7/27/2019 Tutorial2 Sol 250102
1/12
S72-238 WCDMA systemsTutorial 2 25.01.2002.
Solutions1.Two spreading signals in vector form are
; =S and
; =S with
the chip length
U .a) Calculate the cross correlation function between the spreading sequence
S T and
signal
S T that is bit sequence ;= spread by
S T .
b) Calculate the cross correlation function between the spreading sequence
S T and
signal
S T that is bit sequence ; = spread by
S T .
c) Calculate the cross correlation function between the spreading sequence
S T and
signal
; =S T spread by
S T .
d) Calculate the cross correlation function between the spreading sequence
S T and
signal
; =S T spread by
S T .
1.
The spreading signal is described as sequence of values o . Our spreading code haslength four and we can express it as sum of four unit pulses
N
S T S N P T N U
, where S N is the
n
element of the spreading signal
vector, and
P T NU is a unit pulse beginning at 0 and ending at
U .
A spread signal is generated by multiplying the bits of the
S with the spreading
sequence:
S
B
S S S
B B B
N N
S T A N P T N 4 S N P T N
U
U
- - - -- , where
S
B
N N stand for the bit
and chip numbers accordingly. S
B
P P are unit pulses with the bit and chip
length accordingly and
S
B
A N S N stand for the bit and chip amplitudes.
The cross-correlation function X Y
2 U between two signals X T and Y T is defined
as
LIM
4
X Y
4
4
2 X T Y T DT 4
U U
l d
.Since we are dealing with a finite length signal the integration can be limited only to
the values which are different from zero for out signal it would be interval K .For our signal when U is integer the correlation function can be written as sum of
every chip of spread signal
S T and reference signal
S T
2 S T S T DT
U
U
U U
U
7/27/2019 Tutorial2 Sol 250102
2/12
Since the signals are compounded from triangular impulses we can calculate the
correlation only for integer values of U and between those 2 U is given by a linear
function connecting two neighboring values.
For integerI N T
U valuesI N T
2 U is calculated as:
I N T I N T
M
2 S M S M
U
U
U U
U
Between these values we use linear function
2 2 2 2U U U U U U .Where stand for the nearest smallest and highest integer respectively. We
evaluate all the correlation functions with the Matlab.
Figures from 1 to 8
-8 -6 -4 -2 0 2 4 6 8-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
time difference t
R
(t)
correlation between s11
and s1
-8 -6 -4 -2 0 2 4 6 8
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
time difference t
R
(t)
correlation between s10
and s1
7/27/2019 Tutorial2 Sol 250102
3/12
-8 -6 -4 -2 0 2 4 6 8-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
time difference t
R
(t)
correlation between s21
and s1
-8 -6 -4 -2 0 2 4 6 8-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
time difference t
R
(t)
correlation between s21
and s1
-8 -6 -4 -2 0 2 4 6 8-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
time difference t
R(t)
correlation between s20
and s1
7/27/2019 Tutorial2 Sol 250102
4/12
-8 -6 -4 -2 0 2 4 6 8-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
time difference t
R
(t)
all the cross correlations
s1 [1 1]
s1 [0 1]
s2 [1 1]
s2 [0 1]
7/27/2019 Tutorial2 Sol 250102
5/12
2.The transmitted signal sequence is ; =A . The spreading sequence is
; =S .
Calculate amplitude of the signal and the interference for every bit if the channel
response isTap amplitude 0.5 0.3 0.2
Delay U in chips 0 1 2
And Rake receiver is tuned to the first channel tap.
2.Since in the exercise there is no indication on Doppler components we assume that
there is not any and describe the signal as a simple sum of delayed components of
transmitted signal.
The received signal is described in the form
L L
,
X T T S T U T B Once again we can use notation for the spread signal as above:
S
B
S S S
B B B
N N
S T A N P T N 4 S N P T N
U
U
- - - --
0 2 4 6 8 10 12 14 16 18 20-0.25
-0.2
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0.2
0.25the multipath components
amplitude
firstsecond
third
7/27/2019 Tutorial2 Sol 250102
6/12
The receiver correlates the received signal with reference spreading signal and
integrates over it.
I
I
N L
N L
4
I
M K L L
4
U
4
I
I
L L L L
,
4 U
L
Z T M S T U S T DT
M S T U S T DT M S T U S T DT
B
B B
v
- - - - - --
Where the first term describes the signal arriving along the path the receiver has
synchronised on and the other term is sum of the signals from other paths.
Contributions of
Taps\Symbols
1 2 3 4
1 -0.5 0.5 -0.5 0.5
2 0.075 -0.15 0.15 -0.153 -0.05 0.1 -0.1 0.1
Interference amp. 0.025 -0.05 0.05 -0.05
total amplitude -0.475 0.45 -0.45 0.45
0 2 4 6 8 10 12 14 16 18 20-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5total received signal
amplitude
7/27/2019 Tutorial2 Sol 250102
7/12
3.Consider the feedback shift-register shown below with mod 2 calculations.
a) What is the polynomial that the shift register represents?
b) Determine the output of this circuit with the initial shift-register load
A A A .
c) Determine the output of this circuit with the initial shift register load
A A A .
d) Does this circuit generate a M-sequence?e) Calculate autocorrelation of the output sequence.
f) Calculate power spectrum of the output sequence.
g) How much the interference from other user will be suppressed when the users have
spreading code described above but the codes have different phases?
Reference:
R.L. Peterson, R. Ziemer, D. Borth: Introduction to spread spectrum
communications 1995. pages 695. chapter 3.
3.a)The shift register generating the spreading sequence is given on the figure below.
Each element introduces a delay $. The coefficient multiplying the delay element is
1 if the feedback is connected to it and 0 otherwise. The generator figure is described
by a polynomial G $ $ $
a2
a1
a0
1DDD123
a2
a1
a0
1DDD123
7/27/2019 Tutorial2 Sol 250102
8/12
b)
At each step the output of the register is feed back to the register accordingly to the
describing polynomial. By using output as input to the system we can calculate the
registers state at each time step and from there the output sequence.
Cycle Register State A $ Output
0 001 1 1
1 101 1 $ 1
2 111 1 $ $ 1
3 110 $ $ 0
4 011 1 $ 1
5 100 $ 0
6 010 $ 0
7 001 1 1
The same output can also be calculated by dividing the generating polynomial G $
with the initial state I
A $
G $
A $
- - - - . For that at each state we multiply the
polynomial G $ with the output of the generator and add to the state of the generator
at that moment. This can be seen in the following table. Where the first line stands for
the generator output at different steps T$ . Rest of the lines describe the polynomials
of the system state and added polynomial T$ G $ .
System output $ $ $
$ $ $
1+ $ $$ $$ $ $
$ $ $$ $ $
$ $ $ $ $
$
Since the system is recursive such division may be continued to infinity.However we see that at the delay $ the system state
A $ is same as initial state
and the division starts to repeat itself.
c)
The output sequence is calculated as above but for different initial state, 011.
Cycle Register State A $ Output
0 011 1 $ 1
1 100 $ 0
2 010 $ 0
3 001 1 14 101 1 $ 1
7/27/2019 Tutorial2 Sol 250102
9/12
5 111 1 $ $ 1
6 110 $ $ 0
7 011 1 $ 1
d)Maximal length sequence or M-sequence is a shift-register sequence that has the
maximal possible period for an R-stage shift register. For a shift register with 3 stages
the maximal cycle period is . From the output of our sequence we see thatthe period of the output sequence is 7 i.e. the shift register generates a M sequence.
e)
Let represent the sequence of the generator output with binary digits
B B B B B
K K from the alphabet \ ^ . Because the code is periodic with
N
. NB B also the signal waveform C T is periodic, with periodC
4 .4 and isspecified by
N C
N
C T A P T N4 d
d
, where NBN
A , and C
P T N4 is an unit
pulse beginning at 0 and ending atC
4 . The waveform C T is deterministic with auto-
correlation function:
4
C
2 C T C T DT 4
U U ,
4
C N M C C
M N
2 A A P T M4 P T M4 DT 4
U U
d d
d d
.
The integral here is nonzero only when C
P T M4 and CP T N4 U overlap.The delay Ucan be expressed as
C
K4F
U U , where C
4F
Ub . The pulsesoverlap only for N K M and N K M .
C
C
G
4
.
C C M
K M
M
4
.
M C
K M
M
4
2 2 K A A P P D . 4
A A P P 4 D . 4
F
F
U
F F
F
U
U U M M U M
M M U M
Let represent the binary sequence ofB s as a vector B .The discrete periodic auto-correlation function is defined as:
.
$
!
N
B N K
N
. .K A A
. .
2 , where
!
. is a number of places in which B
agrees and$
. is a number of places where B disagrees with KB . Equivalently
!
. is a number of zeros and$
. is a number of ones in modulo 2 sum of B and
KB .
C
B B
C C
2 K K K
4 4
F F
U UU
- - 2 2 - -
7/27/2019 Tutorial2 Sol 250102
10/12
Property: a maximal-length sequence contains one more one than zero. The number of
ones in the sequence is
. .
For an M-sequence the periodic auto-correlation function is two-valued and is
given by:
B
K L.
KK L.
.
2 v
C
4F
Ub b
C
C C C
24 . 4 4 .
U U UU
- -- -- - -
C C
4 . 4F
Ub b
C
C C
2 4 . . 4 .
U U
U
- -- -- - -
C C
. 4 4F
U b b
CC
C
. 42
4 . .
UU
- - - .
f)
The power spectrum of the sequence is the Fourier transform of its auto-correlation
function #
2 U .
C M
M
3 F 0 F MF E
d
d Because we have periodic signal the spectrum of it is discrete with values at locations
O
C
F N.4
where N is integer.
We are dealing with a periodic signal that can be described as sum of two signals.
This is illustrated on the figure where one of the signals is constant at the level
.
and the other one is the correlation function to which is added a constant
. . This
other signal has values only in intervalC C
4 4U .
7/27/2019 Tutorial2 Sol 250102
11/12
For calculating the signal spectrum we use the correlation function in the interval
C C
.4 .4 U . The constant function in this interval has value
. and the
other functions can be described as two functions one interval C
4 U U 1f and
other in interval C
4U
, U
I.
C C
C C
4 . 4
4 . . . . 4
U UU
-- -- - - F
C C
.
4 . . . 4
U U
U
--- - -- - - - - F
First we calculate the spectrum value at frequency - that is the mean of the signal.Because the correlation function is described as a sum of two functions the mean will
be sum of the means of these two.
C
4
4
C
4 4
0 DT T DT T DT .4 .
. .
. . .
. .
. . .
- - - - - - - - --
I I
In order to calculate the spectrum at other frequencies we consider a fact that; by
adding any constant value to the signal will change only mean of the signal, spectral
component
0 , the value of other spectral components will not change. Accordingly
to that we can calculate the spectrum of the signal from the signal 2 in the figure.
1N
Tc
NTc
1.0
1N1.0+
1N
a)
b)
c)
7/27/2019 Tutorial2 Sol 250102
12/12
In order to calculate the spectrum of the signal we consider the relationship between
the differentiation and Fourier transform
D
T IFDT
QF & F
C
C C
C
C C
C C
4
I F T I F T
C
4
4
I F T I F T
C C
4
I F 4 I F 4
C
DDE DT E DT
.4 F I DT F I DT
.E DT E DT
.4 F I . 4
.E E E E
.4 F I F I
Q Q
Q Q
Q Q
UU
Q Q
Q
Q Q
- - - - -- - - - - --
- - - -
FF
-
COS
SIN
SINC
C C
I F 4 I F 4
C
C
C
C
C
C
E E.
.4 F I
.F 4
.4 F I
.F 4
.4 F
.F 4
.
Q Q
Q
Q
Q
Q
Q
Q
-- - - -
g)
Because other signals use the same spreading code and are separated only by phase
they will be suppressed by the correlation factor that is to the level
. .
1NT
N+1N
sinc (fT)
1T
1T
c
cc
c
f
2
S (f)c