156. What is the number of roots of the polynomial F(z) = 4z³ – 8z² – z + 2, lying outside the unit circle?

1. 1. 0

   
   
   

   2. 1

   
   
   

   3. 2

   
   
   

   4. 3

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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157. Causal sequence f(n) if f(z) =	4·z– 3
     	(1 – z– 1)2

1. 1. 4(n – 2) u(n – 2)

   
   
   

   2. 4(n – 1) u(n – 1)

   
   
   

   3. 4(n – 3) u(n – 3)

   
   
   

   4. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   In order to solve such type of very confusing objective problems, start with option (A) and carried to the final option until the required value given in question meet.
   (A) 4(n – 2) u(n – 2)
   We know that
   

   x(n) = u(n) ←z→	z	= X(z)
   	z - 1

   
   

   nu(n) ↔ – z	d	X(z)
   	dz

   
   

   = – z	d		z
   	dz		z - 1

   
   

   = – z	(– 1)
   	(z – 1)2

   
   

   z

   (z – 1)2

   
   

   z

   (z – 1)2

   
   

   z-1

   (1 - z– 1)2

   
   

   4(n – 2) u(n – 2) ↔ 4z–2		z-1
   		(1 - z– 1)2

   
   

   ↔	4z–3
   	(1 – z–1)2

   
   As we got the required solution in first attempt, so no need to solve for further options given in the problem.
   Hence, obviously alternative (A) is the correct choice.

   Correct Option: A

   In order to solve such type of very confusing objective problems, start with option (A) and carried to the final option until the required value given in question meet.
   (A) 4(n – 2) u(n – 2)
   We know that
   

   x(n) = u(n) ←z→	z	= X(z)
   	z - 1

   
   

   nu(n) ↔ – z	d	X(z)
   	dz

   
   

   = – z	d		z
   	dz		z - 1

   
   

   = – z	(– 1)
   	(z – 1)2

   
   

   z

   (z – 1)2

   
   

   z

   (z – 1)2

   
   

   z-1

   (1 - z– 1)2

   
   

   4(n – 2) u(n – 2) ↔ 4z–2		z-1
   		(1 - z– 1)2

   
   

   ↔	4z–3
   	(1 – z–1)2

   
   As we got the required solution in first attempt, so no need to solve for further options given in the problem.
   Hence, obviously alternative (A) is the correct choice.

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158. The causal sequence f(n) if f(z) =	1
     		1 -	1	z-1		2
     			2

1. 1. 			1		n	+ n		1		n		u[n]
      			2					2

   
   
   

   2. 			1		n	+	n		1		n		u[n]
      			2				2		2

   
   
   

   3. 			1		n	– n		1		n		u[n]
      			3					3

   
   
   

   4. 			1		n	+ n		1		n		u[n]
      			3					3

   
   
   

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1. View Hint View Answer Discuss in Forum

   We will apply same approach as discussed in above problem.
   

   (A)			1		n	+ n		1		n		u(n)
   			2					2

   
   

   ∴		1		n	←z→		1
   		2					1 -	1	z-1
   								2

   
   

   and n		1		n	←z→		z-1
   		2						1 -	1	z-1		2
   									2

   
   

   Now,			1		n	+ n		1		n		u(n)
   			2					2

   
   

   =	1				+	z-1
   	1 -	1	z-1				1 -	1	z-1		2
   		2						2

   
   
   

   =	1 –	1	z– 1	+ z– 1
   		2
   		1 –	1	z– 1		2
   			2

   
   
   

   =	1 +	1	z– 1
   		2
   		1 –	1	z– 1		2
   			2

   
   Since, (A) is not the correct choice, so we will solve for next option, i.e., (B).
   

   			1		n	+	n		1		n		u[n]
   			2				2		2

   
   

   ↔	1				+		z-1
   	1 -	1	z-1			2		1 -	1	z-1		2
   		2							2

   
   
   

   ↔		1 –	1	z– 1		+ z– 1
   			2
   		1 –	1	z– 1		2
   			2

   
   

   ↔		2 – z– 1 + z– 1
   	2		1 -	1	z-1		2
   				2

   
   

   ↔		2
   	2		1 -	1	z-1		2
   				2

   
   

   ↔		1
   			1 -	1	z-1		2
   				2

   
   Therefore, option (B) is the correct choice.
   

   Correct Option: B

   We will apply same approach as discussed in above problem.
   

   (A)			1		n	+ n		1		n		u(n)
   			2					2

   
   

   ∴		1		n	←z→		1
   		2					1 -	1	z-1
   								2

   
   

   and n		1		n	←z→		z-1
   		2						1 -	1	z-1		2
   									2

   
   

   Now,			1		n	+ n		1		n		u(n)
   			2					2

   
   

   =	1				+	z-1
   	1 -	1	z-1				1 -	1	z-1		2
   		2						2

   
   
   

   =	1 –	1	z– 1	+ z– 1
   		2
   		1 –	1	z– 1		2
   			2

   
   
   

   =	1 +	1	z– 1
   		2
   		1 –	1	z– 1		2
   			2

   
   Since, (A) is not the correct choice, so we will solve for next option, i.e., (B).
   

   			1		n	+	n		1		n		u[n]
   			2				2		2

   
   

   ↔	1				+		z-1
   	1 -	1	z-1			2		1 -	1	z-1		2
   		2							2

   
   
   

   ↔		1 –	1	z– 1		+ z– 1
   			2
   		1 –	1	z– 1		2
   			2

   
   

   ↔		2 – z– 1 + z– 1
   	2		1 -	1	z-1		2
   				2

   
   

   ↔		2
   	2		1 -	1	z-1		2
   				2

   
   

   ↔		1
   			1 -	1	z-1		2
   				2

   
   Therefore, option (B) is the correct choice.
   

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159. Which one of the following gives the average value or expectation of the random function g(x) of the random variable X? (Given: f(x) is the probability density function)

1. 1. E[g(x)] =		∞	g(x)dx
      		– ∞

   
   
   

   2. E[g(x)] =		∞	g(x).f(x)dx
      		– ∞

   
   
   

   3. E[g(x)] =		∞	g*(x)dx
      		– ∞

   
   
   

   4. E[g(x)] =		∞		g(x)		dx
      		– ∞		f(x)

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

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160. Match List-I (Time Domain Property) with List-II (Frequency Domain Property) pertaining to Fourier representation periodicity properties and select the correct answer using the codes given below the Lists:
     

     List-I (Time Domain Property)	List-II (Frequency Domain Property)
     A. Continuous	1. Periodic
     B. Discrete	2. Continuous
     C. Periodic	3. Non-periodic
     D. Non-periodic	4. Discrete

1. 1. A B C D
      3 4 1 2

   
   
   

   2. A B C D
      2 4 1 3

   
   
   

   3. A B C D
      2 1 4 3

   
   
   

   4. A B C D
      3 1 4 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Time domain	Frequency domain
   A. Continuous	Non-periodic
   B. Discrete	Periodic
   C. Periodic	Discrete
   D. Non-periodic	Continuous.

   Correct Option: D

   Time domain	Frequency domain
   A. Continuous	Non-periodic
   B. Discrete	Periodic
   C. Periodic	Discrete
   D. Non-periodic	Continuous.

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