Tutorial2 Sol 250102

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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

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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)


and s1

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-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)


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)


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)


and s1

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-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]

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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

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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

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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

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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

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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 - -

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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 .

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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)

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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

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Tutorial2 Sol 250102