1. The period for the signal
   

   x(n) = e	jnπ	cos		nπ		is—
   	16			17

1. 1. 32

   
   
   

   2. 34

   
   
   

   3. 544

   
   
   

   4. 272

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Given x(n) = e	jnπ	cos		nπ
   	16			17

   
   Fundamental period of
   

   x1(n) = e	jnπ
   	16

   
   

   or x1(n) = cos	nπ	+ j sin	nπ
   	16		16

   
   

   N1 =	2πm	=	2π	· m = 32
   	Ω		π
   			16

   
   Fundamental period of
   

   x2(n) = cos			nπ
   			17

   
   

   i.e., N2 =	2πm	=	2π	· m = 34
   	Ω		π
   			17

   
   Fundamental period of the signal
   x(n) = GCM of N₁ and N₂
   = GCM of 32 and 34
   

   =	32 × 34	= 544
   	2

   
   Hence, alternative (C) is the correct choice.

   Correct Option: C

   Given x(n) = e	jnπ	cos		nπ
   	16			17

   
   Fundamental period of
   

   x1(n) = e	jnπ
   	16

   
   

   or x1(n) = cos	nπ	+ j sin	nπ
   	16		16

   
   

   N1 =	2πm	=	2π	· m = 32
   	Ω		π
   			16

   
   Fundamental period of
   

   x2(n) = cos			nπ
   			17

   
   

   i.e., N2 =	2πm	=	2π	· m = 34
   	Ω		π
   			17

   
   Fundamental period of the signal
   x(n) = GCM of N₁ and N₂
   = GCM of 32 and 34
   

   =	32 × 34	= 544
   	2

   
   Hence, alternative (C) is the correct choice.

---

2. The fundamental frequency ω0 for the given signal
   

   x(t) = 2 + cos		2πt		+ 4 sin		5πt
   		3				3

1. 1. π 6 radian

   
   
   

   2. π 3 radian

   
   
   

   3. π 2 radian

   
   
   

   4. None of these

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Given, x(t) = 2 + cos		2πt		+ 4 sin		5πt
   		3				3

   
   Fundamental period of
   

   x1(t)= T1 =	2π	= 3
   	2π
   	3

   
   Fundamental period of
   

   x2(t)= T2 =	2π	=	6	b
   	5π		5
   	3

   
   

   Ratio =	T1	=	3	=	5
   	T2		6		2
   			3

   
   or 2T₁ = 5T₂
   

   or 2.3 = 5·	6	= 6
   	5

   
   6 is the fundamental period of the signal x(t)
   So, fundamental frequency,
   

   ω0 =	2π	=	2π	=	π
   	T		6		3

   
   Hence, alternative (B) is the correct choice

   Correct Option: B

   Given, x(t) = 2 + cos		2πt		+ 4 sin		5πt
   		3				3

   
   Fundamental period of
   

   x1(t)= T1 =	2π	= 3
   	2π
   	3

   
   Fundamental period of
   

   x2(t)= T2 =	2π	=	6	b
   	5π		5
   	3

   
   

   Ratio =	T1	=	3	=	5
   	T2		6		2
   			3

   
   or 2T₁ = 5T₂
   

   or 2.3 = 5·	6	= 6
   	5

   
   6 is the fundamental period of the signal x(t)
   So, fundamental frequency,
   

   ω0 =	2π	=	2π	=	π
   	T		6		3

   
   Hence, alternative (B) is the correct choice

---

---

3. The fundamental period for the given signal
   

   x(n) = Re		e	jnπ		+ Img		e	jnπ
   			12					18

1. 1. 72

   
   
   

   2. 36

   
   
   

   3. 24

   
   
   

   4. None of these

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Given signal, x(n) = Re		e	jnπ		+ Img		e	jnπ
   			12					18

   
   

   x(n) = cos	nπ	+ j sin	nπ
   	12		18

   
   Fundamental period of
   

   x1(n)= N1 =	2π	m =	2π	· m = 24
   	Ω		17
   			12

   
   Fundamental period of
   

   x2(n)= N2 =	2π	m =	2π	· m = 36
   	Ω		17
   			18

   
   Fundamental period of the signal x[n]. Greatest common mean (GCM) of x₁[n] and x₂[n]
   GCM of 24 and 36 = 72.

   Correct Option: A

   Given signal, x(n) = Re		e	jnπ		+ Img		e	jnπ
   			12					18

   
   

   x(n) = cos	nπ	+ j sin	nπ
   	12		18

   
   Fundamental period of
   

   x1(n)= N1 =	2π	m =	2π	· m = 24
   	Ω		17
   			12

   
   Fundamental period of
   

   x2(n)= N2 =	2π	m =	2π	· m = 36
   	Ω		17
   			18

   
   Fundamental period of the signal x[n]. Greatest common mean (GCM) of x₁[n] and x₂[n]
   GCM of 24 and 36 = 72.

---

4. The signal x(t) = cos t + 2 sin √2t will be—

1. 1. Periodic with period 2π

   
   
   

   2. Non-periodic

   
   
   

   3. Periodic with period √2π

   
   
   

   4. None of these

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Given signal is x(t) = cos + 2 sin √2t
   Where, cos → x₁(t)
   2 sin √2t → x₁(t)
   Fundamental period of
   

   x1(t)= T1 =	2π	= 2π
   	20π

   
   

   x2(t)= T2 =	2π	= √2π
   	√2

   
   

   The ratio,	T1	=	2π	= √2
   	T2		√2π

   
   which cannot be expressed as ratio of integers, i.e., non-periodic.
   Hence alternative (B) is the correct choice.

   Correct Option: B

   Given signal is x(t) = cos + 2 sin √2t
   Where, cos → x₁(t)
   2 sin √2t → x₁(t)
   Fundamental period of
   

   x1(t)= T1 =	2π	= 2π
   	20π

   
   

   x2(t)= T2 =	2π	= √2π
   	√2

   
   

   The ratio,	T1	=	2π	= √2
   	T2		√2π

   
   which cannot be expressed as ratio of integers, i.e., non-periodic.
   Hence alternative (B) is the correct choice.

---

---

5. The trigonometric Fourier series of an even function of time does not have the—

1. 1. dc term

   
   
   

   2. cosine terms

   
   
   

   3. sine terms

   
   
   

   4. odd harmonic terms

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Trigonometric Fourier series of an even function of time have dc term and cosine terms.

   Correct Option: C

   Trigonometric Fourier series of an even function of time have dc term and cosine terms.

---