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INSTITUTE OF TELECOMMUNICATIONS SYSTEMS – UNIVERSITY OF STUTTGART

Lab Course in Multimedia Communications
Project 4: Simulation of Fixed and Mobile Communications Systems
Miguel Lozano Medina and Ángel Martín Rodríguez 06/11/2009 Group: A2

Lab Course in Multimedia Communications QUESTIONS and TASKS

T 4.4

The bit string is: 0110011110 Q 4.1

Q 4.2

Project 4: Simulation of Fixedand Mobile Communications System

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Lab Course in Multimedia Communications

E S = ∑ P[ Ai ] Ai
i =1

N

2

N =2   • BPSK ⇒   P( A1 ) = P( A2 ) = 

 1 1 1  ⇒ ES = a 2 + a 2 = a 2  2 2 2 

EB =

ES

β
• BPSK ⇒ {β = log 2 2 = 1} ⇒ E B = a2 = a2 1

Q 4.3 The advantage is that is easier to detect the error bit because the adjacent symbols only differ inone bit, so the error has to be in one neighbour’s symbol.

Q 4.4

E S = ∑ P[ Ai ] Ai
i =1

N

2

 N =4  1  1  ⇒ E S = 4 ( a 2 + a 2 ) 2 = 2a 2 • QPSK ⇒   4  P( Ai ) = 4    EB = ES

β

• QPSK ⇒ {β = log 2 4 = 2} ⇒ EB =
Q 4.5 Bit stream: “0110011110”

2a 2 = a2 2

Project 4: Simulation of Fixed and Mobile Communications System

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Lab Course inMultimedia Communications

BITS 01 10 11

SYMBOL STATE a + ja − a − ja a − ja

T 4.5 This is the scheme described.

Project 4: Simulation of Fixed and Mobile Communications System

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Lab Course in Multimedia Communications

We can check in the pictures that the two diagrams are the same Q 4.6 BITS 000 010 SYMBOLS STATE

a
ae
j

π
4

110 111

ja

ae
011 001

j3π 4

−a
ae
j 5π 4

101 100

− ja

ae

j

7π 4

Project 4: Simulation of Fixed and Mobile Communications System

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Lab Course in Multimedia Communications

Q 4.7

Q 4.8

E S = ∑ P[ Ai ] Ai
i =1

N

2

 N =8  1  1  ⇒ ES = 8 a 2 = a 2 • 8 − PSK ⇒  P( Ai ) =  8   8 

EB =

ES

β
a2 • 8 − PSK ⇒ {β = log 2 8 = 3} ⇒ E B = 3

Q 4.9

•a

σ

= 2γ S ⇒ σ =

a 2γ S

= {a = 1} =

1 2γ S

• γ S = 10

γ S dB
10

Project 4: Simulation of Fixed and Mobile Communications System

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Lab Course in Multimedia Communications

γSdB

sigma for a=1

Peb (calculated) 2,5e-2 1e-2 5e-3 2e-3 9e-4 1,9e-4 3e-5

3 dB 4 dB 5 dB 6 dB 7 dB 8 dB 9 dB

0,5 0,446 0,398 0,355 0,316 0,282 0,251

3 dB 1,00E+00

4dB

5 dB

6 dB

7 dB

8 dB

9 dB

1,00E-01

1,00E-02 Peb 1,00E-03 1,00E-04 1,00E-05

γSdB

T 4.6

The first number of this picture indicates the “BER”, the second one “the number of errors” and the third one the “number of bits transmitted”.

Project 4: Simulation of Fixed and Mobile Communications System

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Lab Course in Multimedia Communications

Theobtained results are the same that we expected because they correspond with the chosen parameters. If we make delays in the ERROR RATE CALCULATION, we will obtain errors in the display. T 4.7 We implemented a communication system based in BPSK with Error Rate Calculation.

T 4.8 Table 4.4: Bit Error Ratio with Gaussian Channel and BPSK Mapping γSdB sigma for a=1 3 dB 4 dB 5 dB 6 dB 7 dB 8 dB 9 dB0,5 0,446 0,398 0,355 0,316 0,282 0,251 Peb (calculated) 2,5e-2 1e-2 5e-3 2e-3 9e-4 1,9e-4 3e-5 Peb (simulation) 2,21e-2 1,29e-2 5,72e-3 2,71e-3 8,02e-4 1,74e-4 3,25e-5 206 202 201 200 200 200 200 No. of errors

We can see that the obtained results in the simulation are approximately the same that we calculated.
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LabCourse in Multimedia Communications

Q 4.10
For QPSK Pe = P1 + P2 + P3 + P4 • P1 = (1 − P (Re[a | a ] ∧ Im[a | a ]))P(Re[a ] ∧ Im[a ]) =

= (1 − P[Re{nv } ≥ − a ∧ Im{nv } ≥ − a ])P (Re[a ] ∧ Im[a ]) =

  a   a  1 1    a    a  = 1 − Q − Q −   = 1 − 1 − Q  1 − Q    =       σ   σ  4 4    σ    σ     1   a  1   a   a   = 1 −...

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