1

ECE 480

Wireless Systems

Lecture 14

Problem Session

26 Apr 2006

2

Problem 3.1

Consider-ray channel consisting of a direct ray plus a ground – reflected ray where the transmitter is a fixed base station at height h and the receiver is mounted on a truck (also at a height, h. The truck starts next to the base station and moves away at velocity . Assume that signal attenuation on each path follows a free – space path – loss model. Find the time – varying channel impulse at the receiver tot transmitter – receiver separation d = t sufficiently large for the length of the reflected ray to be approximated by

hr r ' d

d

22

3

n

N tj t

n nn

c t , t e t

0

Solution

j t j tc ,t t e t t e t 0 1

0 0 1 1

LOS Reflectedh

r r ' dd

22

Equivalent low-pass channel impulse response

4

G

t d td

0 4

l

Parameters of the LOS term

t rG G Gl

j tt e t 0

0 0LOS

c Dt f t 00 02

dt d t

c 0

D Dtf t d t 0 0

2

5

D

vf t cos t

0 0

t 0 0

6

j tt e t 1

1 1

Parameters of the Reflected Term

l lR G R Gt d t

r r ' hd

d

1 24 24

R = Ground reflection coefficient

sin ZR

sin Z

c Dt f t 11 12

h

dr r ' dtc c

2

1

2

7

D Dtf t d t 1 1

2

D

vf t cos t

1 1

ht arctan

d 1

2

8

Problem 3.7

Suppose we have an application that requires a power outage probability of 0.01 for the threshold P 0 = - 80 dBm. For Rayleigh fading, what value of the average signal power is required?

r

x xP

zr

p x e e xP

2

22

2

1 10

2

Solution

. 0 01

r

r

P

Poutage

P

P

r

P e .

Pe . .

P

0

0

0

1 0 01

0 99 0 01

9

r

r

r

P P

log P log P d Bm

P d Bm dBm

0

10 10 0

100

10 10 100

20 80 60

10

Problem 3.9

This problem illustrates that the tails of the Rician distribution can be quite different than its Nakagami approximation. Plot the cumulative distribution function (cdf) of the Rician distribution for K = 1, 5, 10 and the corresponding Nakagami distribution with

Km

K

21

2 1

In general, does the Rician distribution or its Nakagami approximation have a larger outage probability p ( < x) for x large?

11

Solution

zR

r r r

z K K z K Kp z exp K I z z

P P P

2

0

2 1 1 12 0

For the Rician distribution:

z

zR zRF z p z0

For the Nakagami-m distribution

m m

zN mr r

m z mzp z exp m .

m P P

2 1 220 5

z

zN zNF z p z0

12

Problem 3.13

Derive a formula for the average length of time that a Rayleigh fading process with average power stays above a given target fade value P 0. Evaluate this average length of time for = 20 dB, P 0 = 25 dB and f D = 50 Hz.

rP

rP

Solution

For Rayleigh fading (K = 0):

Z Z DL below L above f e 2

2r

Z

P

Z = Target level = 25 dB

13

Z

Z

p z Zt above

L above

p z Z p z Z e e 2 2

1 1 1

Z Z DL below L above f e 2

2

Z

DD

et above

ff e

2

2

1

22r

Z

P

r

r r

P dB

r

P dB log P

P

10

20

10 10

10

10 10 100

14

Z dB

Z dB log Z

Z .

10

25

20 20

20

10 10 17 783

r

Z ..

P

17 7831 778

100

Z

D

t abovef

1

2

15

16

Problem 3.15

Consider the following channel scattering function obtained by sending a 900 MHz signal sinusoidal input into the channel:

S , Hz

. s . Hz

else

1

2

70

0 022 49 5

0

where 1 and 2 are determined by path loss, shadowing, and multipath fading. Clearly this scattering function corresponds to a two – ray model. Assume the transmitter and receiver used to send and receive the sinusoid are located 8 m above the ground.

17

a. Find the distance and velocity between the transmitter and receiver

b. For the distance computed in part (a), is the path loss as a function of distance proportional to d – 2 or d – 4?

c. Does a 30 KHz voice signal transmitted over this channel experience flat or rather frequency – selective fading?

18

Solution

8 m

d m

8 m

Distance traveled by LOS component = d

Distance traveled by first multipath component

dd

22

1 2 82

(a)

19

0.022 us

0 t0 t1

t

dd . .

22 6 82 8 0 022 10 3 10 6 6

2Square both sides

dd .

dd . d .

222

22 2 2

2 8 6 62

4 8 13 2 6 64

20

d d . d .

. d .

d . m

2 2256 13 2 43 56

13 2 212 44 0

16 1

DD

fv cosf v

cos

LOS component: = 0 o , f D = 70 Hz

c. m

f

8

6

3 100 3333

900 10

Df . mv .

cos s

70 0 333323 33

1

21

Multipath component: = 45 o , f D = 49.5 Hz

Df . . mv .

cos . s

49 5 0 333323 33

0 707

b. t rc

h hd . m

.

4 4 8 8

768 10 3333

d c >> d: power fall – off is proportional to d - 2

c. f = 30 khz . sBW ,

1 1

33 3330 000

mBW T 1

We have flat fading

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Problem 3.17

Let a scattering function S c ( , ) be nonzero over 0 0.1 ms and – 0.1 0.1 Hz. Assume that the power of the scattering function is approximately uniform over the range where it is nonzero.

a. What are the multipath spread and the Doppler spread of the channel?

b. Suppose you input to this channel two identical sinusoids separated in frequency by f. What is the minimum value of f for which the channel response to the first sinusoid is approximately independent of the channel response to the second sinusoid?

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c. For two sinusoidal inputs to the channel u 1 (t) = sin 2 f t and u 2 (t) = sin 2 f (t + t) find the minimum value of t for which the channel response to u 1 (t) is approximately independent of the channel response to u 2 (t)?

d. Will this channel exhibit flat fading or frequency – selective fading for a typical voice channel with a 3 KHz bandwidth? For a cellular channel with a 30 KHz bandwidth?

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a. What are the multipath spread and the Doppler spread of the channel?

Solution

T m range of definition 0.1 msec = 100 sec

d

. .BWB . Hz

0 1 0 10 1

2 2

25

b. Suppose you input to this channel two identical sinusoids separated in frequency by f. What is the minimum value of f for which the channel response to the first sinusoid is approximately independent of the channel response to the second sinusoid?

f 2 must be out of the bandwidth of f 1

m

BW KHzT . msec

1 1

100 1

f > 10 KHz for independence

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c. For two sinusoidal inputs to the channel u 1 (t) = sin 2 f t and u 2 (t) = sin 2 f (t + t) find the minimum value of t for which the channel response to u 1 (t) is approximately independent of the channel response to u 2 (t)?

d

t sB . Hz

1 1

100 1

d. Will this channel exhibit flat fading or frequency – selective fading for a typical voice channel with a 3 KHz bandwidth? For a cellular channel with a 30 KHz bandwidth?

3 kHz < B c : Flat fading

30 kHz > B c : Frequency selective fading

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Problem 4.2

Consider an AWGN channel with bandwidth 50 MHz, received signal power 10 mW, and noise PSD N 0 /2 where N 0 = 2 10 – 9 W/Hz. How much does capacity increase by doubling the received power? How much does capacity increase by doubling the channel bandwidth?

Solution

PC B log

N B

2

0

1

.C log

. mbps

62 9 6

0 0150 10 1

2 10 50 10

6 875

28

PC B log

N B

2

0

1

.C log

. mbps

62 9 6

0 0250 10 1

2 10 50 10

13 152

Double Power

P ..

N B 9 6

0

0 010 1

2 10 50 10

P ..

N B 9 6

0

0 020 2

2 10 50 10

29

.C log

. mbps

62 9 6

0 01100 10 1

2 10 100 10

7 039

Double Bandwidth

Noise level is increased as well

P ..

N B 9 6

0

0 010 05

2 10 100 10

30

Problem 4.4

Consider a flat fading channel of bandwidth 20 MHz and where, for a fixed transmit power , the received SNR is one of six values: 1 = 20 dB, 2 = 15 dB, 3 = 10 dB, 4 = 0.5 dB, 5 = 0 dB, and 6 = - 5 dB. The probabilities associated with each state are p 1 = p 6 = 0.1, p 2 = p 4 = 0.15, and p 3 = p 5 = 0.25. Assume that only the receiver has CSI.

a. Find the Shannon capacity of this channel

b. Plot the capacity vs. outage for 0 P out 1 and find the maximum average rate that can be correctly received (maximum C out)

P

31

Solution

i ii

C B log p

6

21

1

Must convert to power

dB

dB

SNR

SNR log SNR

SNR

10

10

10

10

1 = 100, 2 = 31.623, 3 = 10, 4 = 3.1623, 5 = 1, and 6 = 0.3162.

i ii

C B log p

6

21

1

32

i ii

C B log p

log . log . .

, , log . log . .

log . log . .

. mbps

6

21

2 2

2 2

2 2

1

1 100 0 1 1 31 623 0 15

20 000 000 1 10 0 25 1 3 1623 0 15

1 1 0 25 1 0 3162 0 1

5 766

a. Ergodic Capacity

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b. Capacity vs. Outage

out r minP P

o out minC p B log 21 1

min out odb p C 20 1 0

min out

min

o

d B db , p .

Assume dB

C . , , log

. mbps

2

15 20 0 9

20

1 0 9 20 000 000 1 100

13 29

34

1 = 100, 2 = 31.623, 3 = 10, 4 = 3.1623, 5 = 1, and 6 = 0.3162.

p 1 = p 6 = 0.1, p 2 = p 6 = 0.15, and p 3 = p 5 = 0.25.

min out

min

o

d B db , p .

Assume dB

C . , , log

. mbps

2

5 10 0 5

10

1 0 5 20 000 000 1 10

34 59

min out

min

o

d B db , p .

Assume dB

C . , , log .

. mbps

2

10 15 0 75

15

1 0 75 20 000 000 1 31 623

25 14

35

1 = 100, 2 = 31.623, 3 = 10, 4 = 3.1623, 5 = 1, and 6 = 0.3162.

p 1 = p 6 = 0.1, p 2 = p 6 = 0.15, and p 3 = p 5 = 0.25.

min out

min

o

d B db , p .

Assume dB

C . , , log

. mbps

2

5 0 0 1

0

1 0 9 20 000 000 1 1

18 00

min out

min

o

d B db , p .

Assume dB

C . , , log .

. mbps

2

0 5 0 35

5

1 0 35 20 000 000 1 3 1623

26 74

36

min out

min

o

db , p .

Assume dB

C . , , log .

. mbps

2

5 0 0

5

1 0 0 20 000 000 1 0 3162

7 93

37

Problem 4.5

Consider a flat fading channel in which, for a fixed transmit power , the received SNR is one of four values: 1 = 30 dB, 2 = 20 dB, 3 = 10 dB, and 4 = 0 dB, and the probabilities associated with each state are p 1 = 0.2, p 2 = 0.3, p 3 = 0.3, and p 4 = 0.2. Assume that both transmitter and receiver have CSI.

a. Find the optimal power adaptation policy for this channel and its corresponding Shannon capacity per unit Hertz (C/B)

b. Find the channel inversion power adaptation policy for this channel and associated zero – outage capacity per unit bandwidth

P

P i

P

38

c. Find the truncated channel inversion power adaptation policy for this channel and associated outage capacity per unit bandwidth for three different outage probabilities P out = 0.1, P out = 0.25, and P out (and the associated cutoff 0) equal to the value that achieves maximum outage capacity

Solution

a. Assume that all channels are used

Convert to magnitudes

i dB

i

1010

1 = 1000, 2 = 100, 3 = 10, and 4 = 1

39

ii i

. . . .p

.

..

4

10

0

1 1 0 2 0 3 0 3 0 21 1

1000 100 10 1

1 2332

10 8109

1 2332

..

0 4

1 1 1 11 2332 1 0

0 8109 1

Our assumption is valid

40

P i.

P .

1 11 2322

0 8109 1000 = 1000:

P i.

P .

1 11 2232

0 8109 100 = 100:

P i.

P .

1 11 1332

0 8109 10 = 10:

P i.

P .

1 10 2322

0 8109 1 = 1:

P i

P i

0

1 1

41

ii

i

Clog p

B

log . log . log . log .. . . .

. bps / Hz

4

21 0

2 2 2 2

1000 100 100 10 2 0 3 0 3 0 2

0 8109 0 8109 0 8109 0 8109

5 2853

b. Channel inversion power adaptation policy

E

1

1

P i

P i

i

i i

p . . . .E .

4

1

1 0 2 0 3 0 3 0 20 2332

1000 100 10 1

..

E

1 14 2882

0 23321

42

P i

P i

P i ..

P 4 2882

0 00431000

= 1000:

P i ..

P 4 2882

0 0429100

= 100:

P i ..

P 4 2882

0 428810

= 10:

P i ..

P 4 2882

4 28821 = 1:

43

Clog log .

Bbps

.Hz

2 21 1 4 2882

2 4028

c. Truncated power capacity

The control policy is the same as for the channel inversion case

outC P B log p

E

0

2 0

11

1

outP p 0

44

N

ij

i j i

pE

0 0

1

P out = 0.1: p . 0 0 9

ii i

. . .p

.

..

4

10

0

1 1 0 2 0 3 0 31 1

1000 100 10

1 0332

10 9679

1 0332

Must omit channel four : Otherwise, p . 0 1 0 9

45

i

i i

p . . .E .

0

3

1

1 0 2 0 3 0 30 0332

1000 100 10

Clog log . .

Bbps

.Hz

2 21 1 30 1205 0 8

3 9678

..

E

1 130 1205

0 03321

46

P out = 0.25: p . 0 0 75

ii i

. .p

.

..

4

10

0

1 1 0 2 0 31 1

1000 100

1 0032

10 9968

1 0332

Must omit channels 3 and 4 : Otherwise,

p . . 0 0 8 0 75

47

i

i i

p . .E .

0

3

1

1 0 2 0 30 0032

1000 100

Clog log . .

Bbps

.Hz

2 21 1 312 5 0 5

4 1462

..

E

1 1312 5

0 00321

Max value

48

Problem 4.7

Assume a Rayleigh fading channel where the transmitter have CSI and the distribution of the fading SNR p () is exponential with mean = 10 dB. Assume a channel bandwidth of 10 MHz.

a. Find the cutoff value and the corresponding power adaptation that achieves Shannon capacity on this channel

b. Compute the Shannon capacity of this channel

c. Compare your answer in part (b) with the channel capacity in AWGN with the same average SNR

d. Compare your answer in part (b) with the Shannon capacity when only the receiver knows [i]

0

49

e. Compare your answer in part (b) with the zero – outage capacity and outage capacity when the outage probability is 0.05

f. Repeat parts (b), (c), and (d) – that is, obtain the Shannon capacity with perfect transmitter and receiver side information, in AWGN for the same average power, and with just receiver side information – for the same fading distribution but with mean = - 5 dB. Describe the circumstances under which a fading channel has higher capacity than an AWGN channel with the same average SNR and explain why this behavior occurs.

50

Solution

a. Find the cutoff value and the corresponding power adaptation that achieves Shannon capacity on this channel

0

P

P

0

1 1

p d

00

1 11

For a Rayleigh fading channel, the probability p () is exponential

p e e

101 1

10

51

e d

ee d d

ee d

e expint

0

0 0

0

0

0

10

0

1010

0

10

0

10

010

0

1 1 11

10

1 11

10 10

1 11

10

1 11

10 10

52

clearclcx=[0.767:0.0001:0.768];y=x;n=length(x);for i=1:1:n y(i)=(1/(x(i)))*exp(-x(i)/10)-(1/10)*expint(x(i)/10);endyx y =

1.0009 1.0008 1.0006 1.0005 1.0003 1.0001 1.00000.9998 0.9997 0.9995 0.9994

x =0.7670 0.7671 0.7672 0.7673 0.7674 0.7675 0.76760.7677 0.7678 0.7679 0.7680

0 = 0.7676

53

C B log p d

0

20

b. Compute the Shannon capacity of this channel

.

Clog e d

B .

10

2

0 7676

1

10 0 7676

.

Clog e d . bps / Hz

B ln .

10

0 7676

12 9792

10 2 0 7676

54

clearclcx=[0.7676:0.01:1];n=length(x);for i=1:1:n I1(i)=log(x(i)/0.7676)*exp(-x(i)/10)*0.01;endSum1=sum(I1(i)) x1=[1:0.1:10];n1=length(x1);for i=1:1:n1 I2(i)=log(x1(i)/0.7676)*exp(-x1(i)/10)*0.1;endSum2=sum(I2(i))Sum = Sum1+Sum2

55

c. Compare your answer in part (b) with the channel capacity in AWGN with the same average SNR

Clog . bps / Hz

B 2 1 10 3 4594

d. Compare your answer in part (b) with the Shannon capacity when only the receiver knows [i]

Clog e d . bps / Hz

B

10

0

11 2 9070

10