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Dr. Uri Mahlab

1


Communication system Information source

Transmitter Transmitter

Channel

Receiver Receiver

Dr. Uri Mahlab

Decision

2


Block diagram of an Binary/M-ary signaling scheme Timing

Information

source

X)t )

Pulse generator

HT)f) XT)t ) Hc)f) Trans filter

channel Channel noise

Output

A/D

Y(t)

Clock recovery network

Dr. Uri Mahlab

n)t)

+

+

Receiver filter

HR)f)

3


Block diagram Description pg (t)

{dk}={1,1,1,1,0,0,1,1,0,0,0,1,1,1}

Tb

Timing X)t )

Information

source

Pulse generator

Tb

HT)f)

XT)t)

Trans filter

pg (t ) For bk=1

Tb

For bk=0

Tb

+ a ak =  − a Dr. Uri Mahlab

if d k = "1"; if d k = "0"; 4


Block diagram Description )Continue - 1) {dk}={1,1,1,1,0,0,1,1,0,0,0,1,1,1}

Tb

Timing X)t )

Information

source

Pulse generator

pg (t ) For bk=1 For bk=0

HT)f)

XT)t)

Trans filter

pg (t ) Tb Tb

Transmitter filter Dr. Uri Mahlab

Tb Tb 5


Block diagram Description )Continue - 2) {dk}={1,1,1,1,0,0,1,1,0,0,0,1,1,1}

Tb

Timing

HT)f)

X)t ) Information

source

100110

Trans filter

Pulse generator

X (t ) =

∑a

k =−∞

Tb

k

p g (t − kTb )

2Tb

5Tb 3Tb

XT)t)

4Tb

Dr. Uri Mahlab

6Tb

t 6


Block diagram Description )Continue - 3) {dk}={1,1,1,1,0,0,1,1,0,0,0,1,1,1}

Tb

X (t ) =

∑a

k =−∞

Timing

100110

source Tb

Tb

2Tb

5Tb

6Tb

t

5Tb

6Tb

t

4Tb

2Tb 3Tb

XT)t)

Trans filter

Pulse generator

3Tb

p g (t − kTb )

HT)f)

X)t ) Information

k

4Tb

Dr. Uri Mahlab

7


Block diagram Description )Continue - 4) Timing X)t ) Information

source

HT)f) Trans filter

Pulse generator

HR)f) +

Receiver filter

Channel noise n)t) Tb

2Tb 3Tb

Tb

6Tb

t

5Tb

6Tb

t

4Tb

2Tb 3Tb

5Tb

4Tb

t Dr. Uri Mahlab

8


Block diagram Description )Continue - 5) Tb

2Tb 3Tb

Tb

6Tb

t

5Tb

6Tb

t

4Tb

2Tb 3Tb

5Tb

4Tb

t

Y (t ) = ∑ Ak pr (t − t d − kTb ) + n0 (t ) k

Dr. Uri Mahlab

9


Block diagram of an Binary/M-ary signaling scheme Timing

Information

source

X)T )

Pulse generator

HT)f) Trans filter

Xt)T)

Hc)f) channel Channel noise

A/D

Output

Y(t)

+

+

n)t)

Receiver filter

HR)f) Clock recovery network

Y (t ) = ∑ Ak pr (t − t d − kTb ) + n0 (t ) k

Dr. Uri Mahlab

10


Block diagram Description Tb

2Tb 3Tb

Tb

5Tb

6Tb

5Tb

6Tb

4Tb

2Tb 3Tb

4Tb

t t t t

1

0

0

0

1

0

1

0

0

1

1

0

Dr. Uri Mahlab

11


Block diagram of an Binary/M-ary signaling scheme Timing

Information

source

X)t )

Pulse generator

HT)f) XT)t ) Hc)f) Trans filter

channel Channel noise

Output

A/D

Y(t)

Clock recovery network

Dr. Uri Mahlab

n)t)

+

+

Receiver filter

HR)f)

12


Explanation of Pr)t) Pg)t)

HT)f)

Hc)f)

Trans filter

channel

HR)f)

Pr)t)

Receiver filter

Y (t ) = ∑ Ak pr (t − t d − kTb ) + n0 (t ) k

Pg)f)

HT)f) Hc)f) HR)f) Dr. Uri Mahlab

Pr)f) p r (0) =131


The output of the pulse generator X)t),is given by

X( t ) =

∑a p (t −k T )

k =−∞

k

g

b

Pg)t) is the basic pulse whose amplitude ak depends on .the kth input bit

Dr. Uri Mahlab

14


The input to the A/D converter is

Y (t ) = ∑ Ak pr (t − t d − kTb ) + n0 (t ) k

For tm =mTb+td and td is the total time delay in the system, we get.

t t

Y( t m )

t2 t1

t3 tm

Dr. Uri Mahlab

t 15


The output of the A/D converter at the sampling time

tm =mTb+td Y (t ) = ∑ Ak pr (t − t d − kTb ) + n0 (t ) k

Y ( t m ) = A m + ∑A k p K ≠m

Tb

Y( t m )

b

2Tb

5Tb 3Tb

t2 t1

r

[( m −k ) T ] + n ( t ) 4Tb

t3 tm

Dr. Uri Mahlab

0

6Tb

m

t t

16


Y ( t m ) = A m + ∑A k p K ≠m

r

[( m −k ) T ] + n ( t ) b

0

m

ISI - Inter Symbol Interference

Y( t m )

t2 t1

t3 tm

Dr. Uri Mahlab

t

17


Explanation of ISI HT)f) Hc)f) HR)f)

Pg)t)

Trans filter

channel

Receiver filter

t

t Fourier Transform

Pr)t)

Fourier Transform

BandPass Filter

f

f

Pg)f)

HT)f) Trans filter

Hc)f) channel

Dr. Uri

HR)f) Receiver Mahlabfilter

Pr)f) 18


Explanation of ISI - Continue t

Fourier Transform

BandPass Filter

Fourier Transform

t

f

f

Tb

2Tb

5Tb 3Tb

4Tb

Dr. Uri Mahlab

6Tb

t

19


-The pulse generator output is a pulse waveform

X (t ) =

∑a

k =−∞

k

p g (t − kTb )

p g ( 0) = 1

ak

a If kth input bit is 1  =  − a if kth input bit is 0 

-The A/D converter input Y(t)

Y (t ) = ∑ Ak pr (t − t d − kTb ) + n0 (t ) k

Dr. Uri Mahlab

20


T y p e o f Im p o rta n t P a ra m e te rs In v o l v e d In T h e D e s ig n O f a P A M S y s te m D a ta ra te

E rro r ra te

T ra n s m i t te d power

N o is e power

Dr. Uri Mahlab

N o is e S p e c tra l d e n s it y

S y s te m c o m p le x it y

21


5.2 BASEBAND BINARY PAM SYSTEMS D e s ig n o f a b a s e b a n d b in a r y P A M s y s t e m P u ls e s h a p e s p g (t)

p r( t )

H R (f)

H T(t)

- minimize the combined effects of inter symbol interference and noise in order to achieve minimum probability of error for given data rate. Dr. Uri Mahlab

22


5.2.1

Baseband pulse shaping

The ISI can be eliminated by proper choice of received pulse shape pr )t).

1 for n = 0  p r ( nTb ) =   0 for n ≠ 0 Doe’s not Uniquely Specify Pr)t) for all values of t.

Dr. Uri Mahlab

23


To meet the constraint, Fourier Transform Pr(f) of Pr(t), should satisfy a simple condition given by the following theorem

Theorem

if

k Pr (f + ) = Tb for ∑ Tb k =−∞

Then

1 for n = 0 p r ( nTb ) =  0 for n ≠ 0

Proof p r (t) =

∫p

r

f <1 / 2Tb

(f ) exp( j2πft )df

−∞

p r ( t ) = ∑k =−∞ ∞

( 2 k +1) / 2 Tb

∫p

(f ) exp( j2πft )df

r (2k− 1) / 2Uri Tb Dr.

Mahlab

24


( 2 k +1) / 2Tb

pr ( nTb ) =∑ k

p r ( nTb ) = ∑ k p r ( nTb ) =

r ( 2 k −1) / 2Tb

1 / 2 Tb

−1 / 2 Tb

1 / 2 Tb

∫p

−1 / 2 Tb

( f ) exp( j 2πfnTb t ) df

p r (f '+

k ) exp( j2πf ' nTb )df ' Tb

(∑ p r (f + k

k )) exp( j2πfnTb )df Tb

1 / 2 Tb

sin( nπ) p r ( nTb ) = ∫ Tb exp( j2πfnTb )df = nπ 1 / 2 Tb

Which verify that the Pr(t) with a transform Pr(f) Satisfy ZERO ISI Dr. Uri Mahlab

25


The condition for removal of ISI given in the theorem is called Nyquist (Pulse Shaping) Criterion

Y ( t m ) = A m + ∑ A k Pr ((m − k )Tb ) + n 0 ( t m ) k ≠m

1 for n = 0  p r ( nTb ) =   0 for n ≠ 0

1

-2Tb -Tb

Tb

2Tb

sin( nπ) p r ( nTb ) = nπ

Dr. Uri Mahlab

26


The Theorem gives a condition for the removal of ISI using a Pr(f) with a bandwidth larger then rb/2/. ISI canâ&#x20AC;&#x2122;t be removed if the bandwidth of Pr(f) is less then rb/2.

Pg)f) Tb

Pr)f)

HT)f) Hc)f) HR)f)

2Tb

5Tb 3Tb

4Tb

Dr. Uri Mahlab

6Tb

t

27


Particular choice of Pr(t) for a given application

p r( t ) R a te o f d e c a y o f p r( t )

S h a p i n g f i lt e r s

The smallest values near Tb, 2Tb, â&#x20AC;Ś In such that timing error (Jitter) will not Cause large ISI

Shape of Pr(f) determines the ease with which shaping filters can be realized.

Dr. Uri Mahlab

28


A Pr(f) with a smooth roll - off characteristics is preferable over one with arbitrarily sharp cut off characteristics. Pr(f)

Pr(f)

Dr. Uri Mahlab

29


In practical systems where the bandwidth available for transmitting data at a rate of rb bits\sec is between rb\2 to rb Hz, a class of pr(t) with a raised cosine frequency characteristic is most commonly used. A raise Cosine Frequency spectrum consist of a flat amplitude portion and a roll off portion that has a sinusoidal form.

Tb , f ≤ rb / 2 −β  rb rb rb  2 π Pr (f ) = Tb cos ( f − +β), −β < f ≤ +β 4β 2 2 2  0, f < rb / 2 +β  FT −1 {Pr (f )} =

cos 2πβt = Pr ( t ) = 1 − ( 4βt ) 2

 sin πrb t    πr t   b  

Dr. Uri Mahlab

30


raised cosine frequency characteristic

Dr. Uri Mahlab

31


Summary The BW occupied by the pulse spectrum is B=rb/2+β . The minimum value of B is rb/2 and the maximum value is rb. Larger values of β imply that more bandwidth is required for a given bit rate, however it lead for faster decaying pulses, which means that synchronization will be less critical and will not cause large ISI. β =rb/2 leads to a pulse shape with two convenient properties. The half amplitude pulse width is equal to Tb, and there are zero crossings at t=3/2Tb, 5/2Tb…. In addition to the zero crossing at Tb, 2Tb, 3Tb,…... Dr. Uri Mahlab

32


5.2.2

Optimum transmitting and receiving filters The transmitting and receiving filters are chosen to provide a proper

H T ,H p u ls e s h a p i n g

R

n o is e im m u n ity Dr. Uri Mahlab

33


-One of design constraints that we have for selecting the filters is the relationship between the Fourier transform of pr(t) and pg(t).

p g ( f ) H T ( f ) H R ( f ) = K c Pr ( f ) exp(â&#x2C6;&#x2019;2 j 2Ď&#x20AC;ft d ) Where td, is the time delay Kc normalizing constant. In order to design optimum filter Ht(f) & Hr(f), we will assume that Pr(f), Hc(f) and Pg(f) are known.

Portion of a baseband PAM system

Dr. Uri Mahlab

34


Pg)f)

HT)f) Hc)f) HR)f)

Pr)f)

If we choose Pr(t) {Pr(f)} to produce Zero ISI we are left only to be concerned with noise immunity, that is will choose

{ H T (f )}

and { H R (f )} â&#x2021;&#x2019; minimum of noise effects

Dr. Uri Mahlab

35


Noise Immunity Problem definition:

For a given : •Data Rate - rb •Transmission power - ST •Noise power Spectral Density - Gn(f) •Channel transfer function - Hc(f) •Raised cosine pulse - Pr(f) Choose

{ H T (f )}

and { H R (f )} ⇒ minimum of noise effects Dr. Uri Mahlab

36


Error probability Calculations At the m-th sampling time the input to the A/D is:

Y ( t m ) = A m + ∑ A k Pr ((m − k )Tb ) + n 0 ( t m ) k ≠m

We decide:

"1" if

Y( t m ) > 0

"0" if

Y( t m ) ≤ 0

Perror = Pr ob[ Y( t m ) > 0 "0" was sent ] • Pr ob( To sent "0" ) + Pr ob[ Y( t m ) < 0 "1" was sent ] • Pr ob( To sent "1" ) Dr. Uri Mahlab

37


Y( t m ) = + A + n0 ( t m ) if

"1"

Y( t m ) = −A + n0 ( t m ) if

"0"

A=aKc

• Pr ob( To sent "0" ) = Pr ob( To sent "1" ) = 0.5

Perror

1 = { Pr ob[ n 0 ( t m ) < −A ] + Pr ob[ n 0 ( t m ) > A ]} 2

The noise is assumed to be zero mean Gaussian at the receiver input then the output should also be Zero mean Gaussian with variance No given by: ∞

N 0 = ∫ G n (f ) H R (f ) df −∞

2

Dr. Uri Mahlab

38


1 (e − [ n ] 2 / 2 N0 ) 2π N 0

0

1 (e − [ n − A ) ] 2 / 2 N0 ) 2π N 0

A ∞

Perror = Pr ob[ n > b] = ∫ b

y(tm)

1 ( −[ z ) ] 2 / 2 N 0 ) e dz 2 πN 0

b

Dr. Uri Mahlab

39


1 (e − [ y ( t m )+ A ) ] 2 / 2 N0 ) 2π N 0

Perror

1 (e − [ y ( t m )− A ) ] 2 / 2 N0 ) 2π N 0

-A = Pr ob[ Y( t m ) > 0]

A

y(tm) Pr ob[ Y( t m ) < 0]

0 Dr. Uri Mahlab

y(tm) 40


1 (e − [ y ( t m )+ A ) ] 2 / 2 N0 ) 2π N 0

-A

1 (e − [ y ( t m )− A ) ] 2 / 2 N0 ) 2π N 0

y(tm)

A

VTransmit

Vreceived Dr. Uri Mahlab

41


Pe = 1 / 2

(

x >A ∞

=∫ A

Pe =

)

1 2 exp − x / 2 N 0 dx = 2πN 0

(

)

1 2 exp − x / 2 N 0 dx 2πN 0 ∞

A / N0 ∞

Q( u ) = ∫ u

z=x / N0

=

  1 A  exp − z 2 / 2 dz =Q  N  2π 0  

(

(

)

)

1 exp − z 2 / 2 dz 2π Dr. Uri Mahlab

42


Q( u ) = ∫ u

(

)

Q(u)

1 exp − z 2 / 2 dz 2π

=∫

dz=

u

Pe =

A / N0 ∞

Q( u ) = ∫ u

U  A  1 2  exp − z / 2 dz =Q  N  2π 0  

(

(

)

)

1 2 exp − z / 2 dz 2π

Dr. Uri Mahlab

43


A = Signal to Noise Ratio N0

Perror decreases asA / N 0

increase

Hence we need to maximize the signal to noise Ratio Thus for maximum noise immunity the filter transfer functions H T(f) and HR(f) must be xhosen to maximize the SNR

Dr. Uri Mahlab

44


Optimum filters design calculations We will express the SNR in terms of HT(f) and HR(f) We will start with the signal: ∞

X (t ) =

∑a

k =−∞

G X (f ) =

p g (f ) Tb

k

p g (t − kTb )

{ }=

E ak

2

2

a p g (f )

2

Tb

The psd of the transmitted signal is given by:: 2

G X (f ) = H T (f ) ⋅ G X ( f )

Dr. Uri Mahlab

45


And the average transmitted power ST is

a2 ST = Tb

∫ P (f )

2

g

2

⋅ H T (f ) df

−∞

A2 = ST = 2 K c Tb

2

A k =K c a k A =K c a

=

∫ Pg (f ) ⋅ H T (f ) df 2

−∞

2

A2 =

ST K c Tb

∫ P (f ) g

2

2

⋅ H T (f ) df

−∞

The average output noise power of n0(t) is given by:

No =

∫G

2

n

(f ) H R (f ) df

−∞

Dr. Uri Mahlab

46


The SNR we need to maximize is ST Tb A2 = 2 ∞ No  ∞  Pr (f ) 2 df   ∫ G n (f ) H R (f ) df ⋅ ∫ 2 −∞  − ∞ H c (f ) H R ( f ) where Pr (f ) = H c (f )H R (f )H T (f )

Or we need to minimize 2 ∞  ∞   Pr (f ) 2 2 min  ∫ G n (f ) H R (f ) df ⋅ ∫ df   = min{ γ } 2   −∞ − ∞ H c (f ) H R ( f )

Dr. Uri Mahlab

47


Using Schwartz’s inequality ∞

∫ V(f ) df ⋅

−∞

2

∫ W (f ) df ≥ 2

−∞

2

∫ V(f ) W (f )df

−∞

The minimum of the left side equaity is reached when

V(f)=const*W(f) If we choose :

V (f ) = H R (f ) G n W (f ) =

1/ 2

(f )

Pr (f ) H R ( f ) H c (f )

Dr. Uri Mahlab

48


Îł2 is minimized when K Pr (f )

2

H R (f ) = 2

H T (f ) =

H c (f ) G n 2

1/ 2

(f )

K c Pr (f ) G n

1/ 2

(f )

K H c (f ) Pg (f )

2

K â&#x2C6;&#x2019; an arbitrary positive constant The filter should have alinear phase response in a total time delay of td

Dr. Uri Mahlab

49


Finally we obtain the maximum value of the SNR to be:

 A2  ST Tb   = 2 1 / 2 ∞  N o  max  Pr (f ) G n (f )  df  ∫ H c (f ) −∞ 

Perror

 = Q  

 A 2      N  o  max  Dr. Uri Mahlab

50


For AWGN with

G n (f ) = η / 2

and pg(f) is chosen such that it does not change much over the bandwidth of interest we get.

H R (f )

2

H T (f )

2

= K1 =K 2

p r (f ) H c (f ) p r (f ) H c (f )

Rectangular pulse can be used at the input of HT(f).

 1 for t <τ/ 2; τ< <Tb p g ( t ) = 0 elsewhere  Dr. Uri Mahlab

51


5.2.3 Design procedure and Example The steps involved in the design procedure. Example:Design a binary baseband PAM system to transmit data at a a bit rate of 3600 bits/sec with a bit error probability less than 10 −4. The channel response is given by: 10 −2 for f  2400 Hc ( f ) =  0 _ elewhere The noise spectral density is Gn ( f ) = 10 −14 watt / Hz Dr. Uri Mahlab

52


Solution: rb = 3600bits / sec pe ≤ 10 −4 B = 2400 Hz Gn ( f ) = 10 −4 watt / Hz

If we choose a braised cosine pulse spectrum with β = rb / 6 = 600  1  3600 , f  1200  1 π pr ( f ) =  cos 2 ( f − 1200),1200 ≤ f  2400  3600 2400   0, f ≥ 2400 Dr. Uri Mahlab 53


We choose a pg(t)

1, t  1200 p g (t ) =  0, elsewhere;τ = Tb / 10 = (0.28)(10 − 4 ) sin πfτ pg ( f ) = τ ( ) πfτ p g (0) = τ , p g (2400) = 0.973τ ≈ τ ⇓

H T ( f ) = K1 p r ( f ) H R ( f ) = Pr ( f )

1/ 2

1/ 2

We choose K1 = (3600)(103 ) ⇓ p g ( f ) H T ( f ) Dr. H c (Uri f ) Mahlab H R ( f ) = pr ( f )

54


Plots of Pg(f),Hc(f),HT(f),HR(f),and Pr(f).

Dr. Uri Mahlab

55


−4 P ≤ 10 To maintain a e

( A2 / N 0 ) max Q( ( A2 / N 0) max ≤ 10 − 4 ( A2 / N 0 ) max ≥ 3.75 ( A2 / N 0 ) max ≥ 14.06 ⇓ 2

 Pr ( f ) G ( f )    10 −14   ∞ 1 A ST = ( ) max  ∫ df  = (3600)(14.06) − 4   ∫ Pr ( f )df  Tb N 0 Hc ( f )  −∞   10  −∞  2

1/ 2 n

For Pr(f) with raised cosine shape

∫ P ( f ) df r

=1

−∞

And hence ST = (14.06)(3600)(10 −10 ) ≈ −23dBm Dr. Uri Mahlab Which completes the design.

56

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