7.Module 5.ppt

3.1
Chapter 3
Data and Signals
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

3.2
3-4 TRANSMISSION IMPAIRMENT
Signals travel through transmission media, which are not
perfect. The imperfection causes signal impairment. This
means that the signal at the beginning of the medium is
not the same as the signal at the end of the medium.
What is sent is not what is received. Three causes of
impairment are attenuation, distortion, and noise.
 Attenuation
 Distortion
 Noise
Topics discussed in this section:

3.3
Figure 3.25 Causes of impairment

3.5
Attenuation
 Means loss of energy -> weaker signal
 When a signal travels through a
medium it loses energy overcoming the
resistance of the medium
 Amplifiers are used to compensate for
this loss of energy by amplifying the
signal.

3.6
Measurement of Attenuation
 To show the loss or gain of energy the
unit “decibel” is used.
dB = 10log10P2/P1
P1 - input signal
P2 - output signal

3.8
Suppose a signal travels through a transmission medium
and its power is reduced to one-half. This means that P2
is (1/2)P1. In this case, the attenuation (loss of power)
can be calculated as
Example 3.26
A loss of 3 dB (–3 dB) is equivalent to losing one-half
the power.

3.9
A signal travels through an amplifier, and its power is
increased 10 times. This means that P2 = 10P1 . In this
case, the amplification (gain of power) can be calculated
as
Example 3.27

3.10
One reason that engineers use the decibel to measure the
changes in the strength of a signal is that decibel
numbers can be added (or subtracted) when we are
measuring several points (cascading) instead of just two.
In Figure 3.27 a signal travels from point 1 to point 4. In
this case, the decibel value can be calculated as
Example 3.28

3.11
Figure 3.27 Decibels for Example 3.28

3.12
Sometimes the decibel is used to measure signal power
in milliwatts. In this case, it is referred to as dBm and is
calculated as dBm = 10 log10 Pm , where Pm is the power
in milliwatts. Calculate the power of a signal with dBm =
−30.
Solution
We can calculate the power in the signal as
Example 3.29

3.13
The loss in a cable is usually defined in decibels per
kilometer (dB/km). If the signal at the beginning of a
cable with −0.3 dB/km has a power of 2 mW, what is the
power of the signal at 5 km?
Solution
The loss in the cable in decibels is 5 × (−0.3) = −1.5 dB.
We can calculate the power as
Example 3.30

3.15
Distortion
 Means that the signal changes its form or
shape
 Distortion occurs in composite signals
 Each frequency component has its own
propagation speed traveling through a
medium.
 The different components therefore arrive
with different delays at the receiver.
 That means that the signals have different
phases at the receiver than they did at the
source.

3.18
Noise
 There are different types of noise
 Thermal - random noise of electrons in the
wire creates an extra signal
 Induced - from motors and appliances,
devices act are transmitter antenna and
medium as receiving antenna.
 Crosstalk - same as above but between
two wires.
 Impulse - Spikes that result from power
lines, lighning, etc.

3.20
Signal to Noise Ratio (SNR)
 To measure the quality of a system the
SNR is often used. It indicates the
strength of the signal wrt the noise
power in the system.
 It is the ratio between two powers.
 It is usually given in dB and referred to
as SNRdB.

3.21
The power of a signal is 10 mW and the power of the
noise is 1 μW; what are the values of SNR and SNRdB ?
Solution
The values of SNR and SNRdB can be calculated as
follows:
Example 3.31

3.22
The values of SNR and SNRdB for a noiseless channel
are
Example 3.32
We can never achieve this ratio in real life; it is an ideal.

3.23
Figure 3.30 Two cases of SNR: a high SNR and a low SNR

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