Microsoft Word - hw5.doc
Problem 1:
a.) What is the minimum E
Ieff (bit energy over effective noise energy) to yield
P(e)<1% in the asynchronous CDMA system (additive white Gaussian noise,
no multipath)?
.) If the spectral efficiency is 1chip/sec/Hz, bandwidth is 3MHz, chip energy
Ec=0.1No, data rate per user=10kbit/sec/user, and P(e)<1%, how many users can
e supported in a single cell?
(Note that No is noise power spectral density (unit is Energy per sec per Hz) and
therefore its unit is energy.)
(Hint: if processing gain is N, for each bit, after the despreader,
the desired signal energy is N*N*Ec
the energy for one interferer is N*Ec
the noise energy is N*No)
Problem 2:
The minimum E
Ieff (bit energy over noise energy) per user in the uplink of an
asynchronous CDMA system is 2. The spectral efficiency is 1chip/sec/Hz. The
andwidth of the available spectrum is 2MHz. Data rate per user 10kbit/sec/user.
The received chip energy at the base station for a user is Ec=No(10^9)/(r^3)
where r is the distance (in meters) between the user and the base station.
(Note that No is noise power spectral density (unit is Energy per sec per Hz) and
therefore its unit is energy. )
Consider the case that all users have the same distances from the base station.
1) What is the maximum range of coverage, if the number of user Nu=100?
2) What is maximum number of users Nu and the co
esponding processing gain
N for r=100m?
3) What is maximum number of users Nu and the co
esponding processing gain
N for r=1km?
4) What is maximum number of users Nu and the co
esponding processing gain
N for r=3km?
5) What is maximum number of users Nu and the co
esponding processing gain
N for r=5km?
Problem 3:
The minimum E
Ieff (bit energy over noise energy) per user in the uplink of an
asynchronous CDMA system is 2. The spectral efficiency is 1chip/sec/Hz. The
andwidth of the available spectrum is 2MHz. Data rate per user =1kbit/sec/user.
The received chip energy at the base station for the j-th user is
Ec=No(10^9)/(r(j)^3) where r(j) is the distance (in meters) between the j-th user
and the base station. Consider the case that r(j)=100+(j-1)*30. Want is maximum
number of users Nu and the co
esponding processing gain N?
(Note that No is noise power spectral density (unit is Energy per sec per Hz)
and therefore its unit is energy.)
(hint: The users with larger distances from the base station have smaller E
No.
Therefore, make a reasonable guess of Nu and calculate the E
No for the
farthest user to check the minimum E
No requirement. By trial and e
or, you
will be able to derive the co
ect value of Nu.)
Problem: 4 (CDMA_asynchronous_cross_co
elation)
∑ ∫∑
∞
−∞=
+∗
+
+=
+−−==
k
ic
NTm
mNT c
j
nkkc
m
n
Nm
mNn
m
n
m dtkTtgnTtgeabxEyyY c
c
i )()(,
)1()(
)1(
1
)()( τθ
Part a: Let m=0, N=3 (i.e., the sum for n is from 1 to 3). Expand the above equation
for k=1 to 3 (i.e., you will have 9 terms). Simplify the expression by
combining the terms with similar integrals (combine terms with n=k=1,2,3;
(n-k)=1; (n-k)=2; (n-k)=-1; (n-k)=-2). Leave the integral intact (don’t perform
the integral). Find the expectation and variance for each term.
Part b: Repeat Part a by letting xk=d, bk=ak, θi=0, τi=0. (Now, combine terms with
n=k=1,2,3; (n-k)=1 or -1; (n-k)=2 or -2).
Microsoft Word - cdma asyn.doc
1
Asynchronous CDMA (Up Link)
After studying this note, students will be able to
1. Formulate asynchronous CDMA systems in a single- path channel with additive white
Gaussian noise
2. Understand the statistical properties of Multiuser Interference in asynchronous CDMA
systems.
3. Understand the statistical properties of noise.
4. Understand Self Noise.
5. Perform Bit-E
or-Rate Analysis of Asynchronous CDMA.
6. Design an optimum Chip-shape Function
7. Compute processing gain, capacity, range, BER, etc.
In the uplink of a CDMA system, the base station receives different signals from different
mobiles. These signals cannot be synchronized because they are processed and transmitted by
different transmitters at different locations. Thus, the orthogonal Walsh codes cannot be used for
spreading and dispreading. Instead, PN codes are employed in the uplink of an asynchronized
CDMA system to minimize multiuser interferences.
2
First Part: Transmitter and Receive
I. Transmitters
Nu users simultaneously transmit signals to the base station. The base station needs to
detect the data transmitted by all users. In the conventional approach, the base station considers
one user as the desired signal at a time while all other users are treated as undesired interferers.
Many multiuser detection algorithms have been developed recently. They are beyond the scope
of this class. Brief discussions on multiuser detection are presented in the “Capacity” note in
Lecture 6.
A. Desired User
The CDMA signal of a certain user over the (m+1)th bit (N-chip) interval is
∑
+
+=
−=
Nm
mNk
ckc
mm kTtgaEdts
)1(
1
)()( )()(
Here,
d(m) is the data value (+1 or -1) for the mth bit interval. Note that d(m) is constant over a period of
N chips;
Tc is the chip interval and N Tc is the bit interval (N chips per bit);
Ec is the energy per chip and Eb=NEc is the energy per bit;
g(t) is the chip pulse shape and its energy Eg is normalized to 1;
ka is the k
th chip of the complex PN code sequence of the desired signal. Real and imaginary
parts of ka are independently generated and take values ±1 / 2 . Note that 1.ka =
3
Without loss of generality and for convenience, we will omit the superscript “(m)” in following
text. Therefore, the (m+1)th bit of the CDMA signal of the desired user is
( 1)
1
( ) ( )
m N
c k c
k mN
s t d E a g t kT
+
= +
= −∑ XXXXXXXXXX)
B. Non-desired Users
There are Ni non-desired users. The transmitted signal of the ith a
itrary user (i=1,2,…,Ni
and Ni=Nu-1) is
XXXXXXXXXXiji i i ic k k c i
k
s t E x b g t kT e θτ
∞
=−∞
= − +∑ XXXXXXXXXX)
where g(t) is the chip pulse shape function and
i
cE is the energy per chip for the i
th user
i
kx is the i
th interferer's data sequence. It takes values ±1 and remains constant for N chips (one
bit period) at a time.
i
kb is the k
th chip of the ith interferer's complex PN code sequence. It has same properties as ak for
desired signal. Real and imaginary parts of ikb are independently generated and take
values ±1 / 2 . Note that 1.kb =
τi is the time offset for the i
th interferer's signal from desired signal. Noting that the sum is from -
∞ to +∞, without loss of generality, the time offset iτ can be considered to be in the range
etween 0 and cT . That is 0 i cTτ≤ ≤ .
θi is the phase offset for the i
th interferer's signal from desired signal
4
Here, we have used the desired signal in (1) as the time and phase references. Not that icE ,
i
kx ,
i
kb , τi , and θi are treated as random variables because there are infinite number of scenarios in
practical situations.
II. Additive Gaussian Noise Channel
If these signals are received together with complex Gaussian white noise n(t) with power spectral
density No Watts/Hz,
1
( 1)
1 1
( ) ( ) ( ) ( )
( ) ( ) ( )
i
i
i
N
i
i
Nm N
ji i i
c k c c k k c i
k mN i k
t s t s t n t
d E a g t kT E x b g t kT e n tθτ
=
+ ∞
= + = =−∞
= + +
= − + − + +
∑
∑ ∑ ∑
(3)
III. Receiver
A. Despreading
The optimum receiver processing is to co
elate the received signal with the "replica",
∑
+
+=
−
Nm
mNk
ck kTtga
)1(
1
* )( , XXXXXXXXXX)
and compare the real part of the result with the threshold zero. In practice, a synchronization
timing t0 has to be searched for initially in the receiver. For convenience and without loss of
generality, reset t0 to be zero. Then, we compute the quantity for the (m+1)th bit
( 1)( 1) *
1
( ) ( )c
c
m Nm NT
k cmNT
k mN
Y r t a g t kT dt
∗++
= +
= −
∑∫ XXXXXXXXXX)
5
Interchange the order of integral and summation in (5) and substitute (3) & (4) into (5). In
practice, the received signal will contain the desired signal, interference and noise. We have
XXXXXXXXXX) *
1
XXXXXXXXXX)
1
XXXXXXXXXX)
1 1
( ) ( ) ( )
( ) ( ) XXXXXXXXXXdesired term )
XXXXXXXXXXinterfere
c
c
c
c
i
c
c
m N m NT
k cmNT
k mN
m N m NT d
k cmNT
k mN
N m N m NT i
k cmNT
i k mN
Y a r t g t kT dt g g
a s t g t kT dt Y
a s t g t kT dt
+ +∗
= +
+