33494836-umts
TRANSCRIPT
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To our parents, professors and all
Who care about
us
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Acknowledgment
First of all we thank ALLAH for teaching us to love
each other and to co-operate in the development of
the project.
We wish to express our deep gratitude toDr. Ahmed sultan (Our supervisor) for his great
support and for teaching us to be self motivated.
As always, the completion of a project would not
be possible without the support of many people. We
would like to thank Walid Abd El-kader, Asser
Osama and Moustafa Shama from ComputerScience Department who helped us with C Coding.
Also, we would like to thank Texas Instruments for
their technical and financial support.
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Team Members
Ahmed Hosny El-Samadony
Bassem Salah Mohammed
011-4583927
Mohammed Amir [email protected]
010-7189710
Mohammed Mokhtar Gaber
012-4259823
Mostafa Anwar Sef [email protected]
010-1254453
Mostafa El-Sayed El-Ashry
Ramy Elarabi [email protected]
018-3276205
Sherif Samir Hassan
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5.1.9. Car Penetration Losses(50)
5.1.10. Transmit Antenna Gain.(51)5.1.11. Cable and Connector Loss.(51)
5.2. Propagation Model...(51)
5.2.1 Okumura-Hata Model(51)5.3. Uplink Balancing(54)
Physical Layer:
Chapter 1: WCDMA Physical layer implementation..(57)
1.1. Introduction ...(57)
1.2. Downlink Transmitter....(57)
1.3. Downlink Receiver.(60)
1.3.1. Received Signal...(60)
1.3.2. Demodulation......(60)
1.3.3. Path Searcher...(61)
1.3.3.1. Multipath..(61)
1.3.3.2. Path Searcher Construction and Analysis.(63)
1.3.3.3. path searching algorithm......(69)
1.3.3.4. False Alarm and Detection Probabilities..(70)
1.3.3.5. Averaging.(73)
1.3.4. Rake Receiver.(77)
1.3.4.1. Channel Estimator(77)
1.3.4.2. Summing over N chips.(79)
1.3.4.3. Decision Level & Probability of Error.(80)
1.3.4.4. Maximal Ratio Combining (MRC)..(83)
Chapter 2: Forward Error Correction (FEC)(83)
2.1. Introduction .. (83)
2.2. Convolutional codes . (86)
2.3. Viterbi Algorithm for detection (90)
2.4. Coding gain .. (91)
Chapter 3: Estimation BER via Simulation.(94)
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Chapter4:DSP(97)
4.1. Processor Supported .. (97)
4.1.2. Operating Systems Supported ........ (97)
4.1.3. Code generator...... (97)
4.1.4. Files.... (97)4.2. Code sequence....(99)
4.3. System components....(99)
4.3.2. Walsh........(99)
4.3.3. Deinterleaver.....(99)
4.3.4. BranchMetrics.(100)
4.3.5. VCP....(101)
References..(102)
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UMTS
Dimensioning
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Chapter 1
Blocking in multiservice systems
Assume a communication system with a total bandwidth of 10MHz.There are
two offered services; one which needs 2MHz, and another with 4MHz
requirement.
Now we will construct a state diagram for this multiservice system. Each state
is characterized by a pair of numbers. The first number gives the number
users in the system utilizing the 2MHz-band width service, where as the
second gives the number of users in the system using the 4MHz-band width
service. For instance, state (2, 1) means two 2MHZ-service users and one
4MHz-service user.
Lets start the system in state (0, 0), i.e., no users. Assume that the average
rate of arrival of users demanding the first 2MHz service ,
(customers/second). Parameter is the average number of users serviced in
unit time. The reciprocal of 1, i.e. (second /customers), is the mean
service time for first service (the 2MHz service). Define as the traffic
(in erlangs) as associated with first service. For the second 4MHz service, we
have , , and a2 defined in a similar way.Since we are in the state (0, 0), the probability of going to state (1, 0)is
proportional to , where as the probability of a transition to (0, 1) isproportional to .
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On other hand, the probability of returning to (0, 0) from (1, 0) is
proportional to , and from (0,1) is proportional to .
Lets investigate this in more detail. A typical assumption in queuing system
(any system where customers or service requesters arrive into a queued
demanding a certain service.) is that the arrival follows Poisson distribution.
Thus, the probability of k users arriving into a queue, with a mean rate of ,
in a period of time of length t is:
! For an infinitesimal period t
1 2 . !
1 1 is negligible compared to 1, as, , , are very smallcompared to .
The probability, therefore, is proportional to . Moreover, it is assumed that
only one user arrive at any instant of time. Note that the inter-arrival time
probability density function, if the arrival is Poisson, is exponential. Why?
If f(t) is the inter-arrival pdf, then
gives the probability of one user
arriving after timet from the last arrival. That is, we have zero arrival fortimet and then one arrival at following period t
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0| 1|
Where 1, then: And
is the mean time between arrivals.Similar argument can be made regarding service times, etc.
Back to our queue: The system moves to state (2, 0) from (1, 0) with a
probability proportional to . To come back to (1, 0), we have twopossibilities; either the 1st user would finish or the second. This mean that the
probability of coming back is proportional to 2 . All this is applicate totransitions between (0, 2) and (0, 1).
2
2
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Now, note that we cant go to a state (0, 3). The second service demands
4MHz. Two already-existing users consume 8MHz. The total band width is
10MHz.
Therefore, there is no (0, 3) state. Using this idea, we can now complete the
state diagram
Note that in states (1, 2),(3, 1),and(5, 0) all the resources (i.e., the full
10MHz) are consumed.
The blocking probability of 1stservice is: P(5,0)+P(3,1)+P(1,2) ,where P(i,j) is
the probability of being in state (i,j).
Why? Because if:
The system is in any of these states, and the first 2MHz service is demanded,
it will not be granted.
On the other hand, the blocking probability of 2nd service is
P(0,2)+ P(1,2)+ P(2,1)+ P(3,1)+ P(4,0)+ P(5,0)
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Because if
The system is in any of these states, and the second 4-MHz service isrequested, the demand will be blocked.
The blocking probability is, thus, service specific.
The challenge now to calculate blocking probabilities is to compute state
probabilities. Though this may be feasible here, if we increase the number of
services, we would get a highly sophisticated state diagram. (If a service
takes one dimension in this diagram, n services take n dimension.)
Fortunately there is a method to collapse the whole state into a simple one
dimension diagram.
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Chapter 2
Kaufman-Roberts Technique
Kaufman and Roberts solved the problem of calculating state probabilities by
assuming a basic service unit, r.
(In the example above, r would be in units of bandwidth.)
Assume that all the services require resources that are integer multiples of r.
So service t demands trr resources, where tr is an integer .In our example
, r can be taken 2MHz .As are result,1r=1 and 2r=2, because first service
demands 2MHz and 14
=r
Hz, whereas the second service demands 4MHz
and 22
42==
r
Hz.The states of the system are numbered according to
how many basic units are being utilized in each state. The final state is the
total resource divided by basic unit .so in our example, we have states 0, 1,
2, 3, 4 and 5.
How can we construct the reduced state diagram?
State l now means one basic unit of resources is being utilized. This is 2MHz.
The only way that we can get there from state 0 is to have a customer
demanding 1st 2MHz service arrives into the system. Note that the states now
are characterized by single numbers.
What about state 2? State 2 means two basic resources are being used
.This could be two 2MHz a service customer or one 4MHz service customer.
Here is the state diagram.
1
1
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Similarly, State 3 may mean 3 2MHz service customer, or one 2 MHz
customer and one 4MHz customer.
If one 4MHz-service user leaves the system, two out of the three utilized
basic service units are set free. For two basic units to be released
simultaneously, the probability is proportional to2
2 .
We have three possibilities so total probability is proportional to
2
3 2 .
32
3 2
A B C
3 possibilities
B CA
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5 4 3 2 52 2 32
Full state diagram
Blocking probability of service 1 is p(5),Whereas blocking probability is
P(4)+P(5) for 2nd 4MHz bandwidth service, because in both states 4 and 5
no 4Mhz service can be admitted due to non-availability of required
resources.
Now how can we calculate state probability?
We will initially compute unnormalized probabilities with P(0) set to 1.
We assume steady state operation where the probability of getting into a
state equals that of exiting the state.
For example,
( ) ( ) ( ) ( )2100 2121 PPPP +=+
=Probability of exiting
state 0
Probability of entering
state 0
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Interestingly, the equations can be written as
( ) ( ) jar
rjPjPtt
t
t = P(0)=1
P()=0 if
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And so on. Recursively we can get each probability. In order to normalize
(j) = so that = 1Where
(j) is the normalized probability of state j
If maximum state is N, the blocking probability of service t is:
1
Utilization, u is the average number of basic units utilized
U = = = = =
1 1 1
Where is the blocking probability for servicet.
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Chapter 3
Uplink capacity calculation
Assume
is signal to interference-plus-noise ratio of user i at the base
statio (BS)
= = =
Where:
: Energy per bit : Interference plus noise spectral density : Received power of user i : Rate of user i (bits per second)
: Total power at receiver of BS = N + Iown + IotherW : Band width = chip rate = 3.84 * 106 Hz
is the sum of powers of noise (N), users controlled by BS (Iown), and
users controlled by other BSs (Iother).
Note that for calculation Pi is subtracted from Itot as a user does notinterfere with her or himself.
=
In the above formula we assume that user i is fully active. If the user is
active for only a fraction vi of the time
=
= = Itot Pi
1
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Let This is the loading by user
At BS the ratio is called rise over thermal
If , i.e., interference from other cells is some factor of
1
1 1
Since
Where : the total loading
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The uplink is interference- limited. In actual systems is kept below, say 0.8.
In general, should be less than
1
1 But
1
1 1 11
1 1
1 Now an additional user x would increase by . If loading plus
1 exceed , the request is blocked. Then condition ofblocking is1
is taken as a log normally distributed random variable. This means thatlog is Gaussian.Blocking
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log log 1
Blocking probability, givena current load i for a demanded x, equals to
Q [| | ], where S is the standard deviation of log (Iother).
Before proceeding, & because it is sometimes needed, lets find statistics of
log (Iother) in terms of those of Iother. (Log=loge)
Let x= log (Iother) Iother=
x= (log (Iother)) =
Var(x) =
S2 std(x) = =xf(x) =
exp (-
)
E (Iother) =
exp (- ) dx= exp (- ) dx= exp (-
) dx=exp (
)E (Iother) =exp (
)E (I2other) =exp (2 2 )| E (Iother) |
2=exp ( 2 )
log
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= log(1+||)=== log [E(Iother)] - log(1+||)
In order to calculate service blocking probability, we consider loading i as a
resource. We quantizei using, for example, the loading of the least
demanding service .If the least demanding services index is 1, then
1
For other services, we use approximation t
rt=round( )=round(
)
The system states are
What is the probability of the Q function defined above??
It is the probability of blocking if servicet is requested while the system is in
state (j) & all the users are active.
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(The definition of i sets activity factor v to one). We will call this probability
(t, j).
The difference between (t, j) & B(t, j) is that in the former all the users are
assumed active & this the true loadingwhich is felt by BS is j .
In the latter, the loading is j but true loading is less because some users are
inactive.
How can we get B (t, j) from (t, j)??P (j) = (1-B (t, j-rt))If Pj (t
*
) is the probability of arriving at state j by requesting servicet*
Pj (t*)=
, Define occupancy (c, j) as the probability of having loading j , while the true
loading due to active users is c .(c, j) = [(c-rt, j- rt) vt+ (c, j- rt) (1- vt)]where: t is the sum over services.
The first term is due to a service requestor who intends to be active. This
requestor would increase both loading&actual loadingby rt.
The second term is due to a service requestor who intends to be inactive.
This requestor would increase loadingby rt, but the actual loading would
remain at c due to users inactivity.
Note that (0, 0) =1 as no loading means no active loading. Nowwe can
relate B (t, j) &
(t, j)
B (t, j) = c,jt,c Note that in (c, j), c can be anything from 0 (no users active) to j (all users
active).
Also (t, 0) = B (t, 0)So here is how the iterations would work to get state probabilities
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- (t, j) are calculated.- B (t, j) & (c, j) are initialized.
(0, 0) =1
B (t, 0) = (t, 0)
For every j: p (0) =1, p (x0) =0
P (j) = (1-B (t, j-rt))Pj (t) = ,
(c, j) = [(c-rt, j- rt) vt+ (c, j- rt) (1- vt)B (t, j) = c,jt,c
After getting the state probabilities, we normalize
Blocking for service t is:
, This calculation depends on statistics of .This would be a tricky part. Forthe time being we can use:
3 10 , 0.57For 0.8
1.2 10
, 0.57
For 0.6
Called coefficient of variation
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5 10 , 0.67For
0.4
1.7 10 , 0.88For 0.2
3.1. UMTS uplink:
(t, j) is the probability of blocking if the system is in state j in which all the
users are active, and the service requestor demands service t.
Condition for blocking: 1
(t, j) probability that Iother exceeds
If Iother follows the lognormal distribution:
, , where w=3.84*106 (Bandwidth) 1.38 10 3004.1410 174
Load due to
requested
service
current
load
12 2
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We will assume Bs receiver noise figure=5 dB. Thus, receiver noise density=-
169 dBm/Hz=1.26*10-20 w/Hz. (w in units is watts.)
If we are given and coefficient of variation of .log log log1
log log1 .(from page 4 uplink capacity calculation)
Note that in the equation for (t, j), Iother and N are in power units. It is
correct to use power density units for both. The value of provided inpage 7 in uplink capacity notes are in watts/Hz.
After (t, j) is calculated, the blocking probability of each service can be
obtained by following the procedure in pages 6 and 7 in uplink capacity
notes.
3.2. The problem with Iother:
Iother is the interference from users not controlled by the Bs under study.
Users controlled by certain Bs generate Iown in the Bs receiver. They also
produce Iother in receivers of the neighboring BSs. The distribution and
statistics of Iother are, therefore, a function of the admission control algorithm
and its parameters such as service load and max.
One way to solve this problem is to start with a guess for probability
distribution of Iother. This pdf determines the probability that Iother exceeds
some value, and, hence, is employed in the computation of (t, j).
Given (t, j), the Kaufman-Roberts is used to get state probabilities, , iswhich are the probabilities of having a certain load in system. Once we get
state probabilities, we can simulate the system to estimate Iother and its pdf.
That is for each loading value, we do a number of simulations proportional to
its probability. For each loading value, and for each simulation run, wegenerate a vector of services with the following two properties:
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1. The sum of loads is equal to the loading value being examined.
2. The services are distributed according to their probability.
Since some users may be inactive, we use the activity factors to restrict our
calculation of users powers to those users who are active.
Assume, for example, that the service vector corresponding to a certain load
has L entries representing L users with L services. Each entry is an integer
that ranges from one to the number of services. The probability of an entry
being equal to t can be taken as the ratio of the traffic of servicest divided
by the total traffic of all services. For three services with
1 0.6, 2 0.3, 3 0.1,We may use the following code, given
p=[0.6 0.3 0.1]Service_vector=zeros(1,300);For g=1:300
u=rand;if u
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user is considered inactive. The sum of the loads of the active users gives theactual loading on the receiver of the BS under investigation.
We can keep an activity vector of zeros and ones to keep track of which users
are active.
Afterwards, we can generate geographical locations for the users. We need to
specify a probability density function for users distribution in a cell. For
example, if the probability is uniform over the area.
P(r, ) The probability of being in a ring of radius r and thickness
dr = f(r) dr =
Using matlabs rand generates a uniformly distributed random variable from
zero to one, x
To transform f(r) d r = f(x) dx to f(r)
2
That is, to generate the locations of L users, we can use:_ _ 1, ;Remember that this presumes a uniform distribution of users over area.
Parameter a can be taken larger than the radius of the cell. We are
interested in users controlled by a certain BS. For a mobile unit to be
controlled by a BS, the total attenuation, , between it and the BS (including
shadowing and path loss) should be less than that between it and
neighboring BSs. can be estimated by ray-tracing techniques, which take
aa
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into account the particular structure of cells, or is represented by several
theoretical and empirical models. We can initially use
10.
10
Where the exponent m of propagation loss is typically taken from 3 to 5, is the distance to BS in meters, and and are two zero-mean normallydistributed random variables, representing shadowing, with a standard
deviation typically in the range of 6 to 10 dB.
10. 10 is the same as the previous formula. is a new random variable that takes
into account the differences in the pathes connecting the mobile unit to the
various BSs. , and are i.i.d.Note that must be less than for the mobile unit to be controlledby the BS under investigation. This can be done as follows
toto = L;Loss_to_own_over _Loss_to_other=[];While toto>0
Shadowing1=shadowing_std*randn;
Shadowing2=shadowing_std*randn;Shadowing3=shadowing_std*randn;Loss_to_own=10^1.4*(distance_to_BS.^m)*10^((shadowing1+
shadowing2) /10/sqrt(2)); %angle of user withincell
phi_user=2*pi*rand;distance_to_other=sqrt(distance_to_BS.^2+4*radius^2-4*radius*
distance_to_BS*cos(phi_user));Loss_to_other=10^1.4*distance_to_other.^m*10^
((shadowing1+shadowing3)/10/sqrt(2));
Loss_to_own_over_Loss_to_other_X=Loss_to_own/Loss_to_other;If Loss_to_own_over_loss_to_other_X < 1Loss_to_own_over_loss_to_other(L-toto+1)=
Loss_to_own_over_loss_to_other_X ;toto=toto 1 ;
endend
other
Apply Cosine rule
BSother
radius
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Then??
We want to calculate
for a particular simulation run.
Remember that
(1)If we write this equation in terms of the transmitted powers
, , The power received at another BS is
, ,, Sum over users
,, (2)
Loss_to_own_over_loss_to_other
Lets expand equations 1 & 2 from the previous page
Iown= (Iown+Iother+N)i
= (Iown+Iother+N) i ,,What is the difference between Iother & ?Iother is the total interference from users not controlled by the BS underinvestigation.
is the interference caused by the users controlled by the BS underinvestigation on a neighboring BS.
To get an equation for Iother , we will assume that
=Iother6 That is, we assume
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*reciprocity* that on average the interference caused by users controlled
by BS1 on neighboring BS2 is the same as the interference caused by users
controlled by BS2 on BS1.
1
11
6 ,
,
6 ,,1
6 ,,1
1
6 ,,1
6 ,,1 6 ,,
Dont forget to weight summations by activity factor (with ones corresponding
to active users, and zero to inactive).
After running the simulations over the possible loadings, we get a vector ofIother that can be used to estimate its pdf, mean, and variance in a way that is
consistent with admission control algorithm and that is derived from it, rather
than being specified independently.
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3.3. Uplink calculat
3.3.1. Beta generati
32
ons flowcharts:
n:
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3.3.2. Kaufman-Ro
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erts:
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3.3.3. Iother cdf esti
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ator:
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Chapter 4
Capacity Calculation for the Downlink
Let Pit,x be the power transmitted by BSx to the mobile unit of user i, which is
power-controlled by X.
The signal to interference-plus-noise ratio (SINR) at the receiver, ,, is givenby:
, ,, , ,, ,, , , , ,,
Where:
Ri,x is the data rate of the service utilized by user i,
W is the ship rate (3.84e6 Hz),
, is the total path loss between the BS and the mobile unit,N is the thermal noise power,
Psch,x is the synchronization channel(SCH) power ,
SCH is a downlink signal used for cell search,
Ptot,x is the total power radiated by BSX,
Ptot,y is the total power radiated by neighboring BSY,
, is the total path loss between the BSY and the mobile unit of user i, and, is the orthogonality factor which describes the fraction of power which is
seen as interference by the receiver .
The Bs can arrange users signals so that they are perfectly aligned. Since
each users data are multiplied by a code orthogonal to the codes assigned to
the other users, the interference from other users should be suppresses.
However, due to the multipath effect, each mobile unit receives replicas of
the transmitted signal. These replicas are delayed versions of one another.
The Walsh code , when shifted, are no longer orthogonal.
Hence, there will remain residual interference despite the use of orthogonal
codes. The factor , characterizes the interference. A value of zero indicates
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perfect orthogonality, whereas a value of one means complete loss of
orthogonality.
In practice , takes the intermediate values. Note its dependence on i. Thereason behind this is that the orthogonality between different users is
degraded by multipath propagation, which in turn, varies according to the
location of the receiver within the cell relative to the Bs.
The SCH has a term as its own in the formulas for , because it is notcoded, i.e. not orthogonal to users data. Now lets use the formula for , toget another for , :
,, ,
, , 1 , , , , , ,
,,
, 1, , , , , 1 , , , , ,,
Define loading factor , , ,,
, , , ,1 , , , , ,, If we sum over users,
, , , ,Where
,is the power of the common channels emitted by BSx. these
include, for example, the common pilot channel (CPICH), which is anunmodulated code channel scrambled by the primary scrambling code of the
BS. It is used for code synchronization and channel estimation.
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We thus have,
, , ,
,
,
,
,
1 ,
,
,
,
, , ,,
, ,, , 1 ,1 , , , ,
,,
1 , ,A simple admission control algorithm can be to accept a call if , remainsless than a specified value Pmax .
The local blocking probability would then be the probability that Ptot,x,
including the required power for the requested service, exceeds Pmax :
Probability (Ptot,x P max)
The computation of this probability generates the beta matrix (t,j).
Sorrowfully, the calculation is not as easy in for the uplink. The problem is
that the loading factors i,x are mixed with , , , and ,. One wayaround this problem is to replace , , , and , by their average values.Hence, a summation like , , , for example, would become i,x
, and we get the total loading separated. We would use a more
rigorous approach based on ..(Monte Carlo) simulations.
These would resemble a lot the simulations we do to get the statistics of
in the uplink.Actually, we would have to do two simulations. Why? Because Ptot,x depends
on Ptot,y and Ptot,y cant considered independent of Ptot,x (this is exactly the
same as the dependence between and in the uplink).So, in the calculation of the beta matrix, we would assume statistics for ,.Then the beta matrix would be fed to the Kaufman-Roberts recursion to getthe blocking probabilities and the state probabilities. The latter can then be
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used in Monte Carlo simulations to find the distribution and statistics of ,.These can be used to repeat the whole process until the statistics we start
with get close enough to the statistics we estimate after computing the state
and blocking probabilities.
4.1. Calculation of beta Matrix:
Parameters:
Noise Power density=1.26*1020
W/Hz
3 services
Rates: 12.2, 64, 144 kbps
: 5, 4, 3 dB
Radius: 1 Km
Traffic: 30, 5.57, 1.2 mE/per user
: 8 W: 1% , : 24%
Path exponent = 3.8
Standard deviation of shadowing=8dB
,: uniformly distributed over [0.4, 0.6],: Gaussian
, 0.5Standard deviation of , 0.05beta=zeros(num_of_services, max_load);
max _
For iter=1:500 (or any large number)
Generate a long service vector taking into account the different service
probabilities. Assume length is G. You may set G to max_load.
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Generate the corresponding loading vector:
For index=1:G
For requested_service=1:num_of_services
** Append loading to the requested service to the first index
elements of theloading vector.
That is,
Loading_vector_partial=[loading_vector(1:index) loading_requested]
** Generate a distribution of index+1 users around BSX
** Caculate the distance to the 6 neighboring cells.
Assume distance between X and
its neighbors is double the radius.
Assume also uniform 60 degrees.
**Make sure the total path loss between X and mobile units is less
than between Y and the mobile units.
**Do not utilize activity factors because the computation of the beta
matrix assumes total activity.
**Calculate the metric
Metric= , , , 1 , 1 , , , , ,, 1 , ,**Calculate the column of matrix to be updated after this
simulation run
Column=round(sum(loading_vector(1:index))/basic_service_unit)
**if(column num_of_columns_in_beta)
x
60 60
60
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visited(requested_service,column+1)=
visited(requested_service,column+1)+1;
visited is a matrix , which , like beta, is initialized by zeros. The visited matrixkeeps track of how many times each pair of (requested_service,loading) has
been visited.
**If (metric>0)&&(columnnum_of_columns_in_beta)beta(requested_service,column+1)=
beta(requested_service,column+1)+1;
end
end the three loops
end
beta=beta./visited;
beta(:,1)=[0;0;0];
beta(find(isnan(beta)==1))=1;
save ..
Note that the entries in beta would be noisy because they are produced via
Monte Carlo Simulations. Each row of beta must be non-decreasing. We can
have oscillations, however, in our beta matrix. These can be reduced byincreasing the number of simulation runs.
4.2. The aftermath of beta calculation:
After beta is estimated, the Kaufman-Roberts recursion is evoked to estimate
the state probabilities and the blocking probabilities of the different services.
Then we need to check the validity of our guesses for the average value and
the standard deviation of total power transmitted. We do this via Monte Carlo
simulation.
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We simulate the loading states whose probabilities exceed some threshold,
say 1%. For each state we do a number of simulations proportional to the
state probability. For instance,
num_of_simulations = round(10000*state_prob) ;
indices_of_loads_to_be_examined = find(state_prob > 0.01) ;
For index = 1:length(indices_of_loads_to_be_examined)
examined_load =basic_service_unit*
(indices_of_loads_to_be_examined(index)-1) ;
num_of_simulations = round(1000*state_prob
(indices_of_loads_to_be_examined(index))) ;
For simulation_index = 1:num_of_simulations
*Generate a service vector with respective service arrival
probabilities.
*Generate the corresponding loading vector.
*Take from loading vector the first entries whose sum is equal to
examined_load.
*Generate activity vector.
*Generate users (locations, total attenuation between them &
SX &neighboring BSs).
Make sure the total path loss between X & mobile units are less
than
that between the units & neighboring BSs.
*Use assumed statistics of Ptot,y to compute:
Ptot,x=
,,,,,, ,,,,,, (Incorporate activity)
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If Ptot,x Pmax, add Ptot,x to a vector that keeps its values, & which is later used
to get the pdf & other statistics. We assume a statistical similarity between
Ptot,x & Ptot,y .
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4.3. Downlink flowcharts:
4.3.1. Beta generation:
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4.3.2. Ptot,y cdf estimator:
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Chapter 5
Uplink Coverage
The coverage calculations based on allowing some losses along the path
between the MU & BS on one condition that the received signal must be atcertain level.
The output power of the MU is around [21 dBm] where the BS output poweris around [13 dBm]; according to that Uplink coverage calculations are thelimiting factor for coverage. But what will be shown in next sections that theDownlink coverage is taken into account but not separated from the capacityas the BS output power is divided on all MUs.
The main idea in this step is to get the maximum allowable path loss alongthe path between the mobile unit and the base station at certain loadingvalue. Then using a suitable propagation model and the maximum path loss,we get the maximum radius of the cell. After getting the cell area we dividethe total area to be covered by the cell area then we get the total number ofsites in that area.
For multi services we get the maximum path loss for each service separatelythen we use the minimum value of alpha to get number of sites (cellrange(R) = 10^alpha).
Thus minimum value of alpha represents a minimum radius which leads tomaximum number of sites which insures total coverage for all services.
On the uplink, perfect power control can be assumed. This means that all
users receive the same power and create the same amount of interferencefor other users even if the channel conditions are different for each path. Onthe downlink each user is in different RF conditions, in terms of channelconditions, interference, and handover conditions.
Link Budget:
Using the link budget we can get the path loss between the transmitter andthe receiver.
For the uplink, the transmitter is the MU and the receiver is the BS.
The next figure shows the path between Tx and Rx.
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SSRBS = Lpath +Gant Lf+j SSdesign
Where the design criterion, SSdesign, is equal to the sensitivity of the radiobase station, RBSsens, plus a number of margins as
SSdesign = RBSsens + BL + CPL + BPL + PCmarg + -Int + LNFmargWhere:
Lpath is the path loss (on the uplink) [dB].
is the maximum user equipment output power (= 21 or 24) [dBm].
RBSsens is the RBS sensitivity. It depends on the RAB [dBm].
LNFmarg is the log-normal fading margin (this margin depends on
the environment and the desired degree of coverage) [dB].
is the noise rise [dB].Int is the Interference margin [dB].
PCmarg is the power control margin, dependent on channel model [dB].
BL is the body loss (= 0 or 3) [dB].
CPL is the car penetration loss (= 6) [dB].
BPL is the building penetration loss [dB].
5.1. Detailed explanation of link budget terms:
5.1.1. Maximum Path loss:
The maximum path loss allowed, is obtained whenSSRBS = SSdesign
So solving for
we obtain:
+Gant Lf+j -RBSsens -BL -CPL -BPL -PCmarg - +Int LNFmarg5.1.2. Power of User Equipment:
Maximum transmit power is dictated by the UE class. Next Table shows themaximum transmit power for each UE class and the associated tolerances. Avoice-centric (i.e., handheld) UE is usually class 3 or 4.
If the network can mix UE classes, the Link Budget should be drafted for thehighest class.
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5.1.3. RBS Sensitivity:
10 log is the thermal noise power density= -174
is the noise figure. is the user bit rate (information bits per second, excludingretransmission)
We can also get the values from tables using the bit rate and channel model.
Thermal Noise Floor:
The thermal noise floor in this Link Budget is calculated using bit rate, notchip rate,
to ensure consistency with the sensitivity calculation. Thermal noise flooraffects the
sensitivity.
Receiver Noise Figure:
The Node B noise figure should be consistent with the site configuration. Avalue of 4 to
5 would assume that no TMAs are used.
Information Full Rate:
Information Full Rate is the bearer data rate. On the Uplink, the main datarates are
12.2 kbps for speech, and 64 kbps for PS or Circuit Switched (CS) dataservices. Release
99 of the standard supports higher data rates of up to 384 kbps for PS(limited to
64 kbps for CS), but these rates are not commonly implemented.
Required Eb/Nt:
The required Eb/Nt, energy per bit-over-total interference (Nt) ratio isinfluenced by four
factors: the coding of the bearer, channel conditions, the target BLER, andthe quality of
the receiver.
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5.1.4. Interference Margin (Rise over Thermal):
RoT is also known as Interference Marginor Load Margin. The RoT isestimated from
the loading value by means of this equation:
RoT = 10 log10(1/(1 load))The result of this equation will be negative to reflect a loss in the LinkBudget.
5.1.5. Log-normal Fading:
Considering that the received signal at a given location can be represented bya Gaussian
distribution, the LNF can be estimated as a complementary cumulativeprobability
distribution function. For easy implementation in a spreadsheet, this can bedetermined
from the cell edge probability () and the standard deviation in the area (),as shown in the equation :
LNF = NORMINV(, 0, )= Q()
Where:
Cell Edge Confidence:
The cell edge confidence, with a standard deviation, is used to calculate the
LNF margin.
The cell edge confidence represents the probability that coverage will beavailable at
the periphery of the cell, on the basis of a statistical distribution of the pathloss. Instead
of the cell edge confidence, the cell areaprobabilitycan be used to estimatethe network
coverage QoS. The cell area and cell edge can be linked.
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Standard Deviation:
Standard deviation represents the dispersion of the path loss or receivedpower measured
over the coverage area. The morphology of the area manmade structuresand natural
obstacles disperses the signal propagation by altering the line of sight andcausing
diffraction, refraction, and reflection. The standard deviation shows limitedcorrelation
with frequency, but this can vary greatly with morphology. In cities, the
standard
deviation can be estimated as 4 to 12 dB. Table 2.5 shows standarddeviations by land
use density.
5.1.6. Handover Gain:
On the Uplink, the handover gain can be seen as a reduction in the LNF. Thisassumption is valid if we consider that the paths to the different cells servinga call are independent, or have a limited correlation. Path independence forcells from different sites is obvious, because the geographical locations of thereceiver lead to independent path obstructions.
For cells of the same site, as in the softer handover case, paths could havegreater correlation, resulting in a reduced statistical handover gain. The LinkBudget does not take this into account, because it is partly compensated forby possible soft combining. Assuming that both paths are equal, the
combining gain would be 3 dB for both directions.
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5.1.7. Body Loss:
Body loss is affected by the evolution of handsets and how people use them.
With hands free kits, the UE can be located anywhere on the user, notnecessarily close to the head.
This could cause large variations in body losses, but no definitivecharacterization has been done.
For video-telephony applications, body loss can be ignored or reduced ascompared to voice applications, because users will be holding the UEs atarms length away from their bodies.
For PS data usage, the body loss depends on both the UE and theapplication. For mini-browser applications, the UE is held in the hand so the
user can navigate the built-in browser. Body loss is assumed to be similar tothat in video-telephony applications. In contrast, for mobile office applicationson a UE with a Personal Computer (PC)-card, equipment loss is a greaterfactor than body loss. The loss created by this equipment depends on thetype of antenna (fixed integrated, swivel integrated, or external) and thecomputer to which the card is connected.
The 3 dB value in the Link Budget can be considered conservative.Measurements can reveal attenuation from 2 to 5 dB for a UE held at head-level, depending on the UE antenna design and its direction relative to themain server.
5.1.8. Building Penetration Losses:
Building Penetration Loss (BPL) is discussed in Chapter 8. In our sample LinkBudget, BPL is a single number, although in reality the value depends on thearea of expected coverage. Values below 20 dB are usually sufficient to coverthe ground floor area immediately Inside the outermost walls of the building.
Values up to 45 dB would be required to cover 95% of the ground floorspace. This amount of coverage may require other deployment.
Scenarios, such as microcells or an indoor solution, to overcome theinterference created by the external building walls.
5.1.9. Car Penetration Losses:
Car penetration losses depend on car type and construction, as well as localregulations.
Historically, for network design, car penetration losses were set between 3 dBand 6 dB.
For newer car materials, such as heat-efficient glass, this value can beincreased up
to 10 dB.
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5.1.10. Transmit Antenna Gain:
Because a handheld UE typically does not use an external antenna, the gain
is set to null (0 dBi).
5.1.11. Cable and Connector Loss:
Because a handheld UE typically does not use an external antenna, this itemis null. For data cards, external antennas are commonly available. If thisoption is widely used in the network, the loss associated with the cableshould be counted. In addition, the Link Budget also must account for theantenna gain (typically 3 to 6 dBi).
After these calculations we get the path loss then using the propagation
model we get the radius of the cell.
5.2. Propagation Model:
5.2.1 Okumura-Hata Model:
Most popular model Based on measurements made in and around Tokyo in1968 between 150 MHz and 1500 MHz
Predictions from series of graphs approximate in a set of formulae (Hata)
Output parameter : mean path loss (median path loss) LdB
Validity range of the model :
Frequency fbetween 150 MHz and 1500 Mhz
TX height hbbetween 30 and 200 m
RX height hmbetween 1 and 10 m
TX - RX distance rbetween 1 and 10 km
3 types of prediction area :
Open area : open space, no tall trees or building in path
Suburban area : Village Highway scattered with trees and house
Some obstacles near the mobile but not very congested
Urban area : Built up city or large town with large building and houses
Village with close houses and tall.
Definition of parameters :
hmmobile station antenna height above local terrain height [m]
dmdistance between the mobile and the building
h0typically height of a building above local terrain height [m]
hbbase station antenna height above local terrain height [m]
rgreat circle distance between base station and mobile [m]
R=rx 10-3 great circle distance between base station and mobile [km]fcarrier frequency [Hz]
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fc=fx 10-6 carrier frequency [MHz]
free space wavelength[m]
Okumura takes urbanareas as a referenceand applies correctionfactors
Urban areas : LdB = A + B log10 R ESuburban areas : LdB = A + B log10 R C
Open areas : LdB = A + B log10 R D
A = 69.55 + 26.16 log10 fc 13.82 log10 hb
B = 44.9 6.55 log10 hb
C = 2 ( log10 ( fc/ 28 ))2 + 5.4
D = 4.78 ( log10 fc)2 + 18.33 log10 fc+ 40.94
E = 3.2 ( log10 ( 11.7554 hm))2 4.97 for large cities, fc 300MHz
E = 8.29 ( log10 ( 1.54 hm))2 1.1 for large cities, fc< 300MHz
E = ( 1.1 log10 fc 0.7 ) hm ( 1.56 log10 fc 0.8 ) for medium to smallcities
Okumura-Hata model for medium to small cities has been extended to cover1500 MHz to 2000 MHz (1999):
LdB = F + B log10 R E + G
F= 46.3 + 33.9 log10 fc 13.82 log10 hb
Edesigned for medium to small cities
G=
0dBmediumsizedcitiesandsuburbanareas
3
dB
metropolitan
areas
a( )=3.2(log(11.75 ))2-4.97
a(1.5)=0
The cell range is given by: 10Where,
13.82 log 44.9 6.55 log
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Number of sites:
The next step is to get the cell area from the cell radius and the area is
different according to cell type, the following figures show the cell types andthe corresponding cell area:
Then the number of sites
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5.3. Uplink Balancing:
In cases where a definite loading has not been given in uplink or downlink itis possible to minimize the number of sites such that coverage and capacity is
in balance. There are numerous ways to perform this balancing. One methodto balance the coverage and capacity is described in Figure 2. A detailed
example using this flow chart:
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The given input data consists of an unloaded path loss and an average user
profile. First an uplink loading is assumed. This is a first guess and based on
this the coverage and capacity sites are calculated. Unless the system is in
balance these will differ. A low loading will for example give few sitesrequired for coverage but many sites for capacity. The uplink load is varied
until balance is achieved.
Once balance in the downlink is achieved a check is made to see that the
downlink capacity/coverage is sufficient. The number of sites is given from
the uplink balancing and it is therefore possible to calculate the load per cell.
This load figure can be used together with the simulated downlink curves
to check that the downlink range is sufficient to cover the required area. If
not the number of sites is increased until enough coverage is achieved.
In our program (see attached cd in balancing.m), we used the bisectionmethod to obtain the previous algorithm. While the difference between thenumber of base stations of the uplink capacity and uplink coverage is greaterthan zero we increase the maximum loading with some step, and if it goesbelow zero we decrease the maximum loading by the same step, then wedivide the step by two and then we reuse the new maximum loading toobtain the new difference between the number of base stations.
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UMTS
Physical LayerImplementation
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Chapter 1
WCDMA Physical layer implementation
1.1. Introduction:
Specifically, we want to implement the base band section in the mobile unit,
which decodes the frames sent by the base band station (BS). The UMTS
frame is 10 ms long and has 38400 chips. The chip rate is, thus,. 3.84 (mega chip per second), and the chip duration Tc0.26s.
The number of chips per bit, N, varies according to the service provided; . The ratio between the bit duration to the chip duration (N) is alsocalled spreading factor.
1.2. Downlink Transmitter:
The signal transmitted by the BS can be written as:
cos2 sin 2 )
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Explanation of terms:
K: Is an index running over the users controlled by BS. Ko is the
number of the users. The K=0 term is the pilot signal sent by the BS
to help the mobile units to:
1. Synchronize the received signal.
2. Estimate the channel characteristic.
d(K):are data of user K. d(0) is always one.dI(K) and dq(k) are
the data sequences multiplied by cos(2fct) and sin(2fct) respectively.
This means that we have quadrature phase shift key (QPSK)
modulation.
In this scheme, every two bits are taken together with one multiplied
by cos and other by sin.I stands for inphase and q for quadrature (The names given to the
two carrier components).
Ec(K): Is the chip energy of the user K. c(0) is the pilots energy.
Note the dependence of c on K. The BS amplifies the signal of its
users by different factors to take into account their different situations
inside the cell.
For instance, users near cell border need higher transmitted energy to
compensate for significant path or propagation loss than those close toBS.
W (K):is the Walsh code or sequence assigned to user K. W(0)
(The pilot Walsh code ) is always one. The different Walsh sequences
are orthogonal when aligned. The two-chip long Walsh codes are :1 1 1 1 Correlation of W1=(1)(1)+(1)(1)=2
Correlation of W2=(1)(1)+(-1)(-1)=2Cross correlation of W1 and W2=(1)(1)+(1)(-1)=0
They are orthogonal when aligned.
Tc
1
-1
1
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Same as 2*2
Walsh codes of length N can be generated recursively from length-2 Walsh
sequence. If we want the codes of length 4,
For length-8 Walsh sequence:
1 11 1 1 11 11 11 1 1 11 1 1 11 1 1 11 11 11 1 1 11 1
1 11 1 1 11 11 11 1 1 11 11 11 1 1 11 11 11 1 1 11 1
Note that:
1.All the codes are orthogonal on each other.
2. With the exception of the all-ones code, the number of ones in
each code is equal to the number of minus ones. This means that
summing the chips of each code gives zero. SI and Sq:Are two BS specific sequences called Scrambling
codes [1]. Each BS has its own scrambling codes ,i.e., scrambling
codes differentiate between BSs (Whereas the users controlled by a
BS are discriminated by their assigned Walsh sequences). The
scrambling codes belong to the pseudo-random noise family of
codes. These are produced by certain shift registers with feedback.
And have the following two important characteristics:
1. The cross correlation between different codes is very low.2. The correlation between a code and a shifted version of itself is very
low.
The scrambling code [2] is at the chip rate.
1 11 1
1 11 1
1 11 1 1 11 1
Same as 4*4 Same as 4*4
Negative of 4*4Same as 4*4
Same as 2*2
Same as 2*2
Negative of 2*2
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1.3. Downlink Receiver:
1.3.1. Received Signal:
After propagation through the channel, the mobile unit receives (see
transmitted signal)
, , cos2
, , sin2
Index 'l' runs over the paths by which the signal is received at the mobile
unit. 'L' is the number of paths.
' ' is the gain of path ' ' and ' ' is the phase.The phase difference betweenpaths results from path length difference and Doppler shift which is a function
of the angle between received path and direction of motion of the mobile
unit . 'dI,l', 'Wl(k)', 'SI,l', 'Sq,l' are the shifted versions of transmitted 'dI', 'W(k)',
'SI', 'Sq'.
1.3.2. Demodulation:
The first order of business at the receiver is to multiply by locally generated
carriers cos2 and sin2 and then low pass filter (or more accuratelymatch filter).
2cos2 cos2 cos4 cos c os after LPF2cos2 sin2 sin4 sin sin after LPF2sin2 cos2 sin4 sin sin after LPF2sin2 sin2 c os4 cos cos after LPF
Received Signal
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Therefore at point A (figure page 3), we have:
, , cos , , sin
At point B:
, , sin , , cos
1.3.3. Path Searcher:
In order to retrieve the in-phase data sequence dI(k), we multiply the signals
at A and B by an SI sequence synchronized with one of the paths. How canthis synchronization be achieved? Let's digress to investigate this issue
thoroughly.
time
1.3.3.1. Multipath:
Each sample carries users' data, noise, and interference from the different
paths through which the transmitted signal has reached the receiver.
Samples of the
received frame
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The BS sends:
0 1 2 3 4 5 101 102 313 314 38399
What is received is:
Noise
+
+
First significant path
+
Second signification path
+
Third significant path
We get the sum
>38400 samples
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1.3.3.2. Path Searcher Construction and Analysis:
At what sample does the first significant path starts? the second? the third ?
How can we construct a path searcher?
What if , is synchronized with ,?A 'one' from , would be multiplied by a 'one' from ,. A minus 'one' by aminus 'one'. The result would be always one. On the other hand, , ,would have an approximatly equal mixture of ones and minus ones.
If we now sum over the chips within one bit duration, i.e., sum over Nconsecutive chips, , , would result in a near zero value (as thenumber of ones number of minus ones).
Not only that, summation over N chips, and in case we are synchronized,
would zero all users' Walsh sequences (see #2, page 2).
Not only that, , , would also have an approximately equal number ofones and zeroes.
Note that we are here talking about two different paths and v(see #2, page 3).
Taking all this into account, what we get after the summation is:
At 'C' N 0 cos + noise + interferenceAt 'D' N 0 sin + noise + interference Note the role of pilot
The other terms are not
eliminated completely
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thus,
At 'C' = N 0 cos + VIAt 'D'
= N
0
sin + Vq
VI , Vq are Gaussian
(VI) = (Vq) = 0
(VI Vq) = 0
( ) = ( ) = = where V = N I0
is the total (interference plus noise) power spectral density. The power
(variance) of interference plus noise is assumed to be equal on average overthe inphase and quadrature branches. The interference plus noise power per
chip in each branch is .
We get because we have summed over N chips over which interference
plus noise are assumed to be uncorrlated from chip to chip.
What is the probability density function of signal 'Z' at point E
Z =
+
, , 0 0 Why?
Any two jointly Gaussian random Variables have the joint distribution:
Where:
1 is the mean of y1y 2 is the mean of y2y 1 is the standard deviation of y1 2 is the standard deviation of y2 is the correlation coefficient between y1 and y2
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In our case: 0 cos 0 cos 0 sin 0 sin (Note that
, l are treated till now as constants)
0 cos 0 sin 0 0Now we will do a transformation of variables:
cos
sin
What is fz,(Z,)?
, , , , ,, ,,
sin cos cos sin
, , Let
, , , ,
1 2 0
To be accurate, this is the pdf of Z conditioned on / as weconsidered to be a constant.From experiments, has been known to follow a number of distributionsdepending on the mobile environment.
Modified Bessel function of order zero
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If there is no line of sight (LOS) between BS and mobile unit, is mostlyRayleigh. Consequently, is exponentially distributed.
, 0
2 | | Standard deviationof 2That is, 2 is both the mean and the standard deviation of .
,
Our Strategy to solve this integral is to cast it as a pdf function with as a
variable. We can then benefit from the fact of pdfs integrating to one.
0 We need here to have:
This guy should be equal to
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1 0 0 0
0 0 0
. 2 0 0 0
Take out the terms that are not a function of .
exp
This integrates to one. Compare with the enclosed formula in page 8 to
ascertain that this is a pdf proper.
1 0
This derivation is predicated on the assumption that locally generated
scrambling code is synchronized with one of the received paths. If this is notthe case something expected when we are still searching for paths within
samples of the received frame, Remember that since is not synchronized with any path, the summationover chips (which with multiplication is basically a correlation process) would
only produce a noise-like signal.
It can be easily shown that in this case:
1
0 needs to be:
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So what we have now are the pdfs of Z in the two cases of synchronization
and lack of alignment:
| 1 0 | 1 Recall in your mind the meaning of a pdf before looking at the figure below.
gives the probability of variable Z lying within the interval
[ , ]
.
z/v
. |
. |
0
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In what region is it more likely that there is no synchronization? In what
region is it more likely that the locally generated scrambling code is aligned
with one of the received paths?
1.3.3.3. path searching algorithm:
We can now devise the following path searching algorithm
received
samples
sequence ofones and
minus ones ofthe locallygeneratedscrambling
code
Multiply and add over N consecutive chips (in both the in-phase and
quadrature branches).
You then get YI & YQ.
Compute Z=YI2+YQ
2.
Compare Z with a threshold Z* . If Z>Z*, decide that a path has been
found. If Zf(z|sync) and above 2V,
f(Z|sync)>f(Z|un-sync) (choosing the point of intersection is only the only
option. Please see below).
Can things go wrong? OF COURSE. We are dealing with random variables
here. Z can be greater than Z*, not because we have alignment, but because
we have had some excessive noise. Z can be less than Z*
whilesynchronization has been achieved, again because of the noise. The first case
Align scrambling codestarting here
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is called false alarm b
just being fooled by n
1.3.3.4. False Alarm
What is the false alar
Well, it is the probabil
1. We do not hav
2.Variable Z exc
Vf
Note that: here we de(Z*/V) as threshold.
What is the detectionDetection is when theSo, detection probabil
1. We have synch
2.Variable Z exce
70
cause we decided that there is alignme
ise. The second case is called miss.
and Detection Probabilities:
probability?
ity that:
synchronization, and
eds Z* because of noise.
(Z|un-sync)
False alarm probability is this
area
liberately have not chosen the point of i
probability?e is alignment and we decide that there
ity PD is the probability that:ronization, and
eds Z*
.
t but we are
Z/V
tersection
is alignment.
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And so we decide in favor of alignment.
Vf(Z|sync)
Z/V
Detection probability is
this area
Note that PF and PD are linked together
l n
corresponding to z*=0
corresponding to z*=
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Bad news! In a perfect world, we want PD =1 and PF =0. In reality,
If PD =1, PF =1 (because if you set the threshold to zero and perpetually
say sync. achieved then in all cases of lack of synchronization, you woulddecide that there is alignment ).
If PF =0 , PD is also zero ( because you cannot just protect your back by
never saying sync. achieved; you would never detect ).
Sometimes, is determined by specifying some tolerable PF ( = - V ln PF ).
This is similar to CFAR (constant false alarm radar).
What is a good operating point for best PD and PF? Assume PD PF relation as
follows
How can we make PD PF relation look like that ?
Increasing
would do the trick. But note that there is a hefty price.We need to increase 0 and the total power transmitted by BS is limited toa maximum value. Also, increasing 0 would increase V for other BSs.We have so far ignored the other cells. They are alive and kicking, however.
When they increase their power, they harm the performance at their
neighbors.
So what is a feasible solution? AVERAGING.
Using Z*
corresponding to this
point would be great
as PD is close to one ,
and PF is close to zero.
x
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1.3.3.5. Averaging:
Rather than taking Z at point (figure in page 6), we would average over Lbit durations to get Z
Z is now here
Lets call variable here Z
What are the new pdfs of Z in both cases of synchronization and lack of
synchronization?
When there is no alignment,
we will take , , as i.i.d independent and identically distributed.This means that the pdf of Z is the convolution of pdfs of , , . Wecan, thus, use Laplace transform to facilitate computation.
Laplace
we have
L i.i.d !
Similarly,
Similarly ,
f(Z|sync) = !
So in case we sum over L bit durations, and then divide by L
f (Z|sync)= ! f (Z|un-sync)= !
|
F
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PF= fZ|u n syncdZ =
! dZ
Let =xPF= !xexpxdx
= (!) x expxdx
= (
!)(
x expxdx-
x expxdx
)
= (! { (L)-(L) inc ( , L)} where: is gamma function
Where incomplete gamma function is defined as
inc(x, L) = ()
(L) = (L-1)!
PF= 1- inc( , L)Similarly ,
PD= 1- inc(, L)
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f(Z|un-sync)
f(Z|sync)
Z/V
L=10 (very close to what we want)
PD
L=5
L=1
PF
0 4L=5
1!
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In practice, we may oversamplethe received signal in order to achieve finer
alignment resolution.
Normalized output of correlator, each
tested shift is one fourthof a chip.
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G
,
Estimate ,
Estimate ,
E
F
, cos(2fct)
sin(2fct)
1.3.4. Rake Receiver:
Back to the detection circuit
, is aligned with path L* as a result of the work of path searcher andsynchronization circuit.
(To get dq sequence, we multiply by , .)1.3.4.1. Channel Estimator:
The channel estimator [2] is for the estimation of and . ituses averaging to get the estimates. Why we do channel estimation isexplained in page 25.
Channel estimator: averaging for Np chips,
Np is an integer multiple of N
Let:
, ,At the point C we have:
, , , , , , , ,
At point D:
0 ,
, , , 0 , , , ,
(signals at A and B are given in page 4)
A
B
C
D
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78
When we sum over Np chips:
Terms with , , are eliminated. Terms with , , when are eliminated. Terms with , and 0 are eliminated.We are left with:
0 , (elimination is not perfect)and 0 , for inphase and quadrature branches respectively.
Dividing by 0 , we get:
, ,
, , Similarly,
, , , and are zero-mean Gaussian random variables
This is similar to and (page 6), but with N replaced by Np. , , , ,
Now at point E we have:
, , , , ,
, , , , ,
,
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, , , , ,
, , , , ,
, 1.3.4.2. Summing over N chips:
when we sum over N chips:
Terms with , , are eliminated. Terms with , , when are eliminated. Terms with
, and
are eliminated.
There for at point 'G' we get :
N , ( , , + , , ) + , , + , , We then add the output from all significant paths. This receiver is called
Rack receiver [3] and each path processor is a finger. Below is a 3-finger
rack. Note that the difference between ,, ,, , and , , is a mere time shift.
User K* 3-finger Rake [1]
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Let X be the aligned sum of all path processors: , , , , , , , , (additive white Gaussian noise)
(Dont forget that dq is obtained by repeating the above but throughmultiplication by , )1.3.4.3. Decision Level & Probability of Error:
Decision Level:If X>0 then dI=1If X0, and that 'one' is sent, but we decide 'minus one' because X
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Hence = =
In our case, 0. We will choose for a Gaussian variable =
Where is the variance and m is the mean.From page 22
m , , , , given that -1 was transmitted , , , , ,,,, , , ,,,,
,,,,,
,
But , , (page 20), , 2 , , , , , , 2 , , , , , , 2 , 2 , 2 , 2 , , 2 , , 2 ,
, , But
, cos (page 19)
, sin Are we done? Not yet because , , are random variables. We need toaverage over their distributions.
This is an upper bound for the integral
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To put in Gaussian pdf format:
Noting has the same distribution as .
Term for imperfect channel estimation
How can we reduce Pe?
1 To reduce , we want this term to be as close toone as possible. This can be achieved by increasing pilot energy 0, andnumber of averaged chips to estimate channel.
Of course, there are limits. The total power emitted by the BS constrained.Also, the channel varies with time and, hence, must be restricted to thechips transmitted to almost the same channel.
1 To reduce , we want this term to be as small aspossible. Increasing the users energy is an option, but it is limited by 1 BSs
total power, and 2 the the ratio required for the second term to
approach one.
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Note that: the contribution of Rake receiver evident in this formula. If whatis inside the brackets is less than one, raising it to L would reduce it
significantly.
1.3.4.4. Maximal Ratio Combining (MRC):
Why do we estimate the channel in the 1st place?
Assume the output of a branch of a rake finger=Xk+nk.We want to combine
the outputs in a way that maximizes signal to interference plus noise ratio
(SINR).
We want to find wkthat maximizes SINR.
Noise power at the output | | For white noise, .Hence, the noise power
.
SINR| | Assume that we have quantity , is real 0 2 0If we choose as
, we get| |
| | 0
| | Schwarz InequalitySINR| | The maximum SINR is attained when wk=kXk, where k is a factor that is
constant. What this means is that the maximum SINR is achieved by
multiplying Xkby (a scaled version) of itself. Since the branches have either
cos or sin , we need to estimate these quantities & multiply theestimates by the coming stream.
This method of combination is called MRC (Maximal Ratio Combining).
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Chapter 2
Forward Error Correction (FEC)
2.1. Introduction:
Forward Error Correction is the addition of redundant bits to sent data in
order to allow the receiver to detect and correct (up to a limit) errors that
have occurred.
An example that illustrates this concept is to send each data bit thrice and
use Voting at the receiver. So rather than sending 0, we send 000. 1 is
sent as 111.
At the receiver we decide whether each bit is 0 or 1, and then apply the rule
000 0
001 0
010 0
011 1
100 0
101 1
110 1
111 1
Assume that the probability of error is .That is, the probability of sending 1 and deciding 0 at the receiver because
of noise and interference is . And the probability of sending 0 and deciding1 is also
.
What is the probability of error in the 3bits-per-bit system?
Assume that 000 was sent. If two or three bits are decoded as 1 in the
receiver, we would say that 1 was transmitted, while it was zero.
Therefore
1 2 errors
2 out of 3 bitsOne correct
Probability of
error in
coded system
3 bits in
error
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3 1 0.01 , 0.000298
This is a significant reduction in . The price, of course, is reduced rate aswe send each bit thrice.A way of looking at coding is Hamming distance. When we send ones and
zeroes, the hamming distance = 1. If we send three ones (for a one) and
three zeroes (for a zero), the hamming distance between two codewords is
three. The hamming distance is the number of differences between two
codewords. Codewords are sequences sent by the transmitter.
In our example, Single errors are corrected. If we send 000, and decide 010
at the receiver, we would say that 0 has been transmitted. Therefore, the
single error has been corrected. Similarly, if we send 111, and decide 011 at
the receiver, we would say that 1 has been sent.
Again the error can be corrected. If we decide 010, however, we would say
that 0 has been transmitted. This is an erroneous decision. Lets look at the
hamming distance as if it is a true distance
How many bits we can correct?
We can correct up to floor ( bits.
Why do we say ? because in actual cases we use more than twocodewords. The minimum hamming distance, , is the minimum amongpairs of codewords. The error correcting capacity is a function of
.
000
1st
111
2nd
codeword codeword
Hamming distance
Points at a distance of one form 000.They are 001, 010, 100. If we
decide them in the receiver, we would say 0 has been sent, which is
a correct decision.
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The probability of error of a coded system, , can be written as a functionof .When do we commit errors?
1
Where n is the length of a codeword in bits. This sum is often dominated by
the 1st term.
In our previous toy example
n=3 , dmin=3 , 1+ floor( ) = 1+1 = 2
P 0.0003, which is very close to actual = 0.000298
2.2. Convolutional Codes:[1]
Redundancy is added via a shift register
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For each input bit , two outputs are produced. The output does not only
depend on the current bit , but also the two previous bits. The constraint
length of a covolutional code is the number of message bits influencing the
output. We denote the constraint length as k. k=3 in our toy example. The is
often initialized by (k-1) zeroes. Also a (k-1) zeroes are added to the bitsequence to be decoded to return the register to the all-zero condition after
the data frame is finished. The input data sequence is thus of length (L+k-1)
where L is the actual data length. If number of outputs per bit is n (n=2 is
our example), the output number of bits = n(L+k-1). The code rate is
therefore . It is the ratio of number of actual data bits to thelength of output. Typically Lk, and r .The number of states of shift register =
2 4if k=3. We have either
00,01,10 or 11 . What happens if the register is in state 00 , and the input is
0 ?
The two outputs are zero
After shifting , the state would remain 00 .If the input data is one
The two outputs are one.
The next state will be 01.
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a solid arrow means that input is zero,
while a dashed arrow denotes one.
The two bits over an arrow are the
outputs.
We can also draw the following state diagram
means that to move out of 00 the input is one, and two ones result as anoutput. In general, the exponent of is the number of input ones, where as
the exponent of is the number of output ones.
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The above diagram gives all the possible ways of starting at 00 and coming
back to it. Using Masons sum rule of flow graphs, the transfer function from
input to output is 11 2 1 2 4 8
2 4 What is the meaning of this? A one at the input produces a sequence of 5
ones at the output. That is one codeword that differs from the all-zero output
by 5 bits.
If two ones are fed to the register, two codewords are produced with 6 ones
before emerging back with the all-zero output sequence. And so forth.
For this convolutional code, dmin consequently, is equal to 5.
Generally dmin can be estimated by inputting one following by zeroes to the
all-zero initial state. What is dmin for the following convolutional encoder?
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2.3. Viterbi Algorithm For Detection :[1]
Suppose that the covolutional
encoder of page 28 generates
an all-zero sequnce. The received
sequence is 0100010000
Calculate Hamming distances
between output of state transitions
and the received sequence.
j=2
j=3
j=4
j=1
Receivedsequence
Receivedsequence
Receivedsequence
Receivedsequence
State 00
State 10
State 01
State 11
State 00
State 10
State 01
State 11
State 00
State 10
State 01
State 11
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j=5
Matlab Convolutional encoding and Viterbi decoding:
Study the instructions poly2trellis, convenc, vitdec, quantiz
Implement WCDMA channel coding procedure using both hard and soft
decoding.
What is explained above is hard decoding. Soft decoding means retaining the
received input by quantizing it to more than one bit. For example, assume
that the input is normalized to a value between 0 and one. Due to noise, the
input to the receiver can be something like 0.34, 0.719, etc. A hard decoder
would transform any value 0.5 to one, and any value < 0.5 to zero.
Afterwards, Hamming distances are used. A soft decoder would quantizeusing, say, 3 bits. The eight possible values would then be compared with
output of state transitions on the basis of Euclidean distances.
2.4. Coding gain:
Using Coding, is reduced (see hard decoding case). The moststraightforward way of reducing is to increase signal-to-noise ratio. So ifwe get a probability for a coded system, we ask ourselves about the SNRneeded to achieve this
without coding. Using coding allows us to operateat a lower SNR while achieving the same error rate. The ratio of SNR without
coding to SNR with coding for the same probability of error is the coding
gain.
The coding gain (dB) can be approximated as 10 log for harddecoding case and 10 log for soft decoding. The superiority of softdecoding case stems from the exploitation of more information from received
data.
Receivedsequence
State 00
State 10
State 01
State 11
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To see the difference between soft and hard decoding, assume a BPSK
system where the code words are a sequence of ones and minus ones. In the
receiver, we correlate received sequence with all the code words forming
correlation metrics.
Where is the bit of the codeword, and is the received bit plusnoise.
Where is the bit of the codeword, the codeword that has beentransmitted, is the bit energy, here multiplied by code rate to take intoaccount the effect of adding redundancy, and is the noise added to the transmitted bit.
If k=i
If ik
2 Where:
n is the length of codeword,
d is the Hamming distance between codewords i and k.
Lets confine ourselves to codeword I with the minimum distance to codeword
k,
2 2 2
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An error takes place if
0If the noise is Gaussian,
, and
0
2 4 2 2
for
1
~ 10 log
For hard decoding case,
2
~ 2
Note: the added r is to take into account the addition of redundancy.
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Chapter 3
Estimation BER via Simulation
If a frame of data bits has m bits, and if the simulated receiver produces v bits
in error, then represents an estimate of BER. For N simulation runs, theestimate of BER is equal to
where Vk is the number of errors in the kthrun.
The most important question now is how much confidence we have in the
estimated BER in relation to its (unknown) true value.
Let be the true value of BER, be the estimated value, and S be thestandard deviation .A known result from probability theory is that, for large N, the quantity
follows Students T distribution With N-1 degrees of freedom
Lets say that we are interested in an upper bound for .
This upper bound depends on the desired confidence. Suppose that we want
the probability that/ goes below be less than , where is chosen to
be something like 5% or 1%.
Since we specify the probability, and since we know the distribution, we can
find the value of .Note that , and thus .
-x
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The Matlab instruction tinv gives the inverse of students T comulative
distribution function. It needs two arguments: and the degrees of freedom of
the T distribution.
X = - tinv(,N-1)
We have (1- ) confidence that
In other words, we have (1- ) confidence that true BER is less than
Guessing N:
Assume the sequence , , . . . . , , of i.i.d random variables y,0 , 1 1,
01 1 01 1
| | 1 We have N observations of y and we want to estimate p. The typical estimatoris
unbiased estimator
By the central limit theorem, can be approximated as Gaussian.
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The probability that
0.00135 (=Q(3))
That is, we have 99.8
This is equivalent to a
we want the upper bo
= 0.1p
N=Note that if p=10, wspecifications above.
In our case, p is the B
The simulation run ex
m=150.
For the specifications
We need about 6000
of 10-4.
96
exceeds three times the standard de
5% confidence that 3 , where99.865% confidence that 3 .nd to be 0.1 ,
30 1
N= need about 900000 ys if we insist on t
ER to be estimated.
mines the transmission and reception o
above, the required number of runs = runs if BER is expected to be of the ord
p 3 iation is
.ssuming that
e
150 bits, i.e.,
.r of magnitude
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Chapter 4
DSP
4.1.Processor Supported:
TMS320C6416 on TMS320C6416 DSK
4.1.2.Operating Systems Supported:
Windows 2000 with service pack 2
Windows XP
Version 31000 MHz DSP, 125 MHz
EMIF Board 5, CPLD 4
4.1.3.Code generator
Code Composer Studio & c code is used
4.1.4.Files
Main.c
Contains data multiplication by Walsh, Scrambling and branch metrics
calculation.
Scrambling_code_generator.c
Walsh.c
deinterleaver_final 8-6.c
Main.h
Vcp_parameters.h
Contains different configurations of VCP parameters
4.2.Code sequence:
First the scrambling code is generated and packed into words,
scrambling_code_generator(Tx->oversampled_scrambling_code);
Then Walsh code is generated,
walsh_code_generator(Tx->walsh,4);
And the data is obtained using a probe point and multiplied by the scrambling and
the walsh codes,to get the interleaved data (Tx->interleaved_input).
for (bit_index=0;bit_index
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{ Tx->Interleaved_input[bit_index]=0;
for(i=0;i>5)+(bit_index*32);
k=i%32;
scrambling=
(Tx->oversampled_scrambling_code[j]>>k)&0x00000001;
if (scrambling==0)
{
scrambling=-1;
}
chips[i]=scrambling*chips[i];
if((i+1)%4==0 & i!=0)
{
summm=(chips[i]+chips[i-1]+chips[i-2]+chips[i-3]);
downsampled[i/4]=(summm/4)*(Tx->walsh[i/4]);
Tx->Interleaved_input[bit_index]=
(Tx->Interleaved_input[bit_index])+(downsampled[i/4]);
}
}
}
for(i=0;iInterleaved_input[i]=Tx->Interleaved_input[i]/256;
}
Afterwards the the Tx-> interleaved_input is passed to the deinterleaver to get the
encoded data with errors (Tx->Deinterleaved).
deinterleaver(Tx->Interleaved_input,Tx->Deinterleaved);
And the the Tx->Deinterleaved to the BranchMetricsCalculator and the result is
Tx->BM_buffer.
BranchMetricsCalculator(Tx->Deinterleaved,Tx->BM_buffer,2);
The BranchMetrics and the VCP parameters are given to the Vcp therough the Edma
*userData[0] isBM address,*vcpParameters [i]is VCP parameters address while
(i) is used to choose the desired configuration from (vcp_parameters.h), anddecisions is the output data.
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submitEdma(&userData[0], &vcpParameters[i], &decisions,&outParms, &numDec);).
4.3.system components4.3.1.Scrambling:
Generated according to the standard.38400 bit oversampled at rate of 4 packed in 4800 word.
4.3.2.Walsh:
Walsh of length 256 is used but generating 256*256 matrix will bust the memory, so
we resorted to generating only one row at a time.
Algorithm:
Knowing the row no. in the 256 matrix inform us about the place of that row within
the matrix, in the upper or lower half.
If it was in upper half so the right half of the row is the same as the left, if in the lower
half ,the right half will the complement of the left one.
And the left half of the row is just the whole row of the smaller submatrix!!!!.
That row can be generated by the same operation and same for matrix ,128,642
submatrices
Take the 3rdrow of a 4*4 walsh matrix.
1 11 1 1 11 11 11 1 1 11 1
3rdrow is in the lower half so right half -1 -1 is the complement of 1 1 which is
the upper half of the smaller matrix1 11 1, and can be obtained by repeating theleft half of the row that is 1.
4.3.3.Deinterleaver :
Deinterleaving is performed using block interleaving, with 10*15 matrix.
Interleaver:Theory:
Used in conjunction with repetition or coding. Interleaving is a form of time diversity
that is employed to disperse bursts of errors in time. A sequence of symbols is
permuted or interleaved before transmission over a bursty channel. If a burst of errors
occurs during transmission, restoring the original sequence to its original ordering has
the effect of spreading the errors over time. If the interleaver is designed well, then
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the errors have a more random pattern that can more easily be corrected by coding
techniques.
- Block interleaving is known for its ease of implementation, so we used it in our
system.
Operation:
An (I, J) block interleaver can be viewed as an array of storage locations which
contains I columns and J rows. The data are written into the array by columns and
read out by rows, as demonstrated in Figure the first symbol written into the array is
written into the top left corner, but the first symbol read out is from the bottom left
corner.
Figure: Reading and writing from and (I, J) block interleaver
4.3.4.BranchMetrics:
The branch metrics (BM) are calculated by the DSP and stored in the DSP memory
subsystem as 7-bit signed values. Per symbol interval T, for a rate R = k/n and a
constraint length K, there are a total of 2K1+k branches in the trellis. For rate 1/n
codes, only 2n1 branch metrics need to be computed per symbol period