Pulse Analog Modulation MCQ Quiz in తెలుగు - Objective Question with Answer for Pulse Analog Modulation - ముఫ్త్ [PDF] డౌన్లోడ్ కరెన్
Last updated on Mar 13, 2025
Latest Pulse Analog Modulation MCQ Objective Questions
Top Pulse Analog Modulation MCQ Objective Questions
Pulse Analog Modulation Question 1:
The main purpose of modulation is to
Answer (Detailed Solution Below)
Pulse Analog Modulation Question 1 Detailed Solution
- The main purpose of modulation is to transmitting low-frequency information over long distance efficiently
- A low-frequency signal requires a large antenna for efficient transmission
- A low-frequency signal is easily affected by Noise
- Modulation helps to reduce the size of the antenna and transmit the message signal efficiently
Pulse Analog Modulation Question 2:
In a 100% amplitude modulated signal, if the total transmitted power is P, then the Carrier Power will be:
Answer (Detailed Solution Below)
Pulse Analog Modulation Question 2 Detailed Solution
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 μ = 1 and Pt = P
\({P} = {P_c}\left( {1 + \frac{{{1^2}}}{2}} \right)\)
\({P} = {P_c}\left( {\frac{{{3}}}{2}} \right)\)
\({P_c} = \frac{2}{3}P\)
Pulse Analog Modulation Question 3:
Consider the AM signal A(1 + μ cos (ωmt)) cos (ωct). The efficiency η of the AM signal for 75% modulation is approximately:
Answer (Detailed Solution Below)
Pulse Analog Modulation Question 3 Detailed Solution
Concept:
The transmission efficiency of an AM signal is given by the expression:
\(\eta = \frac{{{μ ^2}}}{{2 + {μ ^2}}} \times 100\)
μ = Modulation index
Calculation:
With μ = 0.75, the transmission efficiency will be:
\(\eta = \frac{{{{\left( {0.75} \right)}^2} \times 100}}{{2 + {{\left( {0.75} \right)}^2}}} \)
\(\mu= 21.9\% \approx 22\%\)
Derivation:
The transmission efficiency of an AM wave is defined as the percentage of total power contributed by the sidebands.
Mathematically it is expressed as:
\(\eta = \frac{{{P_\;}_{SB}}}{{{P_t}}} \times 100\%\)
For sinusoidal input
PSB = Sideband power given by:
\({P_{SB}} = \frac{{{P_c}\;{μ ^2}}}{2}\)
Pt = Total power given by:
\({P_t} = {P_c}\;\left( {1 + \frac{{{μ ^2}}}{2}} \right)\),
\(\eta = \frac{{{P_c}{μ ^2}}}{{2\left( {{P_c}\left( {1 + \frac{{{μ ^2}}}{2}} \right)} \right)}} = \frac{{{P_c}{μ ^2}}}{{{P_c}\left( {2 + {μ ^2}} \right)}}\)
\(\eta = \frac{{{μ ^2}}}{{2 + {μ ^2}}} \times 100\)
Pulse Analog Modulation Question 4:
Study the given input signal and match the columns.
1. |
a. PAM |
2. |
b. PWM |
3. |
c. PPM |
Answer (Detailed Solution Below)
Pulse Analog Modulation Question 4 Detailed Solution
PWM
The amplitude and position of pulses are constant in modulated signal (PWM), but the width (or duration) of pulses varies proportionally with the amplitude of an analog useful signal.
The carrier signal is produced by a clock.
PPM
The amplitude and width of pulses are constant in a modulated signal (PPM), but the direction of pulses varies proportionally with the amplitude of analogical useful signal.
PAM
The width and location of pulses in a modulated signal (PAM) are constant, while the amplitude of pulses varies proportionally with the amplitude of an analogical useful signal.
Hence from the above option (4) is correct
Pulse Analog Modulation Question 5:
A high PRF would
Answer (Detailed Solution Below)
Pulse Analog Modulation Question 5 Detailed Solution
- Pulse Repetition Frequency (PRF) of the radar system is defined as the number of pulses that are transmitted per second.
- It is normally measured in pulses per second.
- Radar systems radiate each pulse at the carrier frequency during transmit time (or Pulse Width PW), wait for returning echoes during listening or rest time, and then radiate the next pulse.
- The time between the beginning of one pulse and the start of the next pulse is called pulse-repetition time (PRT) and is equal to the reciprocal of PRF.
- Systems using high PRF i.e. above 30 kHz function better known as interrupted continuous-wave (ICW) radar because direct velocity can be measured up to 4.5 km/s at L band, but range resolution becomes more difficult.
- High PRF is limited to systems that require close-in performance, like proximity fuses and law enforcement radar.
- Higher PRFs produce shorter maximum ranges, but broadcast more pulses.
- High PRF pulse Doppler radar has no ambiguity in Doppler frequency but suffers from range ambiguities.
- A high PRF pulse Doppler radar provides an accurate measurement of target's radial velocity.
- In high PRF, It becomes increasingly difficult to take multiple samples between transmit pulses at these pulse frequencies, so range measurements are limited to short distances.
- Range and velocity can both be identified using medium PRF, but neither one can be identified directly.
- Medium PRF is from 3 kHz to 30 kHz, which corresponds with radar range from 5 km to 50 km. This is the ambiguous range, which is much smaller than the maximum range. Range ambiguity resolution is used to determine true range in medium PRF radar.
- Medium PRF is used with Pulse-Doppler radar, which is required for look-down/shoot-down capability in military systems. Doppler radar return is generally not ambiguous until velocity exceeds the speed of sound.
- A technique called ambiguity resolution is required to identify true range and speed. Doppler signals fall between 1.5 kHz, and 15 kHz, which is audible, so audio signals from medium-PRF radar systems can be used for passive target classification.
- Medium PRF has unique radar scalloping issues that require redundant detection schemes.
Low PRF:
- Systems using PRF below 3 kHz are considered low PRF because direct range can be measured to a distance of at least 50 km. Radar systems using low PRF typically produce unambiguous range.
- Unambiguous Doppler processing becomes an increasing challenge due to coherency limitations as PRF falls below 3 kHz.
- Low PRF radar have reduced sensitivity in the presence of low-velocity clutter that interfere with aircraft detection near terrain.
- Moving target indicator is generally required for acceptable performance near terrain, but this introduces radar scalloping issues that complicate the receiver.
- Low PRF radar intended for aircraft and spacecraft detection are heavily degraded by weather phenomenon, which cannot be compensated using moving target indicator.
Pulse Analog Modulation Question 6:
A signal contains components at 400 Hz and 2400 Hz This signal modulates a carrier of frequency 100 MHz. However, after demodulation, it is found that the 400 Hz signal component is present. The channel bandwidth is 15 kHz. What is the reason for the higher frequency signal not to be detected properly?
Answer (Detailed Solution Below)
Pulse Analog Modulation Question 6 Detailed Solution
In FM broadcasting, pre-emphasis improvement is the improvement in the signal-to-noise ratio of the high-frequency portion of the baseband.
In the process of detecting a frequency modulated signal, the receiver produces a noise spectrum that rises in frequency (a so-called triangular spectrum). This is the reason why a preemphasis is required.
Without pre-emphasis, the received audio would sound unacceptably noisy at high frequencies, especially under conditions of low carrier-to-noise ratio, i.e., during fringe reception conditions.
Preemphasis increases the magnitude of the higher signal frequencies, thereby improving the signal-to-noise ratio.
De-emphasis:
At the output of the discriminator in the FM receiver, a deemphasis network restores the original signal power distribution.
De-emphasis means attenuating those frequencies by the same amount by which they are boosted by the pre-emphasis.
Pulse Analog Modulation Question 7:
The plot of modulation index versus carrier amplitude for an amplitude modulated system yields
Answer (Detailed Solution Below)
Pulse Analog Modulation Question 7 Detailed Solution
Concept:
For an amplitude modulated system, the modulation index is defined as:
\(\mu=\frac{A_m}{A_c}\)
Am = Message signal amplitude
Ac = Carrier Amplitude
Observation: Since the modulation index is related to the carrier amplitude through an inverse relationship, we conclude that the plot of modulation index versus carrier amplitude will result in a hyperbola.
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.
Pulse Analog Modulation Question 8:
In a 50% amplitude modulated signal, if the total transmitted power is P, then the carrier power will be
Answer (Detailed Solution Below)
Pulse Analog Modulation Question 8 Detailed Solution
The transmitted power is given by:
\({P_t} = {P_c}\left( {1 + \frac{{{\mu^2}}}{2}} \right)\)
μ = 0.5
\(\begin{array}{l} {P_t} = {P_c}\left( {1 + \frac{{{\mu^2}}}{2}} \right)\\ {P_t} = {P_c}\left( {1 + \frac{{0.25}}{2}} \right) \end{array}\)
\( {P_t} = {P_c}\left( {1 + \frac{{0.25}}{2}} \right) \)
\(P_t= {P_c}\left( {\frac{2.25}{2}} \right) \)
\(P_c=\frac{2}{2.25}P_t\)
Pulse Analog Modulation Question 9:
In which of the following modulation systems does the increase of modulation index result in increase in bandwidth?
Answer (Detailed Solution Below)
Pulse Analog Modulation Question 9 Detailed Solution
For both frequency and phase modulation, an increase in modulation index results in an increase in transmission bandwidth.
For AM
the Modulation index is given by
\(\mu = \frac{{{A_m}}}{{{A_c}}}\)
Bandwidth = 2 fm which is independent of modulation index
For PM
the Modulation index is given by
\(\beta = {k_p}{A_m}\)
Bandwidth = 2 (1 +β ) fm
bandwidth depends on modulation index β
For FM,
modulation index
\({\rm{\beta }} = \frac{{{{\rm{k}}_{\rm{f}}}{{\rm{A}}_{\rm{m}}}}}{{{{\rm{f}}_{\rm{m}}}}}\).
Bandwidth = 2 (1 +β ) fm
bandwidth depends on modulation index β
Pulse Analog Modulation Question 10:
Double spotting in superheterodyne Receiver is caused by:
Answer (Detailed Solution Below)
Pulse Analog Modulation Question 10 Detailed Solution
- Double spotting in a superheterodyne receiver means that the wanted station is tuned in at two spots on the dial.
- This results in an undesired input frequency to be demodulated by the superheterodyne receivers along with the desired incoming signal. This results in two stations being received at the same time, thus producing interference.
- This is mainly because of poor front-end selectivity of the RF stage, i.e. due to insufficient adjacent channel rejection by the front-end RF stage.
- The concept is understood with the help of the following diagram:
Image frequency is given by fsi = fs + 2If
Where fsi = image frequency
fs = incoming signal
If = intermediate frequency