii) A purely resistive network has more states than a reactive network.

(iii) Only a reactive network can have stated.

(iv) A network with energy storing elements has no states at all.

Which of the following statements are correct:

  1. (i) and (iii)
  2. (i) and (iv)
  3. (ii) and (iii)
  4. (ii) and (iv)

Answer (Detailed Solution Below)

Option 1 : (i) and (iii)

State Variables Question 8 Detailed Solution

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The response of a network consists of two parts:

1) Zero input response

2) Zero state response

  • The response of a given input depends on the zero input response.
  • The zero input response in an RLC network is completely determined once the initial inductor currents and capacitor voltages are known.
  • The Initial capacitor voltages and inductor currents (initial conditions) are termed as the initial states of the system.
  • The capacitor voltage and the inductor currents at a specified time are known as the state variables of the network.
  • States can be considered as variables that carry sufficient information about the history of the system.
  • purely resistive network has no states at all as it has no storage property.
  • Only a network with energy storage elements or reactive elements will have states as they store the information about the past states.

The transfer function Y(s) / U(s) of a system described by the state equations

ẋ(t) = - 2x(t) + u(t) and

y(t) = 0.5x(t), is 

  1. \(\frac{0.5}{(s-2) }\)
  2. \(\frac{1}{(s-2) }\)
  3. \(\frac{0.5}{(s+2) }\)
  4. \(\frac{1}{(s+2) }\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{0.5}{(s+2) }\)

State Variables Question 9 Detailed Solution

Download Solution PDF

Concept:

A standard form of State-space matrix:

\(\dot X = AX + BU\)

It is called a State equation or Dynamic equation

Y = CX + DU 

It is known as the output equation

\(\dot X\): Differential State Vector

Y: Output vector

U: Input vector

A: State matrix

B: Input matrix

C: Output matrix

D: Transition matrix

Transfer Function:

For the state model defined by we have to write the Transfer function for the analysis.

It is given by

\(TF = C{\left[ {sI - A} \right]^{ - 1}}B + D\)

 

The characteristic equation is defined by

|sI – A| = 0

Application:

The state equations in the question can be easily solved as follows:

ẋ(t) = - 2x(t) + 2u(t)

Convert the above equation into Laplace domain, we get

sX(s) = - 2X(s) + U(s)

(s+2) X(s) = U(s)

\(X(s) = \frac{U(s)}{(s+2)}\)    ---------    (1)

y(t) = 0.5x(t)

Y(s) = 0.5 X(s)            ----------    (2)

From (1)  and (2), we get

\(Y(s) = 0.5 \frac{U(s)}{(s+2)}\)

\(\frac{Y(s)}{U(s)} = \frac{0.5}{(s+2)}\)

System transformation function H(z) for a discrete-time LTI system expressed in state variable form with zero initial conditions is

  1. c (zI - A)-1 b + d
  2. c (zI - A)-1
  3. (zI - A)-1 z
  4. (zI - A)-1

Answer (Detailed Solution Below)

Option 1 : c (zI - A)-1 b + d

State Variables Question 10 Detailed Solution

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Explanation:

Transfer function which is the ratio of Laplace output to the Laplace input when the initial conditions are zero in discrete is the same as continuous but in the z-domain.

Transfer function from state model is given as:

\(\dot{X}=AX+BU\) -----(1)

Y = CX + DU -----(2)

Where,

Y is single output 

X is single input

X(0) = 0, as initial condition is zero.

Taking Z transform of equation (1):

zX(z) = AX(z) + BU(z)

[zI - A] X(z) = BU(z)

Taking Z transform of equation (2):

Y(z) = CX(z) + DU(z)

putting value of X(z) from equation (1) in equation (2) we get:

\(T(z)=\frac{Y(z)}{U(z)}\)

T(z) = c[zI - A]-1b + d

So, option (1) is correct answer.

State Variables Question 11:

A second-order system is governed by \(\frac{{{d^2}y}}{{d{t^2}}} + 5\frac{{dy}}{{dt}} + 6y = u\left( t \right)\)

The number of state variables required for representing it in the state space representation is ______

  1. 3
  2. 6
  3. 2
  4. 1

Answer (Detailed Solution Below)

Option 3 : 2

State Variables Question 11 Detailed Solution

Number of state variables = order of the system

= number of independent energy storage elements

= number of poles of the system.

Given system is governed by

\(\frac{{{d^2}y}}{{d{t^2}}} + s\frac{{dy}}{{dt}} + 6y = u\left( t \right)\)

As the system is of second order, the number of state variables required are 2.

State Variables Question 12:

A certain linear time invariant system has the state and the output equations given below

\(\left[ {\begin{array}{*{20}{c}} {{{\dot x}_1}}\\ {{{\dot x}_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}\\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0\\ 1 \end{array}} \right]u\)

\(y = \left[ {1\;\;1} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right]\)

If \(\left[ {\begin{array}{*{20}{c}} {{x_1}\left( 0 \right)}\\ {{x_2}\left( 0 \right)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1\\ { - 1} \end{array}} \right]\)and u(0) = -1, then \(\frac{{dy}}{{dt}}\;at\;t = 0\) is ______

Answer (Detailed Solution Below) 0

State Variables Question 12 Detailed Solution

\(y = \left[ {1\;\;1} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right]\)

y(t) = x1(t) + x2(t)

By differentiating the above equation

\(\frac{d}{{dt}}y\left( t \right) = \frac{d}{{dt}}{x_1}\left( t \right) + \frac{d}{{dt}}{x_2}\left( t \right)\)


\(\left[ {\begin{array}{*{20}{c}} {{{\dot x}_1}\left( t \right)}\\ {{{\dot x}_2}\left( t \right)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}\\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0\\ 1 \end{array}} \right]u\)

At t = 0

\(\left[ {\begin{array}{*{20}{c}} {{{\dot x}_1}\left( 0 \right)}\\ {{{\dot x}_2}\left( 0 \right)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}\\ 0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ { - 1} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0\\ 1 \end{array}} \right]\left[ { - 1} \right]\)
\(= \left[ {\begin{array}{*{20}{c}} 2\\ { - 1} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0\\ { - 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2\\ { - 2} \end{array}} \right]\)

Now, \(\frac{d}{{dt}}y\left( 0 \right) = \frac{d}{{dt}}{x_1}\left( 0 \right) + \frac{d}{{dt}}{x_2}\left( 0 \right)\)

= ẋ1(0) + ẋ2(0)

= 2 – 2 = 0 

State Variables Question 13:

Number of state variables of discrete-time system, described by

\(y[n] - \frac{3}{4}y[n - 1] + \frac{1}{8}y[n - 2] = x[n]\) is

  1. 2
  2. 3
  3. 4
  4. 1

Answer (Detailed Solution Below)

Option 1 : 2

State Variables Question 13 Detailed Solution

Concept:

Number of state variables = order of the system

= number of independent energy storage elements

= number of poles of the system.

Analysis:

\(y[n] - \frac{3}{4}y[n - 1] + \frac{1}{8}y[n - 2] = x[n]\)

Taking Z transform we get:

Y(Z) - 0.75 Z-1 Y(Z) + 0.125 Z-2 Y(Z) = X(Z)

Y(Z) [ 1 - 0.75 Z-1 + 0.125 Z-2 ] = X(Z)

\({Y(Z) \over X(Z)} = \frac{1}{0.125 Z^{-2} - 0.75 Z^{-1} + 1}\)

The number of poles = 2

Hence, number of state variables = 2

State Variables Question 14:

Consider the following state-space representation of a linear time-invariant system.

\(\dot x\left( t \right) = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&2 \end{array}} \right]x\left( t \right),y\left( t \right) = {c^T}x\left( t \right),c = \left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right]and\;x\left( 0 \right) = \left[ {\begin{array}{*{20}{c}} 1\\ 1 \end{array}} \right]\)

The value  of \(y\left( t \right)\) for \(t = {\log _{\rm{e}}}2\) is __________.

Answer (Detailed Solution Below) 5.9 - 6.1

State Variables Question 14 Detailed Solution

\(\dot x\left( t \right) = Ax\left( t \right)\)

Taking Laplace transform

\(sX\left( s \right)\;-\;{x_0}\; = \;Ax\left( t \right)\)

\(x\left( t \right)\; = \;\left\{ {{L^{ - 1}}\;{{\left( {sI\;-\;A} \right)}^{ - 1}}\;x\left( 0 \right)} \right\}\)

\(= \left[ {\begin{array}{*{20}{c}} {{e^t}}\\ {{e^{2t}}\;} \end{array}} \right]\)

\(y\left( t \right)\; = \;{c^T}\;x\left( t \right)\)

\(= \;{e^t}\; + \;{e^{2t}}\)

at \(t = {\log _{\rm{e}}}2,\;y\left( t \right) = 2 + 4 = 6\)

State Variables Question 15:

The minimum number of states required to describe the network shown in the figure is-

control systems7 1

  1. 3
  2. 2
  3. 1
  4. 0
  5. 4

Answer (Detailed Solution Below)

Option 2 : 2

State Variables Question 15 Detailed Solution

Concept:

  • States can be considered as variables that carry sufficient information about the history of the system.
  • A purely resistive network has no states at all as it has no storage property.
  • Only a network with energy storage elements or reactive elements will have states as they store the information about the past states.

 

Number of the states required = Order of the network

Also, the Order of the network = Number of energy storage elements

Since the given circuit has two energy storage elements (L and C), the minimum number of required states will be 2.

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