Base Band ELE

8/13/2019 Base Band ELE

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P-1)

(a) what is the theoretical minimum system bandwidth needed for a 10-Mbits/s signal using

16-level PAM without ISI?

b) How large can the filter roll-off factor be if the allowable system bandwidth is 1.375 MHz?

(a)

16 levels = M = 2k

k = 4 bits / symbol

Rs = R /

4 / = 2.5 M symbols/s

Min BW = = 1.25 MHzb)

W= ( 1+r) Rs

1.375 M Hz = ( 1+r ) 1.25 M Hz

r = 0.1

P-2 )

A binary PCM system using polar NRZ signaling operates just above the error threshold with

an average probability of error equal to 10-6. Suppose that the signaling rate is doubled. Find

the new value of the average probability of error ?

In a binary PCM system, with NRZ signaling, the average probability of error is

Pe =erfc

The signal energy per bit is

Eb= A2Tb

Where A is the pulse amplitude and Tb is the bit ( pulse ) duration. If the signaling rate is

doubled, the bit duration is reduced by half. Correspondingly, Eb is reduced by half.

Let u = / , We may then setPe= 10-6=

erfc (u)

Solving for u, we get


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u= 3.3

When the signaling rate is doubled, the new value of Pe is

Pe =

erfc (

)

Pe =erfc (2.33)

Pe = 10-3

P-3 )

In a binary communication channel, the receiver detects binary pulses with an error

probability Pe. What is the probability that out of 100 received digits, no more than three

digits are in error?

Prob (no more than 3 error) = P(no error)+P(1 error) +P(2 error) +P(3 error),which is

P = ( 1- Pe)100+ 1001 Pe( 1- Pe)99+ 1002 Pe2( 1- Pe)98+ + 1003 Pe2( 1- Pe)97(3.5)= ( 1- 100Pe) + 100 Pe( 1- 99 Pe) + 4950 Pe2 ( 1- 98 Pe) + 161700 Pe3 ( 1- 97Pe)

P-4)

Suppose that baseband PCM data is to be sent using Polar NRZ Signaling defined by.

{() = () = The channel noise is additive, white and Gaussian with spectrum N/2.

a) Design the optimum matched filter receiver?

b) Sketch the impulse response of the optimum matched filter receiver in time domain?

c) Determine the optimum matched filter receiver output in time domain?

d) Design the correlation receiver which is equivalent to optimum matched filter

receiver?

e) Determine the probability of bit error for the optimum receiver if 0 and 1 are

equiprobable?

a)


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b)

c)

-A

t

Tb

0

q(t) = - p(t)

A

tTb0

p(t)

A

Tb0

h(t) = p(Tb- t)

t

Threshold

Device

PCM data

Additive White Gaussian

noisen(t)

Matched

Filter h(t)

t= nTb

y(nTb)

Threshold

Decision

decision = 1 , y(nTb) >

decision= 0 , y(nTb) <


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y(t) = ()( ) y(t) = ()( T b t)

y(Tb) = ()( T b t)

y(Tb) = ()() y(Tb) = A2 = A2Tb

d)

decision = 1 , y(nTb) >

decision= 0 , y(nTb) <

e)

Threshold

Device

PCM data

Additive white gaussian

Noise n t

t= nTb

y(nTb)

Threshold

Decision

0

bT

p(t)

y(t)

Tb

A2Tb

2Tb0 t


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Ep = Eq

2 / 2pE E = 0

Pb= Q + Epq= - p2(t) dt Pb= Q +()

Pb= Q 4)

Pb= Q ) Average energy per bit = Eb =

Ep+

Eq = Ep

Pb= Q )

P-5)

In coherent schemes, a small pilot is added for synchronization. Because the pilot does not

carry information, it causes degradation in Pb.Consider coherent PSK using the following

two pulses of duration Tbeach:

() = 1 cos sin () = 1 cos sin

Where sin is the pilot. Show that when the channel noise is white Gaussian,Pb= Q ()

2max= [ () ()] 2dt = +/


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The energy of P(t) is Tbtimes the power of p(t)

Hence, Ep=( )

Tb+( A2m2/2 ) Tb=

= Eb

Similarly , Eq=

= Eb

Epq= ()() = [ 2 (1-m2) Cos2wct + A2 m2Sin2wct] dt=

+ A2m2Tb

Hence,

2max =4 ( )

=8 ()

and

Pe = Q (max / 2) = Q ( ) P-6)

Find the probability of bit error, PB, for the coherent matched lter detection of the equally

likely binary FSK signals s1(t) = 0:5 cos 2000t ands2(t) = 0:5 cos 2020t where the two-

sided AWGN power spectral density is N0=2 = 0:0001. Assume that thesymbol duration is T

= 0:01 s?

Eb = ST = (0.5)2/2 = 0.00125 Joule

=

1()2() = 0.5 (2 1000)0.5 (2 1010) =

.. 0.5 [(21000) (2 2010)]

= 100 4 = [ 0.935 + 0.005 ] = 0.94

Pb= Q () = Q .(.). = Q (0.612 ) = 0.27

The error is much greater than if the tone spacing required for coherent orthogonal

signaling,

= 50 Hz had been used, instead of 10 Hz specified.


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Base Band ELE

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