Control CDA 29 Octubre 2015 PDF

Preview

Summary

Control CDA 29 Octubre 2015...

Description

CDA SISTEL 29 d’octubre de 2015, 10:00h. Duració de l’examen 1:45 h. Dept. Teoria del Senyal i Comunicacions No es pot fer ús de cap dispositiu electrònic excepte calculadores no programables. Problem 1 Consider a communication channel in which we want to transmit the value of a binary random variable X in the presence of additive noise. The noise is modeled as a binary random variable Y independent from X. We denote by Z the output of the channel, Z=X+Y. The random variable X can take values 0 or 1 with equal probabilities:

⎧⎪ Pr(X = x) = p(x) = ⎨ ⎩⎪

1 2 1 2

x =1 x=0

The random variable Y can take values 0 or 1 with probabilities:

⎧⎪ p y =1 Pr(Y = y) = p(y) = ⎨ (1 − p) y = 0 ⎩⎪ 1.a) Obtain the entropy of the three random variables: H(X), H(Y), and H(Z). 1.b) Obtain the capacity of this channel as C = I(X; Z ) = H ( Z ) − H (Z | X) . Justify the result for p=0 and p=1/2. Now assume the random variable Y can take values 0 or 2 with probabilities:

⎧⎪ p y=2 Pr(Y = y) = p( y) = ⎨ , ⎪⎩ (1 − p) y = 0 1.c) Obtain again the entropy H(Z) and the channel capacity C = I(X; Z ) = H (Z ) − H (Z | X) and justify the result.

Hint: You may use the notation:

hb ( p) = −(1 − p )log 2 (1 − p) − p log 2 p Note that the entropy and the conditional entropy of random variables are defined as:

H (X) ≡ − ∑ p ( x ) log2 p( x ) x

H (X | Y = y) ≡ − ∑ p(x | Y = y) log2 p ( x | Y = y ) = −∑ p(x | y) log2 p ( x | y ) x

x

H (X | Y ) ≡ ∑ p ( y ) H (X | Y = y) = −∑ p ( y ) ∑ p(x | y) log2 p ( x | y) = y

= − ∑ p(x, y) log2 p( x | y) x,y

y

x

Problem 2 We consider a communication system in which two independent transmitters TX1 and TX2 wish to transmit to a single receiver RX by sharing a single frequency band of bandwidth B = 1MHz . The noise spectral −14 density of the receiver is N 0 = 10 J . The power of each of the transmitters is PdBm = 30dBm . The complex channel between TX1 and RX is c1 (t ) and between TX2 and RX is c2 (t) . We first consider that both channels are equal, deterministic and constant, c1 (t) = c2 (t) = a , and produce an attenuation adB = 20 log10 a = −60dB . Obtain the capacity of each of the channels when: 2.a) Each transmitter uses 50% of the frequency band all the time (FDMA). 2.b) Each transmitter uses all the frequency band but only 50% of the time (TDMA). 2.c) Each transmitter uses all the frequency band all the time so that each transmitter interferes with each other at the receiver. Assume the transmitted signal is zero mean Gaussian. In this case consider two cases: i) The receiver decodes first the signal from TX1. Then subtracts the decoded data from the received signal and then decodes the signal from TX2. ii) The received cannot perform interference cancellation and decodes both the signal from TX1 and TX2 from the received sequence directly. In the rest of the problem we continue with the case of 2.c) ii), that is, the received cannot perform interference cancellation and decodes both the signal from TX1 and TX2 from the received sequence directly. But we now consider that both channels are modeled as independent Rayleigh flat fading processes: 2

2

2

2

c1 (t) = a w1 (t)

; w1 (t) = w1 ∼ e− w1 ; w1 ≥ 0

c2 (t) = a w2 (t ) ; w2 (t ) = w2 ∼ e− w2 ; w2 ≥ 0 2.d) Provide an integral expression of the ergodic capacity for each channel. 2

2

2

2

2.f) Making the approximation, P a w1 + BN 0 ≈ P a w1 and P a w2 + BN 0 ≈ P a w2 , compute the 10% outage capacity for each channel.

Problem 1 1.a) H (X) = − 21 log2 12 − 21 log2

=1

1 2

H (Y ) = −(1 − p )log 2 (1 − p) − p log 2 p = hb ( p) H (Z ) = − (1−2p) log2

(1− p) 2

−

(

)

p 2

+ (1−2p) log2

(

p 2

)

p + (1−2p) − 2 log2

p 2

=

= 21 (−(1 − p )log2 (1 − p) + (1 − p) − p log 2 p + p ) + 21 = 12 hb ( p) +1 1.b) I(X; Z ) = H ( Z ) − H (Z | X) = 21 hb ( p) +1− H (Z | X); H (Z | X) = − ∑ p ( x ) ∑ p(z | x) log2 p ( z | x ) = z

x

p(x = 0)∑ p(z | x = 0) log2 p( z | x = 0) + p(x = 1)∑ p(z | x = 1) log2 p ( z | x = 1) = z

z 1 2

((1 − p)log2 (1 − p) + p log 2 p ) + ( (1 − p)log2 (1 − p) + p log 2 p ) = hb ( p) 1 2

I(X; Z ) = 1− 12 hb ( p) For p = 0 we have I(X; Z ) = 1, the channel is ideal. For p=1/2 we have I(X; Z ) = 12 ,when we receive 0 or 2 we know exactly what bit was transmitted, and we have total uncertainty when we receive 1. The probability of the first case is ½ and thus the capacity is 1/2 bit /channel use. 1.c) H (Z ) = − (1−2 p) log2 H (Z | X) = hb ( p)

(1− p) 2

p

− p2 log2 2 − (1−2 p) log2

p

(1− p) 2

− p2 log2 2 = hb ( p) +1

I(X; Z ) = 1 When we receive 0 or 2 we know for sure that a 0 was transmitted, and when we receive 1 or 3 we also know for sure that a 1 was transmitted. This is an ideal channel and the capacity is 1 bit/channel use. Problem 2 2.a) C1 = C2 = 21 B log2 (1+

2

a P 1 BN 0 2

−6

3

−6

3

) = 21 10 6 log2 (1+ 101101061010−14 ) = 3.82Mbps −3

2

2.b)

C1 = C2 = 21 B log2 (1+ 2.c)

2

a P BN 0

−3

10 ) = 3.33Mbps ) = 12 10 6 log2 (1+ 1010610 10 −14

i) 2

C1 = B log2 (1+

a P

−6

2

a P+BN 0

−3

3

10 ) = 10 6 log2 (1 + 10−610103 1010−3 +10 6 −14 ) = 0.992Mbps 10

2

−3

a P 10 C2 = B log2 (1 + BN 0 ) = 10 6 log2 (1 + 101010 6 −14 ) = 6.65Mbps 10 −6

3

ii) 2

C1 = C2 = B log2 (1+

a P 2

a P+BN 0

−3

∞

∞

P a w1 1 C1 = Ew1w2 ⎡ B log2 (1 + P aP2 wa +wBN )⎤ = B ∫ ∫ log2 (1 + P a 2 w + BN )e−w1 e−w2 dw1 dw2 = C2 2 0 2 0 ⎦ ⎣ w =0 w =0 2

2

2.d)

3

−6

10 ) = 10 6 log2 (1 + 10−6101031010−3 +10 6 −14 ) = 0.992Mbps 10

1

2

2.e) 2

C1 (w1 ,w2 ) = B log2 (1+

P a w1 2

P a w2 + BN 0

) ≈ B log2 (1 + ∞

Pout = Pr(

w1 w2

≤ γ out ) = Pr(w1 ≤ γ out w2 ) =

2

P a w1 2

P a w2

) = B log2 (1 +

=

∫

∫ ∫

∞ γ

fw1w2 (w1 ,w2 )dw1dw2 =

∞

e−w2 (1− e−γ out w2 )dw2 = 1−

w2 =0

C1−out = (1 − Pout )B log 2 (1 + γ

∫

w2 =0 out

)

γ out w2

w2 =0 w1 =0 ∞

w1 w2

e−(γ out +1)w2 dw2 =

out w2

∫ ∫

e−w1 e−w2 dw1 dw2 =

w2 =0 w1 =0

γ out γ out +1

) = 136,8KBps = C2−out

⇒

γ out =

1 Pout = 1− Pout 9...