Pulse Modulation MCQ Quiz in मल्याळम - Objective Question with Answer for Pulse Modulation - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

• The Noise affects the amplitude of the modulated signal
• In the Amplitude Modulated system, the message is contained in the amplitude modulations of the carrier; hence the introduction of noise distorts the amplitude and hence the message contained in the signal
• In a frequency modulated system the message is contained in the frequency variations of the carrier, therefore, the introduction of noise does not affect the message contained in the message signal
• Since the amplitude of FM remains constant, the noise can be eliminated using an amplitude limiter at the demodulator.

• Analog modulation is a process in which analog low-frequency baseband signals (like an audio or TV signal) are transmitted over a larger distance without getting faded away, by superimposing over a higher frequency carrier signal such as a radio frequency band.
• Different modulation techniques are explained with the help of the following block diagram:

• If the amplitude of a pulse or duration of a pulse is varied according to the instantaneous values of the baseband modulating signal, then such a technique is called as Pulse Amplitude Modulation (PAM).

The bit rate for a PCM system is given by

R_b = nf_s

Where

R_b = Bit rate

n = number of bits in encoder

f_s = sampling frequency

From Nyquist criteria

f_s = 2f_m

Where f_m = message signal frequency

R_b = nf_s

60 MHz = 6 × 2 × f_m

Concept:

The total transmitted power for an AM system is given by:

\({P_t} = {P_c}\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

Pc = Carrier Power

μ = Modulation Index

Calculation:

Given P_t = 600 W, P_c = 400 W

Putting these values in the total power of AM wave expression, we get:

\(600 = 400\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

\(\frac{6}{4}=\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

\(\frac{6}{4}=\left( {1 + \frac{{{μ^2}}}{2}} \right)\)

\(\frac{μ^2}{2}=\frac{1}{2}\)

μ = 1.00

In Amplitude modulated system

Power in side band is given by

\({P_{SB}} = {P_c}\left( {\frac{{{\mu ^2}}}{2}} \right)\)

Hence

\(\frac{{{P_{SB}}}}{{{P_c}}} = \left( {\frac{{{\mu ^2}}}{2}} \right)\)

\(\frac{{50}}{{400}} = \left( {\frac{{{\mu ^2}}}{2}} \right)\)

Concept:

\(B.W = \frac{{{R_b}}}{{{{\log }_2}M}}\)

Where,

Rb = nf_s

n = no. of bits

M = no. of levels of PAM

Calculation:

f_m = 10 kHz

Nyquist sampling rate = 2f­_m = 20 kHz.

The signal is sampled at 20% above Nyquist rate:

∴ f_s = (1.2 × 20) = 24 kHz.

No. of Quantization levels

= 1024

No. of bits required = log₂ (1024)

= log₂ 2¹⁰ = 10 bits.

∴ \(B.W = \frac{{{R_b}}}{{{{\log }_2}M}}\)

\(= \frac{{10\; \times \;24}}{{ \;{{\log }_2}8}}\)

\(= \frac{{240}}{3}\)

= 80 kHz.

• In PCM an analog signal is sampled and encoded into different levels before transmission
• The bandwidth of PCM depends on the number of levels If each sample is encoded into n bits, then the bandwidth of PCM is nfs
• However, in the case of Delta modulation, each sample is sent using only 1 bit which is +Δ or -Δ Hence there is bandwidth saving in Delta modulation​
• DM has a simple hardware requirement in comparison to PCM.

A comparison of different modulation schemes is as shown in the table below:

Analysis:

A general expression for a phase-modulated wave is:

xPM (t) = A cos [2πfct + kpm(t)]

The instantaneous angle is given as:

ϕi(t) = 2πfct + kp m(t)

Phase deviation is, therefore:

\(\Delta [ϕ_i(t)] = k_p m(t)\)

Conclusion: The phase deviation is proportional to the message signal.

In phase modulation, if the question was asking about the frequency deviation, then it will be proportional to the frequency of the message signal as well. But the phase deviation is proportional to the amplitude of the message signal only as derived above.

Important Points

A wave has 3 parameters Amplitude, Phase, and Frequency. Thus there are 3 types of modulation techniques.

Amplitude Modulation:

The amplitude of the carrier is varied according to the amplitude of the message signal.

Frequency Modulation:

The frequency of the carrier is varied according to the amplitude of the message signal.

Phase Modulation:

The Phase of the carrier is varied according to the amplitude of the message signal.

To reduce the Quantization noise two things are done.

Oversampling and Noise shaping.

Explanation:

• An analog signal first undergoes the process of sampling before it is applied to ADC for conversion into a digital signal.
• Oversampling is a process in which an analog signal is sampled at a sampling frequency that is much greater than the Nyquist rate.

• Oversampling is:

  fs' = K × fs , ( K > 1)

• A Sigma-Delta ADC is an example of an ADC that employs oversampling.
• It is the same as “Delta modulation”. That is 1 sample is represented with 1 bit.
• In oversampling correlation increases and Quantization noise decreases.

Delta Modulation: 

A signal is Delta modulated to convert its analog information into a binary sequence, i.e., 1s and 0s. Here only one bit is required to represent the analog input. An over-sampled input is taken to make full use of the signal correlation. A stair-case approximated waveform will be the output of the delta modulator with the step-size as delta (Δ).

Delta Modulator has two main drawbacks:

1. Slope overload distortion: When the rate of rising of the input signal x(t) is so high that the staircase signal can not approximate it. We will see consecutive pulses of the same polarity in the output for such type of distortion. 

2. Granular noise: When the step size is too large compared to a small variation in the input signal (input is almost constant). Here, the output of the Delta modulator will be a sequence of alternate +ve and -ve pulses. 

∴ We can conclude that if the consecutive pulses are of opposite polarity during any time interval then the signal is almost constant during that interval.