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

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.

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.

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

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.

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

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

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

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

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

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

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.
• 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.

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

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

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

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.

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

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

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

⇒ 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

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 

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

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

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

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.