Chapter 5 Analog Transmission

Digital--tto--analoanalog conversion is the process of changing one signal baseoned o ofn the information in digital data.. of thethe characteristics characteristics of Topics discussed in this section: of an Aspects of Digital-to-Analog Conversion an analog analog Amplitude Shift Keying Frequency Shift Keying Phase Shift Keying Quadrature Amplitude Modulation

5.2 Figure 5.1 Digital-to-analog conversion

Digital-to-analog modulation (or shift keying): changing one of the characteristics of the analog signal …… based on the information of the digital signal (carrying digital information onto analog signals)

Changing any of the characteristics of the simple signal (amplitude, frequency, or phase) would change the nature of the signal to become a composite signal 5.3 Figure 5.2 Types of digital-to-analog conversion

5.4 Asppgects of Digital-to-Analog Conversion

 Data Element vs. Signal Element  Data Rate vs. Signal Rate

 S = N/r

 r=logr = log2 L, where L is the number of signal elements  Bandwidth:

 The required bandwidth for analog transmission of digital data is proportional to the signal rate  Carrier Signal:

 The digital data changes the carrier signal by modifying one of its characteristics

 This is called modulation (or Shift Keying)

 The receiver is tuned to the carrier signal’s frequency

5.5

digital

. er second. of

p second

per

transmission

elements per second or equal to the bit rate. analog

data, the baud rate is less than the

Baud rate is the number of signal In the analog transmission of digital

Note Bit rate is the number of bits 5.6 from

value

the

find

Example 5.1

can

An analog signal carries 4 bits per signal element. If An analog signal carries 4 bits per second, find the bit sent per 1000 signal elements are rate. Solution and N is unknown. rS = 1000, = 4, case, In this We can find the value of N from 5.7 n

s the

r element f a o

e dat

y valu

How man ? firs . d d need

e fin bau w e 0 o d .W 100 s n f o

e n Example 5.2 rat element l unknow d e gna i Solutio In this example,ar S = 1000,the N value = of 8000, L. and r and L An analog signalbau has a bitare rate of carried 8000s by bps each and a signal element? How many 5.8 Amppyg()litude Shift Keying (ASK)

 ASK: varying the peak amplitude of the carrier signal to represent binary 1 or 0

 Other signal characteristics remain constant as the amplitude changes

 During each bit duration, the peak amplitude remains constant

 Transmission medium noise is usuallyy(;p), additive (i.e.; affects the amplitude), therefore, ASK is very susceptible to noise interference

 Binary ASK (BASK) or On/Off Keying (OOK) modulation technique:

 One o f the bi nary bit s i s represent ed b y no volt age

 Advantage: requires less transmission energy compared to two-level techniques

 BASK spectrum is most significantly between [fc -S/2, fc + S/2], where fc is frequency of the carrier signal and S is the baud rate

 The ASK bandwidth is BBASK = (1+d) S , where d is a factor related to the modulation process , of which the value is between 0 and 1

 The min. bandwidth required to transmit an ASK is equal to the baud rate

 The baud rate is the same as the bit rate  N = S

5.9 Figure 5.3 Binary amplitude shift keying

5.10 Figure 5.4 Implementation of binary ASK

5.11 = be r r

can and

carrie 1 e = th d e

ar frequency t (with

Wha

. carrier rate z

kH bit ? our 0 1 = 30 the d that o t h

0 find wit 20 to K means

m AS

g n =250kHz.Wecanusetheformulafor Example 5.3 fro This s . c usin

y We have an availablespan bandwidth of 100frequency kHz and which the bitb rate if we modulated ourSolutio data The middlekHz of theat bandwidth f isbandwidth located at1). 250 5.12 e r

th fo

e h

divid

o t

d bandwidt

nee

W . s availabl

Th . s

direction . h

bot

n i bandwidth n e direction

d each an s in Bandwidth of full-duplex ASK used in Example 5.4 Bandwidth of full-duplex

Example 5.4 communicatio kbps

h 25 ure 5.5 ure g In data communications,wit we normally usebandwidth into full-duplex two links with twoFigure carrier frequencies, 5.5. as shown Thefrequencie in figure showseach direction the is now positions 50of kHz, of which leaves two us with carrier a data rate Fi 5.13 Freqqyuency Shift Key yg()ing (FSK)

 FSK: varying the frequency of the carrier signal to represent binary 1 or 0  Other signal characteristics remain constant as the frequency changes  During each bit duration , the frequency remains constant  FSK is mostly immune to the transmission medium additive noise interference since FSK only cares about Frequency changes  The Binary FSK (BFSK) can be thought of as two ASK signals, each with its own carrier frequency f1 and f2  f1 and f 2 are 2fapartf apart  The BFSK required bandwidth is BBFSK = (1+d) S + 2f  What is the minimum value of 2f ?  The baud rate is the same as the bit rate  N = S

5.14 Figure 5.6 Binary frequency shift keying

5.15 at bit

the

and band

the

frequency 2 f

carrier midpoint

the

The be . 1 f FSK should

using What .

kHz

Example 5.5 300 n

We have an available200 bandwidth of 100rate kHz if we which modulated spans our data from by usingSolutio FSK with d = 1? This problemmodulating is similar250 kHz. to We choose 2Δf Example to be 5.3, 50 kHz; but this means we are 5.16 Figure 5.7 Implementation of BFSK

5.17 Multi-Level FSK (()MFSK)

 More than two frequencies can be used to represent more than one bit each

 For example: four different frequencies can be used to send 2 bits at a time

 However, each adjacent pair of frequencies must be 2fapart

 The MFSK required bandwidth is

BMFSK = (1+d) S + (L-1) 2f

 When d=0, the minimum Band width B MFSK = LSL x S

5.18 . z 1

be MH

10 must . i is h t MHz y id 8 w

d =

an bdidh b requenc f frequencies m MHz 1 r mu i × n ii i 8

cente m carrier =

r h he B e t i = 8. The baudthe rate is S = 3 MHz/3 d is 3

an carr , 

e .

rate Th d Example 5.6 . bandwidth s au Mbaud bd b 1

h he Weneedtosenddata3bitsatatimeatabitrateof3 Mbp Calculate the numbert of levels (different frequencies), Solution We have L == 2 MHz apart (2Δf =The 1 MHz). Figure 5.8 shows thebandwidth allocation of frequencies and 5.19 Figure 5.8 Bandwidth of MFSK used in Example 5.6

5.20 Phase Shift Keying (PSK)

 PSK: varying the phase of carrier signal to represent binary 1 or 0  Other signal characteristics remain constant as the phase changes  During each bit duration, the phase remains constant  PSK is mostly immune to the transmission medium additive noise interference since PSK only cares about phase changes  PSK spectrum and bandwidth requirements are similar to ASK  PSK is better than both ASK and FSK. Why?  2-PSK or Binary PSK (BPSK): uses 2 different phases (usually 0 and 180) each representing 1 bit of data  N = S  4-PSK or Quad PSK (QPSK): uses 4 different phases (e .g .; 45, - 45, 135, and -135) each representing 2 bits of data  N = 2 S  PSK is represented by a Constellation (or Phase-State) diagram  PSK is limited by the ability of the receiver to distinguish small phase difference, therefore, limiting the potential bit rate

5.21 Figure 5.9 Binary phase shift keying

5.22 Figure 5.10 Implementation of BPSK

5.23 Figure 5.11 QPSK and its implementation

5.24 N

i is )

rate

(b d

rat l

gna il i s

h he t

S So . 2

r t a . h n t Example 5.7 o i s MH MHz

6 olut Find the bandwidthMbps for for a QPSK. The signal value of transmitting d at = 0. 12 Si S ForQPSK,2bitsiscarriedbyonesignalelement.This mean × (1/r) = 6 Mbaud.= With a value of d = 0, we have B = S 5.25 Figure 5.12 Concept of a constellation diagram

 Used to define the amplitude and phase of a signal element when using two carriers or when dealing with multi-level shift keying

 A signal element is represented by a dot in the diagram, of which:

 The projection on the X axis defines the peak amplitude of the in-phase component

 The projection on the Y axis defines the peak amplitude of the quadrature component

 The length of the line connecting the point to the origin is the peak amplitude of the signal element

 The angle the line makes with the X axis is the phase of the signal element 5.26 . ls

gna il i s

QPSK diagrams constellation Three

an Example 5.8 Figure 5.13 Figure Show the constellation diagrams forBPSK BPSK, an ASK (OOK), Solution constellation three the shows 5.13 Figure diagrams. 5.27 QQpuadrature Amplitude Modulation (Q (Q)AM)

 QAM: varying both the peak amplitude and the phase of the carrier signal to represent binary combination

 During each bit duration, the phase and amplitude remain constant

 Theoretically, any number of measurable changes in phase and amplitude can be combined to give several variations in the signal

 The number of phase shifts is always greater than the amplitude shifts. Why?

 The greater the ratio of phase to amplitude shifts, the better the noiiise immuni ty

 QAM spectrum and bandwidth requirements are similar to ASK and PSK

5.28

modulation

amplitude

combination of ASK and PSK. combination of

te o Quadrature amplitude modulation is a Nt N 5.29 Figure 5.14 Constellation diagrams for some QAMs

5.30 Examples of 4-QAM and 8-QAM constellations

5.31 Examples of 16-QAM constellations

5.32 e

t ith ra w d bit e e itt th

If ransm t e. l d rc au i bd b s are

bit t itt d ith

3 1600 , 8 s on a c = = t 3

3 n / i 2

o p

nce d 4800 Si . t ace p y s y ll d i t i l If th bit t ua egrees apar q each signal unit. Therefore, the each signal unit. baud rate is The constellation indicates 8-PSK 45 The constellation indicates with the points dtSi d A constellation diagram consists of eight e is 4800 bps, what the baud rate? Solution Example 5.33 ps b 16 = 4. 2 log (1000)(4) = 4000

. on

l i

ut na l il ig Thus, A 16-QAM signal has 4 bits per signal unit since A

Compute the bit rate for a 1000-baud 16-QAM s So Solution Example 5.34 d au b 64 = 6. 2 log 72000 / 6 = 12,000 72000 . l na ig on i s

ut l Thus, A 64-QAM signal has 6 bits per signal unit since A

Compute the baud rate for a 72,000-bps 64- Compute the baud rate for a 72,000-bps QAM iSo Solution l Example 5.35 N 2N 3N 4N 5N 6N 7N 8N Bit Rate N N N N N N N N Baud rate Baud rate 1 2 3 4 5 6 7 8 Bits/Baud Bits/Baud bit t a i t B Units Units Dibit en Tribit Octabit Hexabit Quadbit Septabit Ptbit P K S PSK P - - 2 QAM QAM QAM QAM - - - - K, 4 4 S

, Modulation Modulation QAM QAM QAM QAM - - - - QAM QAM QAM QAM QAM QAM K, F ------PSK 4 PSK, 8 PSK, 4 PSK, 8 - - - - ASK, FSK, 2 4 8 16 32 64 128 256 8 16 64 128 256 Bit and baud rate comparison AS S4 2 32 5.36 5-2 ANALOG-TO-ANALOG CONVERSION

Analog-to-analog conversion is the representation of analog information ask whyinformationwe need to bymodulate an analog signal; it is already analog by an an analog is bandpassanalog..in nature Modulationor if onlyanaloga signalbandpass channel is availableavailable to Modulation is signal.. One to us is needed One may Topics discussedus.. in this section:needed if may Amplitude Modulation if thethe medium Frequency Modulation medium Phase Modulation

5.37 Figure 5.15 Types of analog-to-analog modulation

5.38 Ampp()litude Modulation (AM)

 The amplitude of the carrier (or modulated) signal changes with the amplitude changes of the modulating signal

 The frequency and phase of the modulated signal remain constant

 The modulating signal becomes the envelope (i .e .; the outer shape) of the modulated signal

 The bandwidth of the AM signal is equal to twice the bandwidth of the modulating signal

 The bandwidth of an audio sigg(nal (voice and music ) is 5 kHz

 AM radio stations are assigned 10 kHz band per channel

 Everyygp other band is used as a guard band to prevent interference among adjacent channels

5.39 Figure 5.16 Amplitude modulation

5.40

for

= 2B. required

AM nal: B g si bandwidth

can be determined total

from the bandwidth of the audio from the bandwidth The total bandwidth required for AM

Note 5.41 Figure 5.17 AM band allocation

5.42 Freqqyuency Modulation ( ()FM)

 The frequency of the carrier (or modulated) signal changes with the amplitude changes of the modulating signal

 The peak amplitude and phase of the modulated signal remain constant

 The bandwidth of the FM signal is usually about 10 times the bandwidth of the modulating signal

 The bandwidth of a stereo audio signal (voice and music) is about 15 kHz

 FM radio stations are assigned 200 kHz band per channel

 Every o the r b and is used as a gu ard b and to p reve nt interference among adjacent channels

5.43

. can

)B )B β β

1 for (

2 = FM B

required

signal:

bandwidth

audio

the

total

be determined from the bandwidth te of the audio signal: B o

Nt N The total bandwidth required for FM can 5.44 Figure 5.18 Frequency modulation

5.45 Figure 5.19 FM band allocation

5.46 Phase Modulation (()PM)

 The phase of the carrier (or modulated) signal changes with the amplitude changes of the modulating signal

 The peak amplit ud e and f requency of th e mod ul at ed signal remain constant

 PM is the same as FM except that in FM, the instantaneous change in the carrier frequency is proportional to the amplitude of the modulating signal, while in PM it is proportional to the rate of change of the amplitude

 The bandwidth of the PM signal is usually abut 6 times the bandwidth of the modulating signal 5.47 Figure 5.20 Phase modulation

5.48 )B. β signal:

= 2(1 + PM B modulating signal: and maximum amplitude of the be determined from the bandwidth

Note required for PM can The total bandwidth 5.49