01/06/2025, 13:15 State space Approach to Control System Analysis and Design - - Unit 9 - Week 6

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Week 6 : Assignment 6
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1) Which of the following statements is/are true? 1 point
Course outline
State feedback is equivalent to a cascade compensator.
State feedback is equivalent to a feedback compensator.
About NPTEL ()
All coefficients of characteristic equation of a dynamical system can be changed by state feedback.
All coefficients of characteristic equation of a controllable dynamical system can be changed by state feedback.

https://onlinecourses.nptel.ac.in/noc25_ee81/unit?unit=26&assessment=110 1/13
01/06/2025, 13:15 State space Approach to Control System Analysis and Design - - Unit 9 - Week 6

All coefficients of characteristic equation of a controllable and observable dynamical system can be changed by state feedback.
How does an NPTEL
online course work? () All coefficients of characteristic equation of an observable dynamical system can be changed by state feedback.

Yes, the answer is correct.

Week 0 () Score: 1
Accepted Answers:
Week 1 () State feedback is equivalent to a feedback compensator.
All coefficients of characteristic equation of a controllable dynamical system can be changed by state feedback.
Week 2 () All coefficients of characteristic equation of a controllable and observable dynamical system can be changed by state feedback.

2) The system specified by the pair (𝐴, 𝐵) is controllable. Then the system specified by the pair (𝐴 + 𝐵𝐾 , 𝐵) , where 𝐾 is state 2 points
Week 3 ()
feedback control gain, is ............ .

Week 4 () Stable
Controllable
Week 5 ()
Stabilizable

Week 6 () Uncontrollable
None of the above
Pole Placement Design-I:
Concept of State
Yes, the answer is correct.
Score: 2
feedback (unit?
Accepted Answers:
unit=26&lesson=70)
Controllable
Pole Placement Design- Stabilizable
II: Properties of State
Feedback (unit? 3) The Ackermann's formula can be used to place the poles of .......... . 1 point
unit=26&lesson=71)
a controllable system
Pole Placement Design-
III: Pole placement a single input controllable system
formulae, Selection of a single input system
Closed loop pole
a single input stabilizable system
locations (unit?
unit=26&lesson=72) a stabilizable system

Week 6 Feedback Form :

a controllable and observable system
State space Approach to

https://onlinecourses.nptel.ac.in/noc25_ee81/unit?unit=26&assessment=110 2/13
01/06/2025, 13:15 State space Approach to Control System Analysis and Design - - Unit 9 - Week 6

Control System Analysis No, the answer is incorrect.

and Design (unit? Score: 0
unit=26&lesson=37) Accepted Answers:
a single input controllable system
Quiz: Week 6 :
Assignment 6 4) The states of the systems 𝑥˙= 𝐴𝑥 + 𝐵𝑢 and 𝑧˙ = 𝐹 𝑧 + 𝐺𝑢 are related by the transformation 𝑧 = 𝑇 𝑥 . The state feedback 1 point
(assessment? control for the former system is 𝑢 = 𝐾𝑥 . Then for the later system the state feedback will be
name=110)

Week 7 ()
𝑢 = 𝐾𝑧
𝑢 = 𝐾 𝑇 −1 𝑧
Week 8 ()
𝑢 = 𝐾𝑇 𝑧
Week 9 ()
𝑢 = 𝑇 𝐾𝑇 −1
Week 10 () 𝑢 = 𝑇 −1 𝐾𝑇 𝑧
Yes, the answer is correct.
Week 11 () Score: 1
Accepted Answers:
Week 12 () 𝑢 = 𝐾 𝑇 −1 𝑧

Lecture material () 5)
Given the LTI system
𝑥˙ 1 = 0 1 𝑥1 + 1 𝑢, which of the following state--feedback gain vectors can stabilize it 2 points
𝑥2
[ ˙ ] [ 0 2 ] [ 𝑥2 ] [ 0 ]
DOWNLOAD VIDEOS () if the control is generated according to 𝑢 = −𝐾𝑥?

Problem Solving 𝐾 = [1 2]
Session - Jan 2025 ()
𝐾 = [ −2 −3 ]

𝐾 = [ −4 4]
The given system is not stabilizable by state feedback.

Yes, the answer is correct.

https://onlinecourses.nptel.ac.in/noc25_ee81/unit?unit=26&assessment=110 3/13
01/06/2025, 13:15 State space Approach to Control System Analysis and Design - - Unit 9 - Week 6

Score: 2
Accepted Answers:
The given system is not stabilizable by state feedback.

6) Consider the system defined by 𝑥

˙= 𝐴𝑥 + 𝐵𝑢, where 4 points
0⎡ 1 0 ⎤ 0⎤
⎡

𝐴=⎢ 0⎢
0
⎥
1 ⎥ , and 𝐵 = ⎢ 1 ⎥ . The value of 𝐾 in the state-feedback control 𝑢 = −𝐾𝑥 which places the closed-loop poles at
⎢ ⎥

⎣ −1 −5 −6 ⎦ ⎣ 1 ⎦

𝑠 = −2 ± 𝑗4, 𝑠 = −10 is

[ 28.7831 5.5181 2.4819 ]

[ 2.22 5.25 4.435 ]

[ 30.34 5.5181 20.4819 ]

[ 8.7831 7.5181 −2.4819 ]

No, the answer is incorrect.

Score: 0
Accepted Answers:
[ 28.7831 5.5181 2.4819 ]

7) Consider the inverted pendulum on the motor-driven cart. Its state,input and output variables respectively are: 5 points
𝑇 𝑇
𝑋 = [ 𝑥 𝑥˙ 𝜃 𝜃˙ ] , 𝑢 = 𝑒, 𝑦 = [ 𝑥 𝜃 ] and its 𝐴, 𝐵, 𝐶 and 𝐷 matrices are given by
⎡ 0 1 0 0⎤ ⎡ 0 ⎤
⎢ ⎥
−𝑘 2 𝑚𝑔 ⎢ ⎥
⎢ 0 −𝑀 0⎥ 1
𝐴=⎢ 𝑀 𝑟2 𝑅 ⎥ , 𝐵 𝑘
= 𝑀𝑅𝑟
⎢ ⎥
, 𝐶 = [ 1 0 0 0], 𝐷 = [ 0 ]
⎢ 0 0 0 1⎥ ⎢
0 ⎥
0 0 1 0 1
⎢ ⎥
⎢
𝑘2 𝑀 +𝑚 ⎥ 1
⎣ 0 𝑀 𝑟2 𝑅𝑙 𝑀𝑙 𝑔 0⎦ ⎣ −𝑙 ⎦

where,
k = Motor torque constant
R = Motor resistance
r = Ratio of motor torque to linear force applied to the cart (T = rf)

https://onlinecourses.nptel.ac.in/noc25_ee81/unit?unit=26&assessment=110 4/13
01/06/2025, 13:15 State space Approach to Control System Analysis and Design - - Unit 9 - Week 6

e = Voltage applied to the motor

M = Mass of the cart
m = Mass of the pendulum bob
x = Distance from the reference
𝜃 = Vertical angle made by pendulum
The values of different quantities are: 𝑚 = 0.1 𝑘𝑔, 𝑀 = 1.0 𝑘𝑔, 𝑙 = 𝑙.0 𝑚, 𝑘 = 1 𝑉 𝑠, 𝑔 = 9.8 𝑚𝑠 −2 , 𝑅 = 100 Ω, 𝑟 = 0.02 𝑚.
It is desired to place the dominant poles (in a Butterworth configuration) at 𝑠 = −4 and 𝑠 = −2 ± 𝑗2√3
‾ and to leave the pole at 𝑠 = −25
unchanged. The gain matrix 𝐾 that produces this set of closed-loop poles with the state feedback of the form 𝑢 = −𝐾𝑋 , is ........ .

[ −326.5306 100.234 −812.0906 −242.326 ]

[ 326.5306 226.3265 812.0906 −242.326 ]

[ −326.5306 −226.3265 −812.0906 −242.326 ]

[ 326.5306 226.3265 812.0906 242.326 ]

None of these

Yes, the answer is correct.

Score: 5
Accepted Answers:
[ −326.5306 −226.3265 −812.0906 −242.326 ]

Let 𝐻 (𝑠) = where 𝐵 𝑘(𝑧) is a Butterworth polynomial of order 𝑘. Then which of the following is true:
8) 1 2 points
𝐵 𝑘 (𝑠/𝜔0 )

| 𝐻 (𝑗𝜔)| = 1
[1+(𝜔/𝜔0 ) 2𝑘 ] 1/3

| 𝐻 (𝑗𝜔)| = 1
[1+(𝜔/𝜔0 ) 2𝑘 ] 1/2

| 𝐻 (𝑗𝜔)| = 1
[1+(𝜔/𝜔0 ) 3𝑘 ] 1/2

| 𝐻 (𝑗𝜔)| = 1
[1−(𝜔/𝜔0 ) 2𝑘 ] 1/2

https://onlinecourses.nptel.ac.in/noc25_ee81/unit?unit=26&assessment=110 5/13
01/06/2025, 13:15 State space Approach to Control System Analysis and Design - - Unit 9 - Week 6

Yes, the answer is correct.

Score: 2
Accepted Answers:
| 𝐻 (𝑗𝜔)| = 1
[1+(𝜔/𝜔0 ) 2𝑘 ] 1/2

9) Consider the system defined by 4 points

0 1 0
𝑥˙ = [ 𝑥 + 𝑢.
−6 −5 ] [ 1 ]

The performance specifications for the closed loop system with state feedback 𝑢 = 𝑘𝑥 are:
1. Closed-loop system has a damping coefficient = 0.707
2. Peak time for step input is 3.14 s
Which of the following state feedback gains meets the above--mentioned performance specifications?

[ 4 3]

[ 2 1]

[ −4 −3 ]

[ −1 −8 ]

Yes, the answer is correct.

Score: 4
Accepted Answers:
[ 4 3]

10) Consider the system shown in Figure 1. What is the value of 𝐾 such that the damping ratio of the closed-loop system is equal 4 points
to 0.5. What is the value of the undamped natural frequency 𝜔𝑛 of the closed-loop system?

https://onlinecourses.nptel.ac.in/noc25_ee81/unit?unit=26&assessment=110 6/13
01/06/2025, 13:15 State space Approach to Control System Analysis and Design - - Unit 9 - Week 6

Figure 1: The system for Problem 10

0.1, 2.3

0.7, 1.5

0.4, 2.4

0.9, 3.1
Yes, the answer is correct.
Score: 4
Accepted Answers:
0.7, 1.5
11) Consider the system 4 points
𝑌 ( 𝑠) 10
𝑈 (𝑠) = (𝑠+1)(𝑠+2)(𝑠+3) . Define state variables as
𝑥1 = 𝑦, 𝑥2 = 𝑥˙ 1 , 𝑥3 = 𝑥˙ 2 .
By use of the state-feedback control 𝑢 = −𝐾𝑥, it is desired to place the closed-loop poles at:
𝑠 = −2 + 𝑗2√13
‾‾‾ , 𝑠 = −2 − 𝑗2√13‾‾‾ , 𝑠 = −10.
The necessary state feedback gain 𝐾 = ….

[ −16.2 −5.1 −2.3 ]

[ 15.4 4.5 0.8 ]

https://onlinecourses.nptel.ac.in/noc25_ee81/unit?unit=26&assessment=110 7/13
01/06/2025, 13:15 State space Approach to Control System Analysis and Design - - Unit 9 - Week 6

[ 16.2 5.1 2.3 ]

[ −15.4 −4.5 −0.8 ]

Yes, the answer is correct.

Score: 4
Accepted Answers:
[ 15.4 4.5 0.8 ]
12) Consider the type 1 servo system shown in Figure 2. Matrices A, B, and C are given by: 3 points
0
⎡ 1
0 ⎤ ⎡ 0 ⎤

𝐴 ⎢
= ⎢0 1 ⎥,𝐵 = ⎢0⎥,𝐶 = [1 0 0].
0
⎥ ⎢ ⎥

⎣ 0 −5
−6 ⎦ ⎣ 1 ⎦

The control law 𝑢 = −𝐾𝑥 can place the closed-loop poles at 𝑠 = −2 + 𝑗4 , 𝑠 = −2 − 𝑗4 , 𝑠 = −10 if 𝐾 = … .

Figure 2: The system for Problem 12

https://onlinecourses.nptel.ac.in/noc25_ee81/unit?unit=26&assessment=110 8/13
01/06/2025, 13:15 State space Approach to Control System Analysis and Design - - Unit 9 - Week 6

[ 200 55 8]

[ 100 55 8]

[ −200 −55 −8 ]

[ −100 −55 −8 ]
Yes, the answer is correct.
Score: 3
Accepted Answers:
[ 200 55 8]
13) Consider the discrete-time state equation 3 points
⎡ 1 1 −2 ⎤ ⎡ 1 ⎤

𝑥[𝑘 + 1] = ⎢⎢ 0 1 1
⎥
⎥
𝑥[𝑘] + ⎢⎢ 0 ⎥⎥ 𝑢[𝑘], 𝑦[𝑘] = [ 2 0 0 ] 𝑥[𝑘].
⎣ 0 0 1⎦1 ⎦ ⎣

The state feedback gain in 𝑢[𝑘] = −𝐺𝑥[𝑘] so that the resulting system has all eigenvalues at 𝑧 = 0 , is ..........

[ 1 5 2]

[ −1 −5 −2 ]

[ 1 5 1]

[ 2 5 2]
Yes, the answer is correct.
Score: 3
Accepted Answers:
[ 1 5 2]

14) Refer to the Problem 13. For any initial state, the zero-input response of the feedback system becomes identically zero for 1 point
𝑘 ≥ ….

https://onlinecourses.nptel.ac.in/noc25_ee81/unit?unit=26&assessment=110 9/13
01/06/2025, 13:15 State space Approach to Control System Analysis and Design - - Unit 9 - Week 6

2
4
3
1
Yes, the answer is correct.
Score: 1
Accepted Answers:
3
15) Consider the continuous--time LTI system having system, input and output matrices respectively as 3 points
−2 1 1
𝐴 =
[ 1
,𝐵 = ,𝐶 = [1 2]
0] [ 0 ]

If the closed loop system with the state feedback 𝑢 = −𝐾𝑥is unobservable for 𝐾 = [ 1 𝑘2 ], then 𝑘2 = ….

2
3
5
1

Such a 𝐾 does not exist.

No, the answer is incorrect.

Score: 0
Accepted Answers:
3

16) In Problem 15, the unobservability of the closed loop system is .......... 1 point

due to pole-zero cancellation

due to reason that the open loop system itself is unobservable.
due to the uncontrollability of the open loop system

https://onlinecourses.nptel.ac.in/noc25_ee81/unit?unit=26&assessment=110 10/13
01/06/2025, 13:15 State space Approach to Control System Analysis and Design - - Unit 9 - Week 6

None of the above.

Yes, the answer is correct.

Score: 1
Accepted Answers:
due to pole-zero cancellation

17) An idealized two-car train consists of a pair of masses coupled by a spring, as shown in Figure 3. The wheels of each car are 5 points
independently driven by an electric motor. Assume 𝑅 is the motor resistance, 𝐾 is the spring constant, 𝑘 is the motor torque constant, and
𝑟 = 𝜏/𝑓 is the ratio of the motor torque to the linear force applied to the car. The following are the state variables and inputs:
𝑥1 = 𝑧 1 , 𝑥2 = 𝑧 2 , 𝑥3 = 𝑧˙ 1 , 𝑥4 = 𝑧˙ 2 , 𝑢1 = 𝑒 1 , 𝑢2 = 𝑒 2 where 𝑒 1 and 𝑒 2 are the voltages on the drive motors. 𝐴, 𝐵, 𝐶 and 𝐷 matrices
⎡ 0 0 1 0 ⎤
⎡ 0 0 ⎤
⎢ ⎥ ⎢ ⎥
⎢
0 0 0 1 ⎥ 0 0
𝑘
⎢ ⎥
1 2 0 0 0
for the system are given by: 𝐴 = −𝐾 𝐾 , 𝐵= , 𝐶=[ , 𝐷=[ .
− 𝑘2 𝑘
⎢ 2 ⎥ ⎢ ⎥
0 𝑀𝑅𝑟 0 0 1 0 0] 0]
⎢ 𝑀 𝑀 𝑟𝑅 ⎥ ⎢
𝑅𝑟 ⎥
⎢
𝐾 −𝐾 ⎥ ⎢
𝑘 ⎥
− 𝑘2 𝑅
2
0 ⎣ 0 𝑅𝑟 ⎦
⎣
𝑀 𝑀 𝑟 ⎦

The following are the numerical data:

Trains: 𝑀 = 𝑀 1 = 𝑀 2 = 1.0 𝑘𝑔, 𝐾 = 40 𝑁 /𝑚
Motors: 𝑘 = 2 𝑉 . 𝑠, 𝑅 = 100 Ω, 𝑟 = 2𝑐𝑚
It is desired to bring the two-car train to rest at the origin using only the motor on car 1. The gain matrix 𝐾 in the control law 𝑢 = −𝐾𝑥 which

https://onlinecourses.nptel.ac.in/noc25_ee81/unit?unit=26&assessment=110 11/13
01/06/2025, 13:15 State space Approach to Control System Analysis and Design - - Unit 9 - Week 6

places the poles at 𝑠 = −1 ± 𝑗1 and at 𝑠 = −100 ± 𝑗100 is ...........

[ 12517 −1184 8 −14792 ]

[ −12517 1184 −8 14792 ]

[ 10122 −9122 2 −24497 ]

[ −10122 9122 −2 24497 ]

No, the answer is incorrect.
Score: 0
Accepted Answers:
[ 10122 −9122 2 −24497 ]

18) For the two car train specified in the Problem 17, the control law so that the train maintains a constant velocity, is given by 4 points

𝑢 = 50682𝑧 1 − 51682𝑧 2 − 402(−𝑧 3 + 𝑉 ) + 127917(−𝑧 4 + 𝑉 ) − 127515𝑉

𝑢 = 80682𝑧 1 − 41682𝑧 2 − 302(−𝑧 3 + 𝑉 ) + 12917(−𝑧 4 + 𝑉 ) − 7315𝑉

https://onlinecourses.nptel.ac.in/noc25_ee81/unit?unit=26&assessment=110 12/13
01/06/2025, 13:15 State space Approach to Control System Analysis and Design - - Unit 9 - Week 6

𝑢 = 8982𝑧 1 − 5162𝑧 2 − 302(−𝑧 3 + 𝑉 ) + 7917(−𝑧 4 + 𝑉 ) − 7315𝑉

𝑢 = 50682𝑧 1 − 51682𝑧 2 − 402(−𝑧 3 + 𝑉 ) + 1217(−𝑧 4 + 𝑉 ) − 1515𝑉
Yes, the answer is correct.
Score: 4
Accepted Answers:
𝑢 = 50682𝑧 1 − 51682𝑧 2 − 402(−𝑧 3 + 𝑉 ) + 127917(−𝑧 4 + 𝑉 ) − 127515𝑉

https://onlinecourses.nptel.ac.in/noc25_ee81/unit?unit=26&assessment=110 13/13