1) FM is advantageous over AM as
a. The amplitude of FM is constant. So transmitter power remains unchanged in FM but it changes in AM
b. The depth of modulation in FM can be changed to any value by changing the frequency deviation. So the signal is not distorted
c. There is less possibility of adjacent channel interference due to presence of guard bands
d. All of the above
Answer
Explanation
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ANSWER: All of the above
Explanation: The amplitude of FM is constant. So transmitter power remains unchanged in FM but it changes in AM. The depth of modulation in FM can be changed to any value by changing the frequency deviation. So the signal is not distorted. There is less possibility of adjacent channel interference due to presence of guard bands so interference of noise is minimum in FM.
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2) VCO is used to generate
a. Direct FM
b. Indirect FM
c. SSB-SC
d. DSB-SC
Answer
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ANSWER: Direct FM
Explanation: Voltage controlled oscillator is used to generate Direct FM signal. In this type of modulation, the frequency of the carrier is varied directly by the instantaneous voltage of the modulating signal i.e., the instantaneous frequency of the output signal is directly proportional to the voltage of the input modulating signal. Direct FM modulation is obtained by directly giving the modulating signal as an input to the VCO.
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3) Change in instantaneous phase of the carrier with change in amplitude of the modulating signal generates
a. Direct FM
b. Indirect FM
c. SSB-SC
d. DSB-SC
Answer
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ANSWER: Indirect FM
Explanation: Indirect FM is obtained by modulation of the phase of the carrier i.e., the instantaneous phase of the carrier varies with the amplitude of the modulating signal. Crystal controlled oscillators are used to generate Indirect FM and then the result is passed through a frequency multiplier to convert the signal into wideband FM signal.
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4) Phase-locked loop can be used as
a. FM demodulator
b. AM demodulator
c. FM receiver
d. AM receiver
Answer
Explanation
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ANSWER: FM demodulator
Explanation: PLL or Phase Locked Loop is used as a FM demodulator. It is a high performance FM detector due to its key advantages such as Linearity and low manufacturing costs.
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5) The increase or decrease in the frequency around the carrier frequency is termed as
a. Figure factor
b. Frequency deviation
c. Modulation index
d. Frequency pectrum
Answer
Explanation
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ANSWER: Frequency deviation
Explanation: In FM, as the amplitude of the modulating signal changes, the frequency of the carrier changes from its central frequency. This is known as Frequency Deviation.
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6) Carson's rule is used to calculate
a. Bandwidth of FM signal
b. Signal to noise ratio
c. Modulation index
d. Noise figure
Answer
Explanation
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ANSWER: Bandwidth of FM signal
Explanation: According to Carson s rule, the bandwidth required is twice the sum of the maximum frequency deviation and the maximum modulating signal frequency. Or, B=2(Δf +fm) Hz.
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7) The ratio of maximum peak frequency deviation and the maximum modulating signal frequency is termed as
a. Frequency deviation
b. Deviation ratio
c. Signal to noise ratio
d. Frequency spectrum
Answer
Explanation
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ANSWER: Deviation ratio
Explanation: In Frequency Modulation, Deviation ratio is the ratio of maximum peak frequency deviation and the maximum modulating signal frequency. Mathematically, the Deviation Ratio (DR) is given by DR = Δf (max) /fm (max)
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8) The equation v(t) = A cos [ωct + kp Φ(t)] represents the signal as
a. Phase modulated signal
b. SSBSC signal
c. DSB SC signal
d. None of the above
Answer
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ANSWER: Phase modulated signal
Explanation: A phase modulated wave is given by v(t) = A cos [ωct + kp Φ(t)] A = amplitude Φ(t) = baseband signal ωc = carrier frequency kp = phase sensitivity of the modulator.
Where, A is a constant that represents Amplitude and ωc represents carrier frequency and kp is the phase sensitivity of the modulator.
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9) Calculate the maximum frequency deviation for the FM signal v(t) = 10 cos (6000t+ 5sin2200t)
a. 2200 Hz
b. 6000 Hz
c. 1750 Hz
d. 11000 Hz
Answer
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ANSWER: 1750 Hz
Explanation: A standard FM signal is represented by v(t) = Ac cos(2πfct + kfsin2πfmt) Ac = carrier amplitude fc = carrier frequency kf = modulation index fm = modulating frequency = 2200/2π = 350 Hz kf = frequency deviation/modulating frequency 5 = freq deviation/ 350 Therefore, deviation = 5 * 350 = 1750Hz
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10) Calculate the dissipation in power across 20Ω resistor for the FM signal v(t)= 20 cos(6600t+ 10sin2100t)
a. 5W
b. 20W
c. 10W
d. 400W
Answer
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ANSWER: 5W
Explanation: A standard FM signal is represented by v(t) = Ac cos(2πfct + kfsin2πfmt) Ac = carrier amplitude fc = carrier frequency kf = modulation index fm = modulating frequency kf = frequency deviation/modulating frequency the power dissipated across 20Ω resistor is given by Vrms2/R =(20/√2)2/R = 5W
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11) What is the value of carrier frequency in the following equation for the FM signal? v(t)= 5 cos(6600t+ 12sin2500t)
a. 1150 Hz
b. 6600 Hz
c. 2500 Hz
d. 1050 Hz
Answer
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ANSWER: 1050 Hz
Explanation: A standard FM signal is represented by v(t) = Ac cos(2πfct + kfsin2πfmt) Ac = carrier amplitude fc = carrier frequency kf = modulation index fm = modulating frequency kf = frequency deviation/modulating frequency therefore, fc = 6600/2π = 1050Hz
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12) Calculate the modulation index in an FM signal when fm (modulating frequency) is 250Hz and Δf (frequency deviation) is 5KHz.
a. 20
b. 35
c. 50
d. 75
Answer
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ANSWER: 20
Explanation: Modulation index is the measure of how much the modulation parameter changes from its un modulated value. The modulation index of FM is given by μ = frequency deviation/ modulating frequency = Δf/ fm Where Δf is the peak frequency deviation i.e. the deviation in the instantaneous value of the frequency with modulating signal. fm is the value of modulating frequency μ = 5000/250 = 20
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13) After passing the FM signal through mixer, what is the change in the frequency deviation Δ when the modulating frequency is doubled?
a. Becomes 2 Δ
b. Becomes Δ /2
c. Becomes Δ2
d. Remains unchanged
Answer
Explanation
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ANSWER: Remains unchanged
Explanation: When a carrier is Frequency modulated, the output signal has the frequency deviated above and below the carrier frequency, this is known as Frequency Deviation. When the FM signal is passed through a mixer, the mixer changes the carrier frequency but the deviation remains unchanged.
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14) In frequency modulation,
a. Armstrong method is used for generation
b. Multiple side bands are generated
c. The FM signal has infinite bandwidth
d. All of the above
Answer
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ANSWER: All of the above
Explanation: Armstrong method is used for the generation of FM signal. The FM signal has a high quality of audio signal and reduced noise as compared to the AM wave. In FM, multiple side bands are generated therefore has infinite bandwidth though only the first few sets of sidebands are useful for detection of signal.
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15) Maximum frequency deviation and the maximum bandwidth allowed for commercial FM broadcast is
a. 80KHz, 160Khz
b. 75KHz, 200Khz
c. 60KHz, 170Khz
d. 75KHz, 250Khz
Answer
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ANSWER: 75KHz, 200Khz
Explanation: Maximum frequency deviation allowed for commercial FM broadcast is 75KHz. Therefore approximate bandwidth allowed is 2 * 75 = 150KHz. Guard bands of 25KHz are allowed on the either sides so the channel width becomes 2 (75 + 25) = 200KHz. The guard bands are added to prevent the interference between adjacent channels.
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16) Guard bands are provided in FM signal to
a. Prevent interference from adjacent channels
b. To increase the noise
c. To increase bandwidth
d. None of the above
Answer
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ANSWER: Prevent interference from adjacent channels
Explanation: The guard bands are provided to prevent the interference between adjacent channels in FM signals. Guard bands of 25KHz are allowed on the either sides so the channel width becomes 2 (75 + 25) = 200KHz where 75KHz is the maximum permissible frequency deviation allowed for commercial FM broadcast.
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17) For a FM signal v(t) = 15 cos ( 10 * 108t + 10 sin 1220t), calculate
1. Carrier frequency 2. Modulating frequency
a. 159.1MHz, 194.1Hz
b. 185.5MHz, 200.15Hz
c. 350.1MHz, 200.1Hz
d. 159.1Hz, 194.1Hz
Answer
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ANSWER: 159.1MHz, 194.1Hz
Explanation: Standard expression for FM signal is given by v(t) = A cos ( ωct + mf sin ωmt) comparing with the given equation, Carrier frequency is given by fc = 10 * 108/2Π = 159.1MHz Modulating frequency is given by fm = 1220/2Π = 194.1Hz
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18) For a FM signal v(t) = 25 cos (15 * 108t + 10 sin 1550t), calculate Modulation index Maximum frequency deviation
a. 10, 3000.1Hz
b. 20, 1550.9Hz
c. 10, 2465.9Hz
d. 10, 2000.0Hz
Answer
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ANSWER: 10, 2465.9Hz
Explanation: Standard expression for FM signal is given by v(t) = A cos ( ωct + mf sin ωmt) Comparing with the given equation, Modulation index mf = 10 Maximum frequency deviation is given by mf = Δf/fm Δf = mf * fm Here fm = 1550/2Π = 246.59 Hz Δf = 10 * 246.59 = 2465.9Hz
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19) For a FM signal v(t) = 20 cos ( 10 * 108t + 30 sin 3000t), calculate the power dissipated by the FM wave in a 20Ω resistor.
a. 100 Watts
b. 10 Watts
c. 200 Watts
d. 20 Watts
Answer
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ANSWER: 10 Watts
Explanation: Standard expression for FM signal is given by v(t) = A cos ( Ωct + mf sin Ωmt) Comparing with the given equation, A = 20 The dissipated power is given by P = V2rms/R = (20/√2)2/ 20 = 10Watts
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20) A 100MHz carrier is frequency modulated by 5 KHz wave. For a frequency deviation of 100 KHz, calculate the carrier swing of the FM signal.
a. 2000 KHz
b. 100 KHz
c. 105 KHz
d. 200 KHz
Answer
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ANSWER: 200 KHz
Explanation: Carrier frequency fc = 100MHz Modulating frequency fm = 5 KHz Frequency deviation Δf = 100 KHz Carrier swing of the FM signal = 2 * Δf = 2 * 100 = 200 KHz
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