Control system miscellaneous Easy Questions and Answers | Page - 5

Control system miscellaneous

The number of roots of the equation 2s⁴ + s³ + 3s² + 5s + 7 = 0 that lie in the right half of s plane is

1. zero
1. one
1. two
1. three

View Hint View Answer Discuss in Forum

Routh Hurwitz criterion

Since sign changes twice, therefore 2 roots in right half plane.

Correct Option: C

Routh Hurwitz criterion

Since sign changes twice, therefore 2 roots in right half plane.

Consider the following polynomials :
1. s⁴ + 7s³ + 17s² + 17s + 6.
2. s⁴ + 11s³ + 41s² + 61s + 30.
3. s⁴ + s³ + 2s² + 3s + 2.
Among these polynomials, those which are Hurwitz are

1. 1 and 3
1. 2 and 3
1. 1 and 2
1. 1, 2 and 3

View Hint View Answer Discuss in Forum

NA

Correct Option: C

NA

What is the unit step response of a unity feedback control system having forward path transfer
function G(s) = 80
s(s + 18)

1. Overdamped
1. Critically damped
1. Underdamped
1. Undamped oscillatory

View Hint View Answer Discuss in Forum

T(s) = G(s) = 80
1 + G(s) s² + 18s + 80

C(s) = 80
R(s) s² + 18s + 80

Here , 2ξω_n = 18
and ω_n = √80
∴ ξ = 18 = 1.006
2√80

Since, ξ > 1, the system is overdamped.

Correct Option: D

T(s) = G(s) = 80
1 + G(s) s² + 18s + 80

C(s) = 80
R(s) s² + 18s + 80

Here , 2ξω_n = 18
and ω_n = √80
∴ ξ = 18 = 1.006
2√80

Since, ξ > 1, the system is overdamped.

The correct sequence of steps needed to improve system stability is

1. insert derivative action, use negative feedback reduce gain
1. reduce gain, use negative feedback, insert derivation action
1. reduce gain, insert derivation action, use negative feedback
1. use negative feedback, reduce gain, insert derivation action

View Hint View Answer Discuss in Forum

Since all the coefficients in all the three polynomials are positive, it is required to construct Hurwitz table for further checking.
1. s⁴ + 7s³ + 17s² + 17s + 6
This is seen to be a Hurwitz polynomial as all the terms in the first column have the same sign.

2. s⁴ + 11s³ + 41s² + 61s + 30
This also is a Hurwitz polynomial

3. s⁴ + s³ + 2s² + 3s + 2
This is not a Hurwitz polynomial

Correct Option: D

Since all the coefficients in all the three polynomials are positive, it is required to construct Hurwitz table for further checking.
1. s⁴ + 7s³ + 17s² + 17s + 6
This is seen to be a Hurwitz polynomial as all the terms in the first column have the same sign.

2. s⁴ + 11s³ + 41s² + 61s + 30
This also is a Hurwitz polynomial

3. s⁴ + s³ + 2s² + 3s + 2
This is not a Hurwitz polynomial

Given : KK_t = 99; s = j1 rad/s
The sensitivity of the closed-loop system (shown in the figure) to variation in parameter K is approximately

1. 0.01
1. 0.1
1. 1.0
1. 10

View Hint View Answer Discuss in Forum

Given : KK_t = 99; s = j1 rad/sec.

S_K^M = dM
M = K dM
1 + 10 M dK
K

Now M = G = K = K
(10s + 1)
1 + GH 1 + KK_t 10s + 1 + KK_t
(10s + 1)

∴ dM = 1 × (10s + 1 + KK_t) - KK_t
dK (10s + 1 + KK_t)²

= 10s + 1
(10s + 1 + KK_t)²

⇒ K dM = K
M dK K
(10s + 1 + KK_t)

= (10s + 1)
(10s + 1 + KK_t)²

= (10s + 1)
(10s + 1 + KK_t)

= (10s + 1)
10(10 + s)

With s = j1, K dM = √100 + 1 = 0.1
M dK 10 √100 + 1

Correct Option: B

Given : KK_t = 99; s = j1 rad/sec.

S_K^M = dM
M = K dM
1 + 10 M dK
K

Now M = G = K = K
(10s + 1)
1 + GH 1 + KK_t 10s + 1 + KK_t
(10s + 1)

∴ dM = 1 × (10s + 1 + KK_t) - KK_t
dK (10s + 1 + KK_t)²

= 10s + 1
(10s + 1 + KK_t)²

⇒ K dM = K
M dK K
(10s + 1 + KK_t)

= (10s + 1)
(10s + 1 + KK_t)²

= (10s + 1)
(10s + 1 + KK_t)

= (10s + 1)
10(10 + s)

With s = j1, K dM = √100 + 1 = 0.1
M dK 10 √100 + 1