11. The transfer function C (s)/ R (s) of the system, whose block diagram is given below is—
    
    
    
    

1. 1. G1G2

      1 + G1H1+G2H2 - G1G2H1H2

      
      

   
   
   

   2. G1G2

      1 + G1H1+G2H2 + G1G2H1H2

      
      

   
   
   

   3. G1G2

      1 + G1H1+G2H2

      
      

   
   
   

   4. G1(1 + G2H2) + G2(1 + G1H1)

      1 + G1H1+G2H2 + G1G2H1H2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Here in the given block diagram there are two loop, on simplifying, we get
   

   C(s)	=	G1·G2
   R(s)		(1 + G1H1) (1 + G2H2)

   
   

   =	G1·G2
   	1 + G1H1 + G2H2 + G1G2H1H2

   
   

   Correct Option: B

   Here in the given block diagram there are two loop, on simplifying, we get
   

   C(s)	=	G1·G2
   R(s)		(1 + G1H1) (1 + G2H2)

   
   

   =	G1·G2
   	1 + G1H1 + G2H2 + G1G2H1H2

   
   

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12. For the unity feedback system shown below, the steady state, when the input is
    

    R =	3	-	1	+	1
    	s		s2		2s3

    
    would be—
    
    
    
    

1. 1. 1/4
      

   
   
   

   2. 0
      

   
   
   

   3. ∞
      

   
   
   

   4. 1/2

   
   
   

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1. View Hint View Answer Discuss in Forum

   ess = s → 0Lim sE (s) = s → 0Lim	s·R(s)
   	1 + G(s)H(s)

   
   

   = s → 0Lim	s		3	-	1	+	1
   	1 + 4(s + 1)/s2(s + 2).1		s		s2		2s3

   
   

   = s → 0Lim	s3(s + 2)		3	-	1	+	1
   	s3 + 2s2 + 4s + 4		s		s2		2s3

   
   = 0 + 0 + 1/4 = 1/4

   Correct Option: A

   ess = s → 0Lim sE (s) = s → 0Lim	s·R(s)
   	1 + G(s)H(s)

   
   

   = s → 0Lim	s		3	-	1	+	1
   	1 + 4(s + 1)/s2(s + 2).1		s		s2		2s3

   
   

   = s → 0Lim	s3(s + 2)		3	-	1	+	1
   	s3 + 2s2 + 4s + 4		s		s2		2s3

   
   = 0 + 0 + 1/4 = 1/4

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13. The system shown in the given figure has a unit step input. In order to make the steady state error 0.1, the value of K required is—
    
    
    
    

1. 1. 0.1
      

   
   
   

   2. 0.9
      

   
   
   

   3. 1.0
      

   
   
   

   4. 9.0

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given r (t) = u (t) then
   

   R (s) =	1
   	s

   
   

   G(s) =	K
   	(s + 1)(0.1s + 1)2

   
   H (s) = 1
   e_ss = _{s → 0}Lim sE (s)
   

   E(s)	=	1
   R(s)		1 + G(s) H(s)

   
   

   E(s) =	R(s)
   	1 + G(s) H(s)

   
   

   =	1/s
   	1 + K/(s + 1)(0.1 + 1)2 ×1

   
   

   =	(s + 1)(0·1s + 1)2
   	s{(s + 1) (0.1 + 1)2 + K}

   
   

   ess = s → 0Lim	s·(s + 1)(0·1s + 1)2
   	s{(s + 1) (0.1 + 1)2 + K}

   
   

   0·1 =	1 × 1
   	(1 × 1 × K)

   
   or ·
   1 + ·1K = 1 ·1 K = 1 – ·1 = ·9
   K = 9

   
   

   Correct Option: D

   Given r (t) = u (t) then
   

   R (s) =	1
   	s

   
   

   G(s) =	K
   	(s + 1)(0.1s + 1)2

   
   H (s) = 1
   e_ss = _{s → 0}Lim sE (s)
   

   E(s)	=	1
   R(s)		1 + G(s) H(s)

   
   

   E(s) =	R(s)
   	1 + G(s) H(s)

   
   

   =	1/s
   	1 + K/(s + 1)(0.1 + 1)2 ×1

   
   

   =	(s + 1)(0·1s + 1)2
   	s{(s + 1) (0.1 + 1)2 + K}

   
   

   ess = s → 0Lim	s·(s + 1)(0·1s + 1)2
   	s{(s + 1) (0.1 + 1)2 + K}

   
   

   0·1 =	1 × 1
   	(1 × 1 × K)

   
   or ·
   1 + ·1K = 1 ·1 K = 1 – ·1 = ·9
   K = 9

   
   

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14. Given G(s) =	1 - Is
    	s(s + 2)

    
    The system with the transfer function is operated in a close-loop with unity feedback. The closed loop system is—
    

1. 1. stable
      

   
   
   

   2. unstable
      

   
   
   

   3. marginally stable
      

   
   
   

   4. conditionally stable

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given
   

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

   
   H (s) = 1
   

   T.F. =	C(s)	=	G(s)
   	R(s)		1 + G(s) H(s)

   
   

   C(s)	=	1 -s/s(s + 2)
   R(s)		1 + (1 - s)/s(s + 2)× 1

   
   

   C(s)	=	1 - s
   R(s)		s2 + 2s + 1 - s

   
   

   C(s)	=	1 - s
   R(s)		s2 + s + 1

   
   the C.E. equation = s² + s + 1
   R.H.C.
   s² 1 1
   s¹ 1 0
   s⁰ 1
   since all the element in the first column are positive hence the system is stable. Alternative (A) is the correct answer.
   NOTE : In such type of questions on seeing first time it seems to be unstable system, as we realize that one zero in the R.H.S., so the system will be unstable, but actually not.

   Correct Option: A

   Given
   

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

   
   H (s) = 1
   

   T.F. =	C(s)	=	G(s)
   	R(s)		1 + G(s) H(s)

   
   

   C(s)	=	1 -s/s(s + 2)
   R(s)		1 + (1 - s)/s(s + 2)× 1

   
   

   C(s)	=	1 - s
   R(s)		s2 + 2s + 1 - s

   
   

   C(s)	=	1 - s
   R(s)		s2 + s + 1

   
   the C.E. equation = s² + s + 1
   R.H.C.
   s² 1 1
   s¹ 1 0
   s⁰ 1
   since all the element in the first column are positive hence the system is stable. Alternative (A) is the correct answer.
   NOTE : In such type of questions on seeing first time it seems to be unstable system, as we realize that one zero in the R.H.S., so the system will be unstable, but actually not.

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15. The transfer function of transportation lag is e^–sT. If the lag is small as compare to the time constant of the system, it can be approximated by—
    

1. 1. sT
      

   
   
   

   2. 1 + sT
      

   
   
   

   3. 1 – sT
      

   
   
   

   4. 1/1 + sT

   
   
   

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1. View Hint View Answer Discuss in Forum

   On expanding

   e–sT = 1 + (–sT) +	(–sT)2	+...........
   	L2

   
   e^–sT = 1 – sT neglecting higher order terms.

   Correct Option: C

   On expanding

   e–sT = 1 + (–sT) +	(–sT)2	+...........
   	L2

   
   e^–sT = 1 – sT neglecting higher order terms.

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