Control Systems Test 8

Result

/

Score
-

Rank

Time Taken: -

Question 1/10
1 / -0

A second order system is governed by $\frac{d^{2} y}{d t^{2}} + s \frac{d y}{d t} + 6 y = u (t)$
The number of state variables required for representing it in the state space representation is ______

Correct

Wrong

Skipped
A
2-2

Correct

Wrong

Skipped
B
24-21

Correct

Wrong

Skipped
C
27-26

Correct

Wrong

Skipped
D
29-22

Solutions

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
$\frac{d^{2} y}{d t^{2}} + s \frac{d y}{d t} + 6 y = u (t)$
As the system is of second order, the number of state variables required are 2.
Question 2/10
1 / -0

Identify the matrix that can be a state transition matrix.

Correct

Wrong

Skipped
A
$ϕ = [\begin{matrix} e^{- t} & 0 \\ 1 & e^{- 2 t} \end{matrix}]$

Correct

Wrong

Skipped
B
$ϕ = [\begin{matrix} e^{t} & t \\ 1 & 2 e^{- t} \end{matrix}]$

Correct

Wrong

Skipped
C
$ϕ = [\begin{matrix} e^{t} + e^{- t} & 0 \\ 0 & 2 e^{- t} \end{matrix}]$

Correct

Wrong

Skipped
D
$ϕ = [\begin{matrix} e^{- t} & 0 \\ 0 & e^{- 2 t} \end{matrix}]$

Solutions

State transition matrix, ϕ(t) = e^At
From the properties of state transition matrix,
$ϕ (0) = e^{A (0)} = I$
1. $\begin{matrix} ϕ (t) = [\begin{matrix} e^{- t} & 0 \\ 1 & e^{- 2 t} \end{matrix}] \\ ϕ (o) = [\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix}] \neq I \end{matrix}$
2. $\begin{matrix} ϕ (t) = [\begin{matrix} e^{t} & t \\ 1 & 2 e^{- t} \end{matrix}] \\ ϕ (o) = [\begin{matrix} 1 & 0 \\ 1 & 2 \end{matrix}] \neq I \end{matrix}$
3. $\begin{matrix} ϕ (t) = [\begin{matrix} e^{t} + e^{- t} & 0 \\ 0 & 2 e^{- t} \end{matrix}] \\ ϕ (o) = [\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}] \neq I \end{matrix}$
4. $\begin{matrix} ϕ (t) = [\begin{matrix} e^{t} & 0 \\ 0 & e^{- 2 t} \end{matrix}] \\ ϕ (o) = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = I \end{matrix}$
Hence option (d) can be state transition matrix.
Question 3/10
1 / -0

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

Correct

Wrong

Skipped
A
2-2

Correct

Wrong

Skipped
B
24-21

Correct

Wrong

Skipped
C
27-26

Correct

Wrong

Skipped
D
29-22

Solutions

The minimum number of states required
= number of energy storage elements in the circuit
= order of the circuit
Here in the given question, two energy storage elements are present
Question 4/10
1 / -0

The vector matrix differential equation of a system is given by $\dot{x} = [\begin{matrix} 0 & 1 \\ - 2 & - 3 \end{matrix}] x$
The state transition matrix of the system is-

Correct

Wrong

Skipped
A
$[\begin{matrix} - e^{- t} + 2 e^{- 2 t} & 2 e^{- t} + 2 e^{- 2 t} \\ 2 e^{- 2 t} - 2 e^{- t} & e^{- t} - e^{- 2 t} \end{matrix}]$

Correct

Wrong

Skipped
B
$[\begin{matrix} - e^{- t} + 2 e^{- 2 t} & e^{- t} - e^{- 2 t} \\ 2 e^{- 2 t} + 2 e^{- t} & 2 e^{- 2 t} - e^{- t} \end{matrix}]$

Correct

Wrong

Skipped
C
$[\begin{matrix} 2 e^{- t} - e^{- 2 t} & e^{- t} - e^{- 2 t} \\ 2 e^{- 2 t} - 2 e^{- t} & 2 e^{- 2 t} - e^{- t} \end{matrix}]$

Correct

Wrong

Skipped
D
$[\begin{matrix} - 2 e^{- t} + 2 e^{- 2 t} & e^{- t} - e^{- 2 t} \\ - e^{- t} + 2 e^{- 2 t} & 2 e^{- t} - e^{- 2 t} \end{matrix}]$

Solutions

The state transition matrix $ϕ (t) = L^{- 1} [{(S I - A)}^{- 1}]$
Given, $A = [\begin{matrix} 0 & 1 \\ - 2 & - 3 \end{matrix}]$
$S I - A = [\begin{matrix} s & 0 \\ 0 & s \end{matrix}] - [\begin{matrix} 0 & 1 \\ - 2 & - 3 \end{matrix}]$
$= [\begin{matrix} s & - 1 \\ 2 & s + 3 \end{matrix}]$
|SI - A| = s(s + 3) + 2
= s² + 3s + 2
${(S I - A)}^{- 1} = \frac{1}{s^{2} + 3 s + 2} [\begin{matrix} s + 3 & 1 \\ - 2 & s \end{matrix}]$
$L^{- 1} [{(S I - A)}^{- 1}] = L^{- 1} [\begin{matrix} \frac{s + 3}{(s + 2) (s + 1)} & \frac{1}{(s + 2) (s + 1)} \\ \frac{- 2}{(s + 2) (s + 1)} & \frac{s}{(s + 2) (s + 1)} \end{matrix}]$
$= L^{- 1} [\begin{matrix} \frac{2}{s + 1} - \frac{1}{s + 2} & \frac{1}{s + 1} - \frac{1}{s + 2} \\ \frac{2}{s + 2} - \frac{2}{s + 1} & \frac{2}{s + 2} - \frac{1}{s + 1} \end{matrix}]$
$= [\begin{matrix} 2 e^{- t} - e^{- 2 t} & e^{- t} - e^{- 2 t} \\ 2 e^{- 2 t} - 2 e^{- t} & 2 e^{- 2 t} - e^{- t} \end{matrix}]$
Question 5/10
1 / -0

The state diagram is given below.
The state space representation for the above state diagram is
ẋ(t) = A x(t) + B u(t)
y(t) = C x(t) + D u(t)
Then, which of the following is / are true?

Correct

Wrong

Skipped
A
$A = [\begin{matrix} 0 & 1 \\ - 2 & - 3 \end{matrix}]$

Correct

Wrong

Skipped
B
$B^{T} = [\begin{matrix} 0 \\ 1 \end{matrix}]$

Correct

Wrong

Skipped
C
$C^{T} = [\begin{matrix} 1 \\ 2 \end{matrix}]$

Correct

Wrong

Skipped
D
$A = [\begin{matrix} 0 & 1 \\ - 3 & - 2 \end{matrix}]$

Solutions

State Space Representation:
ẋ(t) = A(t)x(t) + B(t)u(t)
y(t) = C(t)x(t) + D(t)u(t)
y(t) is output
u(t) is input
x(t) is a state vector
A is a system matrix
Application:
ẋ₂ = -2x₂ – 3x₁ + u(t)
ẋ₁ = x₂
y(t) = x₁ + 2x₂
$[\begin{matrix} {\dot{x}}_{1} \\ {\dot{x}}_{2} \end{matrix}] = [\begin{matrix} 0 & 1 \\ - 3 & - 2 \end{matrix}] [\begin{matrix} x_{1} (t) \\ x_{2} (t) \end{matrix}] + [\begin{matrix} 0 \\ 1 \end{matrix}] u (t)$
$y (t) = [\begin{matrix} 1 & 2 \end{matrix}] [\begin{matrix} x_{1} (t) \\ x_{2} (t) \end{matrix}]$
$A = [\begin{matrix} 0 & 1 \\ - 3 & - 2 \end{matrix}], B = [\begin{matrix} 0 \\ 1 \end{matrix}], C = [\begin{matrix} 1 & 2 \end{matrix}]$
Question 6/10
1 / -0

Given the homogeneous state space equation $\dot{x} = [\begin{matrix} 0 & 1 \\ - 1 & - 2 \end{matrix}] x$ and the initial state value $x (0) = [\begin{matrix} 10 \\ - 10 \end{matrix}]$
The steady state values of $x_{s s 1} = lim_{t \to \infty} x_{1} (t)$ and $x_{s s 2} = lim_{t \to \infty} x_{2} (t)$ are
Correct

Wrong

Skipped
A
0, 0

Correct

Wrong

Skipped
B
10, 0

Correct

Wrong

Skipped
C
10, -10

Correct

Wrong

Skipped
D
0, -10
Solutions

From the given state space representation,
$A = [\begin{matrix} 0 & 1 \\ - 1 & - 2 \end{matrix}]$
$[s I - A] = s [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] - [\begin{matrix} 0 & 1 \\ - 1 & - 2 \end{matrix}]$
$= [\begin{matrix} s & - 1 \\ 1 & s + 2 \end{matrix}]$
${[s I - A]}^{- 1} = \frac{1}{s (s + 2) + 1} [\begin{matrix} s + 2 & 1 \\ - 1 & s \end{matrix}] = \frac{1}{{(s + 1)}^{2}} [\begin{matrix} s + 2 & 1 \\ - 1 & s \end{matrix}]$
$e^{A t} = L^{- 1} {[s I - A]}^{- 1} = L^{- 1} [\begin{matrix} \frac{s + 2}{{(s + 1)}^{2}} & \frac{1}{{(s + 1)}^{2}} \\ \frac{- 1}{{(s + 1)}^{2}} & \frac{s}{{(s + 1)}^{2}} \end{matrix}]$
$= L^{- 1} [\begin{matrix} \frac{1}{s + 1} + \frac{1}{{(s + 1)}^{2}} & \frac{1}{{(s + 1)}^{2}} \\ \frac{- 1}{{(s + 1)}^{2}} & \frac{1}{{(s + 1)}^{2}} - \frac{1}{{(s + 1)}^{2}} \end{matrix}]$
$= [\begin{matrix} e^{- t} + t e^{- t} & t e^{- t} \\ - t e^{- t} & e^{- t} - t e^{- t} \end{matrix}]$
$x (t) = e^{A t} x (0)$

Sign up to your account

Log in to your account

Sign Up with Google Login with Google

New to Mockers? Sign Up

Already have an account? Log in

By clicking Login, you agree to our Terms of Use & our Privacy Policy.
Most Loved Free Mock Test Platform

I agree to the Terms of Use & Privacy Policy
Sign up to your account

Log in to your account

Varify Mobile Number

Already have an account Log in

New to Mockers? Sign Up

By Log in/Sign Up, you agree to our Terms of Use & our Privacy Policy.

Secure your account:
Enter OTP now.

Please enter verification code sent to

Didn’t receive the code? Resend

By Log in/Sign Up, you agree to our Terms of Use & our Privacy Policy.

x Profile Completed: 0% Complete Profile ➔

×

Open Now

Question 1/10

2-2

24-21

27-26

29-22

Question 2/10

$ϕ = [\begin{matrix} e^{- t} & 0 \\ 1 & e^{- 2 t} \end{matrix}]$

$ϕ = [\begin{matrix} e^{t} & t \\ 1 & 2 e^{- t} \end{matrix}]$

$ϕ = [\begin{matrix} e^{t} + e^{- t} & 0 \\ 0 & 2 e^{- t} \end{matrix}]$

$ϕ = [\begin{matrix} e^{- t} & 0 \\ 0 & e^{- 2 t} \end{matrix}]$

Question 3/10

2-2

24-21

27-26

29-22

Question 4/10

$[\begin{matrix} - e^{- t} + 2 e^{- 2 t} & 2 e^{- t} + 2 e^{- 2 t} \\ 2 e^{- 2 t} - 2 e^{- t} & e^{- t} - e^{- 2 t} \end{matrix}]$

$[\begin{matrix} - e^{- t} + 2 e^{- 2 t} & e^{- t} - e^{- 2 t} \\ 2 e^{- 2 t} + 2 e^{- t} & 2 e^{- 2 t} - e^{- t} \end{matrix}]$

$[\begin{matrix} 2 e^{- t} - e^{- 2 t} & e^{- t} - e^{- 2 t} \\ 2 e^{- 2 t} - 2 e^{- t} & 2 e^{- 2 t} - e^{- t} \end{matrix}]$

$[\begin{matrix} - 2 e^{- t} + 2 e^{- 2 t} & e^{- t} - e^{- 2 t} \\ - e^{- t} + 2 e^{- 2 t} & 2 e^{- t} - e^{- 2 t} \end{matrix}]$

Question 5/10

$A = [\begin{matrix} 0 & 1 \\ - 2 & - 3 \end{matrix}]$

$B^{T} = [\begin{matrix} 0 \\ 1 \end{matrix}]$

$C^{T} = [\begin{matrix} 1 \\ 2 \end{matrix}]$

$A = [\begin{matrix} 0 & 1 \\ - 3 & - 2 \end{matrix}]$

Question 6/10

0, 0

10, 0

10, -10

0, -10

Question 1/10

2-2

24-21

27-26

29-22

Question 2/10

ϕ=[e−t01e−2t]

ϕ=[ett12e−t]

ϕ=[et+e−t002e−t]

ϕ=[e−t00e−2t]

Question 3/10

2-2

24-21

27-26

29-22

Question 4/10

[−e−t+2e−2t2e−t+2e−2t2e−2t−2e−te−t−e−2t]

[−e−t+2e−2te−t−e−2t2e−2t+2e−t2e−2t−e−t]

[2e−t−e−2te−t−e−2t2e−2t−2e−t2e−2t−e−t]

[−2e−t+2e−2te−t−e−2t−e−t+2e−2t2e−t−e−2t]

Question 5/10

A=[01−2−3]

BT=[01]

CT=[12]

A=[01−3−2]

Question 6/10

0, 0

10, 0

10, -10

0, -10

$ϕ = [\begin{matrix} e^{- t} & 0 \\ 1 & e^{- 2 t} \end{matrix}]$

$ϕ = [\begin{matrix} e^{t} & t \\ 1 & 2 e^{- t} \end{matrix}]$

$ϕ = [\begin{matrix} e^{t} + e^{- t} & 0 \\ 0 & 2 e^{- t} \end{matrix}]$

$ϕ = [\begin{matrix} e^{- t} & 0 \\ 0 & e^{- 2 t} \end{matrix}]$

$[\begin{matrix} - e^{- t} + 2 e^{- 2 t} & 2 e^{- t} + 2 e^{- 2 t} \\ 2 e^{- 2 t} - 2 e^{- t} & e^{- t} - e^{- 2 t} \end{matrix}]$

$[\begin{matrix} - e^{- t} + 2 e^{- 2 t} & e^{- t} - e^{- 2 t} \\ 2 e^{- 2 t} + 2 e^{- t} & 2 e^{- 2 t} - e^{- t} \end{matrix}]$

$[\begin{matrix} 2 e^{- t} - e^{- 2 t} & e^{- t} - e^{- 2 t} \\ 2 e^{- 2 t} - 2 e^{- t} & 2 e^{- 2 t} - e^{- t} \end{matrix}]$

$[\begin{matrix} - 2 e^{- t} + 2 e^{- 2 t} & e^{- t} - e^{- 2 t} \\ - e^{- t} + 2 e^{- 2 t} & 2 e^{- t} - e^{- 2 t} \end{matrix}]$

$A = [\begin{matrix} 0 & 1 \\ - 2 & - 3 \end{matrix}]$

$B^{T} = [\begin{matrix} 0 \\ 1 \end{matrix}]$

$C^{T} = [\begin{matrix} 1 \\ 2 \end{matrix}]$

$A = [\begin{matrix} 0 & 1 \\ - 3 & - 2 \end{matrix}]$