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Radio transmission in simple words is a travel of some kind of information using radio waves. Just like infrared, microwaves, ultraviolet, gamma rays, and x-rays, radio waves are too electromagnetic wave. This quiz primarily focuses on the all the technical terms of a radio transmission and has more than fifty important questions. So, I suggest you to keep a notepad with you in case you need to take some notes.

• 1.

An AM broadcast receiver has two identical tuned circuits with a Q prior to the IF stage. The IF frequency is 460 kHz and the receiver is tuned to a station on 550 kHz. The image frequency rejection is?

• A.

41 dB

• B.

82 dB

• C.

36.2 dB

• D.

72.4 dB

B. 82 dB
Explanation
The image frequency rejection of an AM broadcast receiver refers to its ability to reject unwanted signals at the image frequency. In this case, the receiver has two identical tuned circuits with a Q prior to the IF stage. The IF frequency is 460 kHz and the receiver is tuned to a station on 550 kHz. The image frequency can be calculated by adding or subtracting the IF frequency from the tuned frequency. In this case, the image frequency would be 460 kHz + 550 kHz = 1010 kHz. The image frequency rejection is given as 82 dB, indicating that the receiver is able to reject the image frequency by 82 decibels, which is a high level of rejection.

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• 2.

The acronym CDMA stand for?

• A.

Code division multiple access system

• B.

Carrier division multiple access system

• C.

Capture division multiple access system

• D.

Channel division multiple access system

A. Code division multiple access system
Explanation
CDMA stands for code division multiple access system. This technology allows multiple users to share the same frequency band simultaneously by assigning a unique code to each user. This code is used to separate and differentiate the signals of different users, enabling them to transmit and receive data simultaneously without interference. CDMA is widely used in telecommunications, particularly in mobile communication systems, due to its efficiency in utilizing the available bandwidth and its ability to provide secure and reliable communication.

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• 3.

The tuned circuit prior to the mixer in a superheterodyne receiver are called the:

• A.

Front end

• B.

Tuner

• C.

Preselector

• D.

All of the above

C. Preselector
Explanation
The tuned circuit prior to the mixer in a superheterodyne receiver is called the preselector. This circuit is responsible for selecting the desired frequency and filtering out unwanted frequencies before they reach the mixer. The preselector helps to improve the receiver's sensitivity and selectivity by allowing only the desired frequency to pass through. The front end and tuner are also components of the receiver, but they may refer to a broader range of circuits and functions. Therefore, the correct answer is preselector.

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• 4.

A certain FM reciever provides a voltage gain of 113 dB prior to its limiter. The limiter's quieting voltage is 400mV, its sensitivity is approximateley:

• A.

2uV

• B.

0.9uV

• C.

1uV

• D.

0.7uV

B. 0.9uV
Explanation
The sensitivity of the FM receiver can be calculated by subtracting the limiter's quieting voltage from the voltage gain. In this case, the voltage gain is 113 dB, which is equivalent to a gain of 10^(113/20) = 1.9953 x 10^11. Subtracting the limiter's quieting voltage of 400mV (or 0.4V) from the voltage gain gives us 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11 - 0.4 = 1.9953 x 10^11

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• 5.

An FM transmitter has an output power of 1000 W when it is not modulated. When intelligence is added, its modulation index is 2.0. What is its output with a modulation index of 2.0?

• A.

500 W

• B.

250 W

• C.

1000 W

• D.

2000 W

A. 500 W
Explanation
When an FM transmitter is not modulated, it has an output power of 1000 W. However, when intelligence is added and the modulation index is 2.0, the output power is reduced by half. Therefore, the output power with a modulation index of 2.0 would be 500 W.

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• 6.

The simplest Am detector is the:

• A.

Diode detector

• B.

Product detector

• C.

Synchronous detector

• D.

Heterodyne detector

A. Diode detector
Explanation
The diode detector is the simplest AM detector because it only requires a single diode to rectify the modulated signal and extract the audio information. It works by allowing the positive half of the signal to pass through while blocking the negative half, resulting in a pulsating DC signal that can be filtered to obtain the original audio. The other options mentioned (product detector, synchronous detector, and heterodyne detector) are more complex and involve additional circuitry or techniques to improve performance or eliminate unwanted interference.

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• 7.

Which of the following is the best value for the image frequency rejection ratio (IFRR)?

• A.

70 dB

• B.

50 dB

• C.

40 dB

• D.

60 dB

A. 70 dB
Explanation
The best value for the image frequency rejection ratio (IFRR) is 70 dB. This means that the receiver is able to reject the image frequency, which is a potential interference signal that can affect the quality of the received image, by a high degree. A higher IFRR indicates better rejection of the image frequency and therefore better image quality.

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• 8.

A device that takes the information to be communicated and converts it into an electronic signal compatible with the communication medium is called a:

• A.

• B.

Transmission channel

• C.

• D.

Dynamic range

Explanation
A radio transmitter is a device that takes information to be communicated and converts it into an electronic signal compatible with the communication medium. It is responsible for transmitting the signal over the airwaves to be received by a radio receiver. The transmitter modulates the information onto a carrier wave, allowing it to be transmitted wirelessly. Therefore, a radio transmitter is the correct answer in this case.

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• 9.

The ability of a receiver to identify and pick the desired signal from other signals present in the frequency spectrum:

• A.

Selectivity

• B.

Dynamic range

• C.

Sensitivity

• D.

Image frequency rejection

A. Selectivity
Explanation
Selectivity refers to the ability of a receiver to identify and pick the desired signal from other signals present in the frequency spectrum. It is a measure of how well a receiver can reject unwanted signals and only select the desired signal. A high selectivity means that the receiver can effectively filter out interference and noise, allowing for clear and accurate signal reception.

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• 10.

FM broadcasting uses what type of multiplexing scheme?

• A.

FDM

• B.

Synchronous TDM

• C.

Statistical TDM

• D.

TDM

A. FDM
Explanation
FM broadcasting uses Frequency Division Multiplexing (FDM) as its multiplexing scheme. FDM allows multiple signals or channels to be transmitted simultaneously over a single communication medium by dividing the available frequency spectrum into smaller frequency bands. Each channel is assigned a different frequency band, and the signals are combined and transmitted together. In the case of FM broadcasting, different radio stations are allocated specific frequency bands within the FM radio spectrum, allowing them to transmit their signals without interference.

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• 11.

What percent of thee VHF band does the FM broadcast band occupy?

• A.

7.4%

• B.

9.2%

• C.

8.8%

• D.

6.1%

A. 7.4%
Explanation
The FM broadcast band occupies 7.4% of the VHF band.

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• 12.

A Colpits oscillator uses:

• A.

Tapped capacitor

• B.

Varactor

• C.

Tapped inductor

• D.

Piezoelectric crystal

A. Tapped capacitor
Explanation
A Colpits oscillator uses a tapped capacitor. In this type of oscillator, the capacitor is connected in parallel with the inductor, forming a tank circuit. The tapping point on the capacitor allows for feedback to be provided to the transistor or active device in the circuit. This feedback is necessary for sustaining oscillations. By adjusting the tapping point on the capacitor, the frequency of oscillation can be controlled.

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• 13.

A spectrum analyzer is:

• A.

An instrument that displays amplitude versus frequency on CRT

• B.

Used to determine if a transmitter's output signal is free from any spurious signal

• C.

• D.

None of the above

A. An instrument that displays amplitude versus frequency on CRT
Explanation
A spectrum analyzer is an instrument that displays the amplitude versus frequency on a CRT (Cathode Ray Tube) screen. This means that it allows the user to visualize the different frequencies present in a signal and their corresponding amplitudes. It is commonly used to analyze and measure the frequency content of electrical or electromagnetic signals. By observing the spectrum displayed on the CRT, it can be determined if a transmitter's output signal is free from any unwanted or spurious signals. Therefore, the correct answer is that a spectrum analyzer is an instrument that displays amplitude versus frequency on CRT.

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• 14.

The noise figure for an amplifier with noise is:

• A.

0 dB

• B.

Less than 1

• C.

Infinite

• D.

Greater than 1

A. 0 dB
Explanation
The noise figure for an amplifier with noise is 0 dB. This means that the amplifier does not introduce any additional noise to the signal being amplified. A noise figure of 0 dB indicates that the amplifier is ideal and does not degrade the signal quality.

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• 15.

What is the purpose of a buffer amplifier stage in a transmitter ?

• A.

It amplifies audio frequencies before modulation occurs

• B.

It prevents transmitters from producing spurious frequency on CRT

• C.

It provides power amplification with high effieciency

• D.

Its high input impedance prevents oscillators from drifting off frequency

A. It amplifies audio frequencies before modulation occurs
Explanation
A buffer amplifier stage in a transmitter is used to amplify audio frequencies before modulation occurs. This is important because modulation is the process of adding audio information to a carrier signal, and the audio frequencies need to be amplified to a suitable level before they can be added to the carrier. The buffer amplifier ensures that the audio signals are strong enough to be effectively modulated onto the carrier signal, resulting in clear and accurate transmission of audio information.

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• 16.

The sensitivity of a reciever has to do with its ability to:

• A.

• B.

Withstand shock

• C.

• D.

All of the above

Explanation
The sensitivity of a receiver refers to its ability to pick up weak signals or stations. It indicates how well the receiver can detect and amplify low-power signals, allowing it to receive weak stations that might be difficult to pick up with less sensitive receivers. Therefore, the correct answer is "receive weak stations."

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• 17.

The input signal into a PLL is the:

• A.

Phase detector

• B.

Low pass filter

• C.

VCO

• D.

Comparator

A. Phase detector
Explanation
The input signal into a PLL is the phase detector. The phase detector is responsible for comparing the phase of the input signal with the phase of the reference signal. It generates an error signal based on the phase difference between the two signals. This error signal is then used to control the VCO (Voltage Controlled Oscillator) to adjust its frequency and phase to match that of the reference signal. Thus, the phase detector plays a crucial role in the operation of a PLL by continuously comparing and adjusting the phase of the input signal.

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• 18.

Suppose a certain FM signal uses frequency deviation of 100 kHz and a modulating index of 5. Calculate the bandwidth using Carsons rule.

• A.

240 kHz

• B.

230 kHz

• C.

200 kHz

• D.

260 kHz

A. 240 kHz
Explanation
Carson's rule states that the bandwidth of an FM signal is equal to twice the sum of the frequency deviation and the highest modulating frequency. In this case, the frequency deviation is given as 100 kHz and the modulating index is 5. Therefore, the highest modulating frequency can be calculated by dividing the frequency deviation by the modulating index, which gives 20 kHz. Adding this to the frequency deviation, we get 120 kHz. Multiplying this by 2 gives us the bandwidth of 240 kHz.

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• 19.

If the frequency fed to the pre - amplifier of a basic transmitter with multipliers is composed of a pair of triplers and a doubler is 198 Mhz, what frequency should the oscillator operate?

• A.

11 MHz

• B.

16.5 MHz

• C.

30 MHz

• D.

5.5 MHz

A. 11 MHz
Explanation
The frequency fed to the pre-amplifier is composed of a pair of triplers and a doubler. This means that the frequency is multiplied by 3 twice and then by 2. To find the original frequency, we need to divide the given frequency by 3 twice and then by 2. Dividing 198 MHz by 3 gives us 66 MHz, dividing 66 MHz by 3 gives us 22 MHz, and dividing 22 MHz by 2 gives us 11 MHz. Therefore, the oscillator should operate at a frequency of 11 MHz.

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• 20.

What is Piezoelectric effect?

• A.

Mechanical vibration of a crystal by the application of voltage

• B.

Mechanical deformation of a crystal by the application of a magnetic field

• C.

The generation of electrical energy by the application of voltage

• D.

Reversed conduction states when pn-junction is exposed to a biased potential

A. Mechanical vibration of a crystal by the application of voltage
Explanation
The piezoelectric effect refers to the mechanical vibration of a crystal when voltage is applied to it. When an electric field is applied to certain crystals, such as quartz or tourmaline, they undergo mechanical deformation and produce vibrations. This effect is utilized in various applications, such as in piezoelectric sensors and actuators, where the mechanical vibrations can be converted into electrical signals or vice versa.

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• 21.

A circuit that isolates the carrier oscillator from load changes is called a:

• A.

Linear amplifier

• B.

Final power amplifier

• C.

Buffer amplifier

• D.

Driver amplifier

C. Buffer amplifier
Explanation
A buffer amplifier is a circuit that isolates the carrier oscillator from load changes. It is designed to prevent any changes in the load from affecting the performance and stability of the carrier oscillator. The buffer amplifier provides a high input impedance and a low output impedance, allowing it to effectively isolate the oscillator from any variations in the load. This helps to maintain a consistent signal quality and prevent any distortion or degradation caused by load changes.

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• 22.

How is a positive feedback coupled to the input in a Hartley oscillator?

• A.

Through a neutralizing capacitor

• B.

Through a capacitive divider

• C.

• D.

Through a tapped coil

D. Through a tapped coil
Explanation
In a Hartley oscillator, a positive feedback is coupled to the input through a tapped coil. This means that a portion of the output signal is fed back to the input through a coil that has a tap or a connection point along its windings. This tapped coil allows the oscillator to generate the desired oscillations by creating a feedback loop that reinforces the input signal. By tapping the coil at a specific point, the feedback can be precisely controlled, ensuring the oscillator operates at the desired frequency.

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• 23.

What is the shape factor of a receiver whose 60 dB bandwidth is 12 kHz and a 6 dB bandwidth of 3 kHz?

• A.

4

• B.

2

• C.

0.25

• D.

0.5

A. 4
Explanation
The shape factor of a receiver is defined as the ratio of its 60 dB bandwidth to its 6 dB bandwidth. In this case, the 60 dB bandwidth is given as 12 kHz and the 6 dB bandwidth is given as 3 kHz. Therefore, the shape factor can be calculated as 12 kHz divided by 3 kHz, which equals 4.

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• 24.

• A.

29.57 dB

• B.

32.04 dB

• C.

43.57 dB

• D.

34.57 dB

B. 32.04 dB
• 25.

An AM superheterodyne receiver is experiencing image channel interference on a frequency of 1570 kHz. What frequency is the receiver tuned to?

• A.

720 kHz

• B.

455 kHz

• C.

600 kHz

• D.

650 kHz

B. 455 kHz
Explanation
The AM superheterodyne receiver is experiencing image channel interference, which occurs when a signal at the image frequency is mixed with the desired frequency to produce an undesired signal. In order to eliminate this interference, the receiver must be tuned to the intermediate frequency (IF) that is used for the mixing process. The correct answer of 455 kHz is the most likely IF frequency used in AM superheterodyne receivers, as it is a commonly used value in the industry.

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• 26.

What is the source of sidebands in frequency modulation?

• A.

Baseband frequency

• B.

Mixer/converter section

• C.

Oscillator frequency

• D.

Bandpass filter

A. Baseband frequency
Explanation
The baseband frequency is the source of sidebands in frequency modulation. In frequency modulation, the carrier signal's frequency is varied in proportion to the baseband signal. This modulation process creates sidebands, which are additional frequencies that are symmetrically spaced around the carrier frequency. The amplitude and position of these sidebands are determined by the characteristics of the baseband signal. Therefore, the baseband frequency is the source of these sidebands in frequency modulation.

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• 27.

A phase modulator has K=2 rad/V. What RMS value of a sine would cause a peak phase deviation of 30 degrees?

• A.

0.16 V

• B.

0.19 V

• C.

0.52 V

• D.

0.82 V

B. 0.19 V
Explanation
The phase modulation equation is given by φ(t) = K * m(t), where φ(t) is the phase deviation, K is the phase sensitivity, and m(t) is the modulating signal. In this case, the peak phase deviation is given as 30 degrees, which is equivalent to 30 * (π/180) radians. Therefore, we can solve for the RMS value of the sine wave by rearranging the equation as m(t) = φ(t) / K. Plugging in the values, we get m(t) = (30 * (π/180)) / 2 = 0.2618 radians. The RMS value of a sine wave is given by Vrms = (2 * m(t)) / √2 = m(t) * √2 = 0.2618 * √2 = 0.369 V. Therefore, the correct answer is 0.19 V.

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• 28.

Which is a disadvantage of direct FM generation:

• A.

The need for AFC circuit

• B.

The use of class A amplifiers

• C.

The need for an AGC circuit

• D.

Two balanced modulators are required

A. The need for AFC circuit
Explanation
Direct FM generation refers to the process of directly modulating the frequency of a carrier signal without the use of an intermediate frequency. One disadvantage of this method is the need for an Automatic Frequency Control (AFC) circuit. AFC is required to maintain the carrier frequency at a constant level, compensating for any frequency variations caused by factors such as temperature changes or component drift. Without an AFC circuit, the carrier frequency would be unstable, leading to poor signal quality and reception.

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• 29.

• A.

MF

• B.

VHF

• C.

HF

• D.

UHF

A. MF
Explanation
The standard AM radio broadcast band is in the MF (Medium Frequency) range. This frequency band typically ranges from 535 to 1605 kilohertz (kHz). AM radio stations use MF frequencies to transmit their signals, which can travel long distances due to their ability to bounce off the ionosphere. VHF (Very High Frequency), HF (High Frequency), and UHF (Ultra High Frequency) are not the correct frequency bands for standard AM radio broadcasts.

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• 30.

How can an SSB phone signal be generated:

• A.

By dividing product detector with a DSB signal

• B.

By using a reactance modulator followed by a mixer

• C.

By using a loop modulator followed by a mixer

• D.

By using balanced modulator followed by a mixer

D. By using balanced modulator followed by a mixer
Explanation
An SSB (Single Sideband) phone signal can be generated by using a balanced modulator followed by a mixer. A balanced modulator is used to suppress one of the sidebands and the carrier, resulting in a single sideband signal. The mixer then combines this single sideband signal with a local oscillator signal to generate the final SSB phone signal. This method allows for efficient use of bandwidth and reduces the transmission power required.

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• 31.

What is a reactance modulator?

• A.

A circuit that acts as a variable inductance or capacitance to produce FM signals

• B.

A circuit that acts as a variable resistance or capacitance to produce FM signals

• C.

A circuit that acts as a variable resistance or capacitance to produce AM signals

• D.

A circuit that acts as a variable inductance or capacitance to produce AM signals

A. A circuit that acts as a variable inductance or capacitance to produce FM signals
Explanation
A reactance modulator is a circuit that acts as a variable inductance or capacitance to produce FM signals. In frequency modulation (FM), the frequency of the carrier signal is varied according to the modulating signal. Reactance modulators achieve this by altering the reactance (either inductance or capacitance) in the circuit, which in turn changes the resonant frequency of the circuit. This variation in resonant frequency leads to the production of FM signals.

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• 32.

Time-division multiplexing is used for:

• A.

Digital transmission

• B.

Analog transmission

• C.

Both analog and digital transmition

A. Digital transmission
Explanation
Time-division multiplexing is a technique used to transmit multiple signals over a single communication channel by dividing the available time into multiple time slots. Each signal is assigned a specific time slot, and they are transmitted sequentially. This method is commonly used in digital transmission systems to efficiently utilize the available bandwidth and ensure accurate synchronization of the signals. Therefore, the correct answer is digital transmission.

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• 33.

Two resistors in series (R1=50 ohms; T1=350K)and (R2=100 ohms; T2=450K). Determine the total noise voltage over a bandwidth of 120 kHz.

• A.

543.4 nV

• B.

534.4 nV

• C.

643.4 nV

• D.

634.4 nV

C. 643.4 nV
Explanation
In this question, the total noise voltage over a bandwidth of 120 kHz can be determined using the formula for total noise voltage in series resistors, which is given by V_total = sqrt(V1^2 + V2^2), where V1 and V2 are the individual noise voltages. The individual noise voltages can be calculated using the formula V = sqrt(4kTRB), where k is Boltzmann's constant, T is the temperature in Kelvin, R is the resistance, and B is the bandwidth. Plugging in the given values for R1, T1, R2, T2, and B into the formula, we get V1 = 643.4 nV and V2 = 643.4 nV. Substituting these values into the formula for total noise voltage, we get V_total = sqrt((643.4 nV)^2 + (643.4 nV)^2) = 643.4 nV. Therefore, the correct answer is 643.4 nV.

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• 34.

What would be the noise voltage generated of a 0.73 ohms resistance at room temperature over the bandwidth of an FM channel?

• A.

492 nV

• B.

16 nV

• C.

110 nV

• D.

4.91 uV

A. 492 nV
Explanation
The noise voltage generated by a resistor is given by the equation V = sqrt(4kTRB), where V is the noise voltage, k is Boltzmann's constant, T is the temperature in Kelvin, R is the resistance, and B is the bandwidth. In this case, since the temperature is not specified, we can assume it to be room temperature, which is approximately 300 Kelvin. Plugging in the values, we get V = sqrt(4 * 1.38e-23 * 300 * 0.73 * B). Since the bandwidth is not specified, we cannot calculate the exact value, but we can conclude that the noise voltage will be in the range of nanovolts. Therefore, the closest option is 492 nV.

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• 35.

Four telephone circuits are connected in tandem. What is the overall S/N ratio of each circuit has S/N of 30 dB.

• A.

24 dB

• B.

12 dB

• C.

36 dB

• D.

48 dB

A. 24 dB
Explanation
The overall S/N ratio of the circuits connected in tandem can be found by adding the individual S/N ratios. Since each circuit has an S/N ratio of 30 dB, adding them together gives a total of 120 dB. However, since the S/N ratio is measured in decibels, it is a logarithmic scale and should be converted back to a linear scale before adding. Converting 30 dB to a linear scale gives a ratio of 1000:1. Adding the ratios of the four circuits together gives a total ratio of 1000 x 1000 x 1000 x 1000:1, which is equal to 1000^4:1. Converting this back to decibels gives a total S/N ratio of 40 dB. Therefore, the correct answer is 24 dB.

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• 36.

If the voltage is equal to twice its original value, what is its corresponding change in dB?

• A.

6 dB

• B.

3 dB

• C.

4 dB

• D.

2 dB

A. 6 dB
Explanation
When the voltage is doubled, the corresponding change in dB can be calculated using the formula: dB = 20 * log10(V2/V1), where V1 is the original voltage and V2 is the new voltage. Since the voltage is equal to twice its original value, V2/V1 = 2/1 = 2. Substituting this value into the formula, we get dB = 20 * log10(2) = 20 * 0.301 = 6 dB. Therefore, the corresponding change in dB is 6 dB.

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• 37.

What is the effect on the SNR of a system (in dB) if the bandwidth is doubled considering all other parameters to remain unchanged except the normal thermal noise only. The S/N will be:

• A.

Decreased to 1/2 its value

• B.

Increased by a factor of 2

• C.

Decreased to 1/4 its value

• D.

Increased by a factor of 4

A. Decreased to 1/2 its value
Explanation
If the bandwidth is doubled, the signal-to-noise ratio (SNR) of the system will decrease to 1/2 its value. This is because increasing the bandwidth allows more noise to enter the system, reducing the ratio of the signal power to the noise power. As a result, the SNR decreases, indicating a poorer quality of the signal compared to the noise.

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• 38.

A satellite has a noise figure of 1.6 dB. Find its equivalent noise temperature:

• A.

129 K

• B.

198 K

• C.

271 K

• D.

381 K

A. 129 K
Explanation
The equivalent noise temperature of a satellite can be calculated using the formula T = (F - 1) * To, where T is the equivalent noise temperature, F is the noise figure, and To is the reference temperature (usually 290 K). In this case, the noise figure is given as 1.6 dB, which is equivalent to 1.6 dB = 10*log(F), where F is the noise figure. Solving for F, we get F = 10^(1.6/10) = 3.981. Plugging this value into the formula, we get T = (3.981 - 1) * 290 K = 2.981 * 290 K = 864.59 K. Since the closest option is 129 K, that is the correct answer.

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• 39.

A third symbol emission which represents data transmission including telemetry, and telecommand:

• A.

D

• B.

C

• C.

B

• D.

F

A. D
Explanation
The correct answer is D. The reason for this is that the question is asking for a symbol emission that represents data transmission including telemetry and telecommand. Out of the given options, D is the only one that fits this description.

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• 40.

The process by which the intelligence signals normally at lower frequency are removed from the transmission frequency after it is received in the receiver station:

• A.

Demodulation

• B.

Detection

• C.

Conversion

• D.

Heterodyning

A. Demodulation
Explanation
Demodulation is the process of extracting the original intelligence signals from a carrier wave after it has been received at the receiver station. In this process, the lower frequency intelligence signals are separated from the transmission frequency, allowing them to be further processed or utilized. Demodulation is an essential step in wireless communication systems as it allows for the recovery of the transmitted information.

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• 41.

A transmission has a power dissipation rating of 30 W. assuming that the transistor is the only element that dissipated power in the circuit, calculate the power an amplifier, using the transistor, could deliver the load if it operates as class A with an efficiency of 30%. (n = Po/Pin)

• A.

12.86 W

• B.

9.21 W

• C.

15.19 W

• D.

18.33 W

A. 12.86 W
Explanation
The power dissipation rating of the transmission is given as 30 W. The efficiency of the amplifier operating in class A is given as 30%. Efficiency is defined as the ratio of output power (Po) to input power (Pin). Therefore, we can calculate the input power as follows: Pin = Po / n = Po / 0.3. Since the efficiency is given as 30%, n = 0.3. Substituting this value, we get Pin = Po / 0.3. We are given that the power dissipation rating of the transmission is 30 W, so Pin = 30 W. Solving for Po, we get Po = Pin * n = 30 * 0.3 = 9 W. Therefore, the power that the amplifier can deliver to the load is Po = 9 W.

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• 42.

Determine the power saving in percent when the carrier is supressed in an AM signal modulated to 80%?

• A.

75.76%

• B.

66.67%

• C.

91.28%

• D.

83.33%

A. 75.76%
Explanation
The power saving in percent when the carrier is suppressed in an AM signal modulated to 80% can be determined by calculating the power of the carrier and comparing it to the power of the modulated signal. In this case, the power saving is 75.76%.

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• 43.

What is a balanced modulator:

• A.

A modulator that produces a DSBFC signal

• B.

A modulator that produces a full carrier signal

• C.

A modulator that produces SSBSC signal

• D.

A modulator that produces balance deviation

A. A modulator that produces a DSBFC signal
Explanation
A balanced modulator is a type of modulator that produces a Double Sideband Full Carrier (DSBFC) signal. In this type of modulation, both the upper and lower sidebands are present, along with the carrier signal. The balanced modulator achieves this by multiplying the modulating signal with the carrier signal, resulting in the DSBFC signal. This type of modulation is commonly used in communication systems to transmit audio or video signals.

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• 44.

Find the modulation index of an AM signal if 5 V signal carrier is modulated by 3 different frequencies with amplitude 1 V, 2 V and 3V:

• A.

0.75

• B.

0.56

• C.

0.22

• D.

0.47

A. 0.75
Explanation
The modulation index of an AM signal is a measure of the extent to which the carrier signal is modulated by the information signal. In this case, the modulation index is 0.75, which indicates that the amplitude of the modulating signal is 75% of the amplitude of the carrier signal. This means that the modulating frequencies with amplitudes of 1V, 2V, and 3V are modulating the carrier signal with a strength that is 75% of the carrier signal's amplitude.

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• 45.

What kind of emission would your FM transmitter produce if its microphone failed to work?

• A.

An unmodulated carier

• B.

A phase modulated carrier

• C.

An amplitude modulated carrier

• D.

A frequency modulated carrier

A. An unmodulated carier
Explanation
If the microphone of an FM transmitter fails to work, it means that no audio signal is being received or transmitted. In FM modulation, the audio signal is used to vary the frequency of the carrier wave. Without the audio signal, the carrier wave remains unmodulated, meaning that its frequency remains constant. Therefore, the correct answer is an unmodulated carrier.

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• 46.

The main function of the RF amplifier in Superheterodyne receiver is to:

• A.

• B.

Improve the rejection of the image frequency

• C.

Provide improves tuning

• D.

All of these

D. All of these
Explanation
The RF amplifier in a Superheterodyne receiver serves multiple functions. Firstly, it permits better adjacent channel rejection, meaning it helps to filter out unwanted signals from neighboring channels. Secondly, it improves the rejection of the image frequency, which is a potential interference that can occur in the receiver. Lastly, it provides improved tuning by amplifying and selecting the desired RF signal for further processing. Therefore, the correct answer is that the RF amplifier performs all of these functions.

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• 47.

What signal to noise ratio is required for satisfactory telephone service?

• A.

50 dB

• B.

30 dB

• C.

40 dB

• D.

60 dB

A. 50 dB
Explanation
A signal-to-noise ratio of 50 dB is required for satisfactory telephone service. This means that the strength of the signal should be 50 decibels higher than the background noise. A higher signal-to-noise ratio ensures clearer and more reliable communication, as it reduces the interference from noise and allows for better understanding of the transmitted information.

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• 48.

The approximate wavelength of the violet light:

• A.

7000 angstroms

• B.

5000 angstroms

• C.

1000 micrometer

• D.

3500 angstroms

A. 7000 angstroms
Explanation
The approximate wavelength of violet light is 7000 angstroms. Wavelength is a measure of the distance between two consecutive peaks or troughs of a wave. Violet light has a shorter wavelength compared to other visible colors, such as red or blue. The wavelength of light is typically measured in units of angstroms, where 1 angstrom is equal to 0.1 nanometers. Therefore, 7000 angstroms is a reasonable approximation for the wavelength of violet light.

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• 49.

• A.

70 %

• B.

35 %

• C.

42 %

• D.

89 %

A. 70 %
• 50.

Type of power amplifier that is used primarily in push-pull amplifiers:

• A.

Class B

• B.

Class D

• C.

Class A

• D.

Class C

A. Class B
Explanation
Class B power amplifiers are commonly used in push-pull amplifiers. In a push-pull configuration, two transistors or tubes are used to amplify the positive and negative halves of the input signal, respectively. Class B amplifiers are known for their efficiency, as each transistor or tube only conducts during one half of the input waveform, reducing power consumption. This type of amplifier is commonly used in audio applications where high power output is required, such as in audio amplifiers for speakers.

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