This set of Analog Communications Questions and Answers for Freshers focuses on “Coherent Detection of DSBSC Waves”.
1. Which among the following is the drawback of Pulse Position Modulation (PPM)?
a) Synchronization is required between receiver and transmitter
b) Synchronization is not required between receiver and transmitter
c) Amplitude is constant
d) Instantaneous power of PPM modulated signals is constant
Explanation: In PPM, position of the pulse of the carrier is varied with respect to the position of a reference pulse which requires synchronization between the transmitter and the receiver. It also requires large bandwidth as compared to Pulse amplitude modulation.
2. Calculate the Nyquist rate for the signal :
x(t) = 12cos50πt + 7 cos75π t – 13 cos100πt
a) 300 Hz
b) 600 Hz
c) 100 Hz
d) 150 Hz
Explanation: On comparing with general Equation, Acosωt and substituting ω = 2×π× f. We have f1 = 25 Hz, f2= 37.5Hz, f3=50Hz. Nyquist rate = 2 × fmax = = 2 × 50 = 100 Hz.
3. What is the need of doing Pre emphasis?
a) For boosting of modulating signal voltage
b) For restoring of original signal power
c) For removing amplitude variations due to noise
d) For removing frequency variations
Explanation: Pre-emphasis is generally done for boosting the amplitudes of modulating voltages at higher audio frequencies.
4. Amount of data transmitted for a given time is called _________
Explanation: Amount of data transmitted for a given time is called bandwidth. It is generally expressed in bits per second.
5. Consider an AM broadcast station which transmits modulating frequencies up to 10kHz. If it transmits a frequency of 1000 kHz. Find its maximum and minimum upper and lower sidebands frequencies and also the total bandwidth?
a) 900 KHz, 820 KHz, 1000 Hz
b) 720 KHz, 650 KHz, 1020 Hz
c) 1010 KHz, 880 KHz, 15000 Hz
d) 1010 KHz, 990 KHz, 20000 Hz
Explanation: Maximum Frequency, fm = 1000 + 10 = 1010 kHz. Minimum Frequency, fl = 1000 – 10 = 990 kHz and bandwidth = fm – fl = 1010 – 990 =20000 Hz.
6. A superheterodyne receiver receives signal within frequency range of 120 to 180 MHz. Then the required Intermediate frequency is _________
7. Carson’s rule is used to calculate ________
a) Bandwidth of FM signal
c) Modulation index
d) Figure of merit
Explanation: Carson’s rule states that the required bandwidth is equal to the twice of sum of the maximum frequency deviation and the maximum modulating frequency, B = 2(fd +fm)Hz.
8. What is the bandwidth of a FM wave when maximum allowed deviation is 50KHz and the modulating signal has a frequency of 15KHz?
a) 32.5 KHz
b) 47.8 KHz
c) 53.1 KHz
d) 63.25 KHz
Explanation: According to Carson s rule, B = 2(fd +fm) = 2 (50 + 15) = 32.5 KHz.
9. For signal, m(t) = 50cos(10 × 90t + 30 sin100t), the power dissipated by the 20Ω resistor is ________
Explanation: On Comparing with the equation, m(t) = Acos(ωct + mf sinωmt). We have A = 50, dissipated power,
10. Signal and its Hilbert transform have ________
a) same energy density spectrum
b) same power
c) a phase difference of 60°
d) a phase difference of 120°
Explanation: Properties of Hilbert transform states that the signal and its Hilbert transform :
1. have same energy density spectrum
2. are mutually diagonal
3. have same auto correlation function.
Sanfoundry Global Education & Learning Series – Analog Communications.
To practice all areas of Analog Communications for Freshers, here is complete set of 1000+ Multiple Choice Questions and Answers.