Question is ⇒ Consider the following differences between QPSK and MSK <ol style="list-style-type: decimal; padding-bottom:10px;"><li style="padding-top:10px;">The side lobes in MSK are relatively much smaller in comparison to the main lobe</li><li style="padding-top:10px;">The spectrum of MSK has a main centre lobe which is 1.5 times as wide as the main lobe of QPSK</li><li style="padding-top:10px;">The waveforms of both MSK and QPSK exhibit phase continuity</li></ol> Which of the above statements are correct?, Options are ⇒ (A) 1 only, (B) 1 and 2 only, (C) 1, 2 and 3, (D) 2 and 3 only, (E) , Leave your comments or Download question paper.

Question is ⇒ Consider the following differences between QPSK and MSK <ol style="list-style-type: decimal; padding-bottom:10px;"><li style="padding-top:10px;">The side lobes in MSK are relatively much smaller in comparison to the main lobe</li><li style="padding-top:10px;">The spectrum of MSK has a main centre lobe which is 1.5 times as wide as the main lobe of QPSK</li><li style="padding-top:10px;">The waveforms of both MSK and QPSK exhibit phase continuity</li></ol> Which of the above statements are correct?, Options are ⇒ (A) 1 only, (B) 1 and 2 only, (C) 1, 2 and 3, (D) 2 and 3 only, (E) , Leave your comments or Download question paper. Previous question Next question


Q1.	Consider the following differences between QPSK and MSK The side lobes in MSK are relatively much smaller in comparison to the main lobe The spectrum of MSK has a main centre lobe which is 1.5 times as wide as the main lobe of QPSK The waveforms of both MSK and QPSK exhibit phase continuity Which of the above statements are correct?
A.	1 only [Wrong Answer]
B.	1 and 2 only [Correct Answer]
C.	1, 2 and 3 [Wrong Answer]
D.	2 and 3 only [Wrong Answer]
View Answer 1 -1 Explanation:- Answer : B Discuss it below : !! OOPS Login [Click here] is required to post your answer/result Help other students, write article, leave your comments

Previous question Next question

Also Read Similar Questions Below :

⇒

Match the following:

	List I (μP)		List II (Number of bits in register)
A.	8085	1.	32 bit
B.	8086	2.	16 bit
C.	68000	3.	8/16 bit

A-1, B-2, C-3
A-2, B-3, C-1
A-3, B-2, C-1
A-3, B-1, C-2

⇒ The household radio receiver uses
synchronous
envelope detector
radio detector
none of the above

⇒ A thermal power plant has an efficiency of 0.4. If 0.6 kg of coal is burnt to generate 1 kWh, the calorific value of coal is
about 3500 kcal/kg
about 2500 kcal/kg
about 1500 kcal/kg
about 7500 kcal/kg

⇒ Let JCOKE = 3 and LPEPSI = 5. Consider the statement 8
1
15
10

⇒ In a stop-and-wait ARQ system, the transmitting terminal
waits for positive or negative acknowled-gement from the receiving terminal after sending a block
sends another block if positive acknowledge is received through ACK character
resends the previous block if negative acknowledgement is received through a NAK character
does not wait for acknowledgement after sending a block

⇒ A particular green LED emits light of wavelength 5490, Å, the energy bandgap of the semiconductor material used there is .. h = 6.6 x 10^-34 J sec.
2.26 eV
1.98 eV
1.17 eV
0.74 eV

⇒ Counterpoise is used in coils having
high resistivity
low resistivity
medium resistivity
both (b) and (c)

⇒ To couple a coaxial line to a parallel wire line it is best to use
slotted line
balun
directional coupler
λ/4 transformer

⇒ In figure what is the base current if V_BE = 0
10 mA
9.3 mA
10 μA
9.3 μA

⇒ A certain JK FF has t_pd = 12 nsec. The largest MOD counter that can be constructed from these FFs and still operate up to 10 MHz is
any
8
256
10

⇒ Light dependent resistors are
highly doped semiconductor
intrinsic semiconductor
lightly doped semiconductor
either (a) or (c)

⇒ If reference sound pressure P₀ is 2 x 10^-5 N/m², a sound pressure of 90 dB is equal to
0.632 N/m²
0.707 N/m²
0.835 N/m²
0.925 N/m²

⇒ Emitter follower is in which of following configuration.
common Base
common Emitter
common collector
common Base or common Emitter

⇒ In FM modulation pre-emphasis is done for
high frequency components
low frequency components
middle frequency components
both (a) and (b)

⇒ In figure the output V₀
2V₁ + V₂ - 2V₃
2(V₁ + V₂) - V₃
V₁ + V₂ + V₃
V₁ + 2V₂ - V₃

⇒ Which control action can never be used alone?
Proportional
Derivative
Integral
Both (b) and (c)

⇒ The number of protons in germanium atom is
4
16
28
32

⇒ Which of the following is a valid integer constant?
127
127
127
125 + 3

⇒ The normalised form of 0.0152 x 10¹⁸ is
0.152 x 10¹⁷
1.52 x 10¹⁶
15.2 x 10¹⁵
152 x 10¹⁴

⇒ Calculate the wave impedance for TM mode in rectangular waveguide for dominant mode at 3 GHz having
(120p) Ω
377 Ω
226 Ω
629 Ω

⇒ Two voltages are -5 V and -10 V. In positive logic
-5 V is 1 and -10 V is 0
-10 V is 1 and -5 V is 0
-5 V is 1 in some circuits and 0 in others
-10 V is 1 in some circuits and 0 in others

⇒ In single phase half wave regulator, the average current over one full cycle
is always positive
may be positive or negative
may be negative
is always negative

⇒ Which of the following ICs has only one NAND gate?
7410
7420
7430
7447

⇒ In a network with twelve circuit elements and five nodes, what is the minimum number of mesh equations?
24
12
10
8

⇒ For the circuit of the given figure, the output equatio
Y = ABCD
Y = AB + CD
Y = A + BCD
Y = ABC + D

⇒ If each frequency component of f(t) is shifted by p/2, the resulting signal is
Hilbert transform of f(t)
Inverse Hilbert transform of f(t)
Laplace transform of f(t)
Fourier transform of f(t)

⇒ A wave is propagated in a waveguide at frequency of 9 GHz and separation is 2 cm between walls find cut off wavelength for dominant mode.
4 cm
1 cm
2 cm
8 cm

⇒ 11101₂ x 10001₂ =
111101101
111101100
11110
1100110

⇒ A rectangular wave guide is designed to propagate at the dominant mode TE₁₀ at a frequency of 5 GHz. The cut-off frequency is 0.8 of signal frequency. The ratio of the guide width to height is 2. The dimensions of the guide are
3.75 cm, 1.875 cm
4 cm, 2 cm
8 cm, 4 cm
2.54 cm, 1.27 cm

⇒

Assertion (A): The dual of a circuit can be drawn by window-dot method.

Reason (R): The principle of duality is of great help in circuit analysis.

Both A and R are true and R is correct explanation of A
Both A and R are true and R is not the correct explanation of A
A is true but R is false
A is false but R is true