191. The first two rows of Routh's tabulation of a third order equation are as follows.
     

     s3	2	2
     s3	4	4

     
     This means there are

1. 1. two roots at s = ± j and one root in right half s-plane
      

   
   
   

   2. two roots at s = ± j2 and one root in left half s-plane

   
   
   

   3. two roots at s = ± j2 and one root in right half s-plane

   
   
   

   4. two roots at s = ± j and one root in left half s-plane

   
   
   

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1. View Hint View Answer Discuss in Forum

   Two roots at s = ±j and one in left halfs-plane
   

   s3	2	2
   s3	4	4

   
   Characteristic equation is A(s) = 2s³ + 4s² + 2s + 4 = 0
   ⇒ s³ + 2s² + s + 2 = 0
   ⇒ (s + 2) (s² + 1) = 0
   ⇒ s = – 2, s = ± j
   

   Correct Option: D

   Two roots at s = ±j and one in left halfs-plane
   

   s3	2	2
   s3	4	4

   
   Characteristic equation is A(s) = 2s³ + 4s² + 2s + 4 = 0
   ⇒ s³ + 2s² + s + 2 = 0
   ⇒ (s + 2) (s² + 1) = 0
   ⇒ s = – 2, s = ± j
   

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192. The asymptotic approximation of the log-magnitude vs frequency plot of a system containing only real poles and zeros is shown in the figure given below. Its transfer function is
     

1. 1. 	10(s + 5)
      	s(s + 2)(s + 25)

   
   
   

   2. 	1000(s + 5)
      	s2(s + 2)(s + 25)

   
   
   

   3. 	100(s + 5)
      	s(s + 2)(s + 25)

   
   
   

   4. 	80 (s + 5)
      	s2 (s + 2)(s + 25)

   
   
   

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1. View Hint View Answer Discuss in Forum

   At ω = 0.1,	2 poles
   At ω = 2,	1 pole
   At ω = 5 ,	1 zero
   At ω = 25 ,	1 pole

   
   
   So, transfer function,
   

   T(s) =	k(s + 5)
   	s(s + 2)(s + 25)

   
   

   ∴ 80 = 20 log	5k
   	(0.1)2 × 2 × 25

   
   

   ⇒ 4 = log	5k
   	0.5

   
   ⇒ 10k = 10⁴
   ⇒ k = 10³ = 1000
   

   ∴ T(s) =	1000 (s + 5)
   	s2(s + 2)(s + 25)

   
   

   Correct Option: B

   At ω = 0.1,	2 poles
   At ω = 2,	1 pole
   At ω = 5 ,	1 zero
   At ω = 25 ,	1 pole

   
   
   So, transfer function,
   

   T(s) =	k(s + 5)
   	s(s + 2)(s + 25)

   
   

   ∴ 80 = 20 log	5k
   	(0.1)2 × 2 × 25

   
   

   ⇒ 4 = log	5k
   	0.5

   
   ⇒ 10k = 10⁴
   ⇒ k = 10³ = 1000
   

   ∴ T(s) =	1000 (s + 5)
   	s2(s + 2)(s + 25)

   
   

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193. The unit -step response of a unity feedback system with open loop transfer function
     

     G(s) =	K	is shown in the figure below.
     	(s + 1)(s + 2)

     
     
     The value of K is
     

1. 1. 0.5

   
   
   

   2. 2

   
   
   

   3. 4

   
   
   

   4. 6

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given : G(s) =	K
   	(s + 1)(s + 2)

   
   From curve, steady state error = 1 – 0.75 = 0.25
   

   ∴ E(s) =	R(s)	=		1
   				s
   	1 + H(s)G(s)		1 + 1.	k	× 1
   				(s + 1)(s + 2)

   
   
   

   ⇒	1				= 0.25
   	1 +	k
   		2

   
   

   ⇒ 1 = 0.25 +	0.25k
   	2

   
   

   ⇒ k =	0.75 × 2	= 6
   	0.25

   Correct Option: D

   Given : G(s) =	K
   	(s + 1)(s + 2)

   
   From curve, steady state error = 1 – 0.75 = 0.25
   

   ∴ E(s) =	R(s)	=		1
   				s
   	1 + H(s)G(s)		1 + 1.	k	× 1
   				(s + 1)(s + 2)

   
   
   

   ⇒	1				= 0.25
   	1 +	k
   		2

   
   

   ⇒ 1 = 0.25 +	0.25k
   	2

   
   

   ⇒ k =	0.75 × 2	= 6
   	0.25

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194. The open loop transfer function of a unity feedback system is given by
     

     G(s) =	(e-0.1s)
     	s

     
     The gain margin of this system is
     

1. 1. 11.95 dB

   
   
   

   2. 17.67 dB

   
   
   

   3. 21.33 dB

   
   
   

   4. – 23.9 dB

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given : G(s) =	e-0.1s
   	s

   
   

   ⇒ G(jω) =	e-0.1jω
   	jω

   
   Phase crossover frequency :
   

   -90 - 0.1 ω ×	180	= - 180
   	π

   
   

   ⇒	18	ω = 90
   	π

   
   ⇒ ω = 5π = 15.7 rad/sec
   Gain margin :
   

   a = |G(jω)|at ω = 18.7 =	1	=	1
   	ω		15.7

   
   ∴ Gain margin = 20 log a
   

   = 20 log	1	= -23.9 dB
   	15.7

   
   

   Correct Option: D

   Given : G(s) =	e-0.1s
   	s

   
   

   ⇒ G(jω) =	e-0.1jω
   	jω

   
   Phase crossover frequency :
   

   -90 - 0.1 ω ×	180	= - 180
   	π

   
   

   ⇒	18	ω = 90
   	π

   
   ⇒ ω = 5π = 15.7 rad/sec
   Gain margin :
   

   a = |G(jω)|at ω = 18.7 =	1	=	1
   	ω		15.7

   
   ∴ Gain margin = 20 log a
   

   = 20 log	1	= -23.9 dB
   	15.7

   
   

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195. The characteristic equation of a (3 × 3) matrix P is defined as
     α (λ) = |λI – P| = λ³ + λ² + 2λ + 1 = 0.
     If I denotes identity matrix, then the inverse of matrix P will be
     

1. 1. (P² + P + 2I)

   
   
   

   2. (P² + P + I)

   
   
   

   3. – (P² + P + I)

   
   
   

   4. – (P² + P + 2I)

   
   
   

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1. View Hint View Answer Discuss in Forum

   By Clayey – Hamilton theorem,
   Every square matrix satisfies its own characteristic equation
   α (λ) = λ 3 + λ2 + 2λ + 1 = 0
   α(P) = P³ + P² + 2P + I = 0
   ⇒ I = – P³ + P² + 2P
   Premultiplying by p^-1, we get
   P ^-1 = – [P² + P + 2I]

   Correct Option: D

   By Clayey – Hamilton theorem,
   Every square matrix satisfies its own characteristic equation
   α (λ) = λ 3 + λ2 + 2λ + 1 = 0
   α(P) = P³ + P² + 2P + I = 0
   ⇒ I = – P³ + P² + 2P
   Premultiplying by p^-1, we get
   P ^-1 = – [P² + P + 2I]

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