dc_unit_i_chitode

8/6/2019 DC_Unit_I_chitode

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Digital Communicatio'nThese functions are orthonormal and any of Sl(t),5'2(f)..s3,(i) and s4(t) can beexpressed as a linear combination of above basis functions.

1.6 Bandwidth1.6.1 Definitions o'f Bandwidth

shown in Pig, 1,6.1.

ii) Null-to-null bandwidth: This bandwidth defines width of the main lobe. In'iFig. 1.6.1, the' null-to-null bandwidth is ;. This bandwidth is considered for most ofthe digital communication systems. It contains most of the pOlver of the signal.

iii) Fractional power bandwidth : This bandwidth assures the specified pow'er ofthe signal, For example if 95 o povver bandwidth is 10 Ml-Iz. This means within thebandwidth of 10 1VfHzffhe signal power win be 95 ~~.

Iv) Absolute band-width : This is the total bandwidth of the signal No frequencycomponents are' present outside the' absolute bandwidth, In other words, absolutebandwidth is the bandwidth which contains all the 100 % ) pOVv"e r of the signal.

v) Noise equivalentbandwidth: Fig.. 1.6..2shows theresponse of ideal filter andpl'acticalfilteL The bandwidthISN ~ required to ' make theshaded areas of ideal andpractical filters equal is canednoise equivalent bandwidth.

'F igl. . '1.6,.2 Noise equivalent bandwidth


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Communi,cation 1-33 D'igita~Communication Syste,ms,vi) Bounded PSD : Some times the bandwidth is specified as the range~ of

after'\l\7hich PSD falls below certain attenuation level. For exampleandwidth for 35 dB attenuation. This means range of frequencies after which, the psdchieves attenuation higher than 35 dB.

D'iim,ensionallity Theorem- The dimensionality theorem states that" a real waveform can be completely

by 'N independent pieces of information where N is given by"N ~ 2ETa , ... (1.6.1)

Here N - Dimension of the waveform L T l signal spaceB - Bandwidth of the signal

To - Time over which thewaveform is described (period)pplications of diimensionality theorem

Dimensionality theorem is used to calculate storage space required to storedigital signal.

It is used to estimate bandwidth of the signal from equation (1..6.1) '\I\l haveB;;;;~ 2 To ... (1.6.2)

The ratio T N gives number of independent pieces of Infrornation transmitted overoTo - It is also called symbols rate (rs)'. i.e..

... (1.6.3)

B =. rs2 ... (1.6.4)Thus bandwidth can be obained form dimensionality theorem.

The communication channels are mathematically modelled to evaluate their effectsignal transmission,Threemodels are described next.


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gital Communication 2-2 Baseband Fonnatting Techniques, The samples x i s (t) must represent all the information contained inx(t) .' The sampled signal x1 5 (t) is caned Discrete Time (DT) signaL' It is analyzed

with the help of DTFT and z-transform.,.1.,2Sampling Theorem for Lo,w PassO.:P) Signa!lsA low pass or LP signal contains frequencies from 1 Hz to some higher value.

tatement of sam,pUng theorem1) A bandlimited signaJ of finite energy, w'hich: has no frequency components

h.igher than W hertz, is completely descri:bedby specifying the ualues of the Isignai at instants of time separated by 2~r seconds and

2 ) 1 A bandlimited signal ,o t finite energy, which has no frequency componentshigh,er than W hertz, may be com-pletely recovered from the knowledge of itssamples taken at the rate of 2W samp'les per second.

The first part of above statement tells about sampling of the signal and secondtens about reconstruction of the signal. Above statement can be combined and

tated alternatelv as follows :,;A continuous tim e signal can be comp le te ly r ep re sente d in its samples and rec ov ere d back

the sampling frequency is twice afthe highest frequency content of the signal. i.e.,I s . ; : : . 2W

Here I s is the sampling frequency andW is the higherfrequency content.

roof ofsampllnq theoremThere are-1'\\'0 parts: (1)Representation of :lit) in tenus of its samples.(II) Reconstruction of x(t) from its samples,Part I : Representation of x(t) in its samples x(nTs)


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'ommuniication 2-3 Ba;seband Fo~mattingTechniquesStep 1 : Define x 0 (t)Refer Fig. 2.1.1. The sampled signal X8(t) is given as,

=x8(f) = LX(,t) 3(t- nTs) ,.,. (2.1.1)n=-oo

Here observe that x 8 , (t ) is the product of x 8 and impulse train S(t) as shown in. 2.1.1. In the above equation ,B(t-nTs) indicates the samples placed at +T~,2Ts f

3T" ... and so on.Step 2 :FT of xS(t) Le. X,S, ({)Taking FT of equation (2.1.1).

XS(j) = PI L ~ : t } 1 i ( t ~ n T ' ) }- FT {Product of , : x { t ) and impulse train}We know that FT of product' in time domain becomes convolution in frequency

omain. i.e.,Xc/f) = = PI' (x(f)}*FT{o(f-n:T s)}

By definitions, x(t) ~ FT ) X(j) and... (2.1.2)

00o(t~nT~) ( IT ) f s L0(j-n!s)n='-oo

Hence equation (2.1.2) becomes,oox,(I) =: X(f) *1s Lb(f ~ nfs)

Since convolution is linear,I G I C i '

XB(f) = i" 2:X(f) * o{ ! -nfs)n=-oo

QC= I s L X ( f -r.fs) By shifting property of impulse function= . ' - I s X (j ~2fs)+ I s X C !-Is) +Is X(f)+ fs X(f -Is) + fs X(j - 2 /5)+ - ..


5/58

Baseband FOlrmatt i ing Techniques

( i) The RIiS of above equation shows that Xif)at tIs' 2fs t - + 3 fs I-

(ii) This means X i f J is periodic in. f s '(iii) If sampling frequency is L> 2 " V \ l " then the spectrums X(j) just touch each oilier.

IS placed

fs" 2fs""

Fiig. 2,.1..2, Spectrum, o,f o:riginal signal and sam,plied signalStep 3: Relation betweenX(f) and X.)(j)Important assumption : Let us assume that I s = - 2W, then as per above diagram,or

: : : I s X(f)1 .= -"Xry(j)f s

Step 4 .:Relation between x(f) and x(nTs:)

X,3'(f)X { f )

for - W


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Baseband Formatting TechniquesL..,_bove equation 'f is frequency of CT signal. And.! f:;: Frequency of DT signal

)sin equation (2 .1 .4) . Since ,x(n) =x(nTs) ' i,e, samples of x(t), then we have,

Xs(f) =

Putting above expression in equation (2.1..3),XfJ) =. .l. f.x( n Ts ) e~j21tfnTsI,'n=-= .

Inverse Fourier Transform (1FT) of above equation gives x(t) i.e.,x(t) = I F T ' { ' 1~nT5Vj21rjnT,}', f s I'c n=-oc I ,.,. (2.1.5)

Comments :i) Here x(t) is represented completely in terms of x(n Ts)':ti) Above equation holds for I s : : 2 W . Thismeans if the samples are taken at, the

rate of 2W or higher, x(t) is completely represented by its samples.ill) First part of the sampling theorem is proved by above two comments.Part U: Reconstruction of .x(t) from its samples

Step 1 : The IFT of equation (2.1.5) becomes,x(t) = 1 . ' . { 1 , fx(nTs)e- j21


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Interchanging the .order of,summation and integration,x(t) = < > < > WLx(nT~) .~ J ej2x}(t-nTs) df

~-CJO Is -w

L OC I... r. ., . 1 . ! [ . . e . . . j 2 t i f ~ . . (t-n.1 S ) . ' ] w .(nT )..__. ... B h '2 (t T). s J ' 1[ :,'_ n s ) ..~-= '. -w

.~ x(n T)._!_. sin 2 T C W (t - n T~)

. . . 1 . S + n (t - n T )r1~-"""}s s= fx(nT~) sinn(2Wt-2WnTs)

-_ n,{is t- : 0 uTi;)n=-' = v~ J t,Here I s = 2W, hence T, =l2~- Simplifying above equation,

x(t) = ~ x(nT )sin1t(2W;t-n).' s n:(2Wt-n)n""'-O Q ' .rx:L x(n Ts) sine (2 W 't-n) sin 1' t Bsince' = = sine HnEl '. . . (2 .1 .6)11=-00

Step 2 : Let us interpret the above equation. Expanding we get,x(t) :::;::..+x(- 2Ts)s inc (2 Wt+ 2 )+ x(-T _:Jm c( 2 W t+ l)+x(O) s in c (2 W t)+x(T~)sinc(2 W t-l)+_.

omments :(i) The samples xtnTs) are weighted by sine functions,(ii) The sine function is the, interpolating function. Fig. 2.1.3 shows, how x(t) is

interpolated.


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2,-7 Baseband IFormatting! Techniques

F i'g . 2 .1 ., 3, (a) Sampl,ed v,ersion of si ig;nal x(t)(b) Reconstruction of x{t) from its samptes

Step' 3:: Re,con~truction of x(t) by low pass filterWhen the interpolated signal of equation (2.1.6) is passed through the low pass

lter of bandwidth ~.W~f


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ital communleatlon 2-8 Baseband Formatting Techniques.1.3 Effects of tlndersamplmq ,(Aliasing)While proving sampling theorem we considered that I s =2W. Consider the case of


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D' i ig i ta , I . ;Co,mmunicaUon 2.. Baseband Formatting Tech niques

Figl., 2.,1.5 fs ;;:::W avoids aiiasingi by creati!ng a bandgap,i n Bandlimiting the' signalThe' sampling rate is, I s = 2W. Ideally speaking there should be no aliasing. But

there can be few components higher than 2W. These components create aliasing.Hence a low pass filter is used before sampling the signals as shown in Fig. 2.1.6.Thus the output of low pass filter is strictly bandlirnited and there are no frequencycomponents higher than r w . Then there, wil] be no aliasing.

x'{t}~-__,. Bandl l i imUinglow pa ss filte r SampIer ','"""-,,-xa(1t)i

Fig. 2,."1.6Bandlimi'ting the s ign,a t The bandllimiting LPF is calted p relaUas, filter.2.1.4 Nyquist Rate and Nyquist Int,erval

Nyquist rate : When the sampling rate becomes exactly equal to '2W samples/sec rfor a given bandwidth of W hertz, then it is called Nyquist rate.

Nyquist Interval : It is the time interval between any two adjacent samples when,sampling rate is Nyquist rate.

Nyquist rate - 2W HzN .. 1 - 2w 1-econtd S '"yqUlS t mterva.~; c

. .. (2 .1 [,7 )

..,. (2.1.8)

2.,1.5Reconstruction Filter (Interpolation Filler)DefinitionIn section 2.1.2 we have shown that the reconstructed signal is the succession ofsine pulses weighted by;\l( n Ts)" These pulses are interpolated with the help of a low

pass filter. It is also called re,constrnctionfilli:er or interpolation filter.


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'Di,glital Communication 2 , -10Ideal fUte'f

Fig. 2.1.7 shows the spectrum of sampled signal and frequency response ofrequired filter. When the sampling frequency is exactly 2W, then the spectrums justtouch each other as shown in Fig. 2.1.7. The spectrum of original signal, X(j) can befiltered by an ideal filter having passband from - 1 1 \ 7 ' S ; ; f . : : ; ; w.

'I ! I" " t -- ,~ , - " - , , ,~ !

Fig~2~t,7 Ideal reconstruction fUterNon-ideal filter

As discussed above, an ideal filter of bandwidth IW filters out an original signal.But practically ideal filter is not realizable. It requires some transition band. Hence I smust be greater than 2W .. It creates the gap between adjacent spectrums of X,8(fJ. Thisgap can be used for the transition band of the reconstruction filter. The spectrum Xif)is then properly filtered out from X3(f). Hence the sampling frequency must begreater than 1 2 f t V ' 1:0 ensure sufficient gap for transition band.


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2 -11 Baseband Flormatt~ng Techniiq]lIes

' , . . j ' "I

F ' i g l " 2.1.8 Practical reconstruction filterZero,-O.rder Hold for Practical Reconstruction

The zero-order hold circuit is used forpractical reconstruction. It simply holdthe value x(n) for 'T seconds. Here'T'is the sampling period.The output of the zero-order hold isstaircase signal that approximates x(n).This is shown inFig. 2.1.9.Let the impulse response ofzero-order hold be represented as,

h(t) ::: O


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Digital COimmunica'tion 2 - 1:2 Baseband Fo'rm:atting Techniques

..- h(t}* xo(t)- H(w) X(OJ)(ro)

Here H~(ro). ( O TSD_ 2e- j roT /2 2(0

., roT.roT Sfi-~J-,', 2 '""Y(ro) = 2e 2---", - X (w)(I.)The above eq~ation shows that spectnun X(oo)is changed due to convolution or

passing through zero order hold. These changes are,(i) Th . l i n ' ha hi' f t ' di t ' ti d 1 f T[ , iere 181, 'ear F- -se S', .' corresponmgo'ime '-e ay 0; 2 sec.

. ro Tsm ~-(ii)The main lobe of , r o 2 modifies the shape of X(0 0 ) . The above modifications can be reduced by increasing the sampling

frequency ffis or reducing the time 'T. Sometimes anti-imaging filter is used for compensating the modifications. It's

spectrum is given as,

2. wT',5m20 ,

roT

This filter provides reverse action to that of zero-order hold. Fig. 2.1.10 showstheblock diagram with anit-imaging filter.

y(t)x(n)--.......ero-orderhold Anti-imagingfifterFig. 2..1.10 B.lock diagram of practical reeonstruction

2.1.'7 Sampling Theorem 'in Frequenc,y DomaiinStat-emle,nt:

We have seen that if the bandlimited signal is sampled at the rate of if s >2W) intime domain, then it can be fully recovered from its samples. This is samplingtheorem in time domain. A dual of this also exists and it is called sampling theoremin frequency domain. It states that,


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igftal Comlmunication 2 .. '1 3 :Baseband Formatting Techniques

IExpllanation :-Thus the spectrum, is sampled at. f s < 2~ in the frequency domain. T is the

maximum time limit above which signal xet) goes to zero. l i s ' represents thesampling frequency interval in the frequency spectrum of the signal. Notethat here f~does not represent number of samples taken per second. But itrepresents the frequency interval at which the samples are separated infrequency domain.

Fig. 2.1..11 illustrates the sampling theorem in frequency domain. We can seefrom Fig. 2.1.11 (a) that a rectangular pulse is time limited to ~ secondsi.e., x(t) =A for> ~ st - < ~ . The spectrum of rectangular pulse is shown inFig. 2.1.11 (b). This spectrum X(f) of Fig. 2.1.11 (b ) is sampled at the uniformintervals less than 2~ Hz. The sampled version of this spectrum is shown inFig. 2.1.11 (c) and called X8 (f). Thus each frequency sample of X0 (f) isseparated by ! f s ; Hz with respect to the neighbouring frequency samples.

As shown ill Fig. 2.1.11 (c); since X(f) is sampled in frequency domain, lhlsiscalled sampling theorem in frequency domain. Therefore now we can state samplingtheorem in . frequency domain as,

We have represented x(t) in terms of its samples x(nTs) and an interpolationfunction (sine function) in equation (2.1.6) for time domain sampling theorem.Similarly a continuous frequency spectrum X(f) can be represented :i n terms of itssamples X (k f " , ) and an interpolation function (sine function) as follows :


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Oigitai l COlmmUln i :cat ion 2:-14 Baseband ;Formatting Tec,hniqu ..

x(t).iA lime domain

T. . . . . .

t ' " Frequency domainXs(f). ft" ' '' ' . . ..~f i ' ,; f T "_ '. TF ig ,. 2 .1 .'1 1 ; [a] :S ignal x(t) t ime limiitedto 2

(b) Continuo'us spectrumof x(t)(c) Sampled spectrum x, (1)2.'1...8Sampling of B,andp,ass Signals ,Apl'"il/Ma ~2004,Moy/June 2006

Statement ::In the last sections we discussed sampling theorem for low pass signals. However

when the signal is bandpass innature, then different criteria should be used to samplethe signal. The sampling theorem for bandpass signals can be written as follows.


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ta.1 Comm un lcatien 2 15 Baseband Formattingl Tec ihn liqu . . Spectrum of bandpass signal: Thus if bandwidth is 2W, then minimum

sampling rate for bandpass signal should be 41IVsamples per second"Fig. 2..1.12 sho- IS th e spectrum of bandpass signal.

~f j -f II 0 f c - ' ,I , ,f+W fI~ Ii n 'C C ] II III II J . I,,. . . . . 1 i" " ' 1 :2W I I 2W 'I! i I J:I I I ,

X(f)

.Fig,. 2..1.12 -'Spectrum o'f bandpass sJg:nal iThe spectrum is centred around frequency fc . The bandwidth is 2W. Thus the'

in the bandpass signal are from fe ~ w to Ie +W. That is the' highestpresent in the bandpass. signal is fe + 1 ' \ 1 . Normally the centre frequency

>W .' Inphase and quadrature 'components : This bandpass signal is, first: represented in terms of its inphase and quadrature components.

x J(t) = Inphase component of x(t)xQ (t) = Quadrature component of x{t)

LetandThen we can write x(t) in terms of inphase and quadrature components as,

,... ( 2 .1 .10)

- w 0 WFiig,.2.1.13 Spectrum of ~nplhase and

uadr,ature components of bandp,ass s!i'9inal~(t)

f

The inphase and quadraturecomponents are obtained by multiplyingx(t) by cos (2nJct) and --sin (2nfct) and thensuppressing the, sum frequencies bymeans of low pass filters. Thus inphasexIt) and quadrature' x'Q (.t) componentscontain only low frequency components.The spectrum of these components islimited between -W. to + W. This isshown in Fig. 2.1.13.


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Communication 2 16 Baseband Formatting Techniques Representation in terms of samples After some mathematical

manipulations, on equation (2.1.10)we obtain the reconstruction formula as,

Compare this reconstruction formula with that of low pass signals gIven by(2.1.6), It is clear that . 1 : ( t ) is represented b y x ( 4 ; : ' ) completely.

Here, x (-4;:' ) = x (nT s) = Sampled version of bandpass signal1r; = 4Wnd

Thus if 4W samples per second are taken, then the bandpass signal of bandwidth 2W can be completely recovered from its samples.

Thus, for bandpass signals of bandwidth 2WrMinimum sampling rate - Twice of bandwidth

- 4W samples per second!r.... Examp,le 2.1.1 : Show that a b and lim iled signal o f finite energy which has no

frequency com ponents higher than W Hz is comp le te lu desc rib ed by spec; fying values ofthe signals at instants o f tim e separated by 1 / 2W seconds and also show that if theinstantaneous talues of the signal are separated by intervals larger than 2 t V seconds;they fail to d escribe the signal. A bandpass s~gnal has spectral range that extends from20ta 82 kH z. F ind the ac ce ptab le range o f sampling fre qu ency I s .

.A riliMa -20Q5, Ma IJune-2006, 8 Marks

tep 1: Define x;s(t).Let x(t) be the bandlimited signal whdch has no frequency components higher than .

Hz. Let it be sampled by a sampling function,,=x(t) = : L o ( t - nTs )

The sampling function is the train of impulses with Ts as distance betw-eenccessive impulses. Let x (nTs) be, the instantaneous amplitude of signal x(t) at instant'""Ts . The sampled version of x(t) can be represented as multiplication of(nTs )andB (t ) i.e.


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tal Communication 2 : . . 1 7 Baseband Formlatting Techniquesoo

X 6 (t) - L x (nTs) 3(t - nT~ ) ... ( 2.1 .1 2)tep 2 : Fourier transform of X ; S , ( t ) Le, Xa(j)Fourier transform of this sampled signal can be obtained as,

x , , ( , f ) .s oo- Is L X e f -nis)~ . .. ( 2.1 .1 3)H .f' th l i n ' l' hich i .{ 1re rs 1S me San1P 'g rare W, 1S given as Js . = T .sAnd X (f - nls) ~X(f) at nfs =0.; +hI + 2/5, + 3/s ....Thus the same spectrum X(f) appears at f = 0,f =+ I s , f = 2 1 s etc. This means thatperiodic spectrum with period equal to is is generated in frequency domain becausesampling x(t) intime domain, Therefore equation (2.1.13) can be written as"

X am = Is X(j) +/s X (f I s ) +I s X if f2/t;)+I, X(j+3fs)+ I s X(f=t4hJ+ . .. .. (2 .1 .14)

or 00Xs'(f) - I s X(j) + Lis X (f - nfs)11=-=mO

. ... ( 2 .1 .15)

tep 3 : Relation between XV) and Xo.(fJ.00'

By definition of Fourier transform, X(j)= f x(t) :j2rift d tFor sampled version of x(t), we havet = nTs' Then above equation becomes,

,::::0

X, {-0. = ~ x ' ( n " T ' \ e - j2 'IT;fnTs-oVJ- LJ'sJ " . . . . ( 2 .1 . . 1 1 6 )

It is given that the signal is bandlimited to W Hz and,Ts = = 2~ seconds" .... is ~ i =2Ws ... ( 2..1.17 )

From equation (2.1.14) we know that Xs (n is periodic in fs The spectrum -X(fJnd X,S if) an~ shown in Fig. 2.1.14..


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Communication 2 -18

(a)

(b)

f -w -fs,s==-3W

-w fs

fs+W" " ,+3W-f f

Fiig. 2.1.'14 (a) Spectrum of x(t)(b) Spectrumi of X 0 (f) with fs = 2W

Since I s =2W; I s -,W =W and I s +W =3WThus the periodic spectrums X(f) just touch W f 3W !+ 5W .... etc.Thus there is no aliasing. From equation (2.1.15)we can write,

X(j) = = ;X s (r')- ~ X if - nf~))5 ~n;tO

.. ,"(2.1.18)

With fs =2W inabove equation"X(j) = = 2~ Xo(n - f X i f -nls)

n;:;:-;.Q,

m : O"i.e, ... (2.1.19)

4 : Relation between xU) and x(nT.) or { 2;" )-Putting the value of Xo(f) from equation (2.L16) in the above equation,

X ( f J " . = = _!_ ~ x l(nT ) e- _ i 2 1 t f n T s2W L .J . S1F-='12W


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Communicat ion Baseband Forrma ttin g , T echn iques

X(f) - 1 i:.. ( _ _ ! ! _. . . J e: j1rfnlw2W- 2W~ , ... . . . (2.1.20)x(t) can be recovered from X(f) by taking Inverse Fourier Transform of above

l.e. ... (2.1.21)

This equation shows that x(t) is represented completely "by its samples x ( i W ) fur=< n-: =. That is the sequence x ( 2 ~ ) has all the information contained inx(t).econstruction of ,s ig n al from samples :Consider equation (2.1.21), { l .;1O i ' . ) }... n' cx( t ) . ' = IFT -.r- _ .. .~. .x l - - - . . . e" J 1 i f n / W .2 vV z: 2Wl .~-o:;.. . By definition of Inverse Fourier Transform (IFf) the above equation becomes,

w 00 (' )x(t) = J 2~L x l 2 ~ .e-j'njnIW ej2r r . f t d t-w f'P'-=

Interchanging the order of summation of integration,

l . 1 . t ) = ~ ..x ( 2 ~ ) 2 ~ 1/ 2 1 i f { t - 2 W ) dt11='--= \ ~W

_ ~ ...x(. _.1 1 . J sin(2;r WI - n;r)~CD l 2 W (21 t W t ~ nn)~. ( .. 11 J sin 11 (2Wt- n)= . L X 2W n{2Wt-n)tt;;;-o. .

S . . sin 1lt8 b . bIDee SIDee : ; : e ' aoove equation becomes,1 1 :x ( t ) : : : : t x ( 2~ ).s ine (2 Wt - n : )

rF-O-00< n< oo

This is interpolation formula to reconstruct x(t) from its samples x (n Tf j) '


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D'igita.1 Communication Baseband IF orma ttin g Techniques.' Thus the above discussion shows that the signal can be completely

represented Into and recovered from its samples if the spacing betweenilie1successive samples is 2W seconds. i.e. is = 2W samples per second.

S ,ampUng f requency fo r bandpass si:gnal :The spectra] range of the bandpass signal is 20 to 82 kHz.

Bandwidth ~ 2W = = 82 kHz- 2;0lliz = 62 kHzMinimum sampling rate - 2x Bandwidth

= 2x62kHz= 124 kHz

Normally the range of minimum sampling frequencies is specified for bandpasssignallis. It lies between 4 W to 8W samples per second.

:. Range of minimum sampling frequencies= (2.x Bandwidth) to (4.x Bandwidth)= 2 x 62 kl-lz to 4 x 62 kHz= 124 kHz to 248kHZ

u..... Exam!ple .2.1.2:: F ind the Nyquist rate and Nyq uist interval fo r fo llowing signa ls.i) rna) = 2In cos(4000n t) cos (1000 n t)") ......) s:in500ntH' m' t, = ---- : n : t

Solution : i) ml (t) ~. 2Incos( 40001t t) cos (1000 n t)

1 { 1 }. 2n .2 [cos(4000nt-1000n t) +cos (4000 n f+ lOOOn t ) ] .- ~ [cos 30001[ ; t + ens 5000 IT . : t ]4~ .1- 4 1t [cos 2n I I t +cos 2x 1 2 t]


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gital Communication 2 - 21 Baseband Forrn,atting TechniiquesComparing, we get, fl ~ 1500 Hz and h= 2500 HzHere highest frequency W - 1 2 ; ; ; ;2500 I-Iz

Nyquist rate = 2t.-V = = 2x2500 := 5000 HzNyquist interval 11.= - Z - W - , ~ 2 .2 . . . . , 0 . = 0.2 msec.'" X;),

ii) m2(f)- sin500n tntsin2nft

ntComparing, we get, I= 250 Hz orW = 250' Hz'." '! Nyquist rate - 2 V \ l : : : : 2x250 ::::500 HzNyquist interval 1 1- -2W -.,- 2x250 :;;;;2 msec

. . . . Examp;le 2.1.3 : Fig. 2.1.15 shows the spectrum of a message signaL The signal issampled at the rate o f is =1.5 fmax 1 where fmax := 1 Hz, is maximum signal frequency.S ketch the sped-rum of the sam pled 'version of the signal. If the sam pled signal is passedthrough an ideal low pass filter of bandwidth fmax/ sketch the spectrum of the outputsignal from this filter.

X(f)t

Fig. 2:..1.15 S!pectrum of :m,essag,esi~gnaJX '(f~lution : When x(t) is sampled instantaneously its spectrum is given as,

ooXS(j) = !sX(f}+ 'LfsX(j-n/s)

t / F - - - . G < >NO


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Dilgitat Communication 2 ..22 ,Baseband F,ormatting 'TechniquesHere is ; ;1 . .5 fmax. Then above equation will be r

(X

X5(f)= 1.5 fmax"X( j )+ L 1.5 fmaxX(f-1.5nfmax)~--OmO

With fmax =: 1 Hz, above equation will he..ox;

X,S (f) ~' l.sX(j) + L 1.SX (j -l.5n)rF-o:>n O

Fig. 2.1.16 (a) shows the plot of above-equation.Because of under sampling (fs ; : 1.5 f~ax)' there Is aliasing effect and it is visible in

Fig. 2.1.16 (a), When the sampled signal is passed through an ideal low pass filter ofbandwidth fmax [Fig. 2.1.16 (bj], the spectrum of the corresponding 'Output signal isshown in Fig. 1.1.16(e). It shows dearly that the spectrum of Fig. 2.1.15 andFig. 2.1.16 (c) are not same because of aliasing.

(a)1'.5

~ H(f)1,.

I

-- f-f --'''''-1 Hz 0 I - 1 Hzmax ,'max =, :'

(c)

Fig. 2..1.16 (a) S,pectrum ofthe sa impl :ed signal at f,s =1.5fmax(b) Response of iide;allow pass filter withIH (fU='1 fo ,r ~fmax'


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2 -23 Baseband Formatliin~Techniques

1. S tate and prooethe sam pling theorem in tim e dom ain. W hat is N yquist rate?2. Explain the junction of low pass filter in sampling.3 , S tate and prove the sampling theorem in frequen,cy dom ain. S how that the effix t of sam pling

is to produce double sided spectra around each. harmonic oj sampling frequency.4. A btmdlim ited signal fft) is samp'le d by train oj rectangular pulses of width 1: and period T.

(ae)Find an e xp re ssion fo r th e samp ,l ed s igna l.(b) Determ ine the epecirum of the sam pled signa,l an d sk etch it.

5 . Wha t is aliasing and hot it is re du ce d l'Examples

I. S pecify the Nyquist rate and the Nyquist interval for each of the following signals.(a) x ( t ) ; ; : ; sine (200 t)(b) x{t) =sillc2 (200 t) (e) x(t) =sine (200t) + sinc2 ( ZOO t )

.. . . 1 ... 1 'I[ADs: fa)Rate: 200, Interval : 200' (b) Rate; 400"Interval: 400.',(c)Ra.te: 4 O D , Intervsl: 4 ! o l !

2. Consider th e analog signal x ( 1) "" .3 cos 50 ~t +10 sin 300 1 T ;t - cos 100 r . : t . 1 [ . V U a t is theNyquist rate jar this signal ? [Ans. :::Nyquist rate ;,::300Hz]

3 . . Find the .Ny,quist raft tmdNyquist interval for the following.1i) m(t) ~ 2........-.cos ( 4 0 0 :it t) ..008 ( Z O , o 1 1 ; t) [An R 601]H' Int .~ 1 6667 I". '.' S.:: a te = - , . , : . Z,.. "erv.a . l . = . . 'ms.smxtii) mff)"" nt [Ans, : Rate 1HZt Interval = 1 sec]

4. T he signal x (t) = CDS 200m +O.250os700rr t ie snmpled at the rate o.f400 samples per second.S ampled wanejarm is: then passed through em ideal low pass filter with 200 Hz bandsoidth.Write an expression for filter o utp ut .. S k etc h th e fre.quency spectrum of samp le d wav efo rm .

[Ans. :yH) ""'.'os(2n xl00tJ + 1).25cos ( 2 : n x 50.f)15 . A waveform [20+20sm(500t+ 30)] is to be samp.le.d periodically .a nd r ep roduc ed from these

s amp le v alu es. Find m ax im um allo wable tim e interval beneeen sample oalues. H aw many :sample values are needed to be stored . in order to reproduce 1 sec of this Wlf{)qorm ?

[Arts,. : Ts ...6.283ms, Samples = 160]


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' ig,HaICommunica tion 2 -39 Baseband Formatting Techniques2.3 Uniform 'Quantizatio'n (Linear 'Quanti,zati on )

We know that input sample value is quantized to nearest digital level. Thisuantization can be uniform or nonuniform. Inuniform quantization, the quantizationtep or difference between two quantization levels remains constant over the' complete

range. Depending upon the' transfer- Characteristic "there are three types of01' linear quantizers as discussed next-

The transfer characteristic of the midtread quantizer is shown in Fig. 2.3.1.

Fi:g. 2.3',1 (a.)Quantizatiion characteristic of mridtread quantizer(b) Quanti:zati,on error


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COimmunication. ' 2 ' .. 40 Baseband Forma ttin g Techn iquesAs shown in this Fig. 2.3.,1, when an input is between - '0/2 and + 8/2 then the

uantize! output is zero. Le.,For - 0/2 :::;;.(nT~,) < 5/2 ; xq(nTs) '~ QHere 'Oi'is the step size of the quantizer.forSimilarly other levels are assigned. It is called midtread because quantizer outputzero when x(nT s ) is zero. Fig. 2.3.1 (b) shows the quantization error of midtreaduantizer, Quantization error is given as,, ,

___2.3 .1 )InFig. 2.3.1 (b) observe that when x(nTs) ::::0, xa(nTs) ;= O.Hence quantization errorJ' _

s zero at origin. vV'hen x(nTs) :::::8/2, quantizer output is zero just before this levelence error is 8/2 nearthis level. From Fig. 2.3.1 (b) it is dear that"~ / ' 2 " , < ' < '~/2u , ~ e~u ' ... (2 .3 .2 )

Thus quantization error lies between - 15/2and + 0/2. And maximum quantization. .' tiz .. S (2'3 3')1S, maximum quan. ,,'anon error, ,E max:::: 2 ....1 ". ' '~

.3.2 Midriser 'Quan'ti .zerThe transfer' characteristic of the midriser quantizer is shown in Fig. 2.3.2.In Fig. 2.3.2 observe that,when an jnput is between 0 and S, the output is '0/2.

when an input is between 0 and - 0, the output is - B/2. Le.,o < x(nTtJ < 8 ; Xq(nTs) = 5/2

- 6 < x(nTs) < 0 ; xq'nTs) =- , ,& /2Similarly when an input is between 3 l), and 4 - 6, the output is 7 8/2. This is calledidriser quantizer because, its output is either + 3/2 or - 5/2 when input is zero.Fig. 2.3.2 (b) shows the quantization error inmidriser quantization. When input= a , the quantizer will assign the level of 0/2. .Hence quantization error will be,

For

E- Xq(nTs) - x{nTs)= 3/2 - 0 ::::0/2

Thus the quantization error lies between ~ 8/2 and + ,8/2. i.e.,- 8/2 ~ ~ 5/2 ... (2 . 3.4 )


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, , " t a ' I C ' , . 'tligl ,'. 'OmmUnlCa,IOn 2 -42 Baseband Formatting Techniques.3.3 B,iased QuantizerFig. 2.3.3 shows the transfer characteristic of biased uniform quantizer,

For 0 < x(nTs) < ( 5 ;Similarly, for -3 '~. x(nTs) < 0 ;

Xq(nTs) = 0Xq(nTs) ;: - &

:Fig'. 2.3.3, (a) Biased ,quantizer transfer characterlstlc(b) Quantization error

The midriser and midtread quantizers are rounding quantizers. But biaseduantizer is truncation quantizer. Tills is clear from above diagram. VVhen input isetween 0 and 8, the output is zero. i.e.,


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C,ommuinicat ion . 2 , , 63 Baseband Formatting TechniquesWe know that quantization Ievels. land number of hilts v are related as,

..q - - z v

1024 - 2 . vV = 10 hits

The number of bitsr'sec generated by peM system is called bit rate or signallingte. i.e.,

Signalling rate" r - v I s- lOx 10.8 x 106 bits! sec- 108xl06 bita/sec

The output bit rate does not change if Iinear PCM IS converted into companded!. Companded peM is used to improve the signal to noise ratio.

Innonun.i!.0rm quantization, the step size is not fixed. Itvaries according to certainor as per input signal amplitude. Fig. 2..7.1 shows the transfer characteristic and

ror in nonuniform quantization. (See Fig. 2.7.1 on next page.)In this Fig. 2.7.1. observethat step size is small at low input signal levels. Henceuanfizatienerror is also small at these inputs. Therefore signal to quantization noise

ower ratio is improved at low signal levels, Step size is lhigher at high input levels,ence signal to noise power' ratio remains almost same throughout the dynamic range'quantizer,

Nece.ssity o f Nonuniform QuantizationIn uniform quantization" the quantizer has a linear characteristics as we' have seenFig. 2.32 (a). The step size also remains same throughout the :range of quantizer,herefore over the complete range of inputs, the maximum quantization error alsomains sante. From equation (2.3.5)fue quantization error is given as,

~Maximum quantization error =: Emax = ..... (2.7.1)

From equation (2.6.3) step size '6,' is given as,(), =2.xmaxq


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tal Co 'mmunicatio l'1 l Baseband FonnaffingTechniqu8S;

~ I ; ! t ~ 'I i .Ou~put+ 1 ! ! i I I ~ j ; ! l l i ~ ! ' ; '1 t '~ M - - i - - - - - - - - r - - - " ~ ~ ~ j x. nT' . ~i~~:--~r~m~1~-1-~'T'-'I~-~r-'-. -~~'~~1~~I ' " , I ' q( .s) , : ; ~ I : - - - t ' ; ; 1 : ' , i ;f - - + ~ - - I - - - ! - - - I - l + ~ J - - - t - + - - - ~ - - - : - - - ~ - - - - [ - - - l + j - ~ . - . ~ - - \ - - 1 =- - : - - + - l - j - - J - - .L i ~ l , t , , ; j ~ l ; . ,.., , l, I' ,I - - - r - - - t " - 1 - - - - J - - - r -11 - - 1 - - - ' ,- - T - - ~ - - r - - I - - r - ' - - .~~.~~~-!.- - t - - - t - - t - - -i! - T - - t ( ~ J - - l - t - : F ; ~ t ; : ~ 1 I' 1 ~ I (-"Tl r " ' ; " . j' 'I ~ t! ~ _ r . . ~ _ ~ ~ . . ~ I ~ . ~ : : ~ I, ~ :----i--+---...... _ .. _i I j ! ; : ! ~ :.J 1 ! iSm ail s tep 's Ize at ! l i , : ! !. ~ ~ ~ -. . - - I - - ~ 1 - -L - l -- rl - i- ,. ,u ~ - l o w j ~ " t - 'r ~ V ,e i '- ~ . -r -- -. ~ - .- -- - . -. .. - - ~ - - . ~ - . - - . J ._.l...

~ I ; 1 i 1 ! j ~ . , ! 1 I; ; : ~ ~ : ! j x(nT~,), : i f : r 'I f 1 i ' l i o t ' : I 1 ' . ' . 1 l ! j 1! 1 InputI _ ' _ ~ " N _ + w~_,. ~-;__.-- ---.~-.---,,- -.~':...._ -.-..+------.--,-- .. " 1 - - - ~ - t - - - + - - -----L---~-;- --_ : - - - l L . - - J _ - " - - - I - - - _~ ! : i ! ! ~ , . : ' ! I : l. I _~ J _ !LJ 1 _ ! . i ! i- - - - l - i - - + _ I - - - - - - I . . ' : J + _ + - - , - - - , + + - - I I - i - - - - - - l - + + j - - + J I - - i - - ~ - - t - - -, ! ; r , :!; I:, Ii! 1 I . . , ! ;

F ig . 2 ',.7 .1 fa ) Nonun ifo rm quantization transfer charaetertsfic(b) Qua nti,zai!io n error

If x(t) is normalized, its maximum value i.e, xmax =1 -2q ... (2 .7 .2 )

Let us consider an example of peM system inwhich v = 4 bits.Then number of levels q will be,


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.Com:mUniC8.tion 2 65 Baseband Formatting Techniquesq - 24 =16 levels

., From equation (2.7.2) the step size (3 will be,S 2 2 1, = i= 16 :::8

:.Quantization error is given from equation (2 .7 .1) as,'K

Emax = 1 21 1- --- ..-2x. 8 16

Thus the quantization e'rror_!_ volts of the full rang..e voltag 'e. For simp.,ljcih.',.. '.' ..... 16 . ~ssume that fuJI range 'Voltage is 16 volts. Then maximum quantization error will be- .. - - ..volt. Forllie low signal amplitudes like 2 volts, 3 volts etc." the maximumuantization error of 1 volt is quj-te high i.e. about 30 to 50 Ole]. But for signalmplitudes near 15 volts, 16 volts etc., the maximum quantization error (which isame throughout the range) of 1volt can he considered to be small. This problemrises because of uniform quantization. Therefore nonuniform quantization should be

insuch cases. Another example is discussed next..7.2 Necess.ity of No'nuni'fo,rm Quantization for Speech SignalWe know that speech and musk signals are characterized by large crest factor.

hat is for such signals the ratio of peak to rms value is very high.Peak valueCrest factor = - - . ....(2.7.3)R}dSvalue

- Very high for speech and musicWe know that the signal to noise ratio is given as,

s C ' ' 20 .p.)... x .c ..N- .3.2 x ... B y equation ( 2 . 6 . 2 2 ) ... (2 .7 .4)Expressing indecibels, ( 5 ) .-.. ldB ~N.'If we normalize the signal power i.e, if P =tthen above equation becomes r

( S ) " . . . . ".NidB > (4.8+ 6v) dB .... (2.75)


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BasebandF'ormatting Te'chniques;, ,2VSi'anal Xl (t )Here power P is defined as, P := 0,. - ~~R R

Vs7 g n a l - Mean square value of signal voltage= x2 (t)

. I. . x2 (t)., Normalized power will be, P ~ 1 [with R =1]p "'"2 (t ) ... ( 2.7 .6).Crest factor is . given as"

Peak value xmaxCrest factor := - ----RNlS value [x2 (t)]1/2 ... (2 .7 .7)

Xmax# since P = x2 (t ) ... (2 .7 .8)When we normalize the signal x (t), then

Xmax = 1 ... (2 .7 .9)Putting above value of xmax inequation (2.7'.,8),..

. .'. 1Crest factor =. rn- V P ( 2 ' '7 1 ' 0 " )I; .... ' -, '_- . : _ _":

.For a large crest factor of voice (speech) and musk Signals P should be very verythan one in above equation.

P 1 for large crest factor in equation (2.7.10)Therefore actual Signal to noise ratio will be significantly less than the value that is

ven by equation (2.7.5), since in this equation p;= 1. Consider equation (2.7.4),( . '. - N

S . _. .. ' \ ) '. ! = 3x 22v' x P

I I(3x22v xP)! PJ (3X22v xp)lip~l ... (2 .7 .1 1 )This equation shows that the signal to noise ratio for large crest factor signal

1) will be very very less than that of the calculated theoretical value. Theeoretical value is obtained for normalized po 'V \ rer (P' =1) by equation (2..7.4).Therefore such large crest factor signals (speech and music) should use

onuniform quantization to . overcome the problem just discussed. Signal ito noise ratioduces at low povve r levels (P 1) just now we have seen by equation (2.7..11). That. _


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igitalConimlunicatiion- . '2 .. 6 8 Baseband Fonnatting' TechniiquesJl - Law Companding for Speech Signals

Normally for speech and music signals a ,~- law compression IS used. Thisompression is defined by the following equation,

Z (x) = (Sgn x) In (1+ ~ t xl) , Ix l ~ 1 ... (2.7.12)In (1+ Jl )Fig" 2.7.3 shows the variation of signal to noise ratio with respect to signal level

ithout companding and. with companding.

t :(8)L N : J dBo 3 : 0

Withou t. co rnpand inq

I i-40 -30 -20 -10 0Siignalleve~ dB__'"!Fig,.2.7.3 PCM perf-ormance with ,11 law companding

It can be observed from above Fig. 2.7.3 that signal to noise ratio of PCM remainsalmost constant with companding..7.5 A ,_Law for 'CompandingThe ,A - law provides piecewise compressor Characteristic. It has linear segment for

low level inputs and logarithmic segment for high level inputs. It is defined as,

r Alrl ', .., 1~; , , for OS! x [ - < - , .Z(x) = i 1+ InA . _ A! l+.ln(A!,xl } - for_!_$;lxl~'ll l+.lnA A ... (2.7.13)When A ::;:I, we get uniform quantization. The practical value for A is 87,.56. Both

A =Iaw and u -law companding is used for peM telephone systems.


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ta l.Commun i. ca .tion . 2 -69 Basebarndl FonnattingTechniques. 6 S , i'gna l to Noise Ratito o :f C ,omp ,anded PCMThe signal to noise ratio of companded PCM is given as!

5N [ In (1+~Jri. . . . ( 2 ., 7 . . .14)

Here q =2v is number of quantization levels .. Example 2.7.1 : For a random variable are the mean square value and variance

alw ays equal ? C alculate these quantities jor the quantization no ise or error in PCMsystem.

lution: For a random variable X, the variance O'~ is given as,~2 = X , '2_ m,2Ux '". 'x

Here X 2 . is the mean square valueand m ; is the mean value,

Above equation shows that variance (O '~) and mean sqUlMevalue ( X2 ) will b eif mean (mx) is zero.

From quantization characteristics of Fig. 2.3.2 (b) it is dear that quantizationror (e) has zero mean or average value. And it follows uniform distribution fromS' 8., b bilil d ,.. fun . f ' ,. , " ' i l l ' b2 to +2' Hence pro a ity density ... ction 0 quantization error W . oe ,

o S1 for - -,-< E : s ; -. 228 elsewhere By equation (2.6.9)oMean square value can be calculated as,

GO

X 2 = J x2fx(x)dxPutting valuesin above equation"

S2 1- f;c2~dS6 3


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gital Communication 2,,70 Baseband Fomiatting' Techniques

, 82- 12This is the mean square value of quantization error ..

. Example 2..7.2 : A C ompact Disc (CD) records aud io signals d igitally ~11uszngPC lvL Assume the aud io s ig na l bandw id th to be 15 kHz .(i) Whafis., Nyquist rate ?(ii) If the Nyquist sam ples are quantized into L :::.65,536 levels and then binary coded,determ ine the num ber of binary d igits required to encode a sam ple.(iii) D eterm ine the number of binary digits per second (bits/sec) required to encode theaud io s igna l.(iv} For practical reasons, the signals are sampled at a rate well above Nyquist rate at441,00 samples per second .lfL ~ 65. '536 r determ ine num ber of bits per second requiredto encode the signal and transm ission bandwid th o f encoded signaL

olution : (i) To obtaiin Nyquist rateThe bandwidth of the signal is, W = 15 kHz..

iII'iII Nyquist rate = 2W- 2x15 kHz = 3D kHz

To determine number of bitsNumber of levels, q _: L = 65,536Hence binary digits required to encode each sample will be,

_ 2vqv- log2 q

- log 2 65536 = = = 16 bits)To determine signaUing ratev = 16 bitsIsample are used. The samples are taken at the rate of.30,000 samples/sec. Hence signalling rate will be,

or

r = = vfs= 1 6 x 30/000 = 4801 kbitsl sec


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2 -71 Baseba,;ndFo'rmatting Techniquesv ) Toobtam B T if (s = = 44.1 kHz,

Levels used are q - 65,536v= logz q = 16 bits)~- 44100 samples/sec

For PCM, the transmission bandwidth required to encode the signal will be,B T - ; v J s. . . .

.. .

1 ' ,2 x 16x44100- 352.8kHz

E,xample 2.,7~3:: The output signal to noise ratio of a 10 bit PCA1 'llJilS found to be30 dB. The desired SNR is 42 dB. It was decided to increase the SNR io the desiredvalue by increasing the number of quantization levels. Find th e fractio n.a l inc rea se 1ntransmission ,bandwidth re qu ire d fo r :this increase in SNR.

ol!ution :: f i ) To, obtain number of bits for 42 dBSignal to noise ratio of peM is given as,

: ( } r l d B : ; ; : (4.8 + - 6 v ) dB,!V }, "Above equation shows that signal to noise ratio increases by 6 dB with every bit.

is given thats ', ' df f ' 10 b "= 30 I . B or '., ItsN

, , . d " I ' , .. ....... ,C dB H' .'. ( 5 " \ . ." . . 42 i" ') ~O '1") dBThe desired Signa . to noise rano 15042 '::., ence rise inN jrauo IS ' '~,5 ::::::~L " ',.We know that ( ~ ) ratio increases by 6 dB for 1 bit, Hence 2 bi~ are required to

crease signal to noise ratio by 12 dB.Hence, v := 101+ 2= 12 bits are required.To obtaiin fra,ctionail r:ncrease in bandwidthBandwidth in peM is given as,

1BT = 2vis


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2 -1.2 Baseband Formatting TechiniquesB '1'0' bits)\ u mrsj - ~,xl0xfs ;;::;i s

and By (12 bits)

Fractional increase in BT - 6 f s -5/p x 1 0 . 0 . 01_ 2 '0 OfS fs . /0 - .. 1 ,,0. Example2 .7 , .4 : A telephone signal with cutoff frcqueru:y of 4 kH z is digitized into

8 bit PC M ', sampled at J'.lyquist rate. C a lculate baseband transmission band to id th ' and 'C ;quantization ~. ratio .

,olul;on Given data is,w ::: 4kHzv _. B bits

For PCM the transmission bandwidth is given as,

~. 4.8 + 6x 8= 52.8 dB

BT = vVI = 4x8 = 32 kHzTelephone signal is nonsinusoidal signal. Its signal to quantization noise ratio ~

iven by equation (2.6,25) as,c'--" - 4.8 + 6vN

Revie\'V Questions1. iN ith the help ,o f neat d iagrams, explain the transm itter and receiver o f pulse code

modulation.2, "What is unifo Jm ( linear) quantization ? .3. Expla in q uantiz atio n error and derive an ex pressio n f o rma.x1ml f1 f l signal to n01Se

ratio in PC M system th.at u se s line ar q uan tizatio n ..4. Derive the relations for signalling rate and tranem ission bandwid th in PC M system .S . W i1ai' is the necessity of nonuniform quantization and e xp lain. companding?

1 , . : 4 . 40 A.fB hard d isk is used to store PC t\1 data. The signal is sampled at 8 kllz and theencoded PCM is to haoe an twaage signal ta noise ratio of atleasi 30 db. For how numyminutes the PCM data can be stored on the har:d disk ? [Ans, : 133 min]


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BasebandilReception. Tecbniques----------~~----------------------3Here P~(mi Ix) indicates average probability of symbol error when 'x' is the

bservation vector and message m. is selected. To minimize the error Probabilityv ,E . '. ~(m i, x } given by above equation the optimum decision rule can be stated as,se t m = tru, i f Pim, sent I x) ?P (m k sent / x ) for all k - : : # i ... (4..3.3)

This decision rule can be represented graphically. Let tRr~denote. theN-climensional space of all possible vectors 'X'. And this region be partitionedinto'M' decision regions, R1/R2/ .. R,V1 . The decision rule can then bewritten aSIVector x lies in region R, if ln lfx(x /mk)] is maximum for k ;;;;; .:.(4.3.4)

Here ix(x Imk) is the likelihood functio~ which results when ~ s y m D o ] mk 'is. d 'T'L!~ 1" II d > l'kel'h d and th d i n - . d tansmitted, rrus rue IS cauer maxtmumtJ:, :00' ..... I .. e correspon_, .!g ..eector

hich uses this rille is called max imum likeliho od d etecto r. The decision rule of equation 4.3.4 can be further written alternately as,V ector x lies in region R , if II x - skll is m inimum for k= = i _.. (4.3.5)

Here II x - sk IIis the distance between the received signal point and the message .oint. The maximum likelihood decision rule chooses the message point closest to theceived signal point.

An example of maximum likelihood decision forN = .2 , and M = 4Fig. 4.3.2 shows the

regions andoundaries for N = = 2 and= 4 symbols. Observe thate region is ~1~ q,2 t.e,-dimensional. There areour regions R1F R 2,R 3 , and4 corresponding to four

ml fm2! m3 and' The decision boundaries

re shown by the dottedIf the observation

ector falls in region R 1en message m1is selected. Decis ion . boundar y be tween

H1 and R2Fig. ,4.3,.2 'Decjsion regions for N ;; 2


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4-4 Baseband Reception Techniques.4 Receiving Filter-Matched FUte,rMatched filter is used as a receiver for baseband and bandpass signals. Hence it is

so called receiving filter.

The matched filter is used for detection of signals in baseband and passbandtransmission. Fig. 4.4.1 shows the transmitted digital signal and received noisy signal. In

this figure observe that the transmitted signal sequence is 1 0 0 1 1. Thepulse is checked at the point "T ' of every bit period. Because of noise pulsepresent :in the third bit at the instant IT' of checking, it is detected in error,

Flg.4.4,.1 Error due to noise Requirements of detection receiver

i) Signal to noise ratio of the receiver must be improved.ii) The signal must be checked at the instant in hit periodr when signal to

noise ratio is maximum.iii) The error probability should be minimum.


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4-6 Baseband R.eception TechniquesTherefore the decision boundary will be midway between x01 (T) and xvrz (T). , It -is

iyen as,Decision boundary ~ xOl (T) + x02 (T)2 ... (4.4.1)

Review Question1. Exptain zchat is . ma tched filt er .

4.5 Error Rate due to Noise (Error Pr'obabi.lity).5.1 Errol Conditions in Matched Filter

Suppose that X2 (t) was transmitted, hut xOl (T) is greater than x02 (T). If noisenO (T) is positive and larger in magnitude than the voltage difference~ [ X 0 1 (T) + x,OQ (T)] - x02 (T), then jncorrect decision will be taken. i.e, errorwill be generated if,

(T) xO l (T) + X'02 (T) -. '. C . T ) .no: '; ;::.: 2' x02no (T ) > x 01 (T ) - x02 (T)2 ..,' (45.1)

.'t,

Similarly, let xl (t) is transmitted, but x02{T) is greater than xm (T). Thenincorrect decision will be taken if noise no(T) is less than~ [ X O l (T ) + X 0 2 ( T ) ] - X 0 1 ( T ) . i.e., errorwill be generated it

no (T) ~_xm (T); x02 (T)_ x :Q1 (T)x02 (T) - xOl (T)n o (TJ .~ ~__;._--- 2

The above two error conditions are summarized in thefollowing table :-r - (4. .5.2)

Input x(t) Valiueof "O(t}fo;r er:ro'rin the output P'robabi!lity of errorProbability of error will be obtained byevaluating, 'the pmbability that, . > X o { T ) - x O ( 2 ( r )no(T) - . 2

Pili'"obabi~mtyf error can be obtained byevaluaHngthe pmbabiiiity that, . - . ' X 0 2 ( T } - X 0 1 ( T )110JT)s- 2 '

Table 4.,5.1 : Error conditions :in a matcbed filter


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COlflmunication _ 436 Baseband Reception TechniquesReceiving IFilter-Correlation Receiv,elr

In this .section we will study a little different type of receiver which is calledrela tar. Fig. 4.8.1 shows the block diagram of this correlator.

Locally generateds~gna:1x;(t)

f(t)x(t)Input noisy, signa! -----II~f( t)""X(t)+I1(~)

iFig:.4.8.1 Block dia. 'gram of the correlat'orIn the adjacent figure' f (t) represents input noisy signal, i.e.;r f (t) ~ x (t) + n (t). The'f (t) is multiplied to the locally generated replica of input signal x (t)., This resultmultiplication f (t) . .r (t) is integrated. The output of the integrator is sampled at t =Tend of one symbol period). Then based on this sampled value, decision is made.is how the correlator works. It is called correiaior since it correlates the receio edf (t) with a stared replica of the known signal x (t). In the block diagram of above

igure, the product f (t) x (t) is integrated over one symbol period, i.e, T. Hence output(t) can bewritten as, T

r (t) :::; J It) x (t) dto

At t =T, the above equation will be,T

. Output of correlator ::r (T) :: f t (t) x {t)dtI 0

... ( 4.8 .1 )

Now let us consider thematched filter as shown inFig. 4..8.2 below.

In put n o~ s y_ .- - - t - I J I ' i ! I . M ~ ' ! I ~ . , . e h . : r . e dsi,gnali _- 1'!I~'tf(t), ,I h(t)In the above block diagram

t=:T observe that the matched filter does, r(I)~ r{T) not need locally generated replica ofinput signal x (f). The output of the

matched filter can be obtained byconvolution of input f (t) and itsimpulse response h ( t ) . . i.e.,F.ig. 4.8 ..2 'Block dilag!ram of a matched fitlerreceiiver


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COlmmunication 437 !B-asebandReception Techni,qu~es00'

r (t ) = f (t) * h (t)::: j f (1:)' h (t- 1 : ) d't ... {4.8.2)From equation (4.4,.3) we know that impulse response h (t) of the matched filter is

iven as,h (t) 2k .= - - - x(T ~t)No -- ... (4 .8 .3 )

2k _-h (t ~-'t) = N- x(T =]: + 1 : ' )1 . 0

Putting this value of h (t - 1 ; ) inequation (4.8.2) we get,() =]" f)2k - .r t= [i - ('t N o :t(T - t + 1 : " ) d't-00

Since the integration is performed over one bit period, we can change :integrationfrom 0 to T. i.e.,.

. 2k T _. ._.r (t) ::::: N - - f f (r) x (T - t+r)dto. fJAt t =T, the above equation will be, T(T') ' 2k [- f ( - ) - IT T ) dr -._= -, _. I __ "I 't X \ i - '.-+ 't, 'tNo 0Let us put 't ~ t just for convenience of notation,

i - 1 r! -. f hed '1 ' ()- Zk j : ' ~ ' f (-) () dOutput 0- matche _Iter: T.T -= N - _I ,t x tt-0 - n

2k T~ _ ~ f f (T ) X ('t) dta 0... (4.8.4)

This equation gives the output of matched filter. Observe that this equation and(4.8.1) (which gives output or correlator) are identical. Inabove equation the2k "hich b aliz d 1 Th - imil- ity b -N.. IS present v w . II i. cane nOrnl e to . -~.' e S I anr etween 0

uation (4.8.1) and equation (4.8.4) shows that the matched filter and correlator givesme output. Therefore we can state'}"

T he m atched filter and ' correlaior are two d istinct, independent tech niq ues w hich give theresult" These two techniques are used to synthesize the o ptimum flU eT .


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Ji' - _ C ' t"gi'k1 ,"ommunlca, lion 4 -39 !BasebandRe;ception Techn~quesAnd rj+ a- - {l - a ... (4 .9 ,2 )if"i f bk=lbk ~OAk The signal x (t) is then passed through the transmitting filter" The

transmitting filter combines all the necessary transmitting circuits andsystems. The combined transfer function of the transmitting filter is HT (1).The signal is then passed through the channel having the transfer functionHe (fl. The channel delivers the signal to the receiving filter. It consists of allthe necessary receiving circuits and systems. roe combinedtransfer functionof the receiving filter is HR if). The output ()f the receiving filter is y (t)., This~ ? / ] S ,noisy rep~ica of the transmitted Signal.x(t).,. . ' "The Signal y (t ) is sampled synchronouslywith the transrrutter. The samplinginstants are t ; ; ; ; ; ; i Tb. These sampling instants are synchronous to the dockpulses at the transmitter. The sampled signal y (fi)is then given to thedecision device. The decision device ,compares the input signal 'withthreshold' 'N . Then the decision is taken as follows ;

If Y (t i) > I . . select symbol '1'.If Y (f i)


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Communication . 4-40 Bas eb an d R ec eptio n T 'e ch niq ueEfHence equation (4.9.4) becomes,

! - l P{f) = H T (f) He (j)H R if)G if) ... ( 4.9.,6)The receiving filter output is sampled at ti = i Tb. . From equation (4.9.3), at t ~ i Tb we can write,

co

Y (tj) - !1 L Ak P ( i Tb - k Tb)k",,-oo

- .~ L Ak P [( i -k)Tb]k=-=

... (4..9.7)

Let us rearrange the above equation as follows :y(ti) ~ JlAi P (0) + 1 J 1 L Ak p[(i -k)Tb]

k=-r=k i i

... (4.9.8)

The first term represents the value of , Y (fi) 'when i ~ k : P ' (t) is normalized suchthat p (0)=1. Hence above equation becomes,

OQ

y(it) ~ 1 1Ai +~ L Ak P [C i -k)Tb]k ' = - - o obtl ... ( 4.9.9)And i= 0,+1, T2 ..31 ,

i) The first term in above equation is !lAi.' It is the contribution of the iihtransmitted bit.

1 1 ) The second term represents. the residual effect of all other bits transmittedbefore and after the sampling irtstant ti.efinition of lSIThe presence of outputs due to other bits (symbols) mterfere with the output of

equired bit (symbol). This effect is called Intersymbol Interference (lSI). Here note that .e have not considered the effect of channel noise. Actually, channel noise and IS!oth interfere the transmitted signal,

If the intersymbol interference is absent, then the second t.ermwill not bepresent in equation (4.9.7) i.e.,... (4.9.10)


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441At t:=i Tbl the correct bit is Ai' Observe that it is decoded correctly in absence of

._It is not possible to eliminate the second term of equation (4.9.'7) (and hence lSI)The lSI can be reduced by proper design of pulse spectrum G ( ,f) , transmit filter

T {f), receive filter HR C D and the channel He f!J. We will discuss some ofthesesues in subsequent sections.al andSystemDesignfor lSIElimination

ApriI/Mcy-2005, Nov./Dec.-200S

.10..1 Nyquist Putse Shaping: CriterionD'Qimain Criiterion

From equation 4.9,_9we know that the second term (summation) must be zero toeffect of lSI. This is possible if the received pulse p(t) is controlled such that," { 1 for i =kp [ ; ( i - k)T b . ] , = ". .r , ' 0 f o r ' 1 - ;t::k . .. (4.10 ..1 )

If p(t) satisfies the above condition}" then we get a signal which is free from lSI.from equation (4.9.10)

Hence equation (4.10.1) gives the condition for perfect reception in absence ofoise. Equation (4.10.1) is the condition in time domain, This condition gives moreseful criteria in frequency domain.riterion in Freqtlency Domain

. Let p ( nT b ) represent the impulses at which p(t) is sampled for decision.These samples ere taken at the rate of Tb. Fourier spectrum of these impulses, "15gIven as

ooP a . ( f ) = fb LP(f - nfb) . .. . (4 .10 .2 )This means the spectrums of p(t) are periodic with period fb' Here note that the

ampling frequency (instants) ish!" Here- P3(f) represents the spectrum of P(nTb) andis the spectrum of p(t}. We can thlnk of penT b) as the infinite length of impulses with period T b

which are weighted with amplitudes of p(t). i.e.,.. . ( 4 .10 .3 )

it:-=


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Communication Baseband Reception Techniques

4.10..1:: The output of a d igital computer is at a rate o f 64 kbp If theroll l o f f factors (Oa = 1 1 (ii) (l = = 0.5, (iii) a:= 0.2'5, (io) a = 0, find the bandwidthrequired to transmit the data in each case.

e Data rate is

.', Bandwidthf b - 64 kbpsB _ f b . . : . : : ~ 2 , r k b . pC ' . . .o 2 " _ ' . i

Equation 4.9.18 gives the bandwidth required using raised cosine spectrum i.e.,B ~ Bo( l+ n )

I) For ((.:~ I,bandwidth becomes,32xl03(1 +1) = 64 kHz

li) For Ct = 0.5:. 32x 103 (1 + 0,5)= 48 kHzHi)For a = 0.25 32x 103 (1 + 0.25) =40 kHziv) For Cl = - 0 18- EO ::: 32 kHzThus as rolloff factor decreases bandwidth also decreases .

B -B =B --

'1~attern Analysis The Eye Pattern :is used to study the effect of lSI in baseband digital

transmission ..

April/May- 2005

When the sequence is transmitted over a baseband binary dats transmissionsystem of Fig. 4.11.1 the signal obtained at the output Le, y (t ) is a continuoustime' signal as sho-wn in Fig. 4.11.1. Ideally this signal should go high andlow depending on the symbol that was transmitted. But because of thenature of transmission channel, the signal becomes continuous withincreasing and decreasing amplitudes. Fig.4.1Ll (a) shows the binarysequence that is transmitted and Fig. 4.11.1 (b) showsthe signal y (t) obtainedat the output Fig. 4.11.1 (b) also shows various sampling instants t1,t2tt3 ....etc: Thus based on the signal obtained over the penodT b between twosampling instants, decision is taken by the decision device. If we cut thesignal y (t) shown in Fig. 4.11.1. (b) in each interval (Tl;) and place it over oneanother, then we obtain the diagram as shown in Fig. 4.11.1 (c), Thisdiagram is caned Eye pattern of the signal y (t).


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C,ommlUnicaUon The name 'eye' is given because it looks like an eye'. This pattern can also be

obtained on eRO if we apply y (f ) to one of the input channels and apply anexternal trigger signal of 1/ TbHz" This makes one sweep of beam equal to~Tb'; seconds. Therefore the pattern shown in Fig. 4.11.1 (c) will be obtained.When there are large number of bits of the sequence,. then eye patterns willbe as shown in Fig. 4:.11.1 (d).

y(~)

(a)

(b)I tIIII~Its ~Samplinginstants ~ t1

(c) (d)1IIII TI . . . b~l~:1 'l

I I'~ 'T Il..b I! " '1I 'tI II II I

e ye o pe rn :~ n gdependsoneffeet( ) I f JS~

.Fig. 4.11.1 (a) Binary sequence, transmitted (b) Rece,iived signal by basebandtrarrsmtssion system (c) Eye pattem of s:igna,1n (b) (d) Eye pattern for largenumber of' blts in wav,eform yet).11.1 Performenee of the Data Transmission S,yst:emusing Eye ' Pattern

'. Various important conclusions can be derived from eye ' pattern. Fig. 4.11.2shows various points related to eye pattern,i) The width of the eye opening defines the interval over which thereceived wave can be sampled without error from intersymbol

interference, It is preferable to ' sample the instant at which eye IS openwidest The instant is shown as best sampling time in Fig. 4.11.2..


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-. jta~Communication 4,51 Baseband R.eception Techniques------~~~~~~-~~~-~~---'----------' . ._ii) The sensitivity of the system to timing error is determined by the rate of

closure of the eye as the sampling time is varied.i

Best sampling time Distortion atsamp~~ngimeSlope"" sensitivity totiming error

Margin overnoise

Tlrne lnterva! over whtch thewave can be sampled

Fig.. 4,.11.2 lnterpretatlon of the eye patterniii) The height of the eye opening, at the specified sampli.ng time, 1 S called

margin over the noise.

0.5

'1 .5 .1.0.

0.0 I----~.""."-..--,

-1.0.i-1.5 !

As the effect of inter svmbolinterference mcreases, the eyeopening reduces. If the eye is dosedcompletelvv then it is not possible toJ_) ...avoid errors in the output..

AU the above description is for twolevel (binary) system. If there arelv l-le v els r f\J\.'I-a rv system), then eve

... ..' ..!pattern contains (M -1) evJ.e openings~ . '. ~ Gstacked vert-ically one upon theother. Fig. 4.11.3 shows the eyediagram for 4 level (M = = 4) system.Therefore there are 3 eye openings.

o 1.5 2t(s)0 . 5 1ampling instantFig. 4.11.3 Eye, dia'gram for 4~levelsystem!, .1 2 E 'q uaUzing F Ut.e rsIn the baseband transmission system, channel noise and Intersymbol interference

t together. The optimum linear receiver can be used for such transmission system..1 2 .1 ,Zero FOlrc:ing EquaUz:erThe optimum linear receiver is realized with the help of zero forcing equalizer.

his equalizer forces the lSI to zero. Refer the block diagram of baseband dataansmission. system given is Fig. 4..9.1. TIle signal y(t) at the output of receiving filtergiven as'.... -- --I


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ccmmunlcatlen. . 4-52 Baseband Reception Techniques0tJ

y(t) - f c(t) x(t - r) dt.- DO .Here c(t) is impulse response of receive filter and x(t) is aninput signal to receive

lter ,.It is given as ..x (i) - LAkq(t-kTb)+w(f)

k=-=Here Ak. is the symbol transmitted at t= kTb and w(t) is the channel noise. If y(t) is

ampled at i Tbfy{iT b)= zi + ui

Here zi is the signal component and ni is the noise component. Let Ai be theransmitted symbol, Then error between transmitted and received symbol will be,

ei = y(iT b) - Ai

'M'SE" 1 E - [ 21"':! ... " :;;; " 2 .'.ei jThis mean square. error is minimumfor following condition".

t ; : o d f I . No . . 1 .!LRq(t-t:)+-20{t-'t')J c(-r) d't .~ q(-t)-00 .

JHere Rq(t,'t) is temporal autocorrelation function of the sequence q(kTb) ' Ab~v,e !quation can be solved to get c(t)- It gives impulse response of the zero forcing

qualization Techniquesecessity of Equali:zation ,VVhenthe signal is passed through the channel, distortion is introduced in terms.00amplitude and li) delay. This distortion creates the problems of lSI The detection of

he signal also becomes difficult. This distortion can be compensated with the help ofEqualizers are basically filters, which correct the channel distortion.

ig. 4.13.1 shows channel and equalizer for correction of distortion.


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.~ita!Co:mmlillicati:on 4 - 5,3 B,asebandReception TechniquesI - - IC'ha_'i ," ', , net i ' Eq1 uallze r 'x(t) " ,-"" " y{t)- - - . . . . . H e A f ) 1------1 Heq(f ) 1;----

Fi,g. 4.1.3.'1EquaUzer for correction of d~storUoniintroduced in the channel The transfer- function of distortionless system is given as,

HfJ ~ K e : j2nJto The cascade connection of channel + equalizer shown illabove figure will be

distortionless if,He(l) .,H eq({) :::. Ke-.f21tfto

.' Hence transfer function of the equalizer will be,Ke-j'l.nfto

J - leq (1)==, H c ( n . ... (4.l3.1)The, equation is difficult to realize directly" but approximations are available. It canimplemented with the help of tapped delay line filters"13.1 Tapped Delay Line FilterFig. 4.13.2 shows tapped delay line filter.Samp'ledinputs ignal lx(nT)

_.....-.--1 Delay , x{nT - T)T

De,layTI

x(,nT~2T),' x'~rnT-MlT+2T) 1'01 Ix{nT-MT+T). . .- ,,', ,~, , .eay... - , - _. - T bj- - - '

Fixedwefghts ''N1

1y(nIT)Fig. 4..13.,2 Tapped delay line fUter

The output of above filter is given as,A, :1~l

y{nT) = L wix(nT -iT)1=0

... (4.13.2)

Here Wi is theweight of ith tap.M is the total number of taps

and Tis the symbol duration of the signal,


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4 ..54 Baseband Recepti'on T-ecbnh:luesThe' weights are basically filter coefficients. This filter approximates the equalizer

transfer function of equation (4.13.1). The approximation "vHIbe more accurate if weuse more taps inthe filter. The weights are calculated once as per the characteristics ofcharmel.Hence this is fixed filter.4.13.2 Adaptiv,e EquaUzati!on -Ap,rilfMay- 2005

I

110st of the channels 2Je made up of individual links. For example, In theswitched telephone network, the distortion induced depends upon

" :I) transmission characteristics of individual links andii) number of links iruconnnectionHence, the fixed pair of transmit and receive filters will not serve the equalization,problem completely. The transmission characteristics of the channel keep on changing ..

Hence adaptioe equaliza t ion is used.Basic Princ~ple =

In adaptive equalization, the filters adapt themselves to the dispersive effects. ofthe cham ael. That is the coefficients of the-filters are changed continuously accordingto the received data, The filter coefficients are changed in such a way that thedistortion inthe data is reduced.Types, :

"hen an equalization is done at the transmitting side it is called prechannelequalization. This type of equalization requiresfeedhack to know the amount ofdistortion in the received data. When an equalization is done at the receiving side, ':- "i.s called postchannel equalization. In this easel no feedback is required. The equalizer .is placed after the receiving filter in the receiver.Block diagram IThe adaptive equalizer shown in figure below is 81 tapped-delay-line filter. "" jconsists of set of delay elements and variable multipliers. The sequence x(nT) !Sapplied to the input of the adaptive filter. The output y (nT) of the adaptive filter willbel

MY (nT) - L - W i x(nT -n;i"'" ... ' ( 4.13.3) "The weights ui, on the taps are basically adaptive filter coefficients,

sequence {d (nT)} is transmitted first. This seque.nce is known to the receivervThs 'response sequence , Y (nD is observed. As shown in Fig, 4.13.3, the error sequen "between the two sequences is calculated" i.e.,


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dc_unit_i_chitode

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