Communications

Question 1

What is modulation?

Answer 1

Modulation is the alteration of the amplitude or the frequency of a high frequency wave known as the carrier wave by the information signal in order to carry the information.

Question 2

What is the difference between Amplitude modulation and Frequency modulation?

Answer 2

In frequency modulation, the frequency of the carrier wave is made to change in synchrony with the change of displacement of the information signal while the amplitude is made to change in amplitude modulation.

Question 3

Why is modulation necessary?

Answer 3

1. ) Without modulation, only one radio station can operate in an area because a second station would lead to interference.
2. ) The aerial required to transmit frequencies would be long and inefficient without modulation.

Question 4

What is a bandwidth?

Answer 4

This is the frequency range occupied by the waveform. It is equal to 2Fa.

Question 5

Formula of sidebands of an AM waveform.

Answer 5

Fc ± Fa. ( where Fa is the frequency of the audio signal)

Question 6

Advantages of FM.

Answer 6

• Quality of FM is gen. richer than AM as interfering radiation affects the amplitude of the wave but in FM only variations in freq. matter not amp.
• Quality in terms of Fa is richer as the bandwidth is 200kHz unlike AM which is 9kHz, so Fa is 4.5kHz this is well below 15kHz (highest frequency audible to our ears) there4 such brodcast lack higher frequency & thus quality.

Question 7

Advantages of Am.

Answer 7

• AM can use more stations as its bandwidth is 9kHZ while FM is 200kHz (the LW waveband occupies from 30-300kHz in the em spectrum).
• Requires only one transmitter as LW,MW&SW wavebands are propagated very large distances to cover a large area but FM has only 30km line of sight so more trans. are needed. So AM is cheaper and simpler to broadcast.

Question 8

What is noise?

Answer 8

Unwanted random power that adds itself in a random manner to the signal and causes distortion of the signal.

Question 9

How does noise arise?

Answer 9

From the thermal vibrations of the atoms of the material through which the signal is passing.

Question 10

When a repeater amplifier is used for an analogue signal….

Answer 10

It amplifies the noise as well as the original signal. Several repeater amplification are required for transmission over long distances,the signal will become very noisy.

Question 11

What amplifiers are used for digital signals?

Answer 11

Regenerator amplifiers (filter out the noise as it consists of small fluctuations and they are not required to amplify small fluctuations in amplitude but to prod. high or low voltage.)

Question 12

Advantage of digital signal s.

Answer 12

They can have extra info . ( a code used by the receiving system to correct any errors before passing the info to the receiver).
Generally more reliable and cheaper.

Question 13

How to convert analogue to digital.

Answer 13

Take samples of the analogue waveform at regular intervals (measure its instantaneous voltage) then convert it into a binary number that represents its value.

Question 14

A lower sampling frequency means…

Answer 14

Less information can be gathered from the analogue system.

Question 15

In order to recover the analogue signal from its digital conversion….

Answer 15

Sampling has to occur at a frequency greater than twice the highest frequency component in the original signal.

Question 16

Why is output of DAC not smooth?

Answer 16

The output of DAC is grainy and not smooth because the number of bits limits the number of possible voltage levels. (with 4 bits there are 2⁴=16 levels)

Question 17

Why is repeated amplification needed for high frequency signals in wire pairs?

Answer 17

High attention. Energy is lost as heat in the wires and as radiation since the wires act as aerials. The wires also easily pick up external interference. (Cross-linking)

Question 18

Bandwidth of a pair of wires.

Answer 18

About 500kHz. So it is extremely limited as a means of carrying a large amount of info.

Question 19

Coaxial cables have advantage over wire pairs because..?

Answer 19

• It causes less attenuation of the signal.
• Less prone to interference as the outer conductor shields the inner one.(offer slightly greater security).
• bandwidth of 50MHz so it can carry more info than wire pairs.

Question 20

How do optic fibres carry information?

Answer 20

They carry digital information in the form of pulses of light or infrared radiation. (Pulses are prod by lasers and the light has a very high frequency of the order of 10⁸MHz)

Question 21

Advantages of using optic fibres.

Answer 21

• Wide bandwidth (large transmission capacity).
• Signal power losses are relatively small (longer uninterrupted distances between amplifiers & less cost to install).
• It costs less than metal wire.
• It is lighter than ^ (easier handling & storage).
• High security (don’t radiate energy so no crosstalk).
• Don’t pick up e.m interference (can be used in e.mly noisy areas).
• Ideal for digital transmission as light from laser can be turned on & off.

Question 22

Describe the use of satellites in communication.

Answer 22

A transmitting station T directs a carrier wave of frequency f₁(up) towards the satellite.The satellite receives this signal, amplifies it and changes the carrier wave frequency to a lower value f₂(down) before directing it towards a receiver R back on earth.

Question 23

Reasons why satellite is used and not direct transmission. (Can be done with SW & MW wavebands).

Answer 23

• Long distance comm. on these wavebands is unreliable (requires ionospheric reflection, these layers of ions vary in height and density with time of day).
• Wavebands are already filled by existing broadcasts.
• The available bandwidth are to narrow to carry the amount of info required.

Question 24

Features of polar orbits.

Answer 24

• Relatively low with period of rotation of the order of 90min.
• Apart from comm. it is also used for monitoring the state of the earths surface, weather forecasting, spying etc.
• Low orbit resulting in short time delays between transmission.

Question 25

Features of geostationary satellites.

Answer 25

• Orbit the earth above the equator with a period of 24hrs at a distance of 3.6 x 10⁴km.
• rotates with the same period as Earth.
• It can allow continuous .communication between a ground station and anywhere on the earth that can receive a signal from the satellite.
• Don’t need to be tracked like polar as a network of satellites aren’t needed to maintain continuous links so the link can be switched from one satellite to another.

Question 26

Unit of attenuation.

Answer 26

Bells or decibels.

Question 27

Equation for number of bels or decibels. (Attenuation)

Answer 27

Num of B = log(P₂/P₁)
OR
Num of dB = 10log(P₂/P₁)

2->received.
1->transmitted.

Question 28

Equation for attenuation per unit length.

Answer 28

Att. per unit length = 10log(P₂/P₁) / length of cable

Question 29

Equation of signal to noise ratio.

Answer 29

Signal to noise ratio = 10log(signal power / noise power)