Analog Communication Test Questions - Set 1

ANSWER: Change in amplitude of the carrier according to modulating signal

Explanation:
Amplitude modulation is defined as change in the amplitude of the high frequency carrier wave in accordance with the instantaneous value of the modulating signal.

ANSWER: Selectivity

Explanation:
The ability of the receiver to select the wanted signals among the various incoming signals is termed as Selectivity. It rejects the other signals at closely lying frequencies. Selectivity is determines performance of a radio receiver that how much it responds only to the radio signal it is required to receive.

ANSWER: All of the above

Explanation:
Emitter modulator amplifier for Amplitude Modulation operates in class A mode and has a very low efficiency. The output of the modulator is very small, therefore, it is not suitable for modulation at high level.

ANSWER: All of the above

Explanation:
A super heterodyne receiver mixes the incoming signal frequency with the locally generated signal frequency to convert a received signal to a fixed intermediate frequency (IF). A local oscillator in the receiver generates a signal, which mixes with the incoming signal, and then shifts that to intermediate frequency. The IF signal is filtered and is used to detect the original signal. Super heterodyne receivers have better sensitivity, high selectivity but need an extra circuitry for frequency conversion.

ANSWER: All of the above

Explanation:
The amplitude modulated signal consists of a carrier frequency component along with two sidebands around the main carrier. The frequencies obtained in the spectrum after the amplitude modulation are f_c+f_m and f_c+f_m where f_c is the carrier frequency and f_m is the modulating signal frequency.

ANSWER: 455 KHz

Explanation:
Standard intermediate frequency used for AM receiver is 455KHz. Intermediate frequency (IF) is obtained by mixing the incoming signal frequency with the locally generated signal frequency. The output is the IF which is a fixed frequency from which original information is detected using filters and amplifiers.

ANSWER: Varactor diode

Explanation:
Varactor diodes have variable capacitance as a function of variable input voltage. Varactor diodes work in reverse-bias. They are used for tuning the receivers according to incoming signal. Varactor diodes are used in voltage controlled oscillators and in almost all radio, cellular and wireless receivers. They are also used in RF filters for tuning purposes such as tuning of television sets to electronically tune the receiver to different stations.

ANSWER: SSB

Explanation:
Single Side band Transmission carries only one of the sidebands of the AM wave, which is much less than the total power required by the AM signal or the DSB-SC signal.

The total power in an AM is given by

P_t = P_c ( 1 + m²/2)

Where P_c is the carrier power and m is the modulation index.

If carrier is removed, the carrier power is also deducted and then sideband power remains.

And SSB transmission needs half the power of DSB-SC signal.

ANSWER: 19.8 KHz

Explanation:
Here the f_m = 100Hz to 10 KHz = 10000 - 100 = 9900 Hz

Therefore, Bandwidth = 2 f_m = 2 * 9900 = 19.8KHz

ANSWER: RC < 1/W

Explanation:
The discharging time constant RC must be so that it ensures the capacitor to discharge slowly through the load resistor R between the positive peaks of the carrier wave, but not so long that the capacitor voltage will not discharge at the maximum rate of change of the modulating wave, that is

RC < 1/W

Where, W is the message bandwidth. The result is that the capacitor voltage or detector output is nearly the same as the envelope of the AM wave.

ANSWER: 50% of Modulation Depth

Explanation:
Modulation depth is typically the modulation index expressed as a percentage of the figure. Thus a modulation index of 0.5 is expressed as a modulation depth of 50%. However, often the two terms and figures may be used interchangeably. For a modulation index of 1.0, it means 100% modulation, i.e., the carrier level falls to zero level and rises to twice of its un modulated level.

ANSWER: 6.004 GHz

Explanation:
According to Nyquist sampling criteria, a bandlimited signal should be sampled equal to greater than twice the maximum frequency of the signal.

Here, the spectrum obtained after the conversion has a maximum frequency of 3000MHz + 2MHz = 3002MHz. So the sampling frequency required to prevent aliasing is 6004MHz i.e., 6.004GHz.

ANSWER: To combine multiple data streams over a single data channel

Explanation:
Multiplexing is a process where multiple data streams from different sources are combined and transmitted over a single data channel. Multiplexer or MUX is placed at the transmitting end to combine the signals and a De multiplexer or DEMUX is placed at the receiver that separates the received signals and sends them to their corresponding destinations.

Multiplexing techniques are classified as Time-division multiplexing (TDM) and Frequency-division multiplexing (FDM).

ANSWER: Sampling of signals less than at Nyquist rate

Explanation:
Aliasing refers to the sampling of signals less than at Nyquist rate. Nyquist rate states that the rate of sampling of signal should be greater than or equal to twice the bandwidth of the modulating signal. Aliasing is reduced if the sampling is done at higher rate than nyquist rate of sampling.

ANSWER: Bandwidth

Explanation:
Bandwidth is the amount of data that may be transmitted over a communication channel in a specified time. It is also called transmission data rate measured in bits per second (bps). It is the amount of information that may be carried over a channel from one point to another point in a given time period.

ANSWER: 900 KHz, 888 KHz, 12 KHz

Explanation:
Maximum Frequency f_USB = 894 + 6 = 900 kHz

Minimum Frequency f_LSB = 894 - 6 = 888 kHz

Bandwidth BW = f_USB f_LSB = 900 888 = 12 kHz OR = 2(6 kHz) = 12 kHz

ANSWER: 900 W

Explanation:
The total power in an Amplitude Modulated wave is given by

P_T = P_C (1+ m²2)

Here, P_C = 800W,
m = 0.5

therefore,

P_T = 800 (1+ (0.5)²/2) = 900 W

ANSWER: 1458 W, 2187.5 W, 729.25 W

Explanation:
a) P_C=I²R = (5.4)²*50 = 1458W

b) I_T = I_c√(1+m²/2) = 5.4√(1+1²/2)
=6.614 A

P_T = I_T²R
= (6.614)² * 50
= 2187.25 W

c) P_SB = P_T - P_C
= 2187.25 - 1458 W
= 729.25W (for two bands)

For single band, P_SB = 729.25/2
= 364. 625 W

ANSWER: 67%

Explanation:
P_total = 9.8KW

P_c = 8KW

Power of the signal (P_total) transmitted by a transmitter after modulation is given by

P_total = P_c (1+ m²/2)

Where P_c is the power of carrier i.e., without modulation

M is the modulation index

Therefore,
9.8= 8 (1+ m²/2)
9.8/8=1+ m²/2
m=0.67 = 67%

ANSWER: 54

Explanation:
Medium Frequency (MF) is the band of frequencies from 300 KHz to 3MHz. The lower portion of the MF band (300to 500 kilohertz) is used for ground-wave transmission for reasonably long distances. The upper and lower ends of the mf band are used for naval purpose.

Frequency available in MF band= 3000 - 300 = 2700 KHz

The bandwidth required by 25 KHz signal = 2 * 25= 50 KHz

Therefore the number of channels available = 2700/ 50 = 54