156. Figure shows a feedback system where K > 0. The range of K for which the system is stable will be given by
     
     

1. 1. 0 < K < 30

   
   
   

   2. 0 < K < 39

   
   
   

   3. 0 < K < 390

   
   
   

   4. K > 390

   
   
   

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1. View Hint View Answer Discuss in Forum

   K > 0
   Characteristic equation of the system is
   1 + G(s) H(s) = 0
   

   ⇒ 1 +	K	= 0
   	s(s + 3)(s + 10)

   
   ⇒ s(s² + 13s + 30) + K = 0
   ⇒ s³ + 13s² + 30s + K = 0
   Routh array
   
   For system to be stable, all element of first column must be positive.
   

   ∴	390 - K	> 0
   	13

   
   ⇒ K < 390°
   and K > 0
   Hence range of K is 0 < K < 390
   

   Correct Option: C

   K > 0
   Characteristic equation of the system is
   1 + G(s) H(s) = 0
   

   ⇒ 1 +	K	= 0
   	s(s + 3)(s + 10)

   
   ⇒ s(s² + 13s + 30) + K = 0
   ⇒ s³ + 13s² + 30s + K = 0
   Routh array
   
   For system to be stable, all element of first column must be positive.
   

   ∴	390 - K	> 0
   	13

   
   ⇒ K < 390°
   and K > 0
   Hence range of K is 0 < K < 390
   

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157. The transfer function of a system is given as	100
     	s2 + 20s + 100

     
     This system is
     

1. 1. an overdamped system

   
   
   

   2. an underdamped system

   
   
   

   3. a critically damped system

   
   
   

   4. an unstable system

   
   
   

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1. View Hint View Answer Discuss in Forum

   M(s) =	100
   	s2 + 20s + 100

   
   Comparing with standard form,
   

   M(s) =	ω2n
   	s2 + 2ξωns + ω2n

   
   we get 2ξω_n = 20
   

   &rArrr; ξ =	20	= 1
   	2 × 10

   
   and ω²_n = 100
   &rArrr; ω_n = 10
   For ξ = 1, system is critically damped.
   

   Correct Option: C

   M(s) =	100
   	s2 + 20s + 100

   
   Comparing with standard form,
   

   M(s) =	ω2n
   	s2 + 2ξωns + ω2n

   
   we get 2ξω_n = 20
   

   &rArrr; ξ =	20	= 1
   	2 × 10

   
   and ω²_n = 100
   &rArrr; ω_n = 10
   For ξ = 1, system is critically damped.
   

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158. The system shown in the figure given below is
     
     

1. 1. stable

   
   
   

   2. unstable

   
   
   

   3. conditionally stable

   
   
   

   4. stable for input u₁, but unstable for input u₂

   
   
   

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1. View Hint View Answer Discuss in Forum

   Transfer function for u₁
   

   TF1 =

   (s - 1) / (s + 2)

   =

   (s - 2)

   1 +

   s - 1

   .

   1

   (s + 3)

   s + 2

   s - 1

   
   
   

   TF2(s) =			1
   			s - 1
   	1 +		1		.		s - 1
   			s - 1						s + 2

   
   

   =	1
   	(s + 3)(s - 1)

   
   Hence unstable, as it has pole at right side of s-plane.
   

   Correct Option: D

   Transfer function for u₁
   

   TF1 =

   (s - 1) / (s + 2)

   =

   (s - 2)

   1 +

   s - 1

   .

   1

   (s + 3)

   s + 2

   s - 1

   
   
   

   TF2(s) =			1
   			s - 1
   	1 +		1		.		s - 1
   			s - 1						s + 2

   
   

   =	1
   	(s + 3)(s - 1)

   
   Hence unstable, as it has pole at right side of s-plane.
   

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159. If x = Re G(jθ), and y = lm G(jω) then for ω → 0⁺, the Nyquist plot for
     

     G(s) =	1
     	s(s + 1)(s + 2)

     
     becomes asymptotic to the line

1. 1. x = 0

   
   
   

   2. x = -	3
      	4

   
   
   

   3. x = y – 1 / 6

   
   
   

   4. x =	y
      	√3

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given , G(s) =	1
   	s(s + 1)(s + 2)

   
   

   G(jω) =	1	=	-j(1 - jω)(2 - jω)
   	jω(1 - jω)(2 + jω)		ω(1 - ω²)(4 + ω²)

   
   

   =	j(2 - 3jω - ω2)	=	-3ω - j(2 - ω2)
   	ω(1 + ω²)(4 + ω²)		ω(1 + ω²)(4 + ω²)

   
   

   =	- 3	- j	(2 - ω2)
   	(1 + ω²)(4 + ω²)		ω(1 + ω²)(4 + ω²)

   
   

   ⇒ x =	3
   	(1 + ω²)(4 + ω²)

   
   

   y = -	(2 - ω2)
   	ω(1 + ω²)(4 + ω²)

   
   

   At ω → 0 , x →	3	, y → - ∞
   	4

   
   

   Correct Option: B

   Given , G(s) =	1
   	s(s + 1)(s + 2)

   
   

   G(jω) =	1	=	-j(1 - jω)(2 - jω)
   	jω(1 - jω)(2 + jω)		ω(1 - ω²)(4 + ω²)

   
   

   =	j(2 - 3jω - ω2)	=	-3ω - j(2 - ω2)
   	ω(1 + ω²)(4 + ω²)		ω(1 + ω²)(4 + ω²)

   
   

   =	- 3	- j	(2 - ω2)
   	(1 + ω²)(4 + ω²)		ω(1 + ω²)(4 + ω²)

   
   

   ⇒ x =	3
   	(1 + ω²)(4 + ω²)

   
   

   y = -	(2 - ω2)
   	ω(1 + ω²)(4 + ω²)

   
   

   At ω → 0 , x →	3	, y → - ∞
   	4

   
   

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160. The system	900	is be compensated such that its gain-
     	(s + 1)(s + 9)

     
     crossover frequency becomes same as its uncompensated phase-crossover frequency and provides a 45° phase margin. To achieve this, one may use
     

1. 1. a lag compensator that provides an attenuation of 20 dB and a phase lag of 45° at the frequency of 3 √3 rad/s

   
   
   

   2. a lead compensator that provides an amplification of 20 dB and a phase lead of 45° at the frequency of 3 rad/s

   
   
   

   3. a lag-lead compensator that provides an amplification of 20 dB and a phase lag of 45° at the frequency of √3 rad/s.

   
   
   

   4. a lag-lead compensator that provides an attenuation of 20 dB and phase lead of 45° at the frequency of 3 rad/s
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   NA

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