forouzan chp 4

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McGraw-Hill © The McGraw-Hill Companies, Inc., 2004

Chapter 4

Digital Transmission


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4.1 Line Coding

Some Characteristics

Line Coding Schemes

Some Other Schemes


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Figure 4.1 Line coding


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Figure 4.2 Signal level versus data level


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Figure 4.3 DC component


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Example 1 Example 1

A signal has two data levels with a pulse duration of 1ms. We calculate the pulse rate and bit rate as follows:

Pulse Rate = 1/ 10Pulse Rate = 1/ 10-3-3= 1000 pulses/s= 1000 pulses/s

Bit Rate = Pulse Rate x logBit Rate = Pulse Rate x log22 L = 1000 x logL = 1000 x log22 2 = 1000 bps2 = 1000 bps


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Example 2 Example 2

A signal has four data levels with a pulse duration of 1

ms. We calculate the pulse rate and bit rate as follows:

Pulse Rate = = 1000 pulses/sPulse Rate = = 1000 pulses/s

Bit Rate = PulseRate x logBit Rate = PulseRate x log22

L = 1000 x logL = 1000 x log22

4 = 2000 bps4 = 2000 bps


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Figure 4.4 Lack of synchroniation


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Example ! Example !

In a digital transmission, the receiver clock is 0.1 percent

faster than the sender clock. How many extra bits per

second does the receiver receive if the data rate is 1

Kbps? How many if the data rate is 1 Mbps?

Solution Solution

At 1 Kbps:

1000 bits set1001 bits re!ei"e#1 extra bps

At 1 $bps:

1%000%000 bits set1%001%000 bits re!ei"e#1000 extra bps


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Figure 4.5 Line coding schemes


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"nipolar encoding uses only one

voltage level#

&ote: &ote:


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Figure 4.6 "nipolar encoding


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$olar encoding uses t%o voltage levels $olar encoding uses t%o voltage levels

&positive and negative'#&positive and negative'#

&ote: &ote:


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Figure 4.7 Types of polar encoding


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(n )*+,L the level of the signal is (n )*+,L the level of the signal is

dependent upon the state of the -it#dependent upon the state of the -it#

&ote: &ote:


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(n )*+,( the signal is inverted if a 1 is (n )*+,( the signal is inverted if a 1 is

encountered#encountered#

&ote: &ote:


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Figure 4.8 )*+,L and )*+,( encoding


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Figure 4.9 *+ encoding


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. good encoded digital signal must . good encoded digital signal must

contain a provision forcontain a provision forsynchroniation#synchroniation#

&ote: &ote:


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Figure 4.10 /anchester encoding


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(n /anchester encoding0 the (n /anchester encoding0 the

transition at the middle of the -it istransition at the middle of the -it isused for -oth synchroniation and -itused for -oth synchroniation and -it

representation#representation#

&ote: &ote:


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Figure 4.11 Differential /anchester encoding


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(n differential /anchester encoding0 (n differential /anchester encoding0

the transition at the middle of the -it isthe transition at the middle of the -it isused only for synchroniation#used only for synchroniation#

The -it representation is defined -y theThe -it representation is defined -y the

inversion or noninversion at theinversion or noninversion at the-eginning of the -it#-eginning of the -it#

&ote: &ote:


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(n -ipolar encoding0 %e use three (n -ipolar encoding0 %e use three

levels positive0 ero0levels positive0 ero0and negative#and negative#

&ote: &ote:


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Figure 4.12 ipolar ./( encoding


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Figure 4.13 213


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Figure 4.14 /LT,! signal


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4.2 Block Coding

Steps in Transformation

Some Common lock Codes

Fi 4 15 l k di


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Figure 4.15 lock coding

Fi 4 16 S - tit ti i -l k di


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Figure 4.16 Su-stitution in -lock coding


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Ta-le 4#1 456 encoding Ta-le 4#1 456 encoding

Data Code Data Code

0000 1111011110 1000 1001010010

0001 0100101001 1001 1001110011

0010 1010010100 1010 1011010110

0011 1010110101 1011 1011110111

0100 0101001010 1100 1101011010

0101 0101101011 1101 1101111011

0110 0111001110 1110 1110011100

0111 0111101111 1111 1110111101


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Ta-le 4#1 456 encoding &Continued'Ta-le 4#1 456 encoding &Continued'

Data Code

' ('uiet) 0000000000

* (*#le) 1111111111

+ (+alt) 0010000100

, (start #eliiter) 1100011000

K (start #eliiter) 1000110001

. (e# #eliiter) 0110101101

(et) 1100111001

R (Reset) 0011100111

Figure 4 17 E l f 758T di


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Figure 4.17 Example of 758T encoding


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4.3 Sampling4.3 Sampling

$ulse .mplitude /odulation

$ulse Code /odulation Sampling *ate )y9uist Theorem

:o% /any its per Sample;

it *ate

Figure 4 18 $./


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Figure 4.18 $./


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$ulse amplitude modulation has some $ulse amplitude modulation has some

applications0 -ut it is not used -y itselfapplications0 -ut it is not used -y itselfin data communication# :o%ever0 it isin data communication# :o%ever0 it is

the first step in another very popularthe first step in another very popular

conversion method calledconversion method called pulse code modulation# pulse code modulation#

&ote: &ote:

Figure 4 19 3uantied $./ signal


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Figure 4.19 3uantied $./ signal

Figure 4 20 3uantiing -y using sign and magnitude


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Figure 4.20 3uantiing -y using sign and magnitude

Figure 4 21 $C/


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Figure 4.21 $C/

Figure 4 22


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Figure 4.22


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.ccording to the )y9uist theorem0 the .ccording to the )y9uist theorem0 the

sampling rate must -e at least 2 timessampling rate must -e at least 2 timesthe highest fre9uency#the highest fre9uency#

&ote: &ote:

Figure 4.23 )y9uist theorem


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Figure 4.23 )y9uist theorem


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Example 4 Example 4

What sampling rate is needed for a signal with a

bandwidth of 10,000 Hz (1000 to 11,000 Hz)?

Solution Solution

.he saplig rate ust be ti!e the highest reue! ithe sigal:

aplig rate = 2 x (11%000) = 22%000 saples/saplig rate = 2 x (11%000) = 22%000 saples/s


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Example 6 Example 6

A signal is sampled. Each sample requires at least 12

levels of precision (+0 to +5 and -0 to -5). How many bitsshould be sent for each sample?

Solution Solution

e ee# 4 bits5 1 bit or the sig a# 3 bits or the "alue6

A 3-bit "alue !a represet 23 = 7 le"els (000 to 111)%

hi!h is ore tha hat e ee#6 A 2-bit "alue is oteough si!e 22 = 46 A 4-bit "alue is too u!h be!ause 24

= 186


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Example 8 Example 8

We want to digitize the human voice. What is the bit rate,

assuming 8 bits per sample?

Solution Solution

.he hua "oi!e orall !otais reue!ies ro 0

to 4000 +96

aplig rate = 4000 x 2 = 7000 saples/saplig rate = 4000 x 2 = 7000 saples/s

Bit rate = saplig rate x uber o bits per sapleBit rate = saplig rate x uber o bits per saple

= 7000 x 7 = 84%000 bps = 84 Kbps= 7000 x 7 = 84%000 bps = 84 Kbps


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)ote that %e can al%ays change a )ote that %e can al%ays change a

-and,pass signal to a lo%,pass signal-and,pass signal to a lo%,pass signal-efore sampling# (n this case0 the-efore sampling# (n this case0 the

sampling rate is t%ice the -and%idth#sampling rate is t%ice the -and%idth#

&ote: &ote:


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4.4 Transmission Mode4.4 Transmission Mode

$arallel Transmission

Serial Transmission

Figure 4.24 Data transmission


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g

Figure 4.25 $arallel transmission


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g

Figure 4.26 Serial transmission


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g


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(n asynchronous transmission0 %e (n asynchronous transmission0 %e

send 1 start -it &=' at the -eginningsend 1 start -it &=' at the -eginningand 1 or more stop -its &1s' at the endand 1 or more stop -its &1s' at the end

of each -yte# There may -e a gapof each -yte# There may -e a gap

-et%een each -yte#-et%een each -yte#

&ote: &ote:


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.synchronous here means .synchronous here means

>asynchronous at the -yte level0? -ut>asynchronous at the -yte level0? -utthe -its are still synchronied@ theirthe -its are still synchronied@ their

durations are the same#durations are the same#

&ote: &ote:

Figure 4.27 .synchronous transmission


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(n synchronous transmission0 (n synchronous transmission0

%e send -its one after another %ithout%e send -its one after another %ithoutstart5stop -its or gaps#start5stop -its or gaps#

(t is the responsi-ility of the receiver to (t is the responsi-ility of the receiver to

group the -its# group the -its#

&ote: &ote:

Figure 4.28 Synchronous transmission


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