Transmission impairments,Jitter Wander,G 821 anaysis

1. TRANSMISSION UNITS

2. Signals travel through transmission media, are not perfect. TheSignals travel through transmission media, are not perfect. The
imperfection causes signal impairment. This means that the signalimperfection causes signal impairment. This means that the signal
at the beginning of the medium is not the same as the signal at theat 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.end of the medium. What is sent is not what is received.
 Attenuation
 Distortion
 Noise
TRANSMISSION IMPAIRMENTTRANSMISSION IMPAIRMENT

3. Causes of impairment

4. ATTENUATION
 Means loss of energy
 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.
 To compensate- repeaters are used (most commonly)

5. Figure Attenuation

6. DISTORTION
 The Inaccurate reproduction of a signal caused by
changes in the signal's waveform, either amplitude
or frequency

7. NOISE
 Any random disturbance or unwanted signal
caused by the environment
 Measure of Noise:
 Signal to Noise Ratio :average signal power to the
average noise power

8. Figure :Noise

9. Figure 3.30 Two cases of SNR: a high SNR and a low SNR

10. MEASUREMENT OF ATTENUATION
 To show the loss or gain of energy the unit “decibel”
is used.
 "deciBel" is the 1/10th part of a "Bel"
 originated in line telephony in 1923
dB = 10log10P2/P1
P1 - input signal
P2 - output signal

11. SIGNAL TO NOISE RATIO (SNR)
 To measure the quality of a system the SNR is
often used. It indicates the strength of the signal
w.r.t the noise power in the system.
 It is the ratio between two powers.
 It is usually given in dB and referred to as SNRdB.

12. BASE 10 LOGARITHM RULES
 log (AxB) = log (A) + Log10 (B)
 log (A/B) = log (A) - log (B)
 log (1/A) = - log (A)
 log (1) = 0
 log (2) = 0.3
 log (10) = 1
 log (2 x 10) = log (2) + log (10) = 1 + 0.3
 log (100) = 2
 log (1000) = 3
 log (10000) = 4
 Log(an
)=n X log a

13. DBM TO WATT, MW, DBW
CONVERSION TABLE
P o w e r
( d B m )
P o w e r
( d B W )
P o w e r ( w a t t ) P o w e r ( m W )
- 1 0 0 d B m - 1 3 0 d B W 0 .1 p W 0 .0 0 0 0 0 0 0 0 0 1 m W
- 9 0 d B m - 1 2 0 d B W 1 p W 0 .0 0 0 0 0 0 0 0 1 m W
- 8 0 d B m - 1 1 0 d B W 1 0 p W 0 .0 0 0 0 0 0 0 1 m W
- 7 0 d B m - 1 0 0 d B W 1 0 0 p W 0 .0 0 0 0 0 0 1 m W
- 6 0 d B m - 9 0 d B W 1 n W 0 .0 0 0 0 0 1 m W
- 5 0 d B m - 8 0 d B W 1 0 n W 0 .0 0 0 0 1 m W
- 4 0 d B m - 7 0 d B W 1 0 0 n W 0 .0 0 0 1 m W
- 3 0 d B m - 6 0 d B W 1 μ W 0 .0 0 1 m W
- 2 0 d B m - 5 0 d B W 1 0 μ W 0 .0 1 m W
- 1 0 d B m - 4 0 d B W 1 0 0 μ W 0 .1 m W
- 1 d B m - 3 1 d B W 7 9 4 μ W 0 .7 9 4 m W
0 d B m - 3 0 d B W 1 .0 0 0 m W 1 .0 0 0 m W
1 d B m - 2 9 d B W 1 .2 5 9 m W 1 .2 5 9 m W
1 0 d B m - 2 0 d B W 1 0 m W 1 0 m W
2 0 d B m - 1 0 d B W 1 0 0 m W 1 0 0 m W
3 0 d B m 0 d B W 1 W 1 0 0 0 m W
4 0 d B m 1 0 d B W 1 0 W 1 0 0 0 0 m W
5 0 d B m 2 0 d B W 1 0 0 W 1 0 0 0 0 0 m W
6 0 d B m 3 0 d B W 1 k W 1 0 0 0 0 0 0 m W
7 0 d B m 4 0 d B W 1 0 k W 1 0 0 0 0 0 0 0 m W
8 0 d B m 5 0 d B W 1 0 0 k W 1 0 0 0 0 0 0 0 0 m W
9 0 d B m 6 0 d B W 1 M W 1 0 0 0 0 0 0 0 0 0 m W
1 0 0 d B m 7 0 d B W 1 0 M W 1 0 0 0 0 0 0 0 0 0 0 m W

14. Suppose a signal travels through a transmission medium and its
power is reduced to one-half. This means that P2 is (1/2)P1. Find
the attenuation in dB?
A loss of 3 dB (–3 dB) is equivalent to losing one-half the
power.

15. A signal travels through an amplifier, and its power is increased
10 times. This means that P2 = 10P1 . Find the amplification in
dB?

16. Figure 4
Find the overall attenuation in dB?

17. 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 4 a signal
travels from point 1 to point 4. In this case, the decibel value
can be calculated as

18.  Need for db, when using logarithmic ratio, it is easy to
calculate overall lose or gain of the txn system by
addn/subtraction
 Gain of 3dB means double the power, -3dB means power
halved (+value means gain, -ve value means loss)

19. WORKED OUT EXAMPLES
Input 1W, output 2W
calculate loss/gain in dB?
Network
1 W 2 W
Gain = 10 log (output)/(input) dB
10 log 2/1 dB= 10 (0.3010) dB=3.101 dB
= 3dB
approximately

20.  Find the output power for the given network
•0.1W
Network 13 db gain
Gain = 10 log P2/P1 dB
10 log P2/0.1 dB =13db
i.e., log P2/0.1 = 1.3
Or P2/0.1 = antilog 1.3 or
P2 = 0.1 antilog 1.3
P2 = 2W

21. BASIC DERIVED DECIBEL UNITS
 To know the absolute power of a component we use
dBm/Watt
 Power (in dBm) = 10 log( Power /(1mW))

0 dBm = 1mW.

22. EXAMPLES OF DBM
 An amplifier has an output of 20 W; what is its
output in dBm?
 Power (dBm) = 10 log 20 W/1 mW = 10 log
20x10^3 mW/1mW = +43 dBm.

23. BIT ERROR RATE (BER)
 The BER is the measure of error bits with
respect to the total number of bits transmitted in
a given time.

24. JITTER
Abrupt and unwanted variations of one or more signal characteristics, such
as the interval between successive pulses, the amplitude of successive
cycles, or the frequency or phase of successive cycles.

25. WANDER
 Slow variations in signal timing through a
system are called wander.
 Higher speed variations are termed jitter

26. QUALITY PARAMETERS
 Error Seconds (ES)
 Number of one-second intervals with one or more
errors.
 Severely Error Seconds (SES)
 Number of one-second intervals with an error rate,
worse than 10-3
 Non Severely Error Seconds (NSES)
 Number of one-second intervals with an error rate,
better than or equal to 10-3
.

27. AVAILABLE AND NON-
AVAILABLE TIME
 A period of available time begins with a period of
ten consecutive seconds each of which has a BER
better than 10-3
. These 10 seconds are considered
to be available time.
 A period of unavailable time begins when the bit
error rate in each second is worse than 10-3
for a
period of 10 consecutive seconds. These 10
consecutive seconds are considered to be
unavailable time

29. DIGITAL TRANSMISSION
PERFORMANCE CRITERIA
( GENERAL)
 1 in 106
: Better
 1 in 105
: Good
 1 in 104
: Reasonably good
 1 in 103
: Just Acceptable
 More than 1 in 103
: Unacceptable