| (a) | What is noise in the context of data communications. Why is noise a problem? | [3] |
| Noise is undesirable
signal or energy coming from other sources, other than the transmitter. It is a problem because :
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| (b) | What is attenuation? Why is it undesirable? Explain the chief cause of attenuation in terrestrial microwave communications. | [4] |
| Attenuation is the
weakling of signal strength as it propagates through the medium. The further it goes, the
weaker it gets. Increase in frequency may increase attenuation. It's undesirable because if the signal gets too weak, it is hard to distinguish it from noise. And it is also hard to interpret it The chief cause of attenuation in microwave is rainfall because rainfall absorbs the signal.
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| (c) | Some types of noise can be overcome by careful design. However, there is one type of noise which is unavoidable. What is this type of noise called? Explain why it is unavoidable. | [3] |
| Thermal noise is not
avoidable . Thermal noise is a function of temperature and it is caused by random electron
movement in the medium. Therefore, it's no way we can avoid this.
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| (d) | What is meant by delay distortion and how is this problem overcome? | [2] |
| The propagation
velocity of signal along guided medium is not uniform. Signal tends to travel faster in
the center of the medium and slower at the edge of the medium. Therefore, the arrival of
different components of a signal at destination at different time is called delay
distortion. It can be overcome by equalising the velocity of signal across the band.
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| (e) | The following bytes have been received. They are
encoded using even parity. Which ones have been corrupted by communication noise?
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[6] |
| Bytes 3 and 5 have
parity errors. Bit 5 of byte 4 doesn't match on parity; therefore byte 3 should have been 01001110. Bit 6 of byte 8 doesn't match on parity; therefore byte 5 should have been 10001101. The other bytes should be as they are.
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| Suppose that every fourth byte is inserted as an LRC (longitudinal redundancy check) check character (with even parity). What was the original 6-byte message? What assumptions have you made? | ||
| This assumes that the
noise rate/error rate is low, and that a single error has occurred in each block of four
bytes.
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| (f) | What is the transmission efficiency of the message transmission scheme used in part (e)? | [2] |
| Total bits transmitted
: 64 Total overhead bits : 8 (one per char) + 2*7 (for each character). Efficiency = (64-22)/64=65.6%.
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