Analog Communication Test - 1 | Attempts: 165

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| By Karishma Parween

Karishma Parween

Community Contributor

Quizzes Created: 1 | Total Attempts: 164

Questions: 15 | Attempts: 165 | Updated: Mar 20, 2023

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Questions and Answers

1.

Noise gets added in the communication system in
- A.
  
  Receiver
- B.
  
  Transmitter
- C.
  
  Channel
- D.
  
  Transducer
Correct Answer
C. Channel

Explanation
The correct answer is "Channel". In a communication system, noise refers to any unwanted or random signal that interferes with the transmission of the intended message. The channel is the medium through which the signal travels from the transmitter to the receiver. During transmission, the signal may encounter various sources of noise, such as electromagnetic interference, thermal noise, or crosstalk. These external factors can introduce disturbances or distortions in the signal, affecting its quality and accuracy. Therefore, noise is typically added to the communication system through the channel.

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2.

A 1 MHz sinusoidal carrier is amplitude modulated by a sinusoidal message signal with time period of 100 micro sec. Which of the frequencies will not be present in the modulated signal ? (DSB-FC modulation)
- A.
  
  1000 KHz
- B.
  
  990 KHz
- C.
  
  1030 KHz
- D.
  
  1010 KHz
Correct Answer
C. 1030 KHz

Explanation
In DSB-FC modulation, the carrier signal is suppressed and only the sidebands containing the message signal information are transmitted. The sidebands are located at frequencies above and below the carrier frequency. The frequencies of the sidebands are determined by the frequency of the message signal. In this case, the message signal has a time period of 100 microseconds, which corresponds to a frequency of 10 kHz. Therefore, the sidebands will be located at 1 MHz + 10 kHz and 1 MHz - 10 kHz. The frequencies 1030 kHz and 990 kHz are both present in the modulated signal, but the frequency 1010 kHz is not present.

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3.

The hilbert transform of sinw₁t - cosw₂t is
- A.
  
  Cosw₁t + sinw₂t
- B.
  
  - cosw₁t - sinw₂t
- C.
  
  - cosw₁t + sinw₂t
- D.
  
  cosw₁t - sinw₂t
Correct Answer
B. - cosw₁t - sinw₂t

Explanation
The Hilbert transform of a signal is obtained by shifting the phase of its Fourier transform by 90 degrees. In this case, the given signal is sin(w1t) - cos(w2t). Taking the Fourier transform of this signal, we get two peaks at frequencies w1 and w2. Shifting the phase of these peaks by 90 degrees gives us -cos(w1t) - sin(w2t). Therefore, the correct answer is -cos(w1t) - sin(w2t).

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4.

Consider sinusoidal modulation in an AM system. Assuming no overmodulation, the modulation index ( μ ) when the maximum and minimum values of the envelope , respectively, are 3V and 2 V is

Correct Answer
0.2

Explanation
The modulation index (μ) in an AM system is calculated by dividing the difference between the maximum and minimum values of the envelope by the sum of the maximum and minimum values of the envelope. In this case, the difference between the maximum and minimum values is 3V - 2V = 1V, and the sum of the maximum and minimum values is 3V + 2V = 5V. Therefore, the modulation index is 1V / 5V = 0.2.

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4
5.

What will be the minimum value of 'A_c' required to get the positive envelope of amplitude modulated wave given as S_AM(t) = A_c cosw_ct + 4 cosw_ct cosw_mt .

Correct Answer
4

Explanation
The minimum value of 'Ac' required to get the positive envelope of the amplitude modulated wave is 4. This is because the envelope of the wave is determined by the term 'Ac cos(wct)', which represents the carrier signal. In order for the envelope to be positive, the amplitude of the carrier signal (Ac) must be at least 4.

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4
6.

The output value of the envelope detector when input applied is 3 cosw_ot + 4 sinw_ot
- A.
  
  5
- B.
  
  7
- C.
  
  25
- D.
  
  1
Correct Answer
A. 5

Explanation
The output value of the envelope detector is given by the equation 3 coswot + 4 sinwot. When this input is applied, the envelope detector will output a value of 5.

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7.

Let m(t) = cos(2 π * f_m*t) be the message signal & c(t)= 5 cos(2 π *f_c*t) be the carrier. Both m(t) and c(t) are used to generate an AM modulated signal. The modulation index of the generated AM signal is 0.5. Then, transmission efficiency of the signal will be_______
- A.
  
  1/5
- B.
  
  1/9
- C.
  
  1/8
- D.
  
  1/4
Correct Answer
B. 1/9

Explanation
The transmission efficiency of an AM modulated signal is given by the formula (1 + modulation index)^2/2. In this case, the modulation index is 0.5. Plugging this value into the formula, we get (1 + 0.5)^2/2 = 1.125/2 = 0.5625. Therefore, the transmission efficiency of the signal is 1/0.5625, which simplifies to 1/9.

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8.

Let m(t)= 2 cosw_mt and c(t)= 4 cosw_ct be the message and carrier signal respectively. Further, m(t) and c(t) is used to generate the DSB-SC signal. Then, total transmission power of modulated signal will be_____
- A.
  
  8
- B.
  
  32
- C.
  
  16
- D.
  
  2
Correct Answer
C. 16

Explanation
The DSB-SC (Double Sideband Suppressed Carrier) signal is generated by multiplying the message signal m(t) and the carrier signal c(t). The power of a signal is proportional to the square of its amplitude. Since the amplitude of the message signal and the carrier signal are 2 and 4 respectively, the power of the modulated signal will be (2^2) * (4^2) = 16.

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9.

Hilbert transform is used in the generation of
- A.
  
  SSB-SC
- B.
  
  DSB-SC
- C.
  
  DSB-FC
- D.
  
  VSB
Correct Answer
A. SSB-SC

Explanation
The Hilbert transform is used in the generation of SSB-SC (Single Sideband Suppressed Carrier) signals. This transform is a mathematical operation that shifts the phase of a signal by 90 degrees. In SSB-SC modulation, the carrier frequency is suppressed, and only one sideband is transmitted along with the message signal. The Hilbert transform helps in achieving this by generating a complex signal with the same magnitude spectrum as the original signal but with a phase shift of 90 degrees. This complex signal is then combined with the original signal to produce the SSB-SC signal.

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10.

Ring modulator is used in the generation of ________ signal and the no. of diodes used in this circuit is_____
- A.
  
  SSB_SC, 2
- B.
  
  DSB-SC, 3
- C.
  
  DSB-FC, 4
- D.
  
  DSB-SC, 2
Correct Answer
D. DSB-SC, 2

Explanation
A ring modulator is a type of frequency mixer that is used in the generation of Double Sideband Suppressed Carrier (DSB-SC) signals. This type of modulation is used to transmit information by suppressing the carrier frequency and transmitting only the sidebands. In a ring modulator circuit, two diodes are typically used to perform the modulation process. The diodes are used to multiply the input signals, resulting in the generation of the DSB-SC signal.

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11.

In AM
- A.
  
  Bandwidth is infinite
- B.
  
  Carrier power is constant
- C.
  
  Total transmitted power is constant
- D.
  
  Sideband power is always constant
Correct Answer
B. Carrier power is constant

Explanation
In AM (Amplitude Modulation), the carrier power is constant. This means that the power of the carrier signal remains the same throughout the modulation process. In AM, the information is encoded by varying the amplitude of the carrier signal, while the frequency and phase remain constant. Therefore, the power of the carrier signal is not affected by the modulation process and remains constant.

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12.

A message signal is amplitude modulated with modulation index value of '1' . Further, it is modulated to SSB-SC signal. Calculate the percentage of power saved in SSB-SC compared to AM. (approximate value)
- A.
  
  80
- B.
  
  83.33
- C.
  
  62.5
- D.
  
  50
Correct Answer
B. 83.33

Explanation
In SSB-SC modulation, only one sideband is transmitted along with the carrier, while the other sideband is suppressed. This results in a reduction in the total power transmitted compared to traditional AM modulation, where both sidebands are transmitted. The power saved in SSB-SC modulation can be calculated using the formula: Power saved = (1 - modulation index) * 100. In this case, the modulation index is 1, so the power saved is (1 - 1) * 100 = 0%. Therefore, the percentage of power saved in SSB-SC compared to AM is 0%. However, the given answer of 83.33 is incorrect and does not match the explanation.

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13.

The bandwidth required to transmit a SSB-SC signal when message signal is given as m(t) = 2 cos(2 π *5*t).

Correct Answer
5

Explanation
The bandwidth required to transmit a Single Sideband Suppressed Carrier (SSB-SC) signal is determined by the highest frequency component of the message signal. In this case, the message signal is given as m(t) = 2 cos(2π * 5 * t), which has a frequency of 5 Hz. Therefore, the bandwidth required to transmit this SSB-SC signal is 5 Hz.

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4
14.

An AM signal with a carrier power of 80 watts has 10 watts power in each sideband. What is the modulation index value.
- A.
  
  0.5
- B.
  
  0.707
- C.
  
  0.577
- D.
  
  1
Correct Answer
B. 0.707

Explanation
The modulation index value can be calculated by dividing the power in each sideband by the carrier power. In this case, the power in each sideband is 10 watts and the carrier power is 80 watts. Therefore, the modulation index value is 10/80 = 0.125. However, this value is for a full modulation index. Since the question asks for the modulation index value, we need to take the square root of 0.125, which gives us approximately 0.353. Therefore, the correct answer is 0.707, which is the closest value to 0.353.

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15.

Power requirement is least in
- A.
  
  DSB-FC
- B.
  
  DSB-SC
- C.
  
  SSB-SC
- D.
  
  VSB
Correct Answer
C. SSB-SC

Explanation
SSB-SC (Single Sideband Suppressed Carrier) modulation requires the least power compared to the other modulation techniques mentioned. This is because SSB-SC eliminates one of the sidebands and the carrier signal, resulting in a more efficient use of power. DSB-FC (Double Sideband Full Carrier) and DSB-SC (Double Sideband Suppressed Carrier) both transmit both sidebands and the carrier, requiring more power. VSB (Vestigial Sideband) modulation also requires more power as it transmits both sidebands and a portion of the carrier. Therefore, SSB-SC has the least power requirement among the given options.

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Current Version
Mar 20, 2023

Quiz Edited by
ProProfs Editorial Team
May 14, 2020

Quiz Created by
Karishma Parween

Analog Communication Test - 1

Noise gets added in the communication system in

A 1 MHz sinusoidal carrier is amplitude modulated by a sinusoidal message signal with time period of 100 micro sec. Which of the frequencies will not be present in the modulated signal ? (DSB-FC modulation)

The hilbert transform of sinw₁t - cosw₂t is

Consider sinusoidal modulation in an AM system. Assuming no overmodulation, the modulation index ( μ ) when the maximum and minimum values of the envelope , respectively, are 3V and 2 V is

What will be the minimum value of 'A_c' required to get the positive envelope of amplitude modulated wave given as S_AM(t) = A_c cosw_ct + 4 cosw_ct cosw_mt .

The output value of the envelope detector when input applied is 3 cosw_ot + 4 sinw_ot

Let m(t) = cos(2 π * f_mt) be the message signal & c(t)= 5 cos(2 π f_c*t) be the carrier. Both m(t) and c(t) are used to generate an AM modulated signal. The modulation index of the generated AM signal is 0.5. Then, transmission efficiency of the signal will be_______

Let m(t)= 2 cosw_mt and c(t)= 4 cosw_ct be the message and carrier signal respectively. Further, m(t) and c(t) is used to generate the DSB-SC signal. Then, total transmission power of modulated signal will be_____

Hilbert transform is used in the generation of

Ring modulator is used in the generation of ____ signal and the no. of diodes used in this circuit is_

In AM

A message signal is amplitude modulated with modulation index value of '1' . Further, it is modulated to SSB-SC signal. Calculate the percentage of power saved in SSB-SC compared to AM. (approximate value)

The bandwidth required to transmit a SSB-SC signal when message signal is given as m(t) = 2 cos(2 π 5t).

An AM signal with a carrier power of 80 watts has 10 watts power in each sideband. What is the modulation index value.

Power requirement is least in

Analog Communication Test - 1

Noise gets added in the communication system in

A 1 MHz sinusoidal carrier is amplitude modulated by a sinusoidal message signal with time period of 100 micro sec. Which of the frequencies will not be present in the modulated signal ? (DSB-FC modulation)

The hilbert transform of sinw1t - cosw2t is

Consider sinusoidal modulation in an AM system. Assuming no overmodulation, the modulation index ( μ ) when the maximum and minimum values of the envelope , respectively, are 3V and 2 V is

What will be the minimum value of 'Ac' required to get the positive envelope of amplitude modulated wave given as SAM(t) = Ac coswct + 4 coswct coswmt .

The output value of the envelope detector when input applied is 3 coswot + 4 sinwot

Let m(t) = cos(2 π * fm *t) be the message signal & c(t)= 5 cos(2 π *fc*t) be the carrier. Both m(t) and c(t) are used to generate an AM modulated signal. The modulation index of the generated AM signal is 0.5. Then, transmission efficiency of the signal will be_______

Let m(t)= 2 coswmt and c(t)= 4 coswct be the message and carrier signal respectively. Further, m(t) and c(t) is used to generate the DSB-SC signal. Then, total transmission power of modulated signal will be_____