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(1)

Small-Scale Fading II

(and basics about random processes)

PROF. MICHAEL TSAI 2015/4/24

(2)

Random processes

2

t

t

()

()

t

()

()

One realization of X(t)

( ) is a random variable:

( ) 

(3)

Joint CDF for a random process

• If we sample X(t) at times , … , we can have a joint cdf of samples at those times:

3

  ,…,  , … ,  =    ≤ ,   ≤ , … ,   ≤ 

t

X(t)      

(4)

Stationary Random

Process (Strict-sense)

• A random process X(t) is stationary if for all T, all n, and all sets of sample times , , … ,  we have:

4

   ≤ ,   ≤ , … ,   ≤  =

   +  ≤ ,   +  ≤ , … ,   +  ≤ 

If time shifts does not matter, then it is stationary

(5)

Mean (First Moment)

5

t

t

()

()

t

()

[  ]

Averaging over all realizations

[  ]

 

(6)

Autocorrelation (Second Moment)

• “How similar a random process and a shifted version of itself is”

• Autocorrelation of a random process is defined as:

6

! ,  + ≜      +

t

t

#( + )

$()

Shifted by

×

All possible combinations of realizations

! ,  + For a particular t

=

(7)

For stationary random processes…

• Mean

• Autocorrelation

7

   =    −  =   0 = (

Constant. Does not change with t.

! ,  + =    −    + −  = [ 0  ] ≜ ! ( )

(8)

Two random processes

• Two random processes X(t) and Y(t) defined on the same underlying probability space have a joint cdf:

for all possible sets of sample times , … ,  and ) , … , *) .

• Two random processes are independent if we have

8

  ,…,  + , ,…,+ -, , … , , ., … , ./ =

   ≤ , … ,   ≤ , 0 ) ≤ ., … , 0 /) ≤ ./

Similar to how you can define a joint cdf for two random variables

  ,…,  + , ,…,+ -, , … , , ., … , ./ =

   ≤ ,   ≤ , … ,   ≤   0 ) ≤ ., 0 ) ≤ ., … , 0 ) ≤ .

(9)

Cross-correlation

• The cross-correlation between two random processes X(t) and Y(t) is defined as

• Two random processes are uncorrelated if

for all t and 1.

• If both random processes are stationary, we have

9

! + ,  + ≜    0  +

   0  + =     0  +

! + ,  + =    0  + =   0 0 = !23

(10)

Wide-Sense Stationary (WSS)

• A process is wide-sense stationary if

and

• 45(1) has its maximum value at 1 = .

10

   = (

! ,  + =      + = !

! ≤ ! 0 =   

A random process is always “the most similar” to the version of itself without shifting.

(11)

Ergodicity

11

t

t

()

()

t

()

()

Expectation value over time is the same as expectation over all possible realizations

 .

$ .

(12)

Power Spectral Density

• The power spectral density of a WSS process is

defined as the Fourier transform of its autocorrelation function with respect to 1:

• PSD takes its name from the fact that the expected power of a random process X(t) is the integral of its PSD:

12

8 = 9 !A exp −=2? @

BA

   = ! 0 = 9 8A

BA

(13)

Gaussian random processes

• A random process X(t) is a Gaussian process if, for all values of T and all functions g(t), the random variable

has a Gaussian distribution.

• We usually use this to model the noise for a communication receiver.

• Mean & variance:

13

C = 9 D    @E

 Linear combination of samples

 C = 9 D     @E



FGH C = 9 9 D  D I     I @ @IE

 − [C]

E



(14)

Gaussian random processes

• Samples of a random process, 5 J , J = , … , , are jointly Gaussian random variables, if we let K = L − J .

14

C = 9 D    @E



C = 9 M  − E $   @

 =  $

(15)

Example: white noise

• White noise is a zero-mean WSS random process with a PSD that is constant over all frequencies.

• N is often called as one-sided white noise PSD.

• By inverse Fourier transform, the autocorrelation can be obtained:

15

   = 0 8 = O for some constant N

! = N

2 M

White noise is not correlated with any shifted version of itself.

(Not similar at all after ANY time period)

(16)

16

t

t0

τ0 τ

1 τ

τ 3

2 τ

4 τ

5 τ

6 τ(t0) τ(t1) t1

t2

τ(t2) t3

τ(t3)

hb(t,τ)

Two main aspects

of the wireless

channel

(17)

Doppler Effect

• Difference in path lengths PQ = R STUV = WPX YZ[V

• Phase change P\ = ]^PQ_ = ]^WPX_ YZ[V

• Frequency change, or Doppler shift,

`R = 

]^ P\

PX = W

_ STUV

17

(18)

Example

Consider a transmitter which radiates a sinusoidal carrier frequency of 1850 MHz. For a vehicle moving 60 mph, compute the received carrier frequency if the mobile is moving

1. directly toward the transmitter.

2. directly away from the transmitter

3. in a direction which is perpendicular to the direction of arrival of the transmitted signal.

Ans:

Wavelength=a = cb

d = fg××e h = 0.162 (k)

Vehicle speed l = 60km = 26.82 /o 1. p = q.f.qcos 0 = 160 uv

2. p = q.f.qcos ? = −160 (uv)

3. Since cos w = 0, there is no Doppler shift!

18

`R = 

]^ P\

PX = W

_ STUV

(19)

Doppler Effect

• If the car (mobile) is moving toward the direction of the arriving wave, the Doppler shift is positive

• Different Doppler shifts if different V (incoming angle)

• Multi-path: all different angles

• Many Doppler shifts  Doppler spread

19

(20)

Narrow-band Fading Model

• Sending an unmodulated carrier wave with random phase offset \:

• Received signal becomes

20

I  = xy{exp = 2?b + { } = cos 2?b + {

H  = xy } ~  exp −={ 

O 



exp =2?b

= H€  cos (2?b) − H  sin(2?b)

Sum of many MPC Carrier with frequency b

(21)

21

H€  = } ~  cos { 

O 



H  = } ~  sin { 

O 



{  = 2?b   − {„ − {

= H€  cos (2?b) − H  sin(2?b) H  = xy } ~  exp −={ 

O 



exp =2?b

Doppler Shift Carrier phase shift (same for all MPC) Phase shift due to delay

Since N(t) is large & we assume ~() and {() are independent for different MPC, we can approximate H€() and H() as jointly Gaussian random processes.

(22)

Some assumptions

• No dominant LOS component

• … , `† , ‡R 1 change slowly over time

• ]^`S1 changes rapidly relative to all other phase terms

• \( ) uniformly distributed on [−^, ^].

• … and \ are independent of each other.

22

{  = 2?b   − {„ − {

Very large

(23)

Zero-mean

• Similarly,

• So, E[r(t)]=0, and it is a zero-mean Gaussian process.

• If there is a dominant LOS component, then this is no longer true.

23

 H€  =  } ~



cos {  = }  ~ [cos {()]



= 0

 H  = 0

(24)

Un-correlated

24

 H€  H  =  } ~ˆ‰I{ 



} ~/sin ~/ 

/

= } }  ~~/  cos {  sin {/ 

/



= }  ~  cos {  sin { 



= }  ~  sin 2{ 

 2

= 0

~ and { are not correlated.

= }  ~  ~/  cos {   sin {/ 

Š/,/

+ }  ~  cos {  sin { 



Different MPC’s ~ and { are independent

Uniformly distributed over −?, ? , so =0.

H€() and H  are uncorrelated, and they are Gaussian processes

 they are independent.

(25)

Autocorrelation

25

!‹Œ ,  + =  H€  H€  + = }  ~  cos {  cos {  +



= .5[cos(2?„ )] + .5 cos 4?b  − 4?„ − 2?„ − 2{

 cos {  cos {  + =

= [.5 cos {  + − {  + .5 cos {  + + {  ]

{  + = 2?b  − 2?„  + − { {  = 2?b  − 2?„ − {

Large and uniformly distributed over [−?, ?]

0

!‹Œ ,  + = .5 }  ~ [cos(2?„ )]



= .5 }  ~



2?l cos  a

Only depends on , so WSS!

(26)

Autocorrelation

• Finally,

26

!‹Œ ,  + = !‹Œ =  H€  H  +

= −.5 }  ~ sin 2?l cos

 a

= − H  H€  +

H  = H€  cos 2?b − H  sin 2?b

!‹ =  H  H  + = !‹Œ cos 2?b + !‹Œ sin (2?b )

Also only depends on , WSS!

The received signal, representing how the channel changes over time

(27)

Amplitude distribution - Rayleigh

• ‘ = ’ = ’“] + ’”]

• ’“( ) and ’”( ) are both zero-mean Gaussian random process (so at a given time, two Gaussian random

variables).

• z(t)’s distribution - the amplitude distribution of r(t):

27

Channel path loss

t

• v = 2v

– exp −‹ v

– =‹ v

— exp − v

 , v ≥ 0

This is the famous Rayleigh distribution!

(28)

2-variable joint

Gaussian distribution

• PDF for 2-variable joint Gaussian distribution:

™: X and Y’s correlation (in our case, 0)

• ›5 and ›œ: X and Y’s mean

• 5] and œ]: X and Y’s variance (in our case, both are ])

28

, 0 =

, 0 = 1

2?— exp −1

2  + 0

— The rest of the derivation can be found here:

http://www.dsplog.com/2008/07/17/derive-pdf-rayleigh-random-variable/

(29)

Power distribution:

Rayleigh

• We can obtain the power distribution by making the change of variables ‘] = ’ ] to obtain

29

•ž  = 1

– exp −‹ 

– =‹ 1

 exp − 

 ,  ≥ 0

(30)

Example: Rayleigh fading

• Consider a channel with Rayleigh fading (no LOS!) and average received power Ÿ’ = ] dBm. Find the probability that the received power is below 10 dBm.

• We want to find the probability that  ] <  R¢* =

 *£.

30

 ¤ < 10 = 9 1

100 exp − 

100 @



 = 0.095

(31)

With a LOS component – Ricean (or Rician)

• If the channel has a fixed LOS component then ’“( ) and ’”( ) are no longer zero-mean variables.

• The received signal becomes the superposition of a complex Gaussian component and a LOS component.

31

Rayleigh: lots of random nLOS components

Ricean: Rayleigh + one strong static component

(LOS or strong reflection nLOS)

(32)

Ricean distribution

• Amplitude distribution:

• ]] = ∑,Š §[…]] is the average power in the nLOS MPCs.

• U] = …] is the power in the dominant strong component.

• “(¨): the modified Bessel function of zero-th order.

32

• v = v

—exp −v + I

 © vI

— , v ≥ 0

(33)

Ricean distribution

• The average power in the Ricean fading is

• The Ricean distribution is often described in terms of a fading parameter K, defined by

• K is the ratio of the power in the dominant component to the power in the other random MPCs.

K=0, then Ricean degenerates to Rayleigh

K=∞, then Ricean becomes a non-fading LOS channel.

33

– = 9 v‹ A • v @v = I + 2—



« = I 

(34)

Coherence Time

• Coherence Time:

Coherence time is a statistical measure of the range of time over which the channel can be considered “static”.

• 90% coherence time:

• We can define 50% coherence time in a similar way too.

34

¬S,.­ = ‡’K*J1 4’ 1

4’  < . ­

The first time interval that normalized autocorrelation drops below the threshold.

(35)

Fast and slow fading channel

35

!‹

!‹

b

b

o t

o: symbol period

Slow fading

Fast fading

f

®o

®o: signal bandwidth

®„

®„: Doppler Spread

f

¯ > b

®„

®„: Doppler Spread

f

®„ > ®¯

®„ ≪ ®¯

² ≫ ¯

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