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TRANSMISSION IMPAIRMENTTRANSMISSION IMPAIRMENT SignalsSignals traveltravel throughthrough transmission
transmission media,media, which
which areare notnot perfectperfect TheThe imperfectionimperfection causes
causes signalsignal impairmentimpairment This
This
perfectperfect.. TheThe imperfectionimperfection causes
causes signalsignal impairmentimpairment.
. ThisThis
meansmeans thatthat thethe signalsignal at
at thethe beginningbeginning of
of thethe mediummedium isis notnot thethe samesame asas the
the signalsignal atat thethe end
end ofof thethe mediummedium.
gg^
ff WhatWhat isis sentsent isis not
not whatwhat isis receivedreceived.
. ThreeThree causescauses of
of
impairmentimpairment areare attenuationattenuation,
, distortiondistortion,, andand noisenoise.
Topics discussed in this section:Topics discussed in this section: ^ Attenuation ^ Distortion ^ NoiseNoise
Figure 3 25^ C^
f i^ i^ t Figure^ 3.25^ Causes of impairment
AttenuationAttenuation ^ Means loss of energy -> weaker signal ^ When a signal travels through a^ When a signal travels through amedium it loses energy overcoming theresistance of the mediumresistance of the medium ^ Amplifiers are used to compensate forthis loss of energy by amplifying thethis loss of energy by amplifying thesignal.
Measurement of AttenuationMeasurement
of Attenuation
^ To show the loss or gain of energy theunit “decibel” is used.dB = 10log
P /PdB = 10log P^ /P^1021 P - input signal 1 P - output signal 2
Example 3.26 Suppose a signal travels through a transmission mediumand its power is reduced to one-half. This means that P
2
is (1/2)P. In this case, the attenuation (loss of power)^1 b^ l^ l t dcan be calculated as A loss of 3 dB (–3 dB) is equivalent to losing one-halfthe power.
Example 3.27 A signal travels through an amplifier, and its power isincreased 10 times. This means that P
= 10P. In this 2 1 case, the amplification (gain of power) can be calculatedas
- Figure 3.27 Decibels for Example 3.
Example 3.29 Sometimes the decibel is used to measure signal powerin milliwatts. In this case, it is referred to as dB
and ism
calculated as dB
= 10 log10 Pm m^ , where Pis the powerm^ i^ illi^ tt^ C l
l t^ th^
f^ i^ l^ ith dB in milliwatts. Calculate the power of a signal with dB
=m
− 30. SolutionWe can calculate the power in the signal as
p^ g
DistortionDistortion ^ Means that the signal changes its form orea s t at t e s g a c a ges ts
o^ o shape Distortion occurs in composite signals Each frequency component has its ownpropagation speed traveling through amediummedium. The different components therefore arrivewith different delays at the receiver.with different delays at the receiver. That means that the signals have differentphases at the receiver than they did at thesource.
Figure 3.28^ Distortion
NoiseNoise ^ There are different types of noise^ There^ are different types of noise^ ^ Thermal - random noise of electrons in thewire creates an extra signalwire creates an extra signal^ ^ Induced - from motors and appliances,devices act are transmitter antenna anddmedium as receiving antenna.^ ^ Crosstalk - same as above but betweentwo wirestwo wires.^ ^ Impulse - Spikes that result from powerlines, lighning, etc.lines, lighning, etc.
Figure 3.29^ Noise
Example 3.31 The power of a signal is 10 mW and the power of thenoise is 1^ μ W; what are the values of SNR and SNR
?dB
S l tiSolutionThe values of SNR and SNR
can be calculated asdB follows:follows:
Example 3.32 The values of SNR and SNR
for a noiseless channeldB^ are We can never achieve this ratio in real life; it is an ideal.
