13927252 PCS Communication Systems Linear Modulation

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Principles of Communication

Prof. V. Venkata Rao

4

CHAPTER 4

Linear Modulation

4.1 IntroductionWe use the word modulation to mean the systematic alteration of one waveform, called the carrier, according to the characteristic of another waveform, the modulating signal or the message. In Continuous Wave (CW) modulation schemes, the carrier is a sinusoid. We use c ( t ) and m ( t ) , to denote the carrier and the message waveforms respectively. The three parameters of a sinusoidal carrier thatcan be varied are: amplitude, phase and frequency. A given modulation scheme ca

n result in the variation of one or more of these parameters. Before we look into the details of various linear modulation schemes, let us understand the need f

or modulation. Three basic blocks in any communication system are: 1) transmitter 2) Channel and 3) Receiver (Fig. 4.1).

Fig. 4.1: A basic communication system The transmitter puts the information fromthe source (meant for the receiver) onto the channel. The channel is the mediumconnecting the transmitter and the receiver and the transmitted information tra

vels on this channel until it reaches the destination. Channels can be of two types: i) wired channels or ii) wireless channels. Examples of the first type include: twisted pair telephone

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channels, coaxial cables, fiber optic cable etc. Under the wireless category, wehave the following examples: earths atmosphere (enabling the propagation of gr

ound wave and sky wave), satellite channel, sea water etc. The main disadvantage

of wired channels is that they require a man-made medium to be present betweenthe transmitter and the receiver. Though wired channels have been put to extensive use, wireless channels are equally (if not more) important and have found a large number of applications. In order to make use of the wireless channels, theinformation is to be converted into a suitable form, say electromagnetic waves.This is accomplished with the help of a transmitting antenna. The antenna at thereceiver (called the receiving antenna) converts the received electromagnetic e

nergy to an electrical signal which is processed by the receiver. The question is: can we radiate the baseband1 information bearing signal directly on to the channel? For efficient radiation, the size of the antenna should be 10 or more (preferab

y around 4 ), where is the wave

ength of the signa

to be radiated.Take the case of audio, which has spectra

components a

most from DC upto 20 kH

z. Assume that we are designing the antenna for the mid frequency; that is,10 kHz. Then the

ength of the antenna that is required, even for the 10 situationis,

3 108 c = = 3 103 meters, c being the ve

ocity of

ight. 4 10 f 1010

1

Baseband signals have significant spectral content around DC. Some of the baseband signals

that are of interest to us are: a) Speech b) music and c) video (TV signals). Approximate spectral widths of these signals are: Speech: 5 kHz, Audio : 20 kHz, V

ideo : 5 MHz



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Even an antenna of the size of 3 km, will not be able to take care of the entirespectrum of the signal because for the frequency components around 1 kHz, the l

ength of the antenna would be 100 . Hence, what is required from the point of

view of efficient radiation is the conversion of the baseband signa

into a narrowband, bandpass signa

. Modu

ation process he

ps us to accomp

ish this; besides, modu

ation gives rise to some other features which can be exp

oited for the purpose of efficient communication. We describe be

ow the advantages of modu

ation.

1. Modu

ation for ease of radiation

Consider again transmission of good qua

ity audio. Assume we choose the carrierfrequency to be 1 MHz. The

inear modu

ation schemes that wou

d be discussed short

y give rise to a maximum frequency spread (of the modu

ated signa

) of 40 kHz, the spectrum of the modu

ated signa

extending from (1000 20) = 980 kHz to (10

00 + 20) = 1020 kHz. If the antenna is designed for 1000 kHz, it can easi

y takecare of the entire range of frequencies invo

ved because modu

ation process hasrendered the signa

into a NBBP signa

.

2. Modu

ation for efficient transmission

Quite a few wire

ess channe

s have their own appropriate passbands. For efficient transmission, it wou

d be necessary to shift the message spectrum into the passband of the channe

intended. Ground wave propagation (from the

ower atmosphere) is possib

e on

y up to about 2 MHz. Long distance ionospheric propagation ispossib

e for frequencies in the range 2 to 30 MHz. Beyond 30 MHz, the propagation is

ine of sight. Preferred frequencies for sate

ite communication are around3 to 6 GHz. By choosing an appropriate carrier frequency and modu

ation techniq

ue, it is possib

e for us to trans

ate the baseband message spectrum into a suitab

e s

ot in the passband of the channe

intended. That is, modu

ation resu

ts in frequency trans

ation.

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Princip

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3. Modu

ation for mu

tip

exing

Severa

message signa

s can be transmitted on a given channe

, by assigning to e

ach message signa

an appropriate s

ot in the passband of the channe

. Take theexamp

e of AM broadcast, used for voice and medium qua

ity music broadcast. Thepassband of the channe

used to 550 kHz to 1650 kHz. That is, the width of the passband of the channe

that is being used is 1100 kHz. If the required transmission bandwidth is taken as 10 kHz, then it is possib

e for us to mu

tip

ex, at

east theoretica

y, 110 distinct message signa

s on the channe

and sti

be ab

eto separate them individua

y as and when we desire because the identity of eachmessage is preserved in the frequency domain.

4. Modu

ation for frequency assignment

Continuing on the broadcast situation,

et us assume that each one of the messag

e signa

s is being broadcast by a different station. Each station can be assigned a suitab

e carrier so that the corresponding program materia

can be receivedby tuning to the station desired.

5. Modu

ation to improve the signa

-to-noise ratio

Certain modu

ation schemes (notab

y frequency modu

ation and phase modu

ation) have the feature that they wi

permit improved signa

-to-noise ratio at the receiver output, provided we are wi

ing to pay the price in terms of increased transmission bandwidth (Note that the transmitted power need not be increased). Thisfeature can be taken advantage of when the qua

ity of the receiver output is very important. Having understood the need and the potentia

benefits due to modu

ation,

et us now get into the detai

s of various

inear modu

ation schemes. The

four important types of

inear modu

ation schemes are 1) 2) Doub

e SideBand, Suppressed Carrier (DSB-SC) Doub

e SideBand, Large Carrier (DSB-LC) (a

so ca

ed conventiona

AM or simp

y AM)

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Princip

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3) 4)

Sing

e SideBand (SSB) Vestigia

SideBand (VSB)

We sha

begin our discussion with DSB-SC.

4.2 DSB-SC Modu

ation4.2.1. Modu

ationThe DSB-SC is the simp

est of the four

inear modu

ation schemes

isted above (simp

est in terms of the mathematica

description of modu

ation and demodu

ationoperations). Consider the scheme shown in Fig. 4.2

Fig. 4.2: DSB-SC modu

ation schemem ( t ) is a baseband message signa

with M (f ) = 0 for f > W , c ( t ) is a high

frequency carrier, usua

y with fc >> W . DSB-SC modu

ator is basica

y a mu

tip

ier. Let g m denotes the amp

itude sensitivity (or gain constant) of the modu

ator, with the units per vo

t (we assume that m ( t ) and Ac are in vo

ts). Thenthe modu

ator output s ( t ) is,s ( t ) = g m m ( t ) ( Ac cos ( c t ) ) For convenience, let g m = 1 . Then,s ( t ) = Ac m ( t ) cos ( c t )

(4.1a)

(4.1b)

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As DSB-SC modulation involves just the multiplication of the message signal andthe carrier, this scheme is also kno n as product modulation and can be sho n asin Fig. 4.3.

Fig. 4.3: Product Modulation scheme The time domain behavior of the DSB-SC signal ( ith Ac = 1 ) is sho n in Fig. 4.4(b), for the m ( t ) sho n in Fig. 4.4(a).



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Fig. 4.4: (a) The message signal (b) The DSB-SC signal Note that the carrier undergoes a 180 phase reversal at the zer

cr

ssings

fm ( t ) . This is br

ught

ut m

re clearly in the

scill

grams, sh

wn in Fig. 4.

5 and

Fig. 4.6, where m ( t ) is a sinus

idal signal.

With reference tFig. 4.5, between the p

ints a and b, the carrier in the D

SB-SC signal and the actual carrier (b tt m picture) are in phase whereas between the p

ints b and c, they are 1800

ut

f phase. Fig. 4.6 is an expanded ve

rsi n f the central part f the wavef rms in Fig. 4.5. Here, we can

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Principlesf C

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Prf. V. Venkata Ra

very clearlybserve that t

the left

f b, b

th the carriers are in phase whe

reas t

the right, they are 1800

ut

f phase.

Fig. 4.5: (t p) DSB-SC signal with t ne m dulati n (b tt m) The carrier

Fig. 4.6: Expanded versi

ns

f a part

f the wavef

rms in Fig. 4.5 C

nsider wavef

rms sh

wn in Fig. 4.7. We have

n the t

p, m

dulating t

ne signal and at the b

tt

m, the c

rresp

nding DSB-SC. What d

we

bserve

n the

scill

sc

pe, if we f

eed the X-plates the t ne signal and the Y-plates, the DSB-SC signal? The resultis sh

wn in Fig. 4.8, which can be explained as f

ll

ws.

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At the pint a in Fig. 4.7, the m

dulating t

ne is at its maximum and hence th

e DSB-SC signal has the maximum value. P

int a in Fig. 4.8 c

rresp

nds t

thep

int a in Fig. 4.7. Between the p

ints a and b in Fig. 4.7, the t

ne ampl

itude decreases (reaching the value zer at p int b); hence the maximum value reached by the DSB-SC signal during each carrier cycle keeps decreasing. As the X-plates are being fed with the same t

ne signal, this decrease will be linear andthis c

rresp

nds t

segment a t

b in Fig. 4.8. (N

te that DSB-SC signal is

zerat p

int b). In the time interval between b and c

f Fig. 4.7, the DSB

signal increases and this increase is seen as a straight line between the p ints b and c in Fig. 4.8. Between the p

ints c and e in Fig. 4.7, the t

ne

amplitude varies fr m the m st negative value t the m st p sitive value. C rresp ndingly, the display n the scill sc pe will f ll w the trace c d e sh wnin Fig. 4.8.

Fig. 4.7: (t p) m dulating signal (b tt m) DSB-SC signal

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Fig. 4.8 Displayn the

scill

sc

pe with the f

ll

wing inputs: X-plates: T

ne s

ignal Y-plates: DSB-SC signal Taking the F

urier transf

rm

f Eq. 4.1(b), we have

S (f ) = Ac M ( f fc ) + M ( f + fc ) 2

(4.2)

If we ignore the constant

Ac on the R.H.S of Eq. (4.2), we see that the 2

modulation process has simply shifted the message spectrum by fc . As the frequency translation of a given spectrum occurs quite often in the study of modulation and demodulation operations, let us take a closer look at this. i) Let m ( t) be a real signal with the spectrum M ( f ) shown below (Fig. 4.9(a)). Let fc

be 100 kHz. Assuming Fig. 4.9(b).Ac = 1 , we have S ( f ) as shown in 2



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Fig. 4.9: Frequency translation (a) baseband spectrum (real signal) (b) Shiftedspectrum. Note that S ( f ) = M ( 2 kHz ) + M ( 202 kHz )

f = 102 kHz

=1+ 0=1 and is the point a in Fig. 4.9 ii) Let m ( t ) be a complex signal with M ( f ) as shown in Fig. 4.10(a). The corresponding shifted spectrum (with fc= 100 kHz) is shown in Fig. 4.10(b)



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Fig. 4.10: Frequency translation (a) Baseband spectrum (complex signal) (b) Shifted spectrum. In figures 4.9(b) and 4.10(b), the part that is hatched in red iscalled the Upper Sideband (USB) and the one hatched in blue is called the Lower

Sideband (LSB). Any one of these two sidebands has the complete information about the message signal. As we shall see later, SSB modulation conserves the bandwidth by transmitting only one sideband and recovering the m ( t ) with appropriate demodulation.

Example 4.1

Consider the scheme shown in Fig. 4.11(a). The ideal HPF has the cutoff frequency at 10 kHz. Given that f1 = 10 kHz and f2 = 15 kHz, let us sketchY ( f ) for the X ( f ) given at (b).

4.12



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Fig. 4.11: (a) The scheme of example 4.1 (b) The input spectrum, X ( f ) We haveV ( f ) = X ( f f1 ) + X ( f + f1 ) , which is as shown in Fig. 4.12(a). TheHPF eliminates the spectral components for f 10 kHz. Hence W ( f ) is as shown

in Fig. 4.12(b).Y ( f ) = W ( f f2 ) + W ( f + f2 ) . This is shown in Fig. 4.12(c).



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Fig. 4.12: Spectra at various points in the scheme of Fig. 4.11

4.2.2. Coherent demodulation

The process of demodulation of a DSB SC signal, at least theoretically, is quitesimple. Let us assume that the transmitted signal s ( t ) has been received without any kind of distortion and is one of the inputs to the demodulator as shownin Fig. 4.13. That is, the received signal r ( t ) = s ( t ) . Also, let us assume that we are able to generate at the receiving end a replica of the transmitted carrier (denoted cr ( t ) = Ac cos ( c t ) in Fig. 4.13) hich is the other input to the demodulator.



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Fig. 4.13: Coherent demodulation of DSB-SC The demodulation process consists ofmultiplying these t o inputs and lo pass filtering the product quantity v ( t ). From Fig. 4.13,

e have ' v ( t ) = dg ( Ac m ( t ) cos ( c t ) ) Ac cos ( c

t )

(

)

here d g is the gain constant of the multiplier, called the detector gain constant, in the context of demodulation. For convenience, let us take d g = 1

' v ( t ) = Ac Ac m ( t ) cos2 ( c t )1 + c

s ( 2 c t )

= Ac Ac m ( t ) 2

Assuming that Ac Ac = 2 we have v ( t ) = m ( t ) + m ( t ) cos ( 4 fc t )

(4.3)

The second term on the R.H.S of Eq. 4.3 has the s ectrum centered at 2 fc andwould be eliminated by the low ass filter following v ( t ) . Hence v 0 ( t ) ,the out ut of the demodulation scheme of Fig. 4.13 is the desired quantity, namely,m (t ) .

Let us illustrate the o eration of the detector in the frequency domain. Letm (t )

be

real

with

the

s ectrum

shown

in

Fig.

4.14(a).

Let



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r ( t ) = s ( t ) = 2 m ( t ) cos ( c t ) . Then S ( f ) = M ( f fc ) + M ( f+ fc ) , shown in

Fig. 4.14(b).

Fig. 4.14: S ectra at various oints in the demodulation scheme of Fig. 4.13 (Note that the ositive frequency art of S ( f ) is shown in red and the negative' frequency art in blue). Assuming v ( t ) = s ( t ) cos ( c t ) (Fig. 4.13 ith Ac = 1 ),

then V ( f ) =

1 1 1 S ( f fc ) + S ( f + fc ) . S ( f fc ) and S ( f + fc ) are shown in 2 2 2



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Fig. 4.14(c) and (d) respectively. V0 ( f ) is the sum of the outputs of the lowpass filters shown in Fig. 4.14(c) and (d) which is the desired message spectrum,

M (f ) .

From the discussion of the demodulation process so far, it appears that demodulation of DSB SC is quite simple. In practice, it is not. In the scheme of Fig. 4.13, we have assumed that, we have available at the receiver, a carrier term thatis coherent (of the same frequency and phase) with the carrier used to generatethe DSB

SC signal at the transmitter. Hence this demodulation scheme is known a

s coherent (or synchronous) demodulation. As the receiver and the transmitter are, in general, not collocated, the carrier source at the receiver is different from that used at the transmitter and it is almost impossible to synchronize twoindependent sources. Fairly sophisticated circuitry has to be used at the receiver in order to generate the coherent carrier signal, from an

r ( t ) that has no carrier component in it. Before we discuss at the generationof

the coherent carrier at the receiver, let us look at the degradation caused to the demodulated message due to a local carrier that has phase and frequency differences with the transmitted one.Case i): Constant phase difference between c ( t ) and cr ( t )

Let c ( t ) = cos ( 2 fc t ) and cr ( t ) = cos [ 2 fc t + ] (the am litudequantities,' Ac and Ac can be treated as 1) v ( t ) = m ( t ) cos ( c t ) cos ( c t + )

= m ( t ) cos ( c t ) c s ( c t ) cos sin ( c t ) sin = m ( t )c s2 ( c t ) cos sin ( c t ) cos ( c t ) sin 1 + c

s ( 2 c t ) m ( t ) sin ( 2 c t ) sin = m (t ) cos 2 2



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At the output of the LPF, we will have only the term

m ( t ) cos 2

. That is, the

out ut of the demodulator, v 0 ( t ) , is ro ortional to m ( t ) cos . As long as remains a constant, the demodulator out ut is a scaled version of the actual message signal. But values of close to 2 will force the out ut to near about zero. When = 2 we have zero out ut from the demodulator. This is calledthe quadrature null effect of the coherent detector.Case ii): Constant frequency difference between c ( t ) and cr ( t )

Let c ( t ) = cos ( 2 fc t ) and cr ( t ) = cos 2 ( fc + f ) t . Then, v ( t ) = m ( t ) cos ( 2 fc t ) cos 2 ( fc + f ) t By carryin

g out the analysis similar to case (i) above, we find that v 0 ( t ) m ( t ) cos 2 ( f ) t (4.4a)

Let us look in some detail the implications of Eq. 4.4(a). For convenience, letv 0 ( t ) = m ( t ) cos 2 ( f ) t (4.4b)

Assume f = 100 Hz and consider the s ectral com onent at 1 kHz in M ( f ) . After demodulation, this gives rise to two s ectral com onents, one at 900 Hz andthe other at 1100 Hz, because

cs 2 103 t c s ( 2 100 t ) = 1 c s ( 2 1100 t ) + cos ( 2

900 t ) 2The behavior of the sum of these two components is shown in Fig. 4.15. As can be

seen from this figure, the envelope of sum signal (broken red line) attains thepeak value twice in a cycle of the beat frequency f . Also, it goes through zero twice in a cycle of the beat frequency.

(

)



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Fig. 4.15: Time domain behavior of cos 2 103 t cos ( 2 100 t )

(

)

Let us examine the effect of frequency offset in the frequency domain. Let M ( f) be as shown in Fig. 4.16(a). Assume f = 300 Hz. Then,

V0 ( f ) =

1 M ( f f ) + M ( f + f ) will be as shown in Fig. 4.16(d), which is 2

one half the sum of the spectra shown at (b) and (c). Comparing Fig. 4.16(a) and

(d), we are tempted to surmise that the output of the demodulator is a fairly distorted version of the actual message signal. A qualitative feeling for this distortion can be obtained by listening to the speech files that follow.

Introduction Output 1 Output 2 Output 3 Output 4



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Fig. 4.16: The effect of frequency offset in the demodulation of DSB SC: (a) Typical message spectrum, M ( f ) (b) M ( f + 300 ) (c) M ( f 300 ) (d) 1 M ( f+ 300 ) + M ( f 300 ) 2



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Example 4.2

In this example, we will show that the DSB SC signal can be demodulated with the

help of any periodic function p ( t ) , as long as p ( t ) has a spectral component at fc , the carrier frequency. This component could be due to the fundamental or some harmonic of the fundamental. a) Let s ( t ) = Ac m ( t ) cos ( c t ). Consider the roduct s ( t ) x ( t ) here x ( t ) is ny eriodic sign l ith the eriod T0 =x (t ) = 1 . Th t is, fc

n =

xn e j 2 n fc t

here xn is the nth Fourier coefficient. We ill sho th t if x1 0 , then it is ossible to extr ct m ( t ) from the roduct s ( t ) x ( t ) . b) ' Let y ( t ) be

nother eriodic sign

l ith the eriod T0 = NT0 . We ill sho th

t,

ro ri te filtering of the roduct s ( t ) y ( t ) , ill result in m ( t ) .

)

As cos ( 2 fc t ) =s (t ) x (t ) =

1 j 2 fc t + e e 2

j 2 fc t

, we have

Ac m ( t ) j ( n + 1) c t + xn e 2 n

xn e (n

j n 1) c t

j ( n 1) c t

Ac m ( t ) = x 1 + 2

n, n 1

xn e

j ( n + 1) c t


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+ x1 +

n, n1

xn e

as

x 1 + x1 2

= Re [ x1] , the output, after lowpass filtering would be,

Re [ x1] Ac m ( t ) . (We assume that the LPF will reject all the other

spectral components) b) The product s ( t ) y p ( t ) can be written as



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n j 2 Ac m ( t ) N s (t ) y (t ) = y n e 2 n

+ 1 fc t

+

ynn

n j 2 1 fc t N e

Ac m ( t ) = y N + 2

n, n N

yn

n j 2 + 1 fc t N e

+ yN +

n, nN

yn

n j 2 1 fc t N e

We ssume th t y N 0 . Then, the out ut of the LPF ould beRe [ y N ] Ac m ( t ) . 0,

(Note

th t

y

(t ) s (t )h s

s ectr l

lobes

t

fc 2f , c , etc. We ssume th t the LPF ill extr ct the lobe t f = 0 N N

nd reject others).

Ex m le 4.3


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Consider the scheme sho n in Fig. 4.17. s ( t ) is the DSB SC sign lm ( t ) cos ( c t ) ith

1 , 99 kHz f 101 kHz S (f ) = 0 , outside Let g ( t ) be nother b nd ss sign l ith 1 , 98 kHz f 102 kHz G (f ) = 0 , outside ) b) We ill sho th

t the out ut y ( t ) m ( t ) . We ill sho th

t it ould not be

ossible to recover m ( t ) from v ( t ) if 1 , 98.5 kHz < f 101.5 kHz G (

f ) = 0 , outside

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Princi les of Communic tion

Prof. V. Venk t R o

Fig. 4.17: Scheme of DSB SC demodul tion (ex m le 4.3) ) Let g ( t ) = m1 ( t )cos ( 2 fc t ) here fc = 100 kHz

nd 2 , f 2 kHz M1 ( f ) = 0 , outsi

de From S ( f ) , e see th t 2 , f 1 kHz M (f ) = 0 , outside We h

ve, v ( t ) = m ( t ) m1 ( t ) cos2 ( c t ) 1 + c s ( 2 c t ) = m ( t ) m1( t ) 2 = m ( t ) m1 ( t ) 2 + m ( t ) m1 ( t ) 2 cos ( 2 c t )

We ill ssume th t the LPF rejects the s ectrum round 2 fc , m ( t ) m1 ( t) 2 M ( f ) M1 ( f ) 2

M ( f ) M1 ( f ) ill hve

fl

t s ectrum for f 1 kHz . By using

n ILPF

ith cutoff t 1 kHz, e c n recover m ( t ) from v ( t ) . b) For this c se M1( f ) ould be 2 , f 1.5 kHz M1 ( f ) = 0 , outside M1 ( f ) M ( f ) ill be fl

t only for f 0.5 kHz . Hence m ( t ) c

nnot be

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Prof. V. Venk t R o

Exercise 4.1

A sign l m ( t ) hose s ectrum is sho n in Fig. 4.18( ) is gener ted using the

sign ls m1 ( t ) nd m2 ( t ) . M1 ( f ) nd M2 ( f ) re sho

n t (b) nd (c) res ectively in Fig. 4.18. The sign l s ( t ) = 2 m ( t ) cos 105 t tr nsmittedon the ch

nnel.

) b) Suggest

scheme to obt

in m ( t ) from m1 ( t )

nd m2 (

t ) .m1 ( t ) nd m2 ( t ) re to be recovered from the received sign l r ( t ) = s (t ) . A rt of this receiver is sho n in Fig. 4.18(d). Com lete

(

)

is

the receiver structure by indic ting the o er tions to be erformed by the boxes ith the question m

rk inside.

( )

(b)

(c)

(d) Fig. 4.18: Pro osed receiver structure for the exercise 4.1

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Prof. V. Venk t R o

4.2.3 C rrier recovery for coherent demodul tionAs ex l

ined in det

il in sec. 4.2.2, coherent demodul

tion requires

c

rrier

t the receiving end th t is h se coherent ith the tr nsmitted c rrier. H d the

re been c rrier com

onent in the tr nsmitted sign l, it

ould h ve been

ossible to extr ct it t the receiving end nd use it for demodul tion. But the DSB SC sign

l h

s no such com onent

nd other methods h

ve to be devised to gener

te

coherent c rrier t the receiver. T o methods re in common use for the c rrier recovery ( nd hence demodul tion) from the su ressed c rrier modul tion schemes, n mely ( ) Cost s loo nd (b) squ ring loo .

) Cost

s loo : This scheme is sho n in Fig. 4.19.

Fig. 4.19: Cost s loo The VCO (Volt ge Controlled Oscill tor) is source th t roduces eriodic veform1 hose frequency is controlled by the in ut volt geec ( t ) . The out ut frequency fo of the VCO hen ec ( t ) = 0 is c

lled the f

ree running frequency of the VCO. The frequency ut out by the VCO t ny inst n

t de

ends on the sign

nd m

gnitude of the control volt

ge, ec ( t ) .1

Here, e sh ll ssume th t the VCO out ut is sinusoid l.



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Prof. V. Venk t R o

To underst nd the loo o er tion, let us ssume th t the frequency nd h se ofthe VCO out ut

re the s

me

s th

t of the incoming c

rrier. Then, v1 ( t ) = (

Ac m ( t ) cos ( c t ) ) ( A0 cos ( c t ) ) = A0 Ac m ( t ) cos2 ( c t )1 + 2c s ( c t ) = A0 Ac m ( t ) 2

The output of the LPF1 is v2 (t ) = A0 Ac m ( t ) 2 ; that is, v 2 ( t ) m ( t) , the desired sign l.

Simil r n lysis sho sv 4 (t ) = 0

No su ose th t VCO develo s sm ll h se offset of radians. The Ichannel out ut will remain essentially unchanged but a small voltage will develo at the out ut of the Q

channel which will be ro ortional to sin (If the hase shift i

s rad, then the Q channel out ut is ro ortional to sin ). Because of th

is, e ( t ) is a non

zero quantity given by e (t ) = v2 (t ) v4 (t ) = =2 1 Ac A0 m ( t ) cos sin 4

2 1 Ac A0 m ( t ) sin 2 8

e ( t ) is input to LPF3, which has very narrow passband (Note that LPF1 and

LPF2 should have a bandwidth of at least W Hz). Hence ec ( t ) = C0 sin2 whereC0 is the DC value of2 1 Ac A0 m ( t ) . This DC control voltage ensures 8

that the VCO output is coherent with the carrier used for modulation.

b) Squaring loop

The operation of the squaring loop can be explained with the help of Fig. 4.20.



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Fig. 4.20: Demodulation of DSB SC using a squaring loop Let r ( t ) = s ( t ) =Ac m ( t ) cos ( c t ) . Then, v (t ) = r 2 (t ) =2 Ac 2 m ( t ) 1 + c s ( 2 c t ) 2

m 2 ( t ) will have nonzero DC value which implies its spectrum has an impulse at

f = 0 . Because of this, V ( f ) will have a discrete spectral component at 2 fc.v ( t ) is the input to a very narrowband bandpass filter, with the centre frequency

2 fc . By making the bandwidth of LPF1 very narrow, it is possible to make the VCO to lock on to the discrete component at 2 fc , present in w ( t ) . (The dotted box enclosing a multiplier, LPF and a VCO, connected in the feedback configur

ation shown is called the Phase Locked Loop (PLL)). The VCO output goes througha factor of two frequency divider, yielding a coherent carrier at its output. This carrier is used to demodulate the DSB

SC signal. Note that LPF2 must have ade

quate bandwidth to pass the highest frequency component present in m ( t ) .

Both the Costas loop and squaring loop have one disadvantage, namely, an 1800 phase ambiguity. Consider the Costas loop; if the input to the loop were 4.27Indian Institute of Technology Madras


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to be Ac m ( t ) cos ( c t ) , out ut of the LPF1 ould be th t of LPF2 ould be

1 Ac A0 m ( t ) cos and 2

1 Ac A0 m ( t ) sin with the result that, e ( t ) would be the 2

same as in the case discussed earlier. Similarly, for the squaring loo , v (t )would be the same whether r ( t ) = Ac m ( t ) cos ( c t ) or Ac m ( t ) cos( c t ) . Hence the demodul

ted out ut could be either m ( t ) or m ( t ) . H

o ever, this ill not c use ny roblem for udio tr nsmission bec use m ( t ) nd m ( t ) , ould sound the s me to our e rs. Though DSB SC modul tion schemes l ce the entire tr nsmitted o er into the useful sideb nds, the demodul tionh

s to be coherent. The circuit required to gener

te

coherent c

rrier incre

s

es the cost of the receiver. If only fe receivers re to be built for s eci

fic communic

tion need, the cost m

y not be

m

jor f

ctor. But in

bro

dc

st situ tion, there ould be l rge number of receivers tuned to given st tion nd in th

t scen

rio, it is better m

ke the receiver f

irly che

nd ush the cos

t u of the tr nsmitter, if required. As ill seen l ter, the Envelo e Detector(ED) is f irly che to im lement s com red to coherent detector. But to m keuse of ED, the modul

ted c

rrier should c

rry m ( t ) in its envelo e. Evidentl

y, DSB SC does not s tisfy this ro erty s ex l ined belo . Let

s ( t ) = Ac m ( t ) cos ( c t )

Pre envelo e of s ( t ) = s ( t ) pe = Ac m ( t ) ei c t Complex envelope of s ( t ) = s ( t )ce = Ac m ( t ) Hence the envelope of DSB SC = s (t ) pe = s ( t )ce

m (t )



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Prof. V. Venk t R o

We sh ll no describe modul tion scheme th t h s m ( t ) s its envelo e hichc

n e

sily be extr

cted.

4.3 DSB LC Modul tion (or AM)By dding l rge c rrier com onent to the DSB SC sign l, e ill h ve DSB LC,

hich, for convenience, e shll c

ll sim ly

s AM. By choosing the c

rrier com o

nent ro erly, it is ossible for us to gener te the AM sign l such th t it reserves m ( t ) in its envelo e. Consider the scheme sho n in Fig. 4.21.

Fig. 4.21: Genertion of

n AM sign

l from

DSB

SC sign

l Let

v ( t ) = Ac g m m ( t ) cos ( c t ) . Then, s ( t ) AM = Ac cos ( c t) + v ( t )

= Ac 1 + g m m ( t ) cos ( c t )

(4.5)In this section, unless there is confusion, we use s(t ) in place of [s(t )]AM .We shall assume that m(t ) has no DC component and ( m ( t ) )max = ( m ( t ))min . Letg m m ( t ) be such that g m m ( t ) 1 for all t. Then 1 + g m m ( t ) 0and s ( t ) AM preserves m ( t ) in its envelope because

s ( t ) = Ac 1 + g m m ( t ) e j c t pe s ( t ) ce = Ac 1 + g m m ( t )



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As 1 + g m m ( t ) 0 , we have Envelope of s(t ) = s(t )ce = Ac [1 + gm m(t )] The quantity after the DC block is proportional to m ( t ) . If 1 + gm m ( t ) is not nonnegative for all t , then the envelope would be different

from m ( t ) . This would be illustrated later with a few time domain waveforms of the AM signal. Fig. 4.22(b) illustrates the AM waveform for the case 1 +g m m ( t ) 0 for all t .

Fig. 4.22: (a) An arbitrary message waveform m ( t ) (b) Corresponding AM waveform



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A few time instants have been marked in both the figures (a) and (b). At the time instants when m ( t ) = 0 , carrier level would be Ac which we have assumed tobe 1. Maximum value of the envelope (shown in red broken line) occurs when m (

t ) has the maximum positive value. Similarly, the envelope will reach its minimum value when m ( t ) is the most negative. As can be seen from the figure, theenvelope of s ( t ) follows m ( t ) in a one

to

one fashion. Let g m m ( t ) = x

1. Then s ( t ) is said to have (100x) percentage max modulation. For the case of 100% modulation,gm m ( t ) max = g m m ( t ) min = 1. If g m m ( t ) max > 1 , then we have over

modulation which results in the envelope distortion. This will be illustrated inthe context of tone modulation, discussed next.

Exercise 4.2

For the waveform m ( t ) shown in Fig. 4.23, sketch the AM signal with the percentage modulation of 40. Assume Ac = 1 (the figure has to be shown with referenceto m ( t ) )

Fig. 4.23: Baseband signal for the exercise 4.2



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4.3.1. Tone ModulationLetm ( t ) to be a tone signal; that is, m ( t ) = Am cos ( 2 fm t ) here

fm 1 , we have overmodulation).



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Fig. 4.24: AM with tone modulation ( = 0.5 )

Fig. 4.25: AM with tone modulation ( = 1)



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Fig. 4.26: AM with tone modulation ( = 1.5 )

Fig. 4.27: Envelope of the AM signal of Fig. 4.26 As can be seen from 4.24 and 4

.25, the envelope (shown with a red broken line) is one to one related to the message sinusoid. Note that, for = 1, the carrier amplitude (and hence the envelope) goes to zero, corresponding to the time

instant when the sinusoid is going

through the negative peak. However, when > 1, the one to one relationship between the envelope of the modulated carrier and the modulating tone is no longer maintained. This can be more clearly seen in Fig. 4.27 which shows the output ofthe envelope detector when the input 4.34Indian Institute of Technology Madras


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is the modulated carrier of Fig. 4.26. Notice that the tone signal between ( t1, t2 ) and to the right of t3 of Fig. 4.26 is getting inverted; in other words,the output of the ED is proportional to (1 + cos mt ) hich is not equ l to (

1 + cos m t ) ,

hen > 1.

Fig. 4.28: Oscillogrm hen the CRO in uts

re: X

ltes: tone sign

l Y

ltes:

AM sign l ith = 1 2

Fig. 4.29: Oscillogr m hen the CRO in uts re: X l tes: tone sign l Y l tes:AM sign

l ith ( = 1) 4.35



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Prof. V. Venk t R o

Fig. 4.28 nd Fig. 4.29 illustr te the oscillogr ms hen the X l tes of the CRO

re fed ith the modul

ting tone

nd the Y

ltes ith the AM sign

l ith = 0

.5 nd = 1 res ectively. In Fig. 4.28, A re resents the e k to e k v lue of

the c rrier t its minimum (th t is, A = 2 Ac [1 ] )

here s B is the

e k to e k v lue of the c rrier t its m ximum (th t is, B = 2 Ac [1 + ] ). Hence c

n be c

lcul

ted

s = BA B+A

In Fig. 4.29, s A = 0 e h ve = 1

Exercise 4.3

Picture the oscillogr m hen the X l tes of the CRO re fed ith the modul tingtone nd the Y l tes ith the AM sign l ith = 1.5 .

4.3.2. S ectr of AM sign ls

T

king the Fourier tr

nsform of Eq. 4.5,Ac Ac S ( f ) AM = 2 ( f fc ) + ( f + fc ) + 2 g m M ( f fc )+ M ( f + fc )

(4.8) The plot of [S(f )]AM is given in Fig. 4.30.



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Fig. 4.30: (a) Baseband message spectrum M ( f ) (b) Spectrum of the AM signal Based on Fig. 4.30, we make the following observations: 1) The spectrum has two sidebands, the USB [between fc to fc + W , and ( fc W ) to fc , hatched in

red] and the LSB ( fc W to fc and fc to ( fc + W ) , hatched in blue). 2)If the baseband signal has bandwidth W , then the AM signal has bandwidth 2 W .That is, the transmission bandwidth BT , required for the AM signal is 2 W . 3)Spectrum has discrete components at f = fc , indicated by impulses of area 4)Ac 2 S ( f ) , fc > W (In practice, fc >> W , so that s ( t ) is a narrowband

In order to avoid the overlap between the positive part and the negative part of

signal)



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The discrete components at f = fc , do not carry any information and as such AM does not make efficient use of the transmitted power. Let us illustrate this taking the example of tone modulation.

Example 4.4

For AM with tone modulation, let us find = function of mo

ulation in

ex .

Total si

eban

power , as a Total power

For tone mo

ulation, we

ave s ( t ) = Ac 1 + c s ( m t ) cos ( c t ) Carrier term = Ac cos ( c t ) C rrier Po er = USB term =2 Ac 2

Ac cos ( c + m ) t 2

2 2 Ac 2 8 Ac 2 Po er in USB = 2

=

Po er in LSB = Po er in USB Tot l sideb nd Po er = 2 2 Ac 2 Ac 2 2 = 8 4

2 2 2 2 Ac 2 + 2 Ac Ac 2 Ac 2 Tot l Po er= + = 1 + = 2 4 2 2 4 2 Ac 2 Hence, = 2 4 2 Ac 2 +

(

)

(

)

=

2 2 + 2

4

Calculating t

e value of for a few value of , we

ave4.38In

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0.25 0.03 0.50 0.11 0.75 0.22 1.0 0.33

As can be seen from t

e above tabulation, increases as 1;

owever, even at = 1, only 1/3 of t

e total power is in t

e si

eban

s (or si

e frequencies), t

e remaining 2/3 being in t

e carrier. From t

is example, we see t

at AM is notan efficient mo

ulation sc

eme, in terms of t

e utilization of t

e transmitte

power. T

e complex envelope be

avior of s ( t ) for tone mo

ulation is quite illustrative. s ( t ) = Ac 1 + c s ( m t ) cos ( c t ) A j = Re Ace j c t + c e ( c 2 s ( t ) = Ac + ce+ m ) t

j +e( c

m t )

Ac j m t Ac e e + 2 2

j m t

(4.9)

Let us dr

h sor

iagram, using t

e carrier quantity as t

e reference. T

eterm ofAc j m t e 2

c n be re resented s rot ting vector ith m gnitude rev/sec. Simil rly,

Ac , rot ting counterclock ise t the r te of fm 2j m t

Ac e 2

c n be sho n s vector ith clock ise rot tion l s eed of fm

rev/sec.



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Prof. V. Venk t R o

Fig. 4.31: Ph sor di gr m for AM ith tone modul tion Fig. 4.31 de icts the beh vior of

ll the qu

ntities on the RHS of Eq. 4.9. From Eq.4.9, e find th

t the

com lex envelo e is re l nd is given by Ac 1 + c s ( m t ) . This can als

o be seen from the phasor diagram, because at any given time, the quadraturecomponents of the sideband phasors cancel out where as the in phase componentsadd up; the resultant of the sideband components is collinear with the carrier.The length of the in phase component of s ( t ) depends on the sign of cethe resultant of the sideband phasors. As can be seen from Fig. 4.31 this varies between the limits Ac [1 1, Ac [1

]

to Ac [1 + ] . If the modulation index is less than

]

> 0 and envelope of s ( t ) is

s ( t ) ce = Ac 1 + c s ( m t ) = Ac 1 + c s ( m t )

Phasor diagrams such as the one shown in Fig. 4.31 are helpful in the study of unequal attenuation of the sideband components. We shall illustrate this with anexample.

Example 4.5

Let Ac = 1 , =

1 and let the upper sideband be attenuated by a factor of 2

2. Let us find the expression for the resulting envelope, A ( t ) .



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The phasor diagram for this case is shown in Fig. 4.32.

Fig. 4.32: Phasor diagram for an AM signal with unequal sidebands As can be seen

from the figure, the resultant of the sidebands is no longer collinear with thecarrier. 1 1 s ( t ) = 1 + ( cos ( m t ) + j sin ( m t ) ) + ( cos ( m t) j sin ( m t ) ) ce 8 4 = 1+ 3 1 cos ( m t ) j sin ( m t ) 8 81

2 2 3 1 2 A ( t ) = 1 + cos ( m t ) + sin ( m t ) 8 8

Evidently, it is not possible for us to recover the message from the above A ( t) .

4.4 Generation of AM and DSB SC signals

Let x ( t ) be the input to an LTI system with the impulse response h ( t ) andlet y ( t ) be the output. Then,y (t ) = x (t ) h (t ) Y (f ) = X (f ) H (f )

That is, an LTI system can only alter a frequency component (either boost or attenuate), that is present in the input signal. In other words, an LTI system cannot generate at its output frequency components that are not present inX ( f ) . We have already seen that the spectrum of a DSB or AM signal is



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different from that of the carrier and the baseband signal. That is, to generatea DSB signal or an AM signal, we have to make use of nonlinear or time

varying

systems.

4.4.1 Generation of AMWe shall discuss two methods of generating AM signals, one using a nonlinear element and the other using an element with time varying characteristic.

a) Square law modulator

Consider the scheme shown in Fig. 4.33(a).

Fig. 4.33 (a): A circuit with a nonlinear element (b): v i characteristic of the diode in Fig. 4.28(a) A semiconductor diode, when properly biased has a v icharacteristic that nonlinear, as shown in Fig. 4.33(b). 4.42



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For fairly small variations of v around a suitable operating point, v 2 ( t ) can be written as2 v 2 ( t ) = 1 v1 ( t ) + 2 v1 ( t )

(4.10)

here 1 nd 2 re const nts.' Let v1 ( t ) = Ac cos ( 2 fc t ) + m ( t ) . Then,

2 2 ' ' v 2 ( t ) = 1 Ac 1 + m ( t ) cos ( 2 fc t ) + 1 m(t ) + 2 m2 ( t ) + 2 Ac2 cos2 ( 2 fc t ) 1

(4.11) The first term (on the RHS of Eq. 4.11) is s ( t ) , with the carrieramplitude AM2 2 ' Ac = 1 Ac nd g m = . 1

No the question is: c n e extr ct s ( t ) AM from the sum of terms on the RHS of Eq. 4.11? This can be answered by looking at the spectra of the various terms constituting v 2 (t ) . Fig. 4.34 illustrates these spectra (quantitiesmarked A to E) for the M ( f ) of Fig. 4.14(a). The time domain quantities corresponding to A to E are listed below.

Fig. 4.34: Spectra of the components of v 2 ( t ) of Eq. 4.11



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Component

Spectrum indicated by

(i) Ac cos ( 2 fc t )' (ii) 2 2 Ac m ( t ) cos ( 2 fc t )

A B C D E

(iii) 1 m ( t ) (iv) 2 m 2 ( t ) ' (v) 2 Ac

()

2

cos2 (2 fc t )s ( t ) AM consists of the components (i) and (ii) of the above list. Ifit is possible

for us to filter out the components (iii), (iv) and (v), then the required AM signal would be available at the output of the filter. This is possible by placinga BPF with centre at fc and bandwidth 2W provided ( fc W ) > 2W or fc > 3 W .Usually, this is not a very stringent requirement. However, this scheme suffersfrom a few disadvantages. i) The required square law nonlinearity of a given device would be available only over a small part of the (v i ) characteristic. Hence, it is possible to generate only low levels of the desired output.

ii)

If fc is of the order of 3W, then we require a BPF with very sharp cut off characteristics.

b) Switching modulator

In the first method of generation of the AM signals, we have made use of the nonlinearity of a diode. In the second method discussed below, diode will be used as a switching element. As such, it acts as a device with time varying characteristic, generating the desired AM signals when it is used in the circuit configuration shown in Fig. 4.35.



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Fig. 4.35: (a) Switching modulator (b) Switching characteristic of the diode load combination. The (ideal) transfer characteristics of the diode

load combinatio

n is shown at (b) in Fig. 4.35. This is explained as follows. We have,

v1 ( t ) = c ( t ) + m ( t ) = Ac cos ( 2 fc t ) + m ( t )

If essume th

t m ( t ) 0 , c (t ) 0 0

(Tht is, the diode offers infinite im ed

nce hen reverse bi

sed

nd h

s zero i

m ed nce, hen for rd bi sed. Hence, hether v1 ( t ) is s itched to the out ut

or not de

ends on the c

rrier cycle) We c

n ex

ress v 2 ( t )

sv 2 ( t ) = v1 ( t ) x ( t )

(4.12)

here x ( t ) is the eriodic rectngul

r ulse tr

in of ex

m le 1.1. Th

t is,

1 , x (t ) = 0 ,

if if

c ( t ) > 0 ( ositiveh lf cycleof c ( t ) ) c ( t ) < 0 ( neg tiveh lf cycleofc ( t ) )

But from ex m le 1.1 ( ith f0 = fc ),



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Prof. V. Venk t R o

x (t ) =

n =

xn e j 2 n fc t

here xn =

1 n sin c 2 2

(4.13 )

From Eq. 4.13( ), e find th t xn = 0 , for n = 2, 4 etc. Combining the terms x n

nd xn , e obt

in the trigonometric Fourier series, n

mely, 12 x (t ) =

+ cos ( 2 fc t ) + 2

n=2

( 1)n 1 cos 2 2n 1

( 2 n 1) fc t

(4.13b)

From Eq. 4.12 and 4.13(b), we see that v 2 ( t ) is composed of two components,namely, a) b) The desired quantity:Ac 2 4 m ( t ) cos ( 2 fc t ) 1 + Ac

The undesired terms with i) ii) Impulses in spectra at f = 0, 2 fc , 4 fc etc. Spectral lobes (same in shape as M(f)) of width 2W, centered at 0, 3 fc , 5 fc etc.

As compared to the square law modulator, switching modulator has the following advantages: a) b) Generated AM signals can have larger power levels. Filtering requirements are less stringent because we can separate the desired AM signal if fc > 2W . However, the disadvantage of the method is that percentage modulation has to be low in order that the switching characteristics of the diode are controlled only by the carrier.



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4.4.2. Generation of DSB SCa) Product modulator

Generation of a DSB SC signal involves the multiplication of m ( t ) withAc cos ( c t ) . To gener te this sign l in the l bor tory, ny of the commercilly

v il ble multi liers c n be used. Some of them re: N tion l: Motorol : An logDevices: Signetics: LM 1496 MC 1496 AD 486, AD 632 etc 5596

The o er levels th t c n be gener ted, the c rrier frequencies th t c n be used ill de end on the IC used. The det ils c n be obt ined from the res ective m nu ls. Gener lly, only lo o er levels re ossible nd th t too over limitedc

rrier frequency r

nge.

b)Ring modul

tor

Consider the scheme sho n in Fig. 4.36. We ssume th t the c rrier

Fig. 4.36: Ring modul tor sign l c ( t ) is much l rger th n m ( t ) . Thus c (t ) controls the beh vior of diodes hich ould be cting s ON OFF devices. Consider the c rrier cycle here the



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Prof. V. Venk t R o

termin l 1 is ositive nd termin l 2 is neg tive. T1 is n udio frequency tr nsformer hich is essenti

lly

n o en circuit

t the frequencies ne

r

bout the c

rrier. With the ol rities ssumed for c ( t ) , D1, D4 re for rd bi sed, he

re s D2, D3 re reverse bi sed. As consequence, the volt ge t

oint getsswitc

e

to a an

voltage at point b to b . During t

e ot

er

alf cycle ofc ( t ) , D2 an

D3 are forwar

biase

w

ere as D1 an

D4 are reverse biase

. Asa

result, t

e voltage at a gets transferre

to b an

t

at at point b to a .T

is implies,

uring, say t

e positive

alf cycle of c ( t ) , m ( t ) is switc

e

to t

e output w

ere as,

uring t

e negative

alf cycle, m ( t ) is switc

e

. In ot

er wor

s,v ( t ) can be taken as v (t ) = m (t ) xp (t )

(4.14)

w

ere x p ( t ) is square wave as s

own in Fig. 4.37.

Fig. 4.37: x p ( t ) of Eq. 4.14

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Fig. 4.38: (a) A message waveform m ( t ) (b) v ( t ) of t

e ring mo

ulator Fig.4.38(b) illustrates t

e pro

uct quantity m ( t ) x p ( t ) , for t

e m ( t ) s

own in Fig. 4.38(a). T

e Fourier series expansion of x p ( t ) can be written as

4 xp (t ) =

n = 1, 3, 5, ...

( 1)

n 1 2

n

cos ( n c t ) .When v ( t ) is

ssed through

BPF tuned to fc , the out ut is the desired DSB

SC sign l, n mely, s ( t ) = 4 m ( t ) cos ( c t ) .

Exm le 4.6: Gener

tion of DSB

SC

Consider the scheme sho n in Fig. 4.39. The non line r device h s the in ut out

ut ch r cteristic given byy ( t ) = 0 x ( t ) + 1 x 3 ( t )



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Prof. V. Venk t R o

Fig. 4.39: The scheme for the ex m le 4.6 Let x ( t ) = A cos ( 2 f1 t ) + m (t ) here m ( t ) is the mess

ge sign

l. If the required out ut s ( t ) is

DS

B SC sign l ith c rrier frequency of 1 MHz, let us find the v lue of f1 , ss

uming th t suit ble BPF is v il ble.y ( t ) = 0 A c s ( 2 f1 t ) + m ( t ) + a1 A c s ( 2 f1 t ) + m ( t) 1 23

2 = a1 A3 cs3 ( 2 f1 t ) + m3 ( t ) + 3 A2 cos2 ( 2 f1 t ) m ( t ) + 3 A

cos ( 2 f1 t ) m 2 ( t ) In the equation for the quantity 2 above, the only term on the RHS that can giverise to the DSB SC signal is 3 a1 A2 m ( t ) cos2 ( 2 f1 t ) .3

1 A2 m ( t ) cos2 ( 2 f1 t ) = 3

1 A2 m ( t )

{1 + cos 2 ( 2 f1 ) t } 2

Assume that the BPF will pass only the components centered around 2 f1 . Then, choosing f1 = 500 kHz, we will haves ( t ) = Ac m ( t ) cos ( 2 fc t )

here Ac = 3 1 A2 nd fc = 1 MHz.



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Prof. V. Venk t R o

Exercise 4.4

Consider the circuit configur tion (c lled Co n modul tor) sho n in Fig. 4.40.

Sho

th t the circuit c n

roduce t its out

ut the DSB SC sign l. T1 is the udio frequency tr nsformer here s T2 nd T3 re designed to o er te round the crrier frequency.

Fig. 4.40: Co n modul tor

Exercise 4.5: Bl

nced Modul

tor (BM)

Consider the scheme sho n in Fig. 4.41. This configur tion is usu lly c lled b l nced modul tor. Sho th t the out ut thereby est blishing th t BM is essenti lly

multi lier.

s(t ) is

DSB

SC sign

l,Fig. 4.41: B

l

nced modul

tor (BM) 4.51



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Prof. V. Venk t R o

4.5 Envelo e DetectorAs mentioned e

rlier, the AM sign

l, hen not over modul

ted

llo s the recovery

of m(t ) from its envelo e. A good roxim tion to the ide l envelo e detector

c n be re lized

ith f irly sim

le electronic circuit. This m kes the receiver for AM some h t sim le, there by m king AM suit ble for bro dc st lic tions. We sh

ll briefly discuss the o er

tion of the envelo e detector, hich is to b

e found in lmost ll the AM receivers. Consider the circuit sho n in Fig. 4.42.

Fig. 4.42: The envelo e detector circuit We ssume the diode D to be ide l. Whenit is for

rd bi

sed, it

cts

s

short circuit

nd thereby, m

king the c

ci

tor C ch rge through the source resist nce Rs . When D is reverse bi sed, it cts s n o en circuit nd C disch rges through the lo d resist nce RL . As the o

er tion of the detector circuit de ends on the ch rge nd disch rge of the c citor C , e sh

ll ex l

in this o er

tion ith the hel of Fig. 4.43.



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Prof. V. Venk t R o

Fig. 4.43: Envelo e detector veforms ( ) v1 ( t ) (before DC block) (b) v out( t ) (

fter DC block) If the time const

nts Rs C

nd RL C

re ro erly chosen,

v1 ( t ) follo s the envelo e of s ( t ) f irly closely. During the conduction c

ycle of D , C quickly ch rges to the

e k v lue of the c rrier t th t time inst nt. It ill disch rge little during the next off cycle of the diode. The timeconst

nts of the circuit ill control the ri le

bout the

ctu

l envelo e. CB

is blocking c citor nd the fin l v out ( t ) ill be ro ortion l to m ( t) , s sho n in Fig. 4.43(b). (Note th t sm ll high frequency ri le, t the c

rrier frequency could be resent on v out ( t ) . For udio tr nsmission, this ould not c

use

ny roblem,

s fc is gener

lly much higher th

n the u er limit

of the udio frequency r nge).

4.53Indi

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Prof. V. Venk t R o

Ho do e choose the time const nts? Rs , though not under our control c n be ssumed to be f

irly sm

ll. V

lues for RL

nd C c

n be

ssigned by us. During the

ch rging cycle, e nt the c citor to ch rge to the e k v lue of the c rrier

in s short time s

ossible. Th t is,Rs C


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Fig. 4.44: The scheme for the example 4.7 It is not too difficult to see that w(t ) = w (t ) = As 1 cos 2 (10 + f ) 106 t where f = 0.01 Hz. That is, 2 1 c s 2 ( f ) t cos ( c t ) sin 2 ( f ) t sin ( c t )

2

z ( t ) = w ( t ) + sin ( 2 fc ) t , we have z (t ) = 1 1 cos 2 ( f ) t cos ( c t ) sin 2 ( f ) t 1 sin ( c t ) 2 2 a narrowband signal with the in phase component

z (t )

represents

1 1 cos 2 ( f ) t and the quadrature component sin 2 ( f ) t 1 . Hence, 2 2

2 1 1 2 y ( t ) = cos2 2 ( f ) t + sin 2 ( f )t 1 2 4 1

5 2 = sin 2 104 t 4

(

)

1

Ex m le 4.8

Consider the scheme sho

n in Fig.4.45.

Fig. 4.45: The scheme for the ex m le 4.8 Let us find the out ut y ( t ) hen,



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Prof. V. Venk t R o

)

x ( t ) = 1 + g m m ( t ) cos ( c t ) . g m m ( t ) < 1 nd m ( t ) is b nd

limited to

W Hz and the LPF has a bandwidth of 2W . Assume that fc >> 2W .b)x ( t ) is a DSB SC signal; that is x ( t ) = m ( t ) cos ( c t ) .

)

v ( t ) = x 2 ( t ) = 1 + g m m ( t ) cos2 ( c t ) 2 1 + c s ( 2 c t ) = 1 + g m m ( t ) 2

2

1 + g m m ( t ) + 1 + g m m ( t ) cos 2 t . = ( c) 2 2 The secondterm on the RHS will be eliminated by the LPF. Hence, 1 + g m m ( t ) . As1 + g m t 0 , we have w (t ) = ( ) m 21 + g m m ( t ) . y (t ) = 22

2

2

b)

When x ( t ) = m ( t ) cos ( c t ) ,

e h ve 1 + c s ( 2 c t ) v ( t ) =m 2 ( t ) cos2 ( c t ) = m 2 ( t ) 2 The output of the LPF would be w (t ) =m2 ( t ) 2

As the squaring operation removes the information about the sign of the signal,the output of y ( t ) is y (t ) = m (t ) 2



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Exercise 4.6

Consider the waveform m ( t ) shown in Fig. 4.46. A DSB SC is generated using m

( t ) and a suitable high frequency carrier. Sketch the output of an ideal envelope detector when the input to the detector is the DSB SC signal.

Fig. 4.46: m ( t ) for the exercise 4.6

4.6 Theory of Single SidebandAssume that from a DSB

SC signal, we have completely suppressed one

Ac of the sidebands, say, the LSB. Let S ( f ) DSB = 2 M ( f fc ) + M( f + fc )

where M ( f ) is as shown in Fig. 4.30(a). The resulting spectrum S ( f ) USB will be as shown in Fig. 4.47. Can we get back m ( t ) from the above signal

? The answer is YES. Letv ( t ) = s ( t ) USB cos ( 2 fc t )

1 If S ( f ) = S ( f ) USB , then V ( f ) = 2 S ( f fc ) + S (f + fc )



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Fig 4.47: Spectrum of the upper sideband signal By plotting the spectrum of v (t ) and extracting the spectrum for f W , we see that it is 1 Ac M ( f ) . A similar analysis will show that it is possible to extract 2

m ( t ) from S ( f ) LSB . In other words, with coherent demodulation, itis possible

for us to recover the message signal either from USB or LSB and the transmissionof both the sidebands is not a must. Hence it is possible for us to conserve transmission bandwidth, provided we are willing to go for the appropriate demodulation. Let us now derive the time domain equation for an SSB signal. Let us startwith the two sided spectrum and then eliminate the unwanted sideband. We shallretain the upper sideband and try to eliminate the lower sideband. ConsiderAc M ( f fc ) + sgn ( f fc ) M ( f fc ) 2

But M ( f fc ) , f > fc sgn ( f fc ) M ( f fc ) = M ( f fc ) , f fc Ac M ( f fc ) 1 + sgn ( f fc ) = 2 , f < fc 0

Hence

Th t is, the lo er sideb nd h s been elimin ted from the ositive rt of the s

ectrum.M ( f fc ) m ( t ) e j 2 fc t

4.58



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Prof. V. Venk t R o

sgn ( f fc ) M ( f fc ) here m ( t ) is the HT of m ( t ) . Th t is,

1 m ( t ) e j 2 fc t , j

Ac A 1 M ( f fc ) 1 + sgn ( f fc ) c m ( t ) m ( t ) e j 2 fc t j 2 2 1 M ( f + fc ) , f < fc M ( f + fc ) 1 sgn ( f +fc ) = 2 , f > fc 0

(4.16 )

Simil rly,

Ac A A M ( f + fc ) 1 sgn ( f + fc ) c m ( t ) e j 2 fc t + c m (t ) e j 2 fc t 2 2 2j the upper single sideband signal, namely, s ( t ) USB = Ac m ( t ) cos ( c t ) m ( t ) sin ( c t ) Ac m (

t ) cos ( c t ) , by filtering out the LSB

rt. Then, Ac s ( t ) USB = 2 m ( t ) cos ( c t ) m ( t ) sin ( c t )

(4.16b)

Combining Eq. 4.16(a) and Eq. 4.16(b), we have the time domain equation for

(4.17)

Assume that the USSB signal is obtained from the DSB SC signal,

(4.18)

A few authors take Eq. 4.18 as representative of the SSB signal. Eq. 4.18 has the feature that the average power of the SSB signal is one half the average powerof corresponding DSB signal. We shall make use of both Eq. 4.17 and Eq. 4.18 inour further studies. By a procedure similar to that outlined above, we can derive a time domain expression for the LSB signal. The result would be s ( t ) LSB = Ac m ( t ) cos ( c t ) + m ( t ) sin ( c t ) Ac s ( t ) LSB = 2 m ( t ) cos ( c t ) + m ( t ) sin ( c t ) or

(4.19a) (4.19b)

An SSB signal, whether upper or lower, is also a narrowband bandpass signal. Eq.4.18 can be treated as the canonical representation of USB signal with m ( t )

as the in

phase component and m ( t ) as the quadrature component. Similarly 4.59Indian Institute of Technology Madras


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Eq. 4.19 provides the canonical representation of the LSB signal where m ( t ) is the in

phase component and m ( t ) is the quadrature component.

We have already seen that a narrowband signal can also be expressed in the envelope and phase form. Taking the USB signal, we have for the complex envelope thequantity signal is Ac A ( t ) USB = 2 m2 ( t ) + m2

Ac m ( t ) + j m ( t ) . Hence the envelope A ( t ) of the USB 2

(

)

(t )

(4.20a)

Similarly for the phase ( t ) , we have m (t ) ( t ) = arc tan m (t )

(4.20b)

Expressing the USB signal with the envelope phase form, we haves ( t ) USB = A ( t ) c s ( c t ) + ( t )

(4.21)

where A ( t ) and ( t ) are given by Eqs. 4.20(a) and 4.20(b) res

ectively. The ex ression for s ( t ) LSB is identical to Eq. 4.21 but with ( t ) given by m (t ) ( t ) = arc tan m (t )

(4.22) (4.23)

That is, s ( t ) SSB = A ( t ) cos ( c t + ( t ) )

where ( t ) is given either by Eq. 4.20(b) or Eq. 4.22. Eq. 4.23 indicates that an SSB signal has both am litude and hase variations. (AM and DSB SC signalshave only the am litude of the carrier being changed by the message signal. Note

that AM or DSB

SC signals do not have quadrature com

onents.) As such, SSB signals belong to the category of hybrid am litude and hase modulation.



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Exam le 4.9: SSB with tone modulation

As a sim le a lication of the Eqs. 4.18 and 4.19, let m ( t ) be cos ( m t ) .

Let us find the SSB sign ls. m ( t ) = cos ( m t ) m ( t ) = sin ( m t ) .Therefore,Ac s ( t ) USB = 2 c

s ( m t ) cos ( c t ) sin ( m t ) sin ( c t )

=

Ac c s ( c + m ) t 2

Ac s ( t ) LSB = 2 c s ( m t ) cos ( c t ) + sin ( m t ) sin ( c t )

=Ac c

s ( c m ) t 2

Alternatively,s ( t ) DSB SC = Ac cos ( m t ) cos ( c t ) =Ac c s ( c + m ) t + cos ( c m ) t 2

Extracting the USB, we haveAc s ( t ) USB = 2 cos ( c + m ) t

If we eliminate the USB, thenAc s ( t ) LSB = 2 cos ( c m ) t

Example 4.10

Let m ( t ) = x ( t ) y ( t ) where X ( f ) and Y ( f ) are as shown in Fig. 4.48. An LSB signal is generated using m ( t ) as the message signal. Let us develop the expression for the SSB signal.



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Fig. 4.48: X ( f ) and Y ( f ) of example 4.10

Let us take the SSB signal as s ( t ) LSB = m ( t ) cos ( c t ) + m ( t

) sin ( c t ) We h ve x ( t ) = 2 103 sin c 2 103 t y ( t ) = 103 sin c 2 ( 500 t ) cos 4 103 t m (t ) = x (t ) y (t )

(4.24)

(

)

(

)(4.25a)

What is required is m ( t ) , the HT of m ( t ) . m ( t ) is the product of a lowpass and a bandpass signal. Hence m ( t ) = x ( t ) y ( t ) . (See the note, after example 1.25) But y ( t ) , from the result of example 1.25, is y ( t ) = 103 sin c 2 ( 500 t ) sin 4 103 t That is,

(

)

m ( t ) = 2 106 sin c 2 103 t sin c 2 ( 500 t ) sin 4 103 t s ( t ) LSB is obtained by using Eq. 4.25 in Eq. 4.24.

(

)

(

)

(4.25b)4.62Indian Institute of Technology Madras


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Exercise 4.7

Let M ( f ) be as shown in Fig. 4.49. An upper sideband signal is generated usin

g the signal with this M ( f ) . Compute and sketch the spectrum of the quadrature component of the SSB signal.

Fig. 4.49: Baseband spectrum for the Exercise 4.7

Exercise 4.8

Let m ( t ) = sin c ( t ) . Develop the expression for the following: a) b) USBsignal, in the canonical form. USB signal, in the envelope and phase form

1 t Ans. (b): sin c cos 2 fc + t 4 2

Exercise 4.9Let s ( t ) = m ( t ) cos ( c t ) m ( t ) sin ( c t ) here M ( f ) is

s sh

o n in Fig. 4.49. Let W = 5 kHz nd M ( 0 ) = 1. ) b) c) Sketch the s ectrum of(i) s ( t ) cos ( c t ) nd (ii) s ( t ) sin ( c t ) Sho th t sum of the s ectr

of

rt (

) is ro ortion

l to M ( f ) Sketch the s ectrum of s ( t ) cos (

c t ) s ( t ) sin ( c t ) . Is this rel ted toM ( f ) ? Ex l in.



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Prof. V. Venk t R o

4.7 Gener tion of SSB Sign lsWe sh

ll consider t o bro

d c

tegories of SSB gener

tion, n

mely, (i) frequency

discrimin tion method nd (ii) h se discrimin tion method. The former is b sed

on the frequency dom in descri

tion of SSB,

here s the l tter in b sed on the time dom in descri tion of n SSB sign l.

4.7.1 Frequency discrimin tion methodConce tu lly, it is very sim le scheme. First gener te DSB sign l nd then filter out the un nted sideb nd. This method is de icted in Fig. 4.50.

Fig. 4.50: Frequency discrimin tion method of SSB gener tionv ( t ) is the DSB SC sign l gener ted by the roduct modul tor. The BPF is

designed to su ress the unnted sideb

nd in V ( f ) , thereby roducing the de

sired SSB sign l. As e h ve lre dy looked t the gener tion of DSB SC sign ls,

let us no

look

t the filtering

roblems involved in SSB gener

tion. BPFs

ith bru t ss nd sto b nds c nnot be built. Hence, r ctic l BPF ill h ve them

gnitude ch

r

cteristic H ( f ) ,

s sho n in Fig. 4.51. As c

n be seen from t

he figure, ( H ( f ) is sho n only for ositive frequencies) r ctic l filter,besides the P ssB nd (PB) nd Sto B nd (SB), lso h s Tr nsitionB nd (TB), during hich the filter tr

nsits from

ssb

nd to sto b

nd. (The edges of the PB

nd SB de end on the ttenu tion levels used to define these b nds. It is common r ctice to define the ssb nd s the frequency interv l bet een the 3 dB



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Prof. V. Venk t R o

oints. Attenu tion requirements for the SB de end on the lic tion. Minimum ttenu

tion for the SB might be in the r

nge 30 to 50 dB.)

Fig. 4.51: M gnitude ch r cteristic of

r ctic l BPF. Centre frequency = f0 Bec use of TB ( here the ttenu tion is not to the desired level), rt of the undesired sideb

nd m

y get through the filter. As

rule of the thumb, it is oss

ible to design filter if the ermitted tr nsitionb nd is not less th n 1% of center frequency of b nd ss filter. Fortun tely, quite fe sign ls h ve s

ectr l null round DC nd if it is ossible for us to fit in the tr nsitionb ndinto this g

, then the desired SSB signl could be gener

ted. In order to

ccom

lish this, it might become necess ry to erform the modul tion in more th n onest ge. We sh ll illustr te this ith the hel of n ex m le.

Exm le 4.11

Tele

hone qu

lity s

eech sign

l h

s

s

ectrum in the r

nge 0.3 to 3.4 kHz. We

ill suggest scheme to gener te u er sideb nd sign l ith c rrier frequencyof 5 MHz. Assume th

t b

nd

ss filters

re

v

il

ble, roviding

n

ttenu

tion o

f more th n 40 dB in TB of idth 0.01 f0 , here f0 is the centre frequency ofthe BPF. Let us look t the gener tion of the SSB sign l in one st ge using c

rrier of 5 MHz. When

DSB sign

l is gener

ted, it ill h

ve

s ectr

l null of

600 Hz centered t 5 MHz. Th t is, the tr nsitionb nd is bout 0.01 ercent ofthe c rrier 4.65Indi n Institute of Technology M dr s


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Prof. V. Venk t R o

nd hence it ould not be ossible to design such sideb nd filter. Ho ever, it ould be ossible to gener

te the desired SSB sign

l using t o st

ges of modul

tion. Consider the scheme sho n in Fig. 4.52.

Fig. 4.52: T o st ge gener tion of SSB sign lv1 ( t ) is

DSB

SC sign

l ith the USB occu ying the ( ositive frequency) r

ng

e

( fc1 + 300 )( fc1 3400 )

Hz to

( fc1 + 3400 )

Hz. The frequency r

nge of the LSB isHz to ( fc1 300 ) Hz. Let us extr

ct the u er sideb

nd from v1 ( t )

ith the hel of BPF1. Then the centre frequency of BPF1, ( f0 )1 , is

( f0 )1

=

( fc1 + 300 ) + ( fc1 + 3400 )2

= ( fc1 + 1850 ) idth of the s ectr l null round fc1 = 600 Hz. Hence orfc1 + 1850 600 100

fc1 60,000 1850

58.1 kHzLet us t ke fc1 s 50 kHz. Let M ( f ) be s sho n in Fig. 4.53( ). Then V2 ( f) ill be s sho n in Fig. 4.53(b).

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Prof. V. Venk t R o

Fig. 4.53: ( ) Mess ge s ectrum (b) The s ectrum of the USB sign l ith fc1 = 50kHzv 3 ( t ) is DSB SC sign l ith the c rrier fc 2 , v 2 ( t ) being the modul t

ing sign l.

Then V3 ( f ) ill bes sho n in Fig. 4.54. In this figure,

Fig. 4.54: S ectrum of v 3 ( t ) of Fig. 4.52f1 = ( fc 2 53,400 ) Hz f2 = ( fc 2 50,300 ) Hz f3 = ( fc 2 + 50,300 ) Hz f4= ( fc 2 + 53,400 ) Hz



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Prof. V. Venk t R o

Hence the tr nsition b nd v il ble to design the sideb nd filter is

( f3

f2 ) = 100.6 kHz. With this TB, e c n choose centre frequency of BPF2

less th n or equ l to 10.06 MHz. If e choose fc 2 s 4.95 MHz then e ill h vethe u er sideb nd occu ying frequency r nge (4.95 + 0.0503) = 5.0003 MHz to (4.95 + 0.0534) = 5.0034 MHz. This is ex ctly h t ould h ve h ened if the modul

tion scheme

s

ttem ted in one ste ith 5 MHz

s the c

rrier. Note: As the

s ectr l occu ncy of the USB sign l is from 5.0003 MHz to 5.0034 MHz, theoretic l centre frequency of the BPF2 is 5.00185. With res ect to this frequency, e h ve TB idth 100.6 = 100 = 2.01 ercent centre freq. 5001.85 hich is bout t

ice the ermitted rtio. Hence, it is ossible to choose for fc1

v

lue lo er t

h n 50 kHz.

4.7.2 Ph se discrimin tion methodThis method im lements Eq. 4.17 or Eq. 4.19(

), to gener

te the SSB sign

l. Cons

ider the scheme sho n in Fig. 4.55. This scheme requires t o roduct modul tors,t o h se shifters nd n dder. One of the h se shifter 2 h se shift for

ll the 2

is ctu lly Hilbert tr nsformer (HT); it should rovide

com onents in M ( f ) . This is not such n e sy circuit to re lize. Assuming itis ossible to build the HT, the SSB c n be gener ted for ny fc , rovided the roduct modul tors (multi liers) c n ork t these frequencies.



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Prof. V. Venk t R o

Fig. 4.55: SSB gener tion: Ph se discrimin tion method Inste d of single ideb

nd h

se shifter

cting

s the HT, it is ossible to h

ve

n SSB gener

tor it

h t o Ph se Shifting Net orks, (PSN), one in e ch br nch s sho n in Fig. 4.56.

Fig. 4.56: An ltern te configur tion for the h se discrimin tion schemeH1 ( f )

nd H2 ( f )

re the h

se shifting net orks. Let H1 ( f ) = e H2 ( f )

= ej 2 ( f ) j 1( f )

and

. 1 ( f ) and 2 ( f ) are such that 1 ( f ) 2 ( f ) = for the 2

frequency range of interest. That is, PSN1 and PSN2 maintain a constant difference of . 2



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Princi

les of Communication


Let us ex

lain the o

eration of the scheme shown in Fig. 4.56 by takingm ( t ) to be a tone signal; that is, m ( t ) = Am cos ( m t ) . Let

1 ( f )

f = fm

= 1 and 2 ( f ) . Then, 2

f = fm

= 2

where 2 = 1 +

v1 ( t ) = Am cos ( m t + 1 ) andv 2 ( t ) = Am cos ( m t + 2 ) .

v 3 ( t ) = Am Ac cos ( m t + 1 ) cos ( c t ) v 4 ( t ) = Am Ac cos ( m t +2 ) sin ( c t )

= Am Ac cos m t + 1 + sin ( c t ) 2 = Am Ac sin ( m t + 1 ) sin ( c t )

v 3 ( t ) + v 4 ( t ) = Am Ac cos ( c + m ) t + 1

After coherent demodulation, we will have cos ( m t + 1 ) . We shall assume that the additional

hase shift 1 which is actually fre

uency de

endent will notcause any

roblem after demodulation. As it is not too difficult to design a Hilbert transformer using digital filter design techni

ues,

hase shift method of SSB generation is better suited for digital im

lementation. For a brief discussion on SSB generation using digital signal

rocessing, the reader is referred to [1].



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Princi



Exercise 4.10

There is a third method of generating the SSB signal, known as Weavers

method. This scheme is shown in Fig. 4.57.

Fig. 4.57: Weavers method of SSB generation Let M ( f ) be nonzero only in the

interval fl f fu ; and let f1 =fl + fu . 2

The low

ass filters in the I and Q channels are identical and have a cutoff fre

uency of fc 0 =fu fl . Assume that f2 >> fc 0 . By sketching the s

ectra at 2

various

oints in the above scheme, show that s ( t ) is an SSB signal. What isthe actual carrier fre

uency with res

ect to which, s ( t ) would be an SSB sign

al?

4.8 Demodulation of SSBSSB signals can be demodulated using coherent demodulation as shown in Fig. 4.58.



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Princi



Fig. 4.58: Coherent demodulation of SSB The received SSB signal is multi

lied bythe local carrier which is of the same fre

uency and

hase as the carrier usedat the transmitter. From Fig, 4.58, we have

v ( t ) = s ( t ) SSB Ac cos ( c t ) =Ac m t c s ( c t ) m ( t ) sin ( c t ) Ac cos ( c t ) () 2

=

Ac Ac m ( t ) c s2 ( c t ) m ( t ) cos ( c t ) sin ( c t ) 2

(4.26)

The second term on the RHS of Eq. 4.26 has the spectrum centered at 2 fc whichwill be eliminated by the LPF following v ( t ) . The first term of Eq. 4.26 can be written as,

Ac Ac 1 + c

s 2 ( c t ) m (t ) . 2 2 As m ( t ) cos ( 2 c t ) h

s the s ectrum centered

t 2 fc , even this ill b

e ' Ac Ac elimin ted by the LPF. Hence v 0 ( t ) = m ( t ) . Th t is, 4v0 (t ) m (t )

The difficulty in demodul tion is to h ve coherent c rrier t the receiver. C n e use squ ring loo or Cost s loo to recover the c rrier from s ( t ) SSB ? The answer is NO. Let us look at the squaring loop. From Eq. 4.23,

s ( t ) SSB = s ( t ) = A ( t ) cos c t + ( t )

After squaring, we obtain,



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s 2 ( t ) = A2 ( t ) cos2 c t + ( t ) = A2 ( t ) 1 + cos2 2 ( c t + ( t ) )

{

}

As ( t ) is a function of time, we do not have a discrete com onent at f = 2 fc and hence, carrier acquisition is not ossible. It is left as an exercise to show that Costas loo will not be able to ut out m ( t ) when the in ut to the loo is the SSB signal. Hence, when SSB is used in a communication system, highlystable crystal oscillators are used both at the transmitter and receiver. If this scheme does not work (es ecially, at very high frequencies) a small ilot carrier can be sent along with the SSB signal. This ilot carrier is recovered at the receiver and is used for demodulation. Let us now look at the effects of freq

uency and

hase offset in the carrier used for demodulation. Let the carrier term at the receiver be ' Ac cos 2 ( fc + f ) t . Let the received input tothe demodulator (Fig. 4.58) be 1 Ac m ( t ) c

s ( c t ) m ( t ) sin ( c

t ) 2 Then, v ( t ) = 1 Ac Ac m ( t ) c s ( c t ) m ( t ) sin ( ct ) cos ( c + ) t 2

v 0 ( t ) , the output of the LPF would be

v0 (t )

1 ' Ac Ac m ( t ) c s ( 2 f t ) + m ( t ) sin ( 2 f t ) 4

(4.27)

Assume thatv 0 ( t ) = m ( t ) cos ( 2 f t ) + m ( t ) sin ( 2 f t )

(4.28)

Consider a s ecial case, namely, a frequency com onent at f = 1 kHz in M ( f ) and f = 100 Hz. With these values, Eq. 4.28 becomesv 0 ( t ) = cos 2 103 t cos ( 2 100 t ) + sin 2 103 t sin ( 2 100 t )

(

)

(

)

e j 2 900 t + e = cos ( 2 900 t ) = 2

j 2 900 t

(4.29)



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As this is true of every frequency com onent in M ( f ) , we have the result that when f is ositive and the in ut is a USB signal, there is an inward shift of

M ( f ) by f . (We see from Eq. 4.28, e j 2 1000 t is converted to e j 2 900 t and

e

j 2 1000 t

is converted to e

j 2 900 t

. That is, both the s ectral com onents

have been shifted inward by 100 Hz.) By a similar analysis, we can easily see that if f is negative and the in ut is a USB signal, then, after demodulation, the s ectral com onents in M ( f ) would undergo an outward shift by f . In all, we have four cases to be taken into account and the effects of non zero f onthe resulting out ut after demodulation are summarized below. Case i) f > 0 and the in ut signal is USB: S ectral com onents in M ( f ) will

undergo an inward shift by f Case ii)

f > 0 and the in ut signal is LSB: S ectral com onents in M ( f ) willundergo an outward shift by f

Case iii)

f < 0 and the in ut signal is USB: S ectral com onents in M ( f ) willundergo an outward shift by f

Case iv)

f < 0 and the in ut signal is LSB: S ectral com onents in M ( f ) willundergo an inward shift by f

Let

13927252 PCS Communication Systems Linear Modulation

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Transcript of 13927252 PCS Communication Systems Linear Modulation