Bandpass Modulation MCQ || Bandpass Modulation Questions and Answers

Answer.1. M/2 kHz

Explanation:-

In the M-Array modulation scheme, the minimum bandwidth required for ideal transmission is given by:

${\left( {BW} \right)_{min}} = \frac{{{2R_b}}}{{{{\log }_2}N}}Hz$

Where,

Rb = bit rate in bps

N = number of levels in M-Array scheme

Calculation:

Given that,

Bit rate = M kbps

Number of levels = N = 16

$\therefore {\left( {BW} \right)_{{\rm{min}}}} = \frac{{{2R_b}}}{{{{\log }_2}N}}Hz = \frac{2M}{{{{\log }_2}16}}kHz$

$ = \frac{2M}{{{{\log }_2}{2^4}}}kHz$

$ = \frac{2M}{{4\; \times\; {{\log }_2}2}}kHz$

$ (BW)_{min}= \frac{M}{{2}}kHz$

So. The minimum bandwidth will be M/2 kHz, for ideal transmission.

Answer.3. (i), (ii), (iv), (iii)

Explanation:-

For ASK: BWASK = 2Rb

For M-PSK BW is calculated as:

$Bandwidth = \frac{{2{R_b}}}{{{{\log }_2}M}}$

For QPSK M = 4 bandwidth = Rb

For FSK: BW_FSK = 2Rb + |f₁ − f₂|

f1 and f2 are the two lower and upper frequencies.

for MSK bandwidth B = 1.5 Rb

f1 and f2 are the two lower and upper frequencies.

Observation:

BW_QPSK < BWMSK < BW_ASK< BWFSK

Answer.2. λ/4

Explanation:-

The length of the antenna is inversely proportional to the frequency and directly proportional to the wavelength. The higher the frequency and the shorter the wavelength, the shorter the antenna can be made.

The transmission of E_m field in space is done with the help of antennas. Antenna size depends on the wavelength. The length of the antenna is equal to λ/4.

Answer.3. Frequency shift keying

Explanation:-

• FSK is a system of frequency modulation in which the nominal unmodulated carrier frequency corresponds to the mark condition, and space is represented by a downward frequency shift
• In the FSK generator, the frequency shift may be obtained by applying the varying dc output of the telegraph machine to a varactor diode in a crystal oscillator
• At the receiving end, the signal is demultiplexed and applied to a standard phase discriminator
• This is how a telegraph works with FSK modulation

Answer.4. 222

Explanation:-

Observe the first 5 consecutive 0s

Summation rule: the first 5 consecutive 0’s could start at position 1, 2, 3, 4, 5, or 6

Start at the first position

The other 5 bits can be anything: 2⁵ = 32

Start at the second position

The first bit must be a 1

There are various possibilities that can be included in it.

The remaining bits can be anything: 2⁴ = 16

Start at third position

The second bit must be a 1 due to the above-mentioned reason.

First bit and last 3 bits can be anything: 2⁴ = 16

Starting at 4,5 and 6 positions

Same as starting at positions 2 or 3: 16 each

Total = 32 + 16 + 16 + 16 + 16 + 16 = 112

The five consecutive 1’s ensue the same pattern and have different like 112 possibilities

There would be two cases counted twice (that we thus need to exclude): 0000011111 and 1111100000

Total = 112 + 112 – 2 = 222

Answer.1. Coherent detection

Explanation:-

Coherent detection originates from radio communications, where a local carrier mixes with the received radio frequency (RF) signal to generate a product term. As a result, the received RF signal can be frequency translated and demodulated.

When the receiver uses the carrier’s phase as the major factor for detection then it is called coherent detection and when the carrier is not given the importance it is called non-coherent detection.

Answer.3. QAM

Explanation:-

QAM is a mixture of both ASK and PSK.

2) QAM uses two carrier signals with the same frequency but with which are in quadrature. Quadrature here means out phase by 90°.

3) Since the two signals have the same frequency, they are detected using synchronous detection.

4) Hence, amplitude and the phase of the carrier frequency both vary with the message signal.

Answer.1. Actual Carrier

Explanation:-

• PSK: It stands for phase shift keying.
• In PSK, binary 1 is represented with the actual carrier, and 0 is represented with the 180^o phase shift of carrier.
• 1 is represented as  s(t) = Accos(2πf_ct)
• 0 is represented as  s(t) = – Accos(2πf_ct)
• As in PSK, one bit is transmitted in a specific time constant.
• Similarly, in 16 PSK, log₂(16) = 4 bits are transmitted in a specific time constant which is known as 16-ary PSK.

Answer.4. All of the mentioned

Explanation:-

Coherent Modulation is a technique that uses modulation of Amplitude and Phase of light, as well as transmission across two polarizations to enable transport of more information across the optical fiber.

Some of the examples of coherent modulation techniques are phase-shift keying, amplitude shift keying, frequency-shift keying, and continuous phase modulation.

Answer.2. ASK

Explanation:-

In ASK two states (logic high and logic low) in the envelope (modulated signal) exist. A 1 represents logic high, and a 0 represents logic low. ASK can provide a high data rate but has low noise immunity.