Übung zu Drahtlose Kommunikation

Übung zu Drahtlose Kommunikation

2. Übung

20.11.2018

Exercise 1

Exercise 1

𝑑𝑑

= 2 � 𝐷𝐷

𝜆𝜆

Fraunhofer’s distance

- D – the largest antenna dimension

- defining difference between near-field and far-field effects

𝜆𝜆 = 𝑐𝑐

𝑓𝑓 = 3 � 10⁸

5 � 10⁹ = 6𝑐𝑐𝑐𝑐

⇒ 𝑑𝑑_𝑓𝑓 = 2 � 0,12²

0,06 = 48𝑐𝑐𝑐𝑐

Exercise 1

b) What happens to the near field, when increasing the frequency and at the same time:

1) keeping the antenna length the same?

Frequency increases => wave length is smaller =>

=> Fraunhofer distance increases => Near-field increases 2) adjusting the antenna length according to λ/4?

=> Fraunhofer distance decreases

=> Near-field decreases

𝑑𝑑 _𝑓𝑓 = ^2�( _𝜆𝜆

⁾

= ^2� _𝜆𝜆

= ^𝜆𝜆 ₈ = _8�𝑓𝑓 ^𝑐𝑐

Exercise 2

Exercise 2

𝑐𝑐 = 3� 10⁸ 𝑐𝑐 𝑠𝑠 𝜆𝜆 = 𝑐𝑐

𝑓𝑓 = 3 �10⁸

900 �10⁶ = 1 3 𝑐𝑐 𝑓𝑓_𝑑𝑑 = 𝑣𝑣

𝜆𝜆 = 27,78 �3 = 83,34 𝐻𝐻𝐻𝐻

⇒ 𝑓𝑓^′ = 𝑓𝑓 + 𝑓𝑓_𝑑𝑑 = 900 000 083,34 𝐻𝐻𝐻𝐻

Exercise 2

𝑓𝑓^′′ = 𝑓𝑓^′ + 𝑓𝑓_𝑑𝑑 = 900 000166.68 𝐻𝐻𝐻𝐻

Exercise 3

Exercise 3

𝑃𝑃𝑃𝑃 = 𝑃𝑃𝑃𝑃 𝑑𝑑

+ 10 � 𝑛𝑛 � log 𝑑𝑑

𝑑𝑑

+ 𝐸𝐸 𝑋𝑋

𝑃𝑃𝑃𝑃 = 10 + 10 � 2,5 � log 100 = 60 𝑑𝑑𝑑𝑑

Exercise 3

𝑃𝑃_{𝑅𝑅𝑅𝑅} 𝑑𝑑 = 𝑃𝑃_{𝑇𝑇𝑅𝑅} − (𝑃𝑃𝑃𝑃₀(𝑑𝑑₀) + 10 � 𝛾𝛾 � log 𝑑𝑑

𝑑𝑑₀ + 𝑋𝑋_𝛿𝛿) 𝑃𝑃_{𝑅𝑅𝑅𝑅} 𝑑𝑑 ≥ −80 𝑑𝑑𝑑𝑑𝑐𝑐

𝑃𝑃_{𝑇𝑇𝑅𝑅} − 𝑃𝑃𝑃𝑃₀ 𝑑𝑑₀ − 10 � 𝛾𝛾 � log 𝑑𝑑

𝑑𝑑₀ − 𝑋𝑋_𝛿𝛿 ≥ −80 𝑑𝑑𝑑𝑑𝑐𝑐

−10 − 10 − 50 − 𝑋𝑋_𝛿𝛿 ≥ −80𝑑𝑑𝑑𝑑𝑐𝑐 𝑋𝑋_𝛿𝛿 ≤ 10

Exercise 3

𝑃𝑃_{𝑅𝑅𝑅𝑅} 𝑑𝑑 = 𝑃𝑃_{𝑇𝑇𝑅𝑅} − (𝑃𝑃𝑃𝑃₀(𝑑𝑑₀) + 10 � 𝛾𝛾 � log 𝑑𝑑

𝑑𝑑₀ + 𝑋𝑋_𝛿𝛿) 𝑃𝑃_{𝑅𝑅𝑅𝑅} 𝑑𝑑 ≥ −80 𝑑𝑑𝑑𝑑𝑐𝑐

𝑃𝑃_{𝑇𝑇𝑅𝑅} − 𝑃𝑃𝑃𝑃₀ 𝑑𝑑₀ − 10 � 𝛾𝛾 � log 𝑑𝑑

𝑑𝑑₀ − 𝑋𝑋_𝛿𝛿 ≥ −80 𝑑𝑑𝑑𝑑𝑐𝑐

−10 − 10 − 50 − 𝑋𝑋_𝛿𝛿 ≥ −80𝑑𝑑𝑑𝑑𝑐𝑐 𝑋𝑋_𝛿𝛿 ≤ 10

𝑃𝑃 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑟𝑟𝑣𝑣𝑐𝑐 = 𝑃𝑃 𝑋𝑋_𝛿𝛿 ≤ 10

Φ 𝑥𝑥 = 1

2𝜋𝜋 �

−∞

𝑅𝑅

𝑐𝑐^{−𝑡𝑡22}𝑑𝑑𝑐𝑐 𝑤𝑤𝑤𝑐𝑐𝑐𝑐𝑐𝑐 𝑥𝑥 = 10 − 0 5

𝜱𝜱 𝒙𝒙 = 𝟎𝟎,𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗

Generic normal distribution:

Φ 𝑥𝑥 − 𝜇𝜇

𝜎𝜎 = 1

2 � 1 + 𝑐𝑐𝑐𝑐𝑓𝑓 𝑥𝑥 − 𝜇𝜇 𝜎𝜎 2

Exercise 3

𝑷𝑷_𝒔𝒔 = 𝟎𝟎,𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗

𝑷𝑷_𝑭𝑭 = 𝟏𝟏 − 𝑷𝑷_𝒔𝒔 = 𝟎𝟎,𝟎𝟎𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗

Exercise 3

𝑷𝑷_𝒔𝒔 = 𝟎𝟎,𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗

𝑷𝑷_𝑭𝑭 = 𝟏𝟏 − 𝑷𝑷_𝒔𝒔 = 𝟎𝟎,𝟎𝟎𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗

A B

C

Exercise 3

𝑷𝑷_𝒔𝒔 = 𝟎𝟎,𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗

𝑷𝑷_𝑭𝑭 = 𝟏𝟏 − 𝑷𝑷_𝒔𝒔 = 𝟎𝟎,𝟎𝟎𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗

A B

C

To have a successful communication we need to have either all three links functional OR two links functional and one nonfunctional

𝐴𝐴𝐴𝐴 ∧ 𝑑𝑑𝐴𝐴 𝐴𝐴𝑑𝑑 ∧ 𝑑𝑑𝐴𝐴

𝐴𝐴𝐴𝐴 ∧ 𝐴𝐴𝑑𝑑 ∨ 𝐴𝐴𝑑𝑑 ∧ 𝑑𝑑𝐴𝐴 ∧ 𝐴𝐴𝐴𝐴

Exercise 3

𝑷𝑷_𝒔𝒔 = 𝟎𝟎,𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗

𝑷𝑷_𝑭𝑭 = 𝟏𝟏 − 𝑷𝑷_𝒔𝒔 = 𝟎𝟎,𝟎𝟎𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗

A B

C

To have a successful communication we need to have either all three links functional OR two links functional

𝐴𝐴𝐴𝐴 ∧ 𝑑𝑑𝐴𝐴 𝐴𝐴𝑑𝑑 ∧ 𝑑𝑑𝐴𝐴

𝐴𝐴𝐴𝐴 ∧ 𝐴𝐴𝐴𝐴 ∨ 𝐴𝐴𝑑𝑑 ∧ 𝑑𝑑𝐴𝐴 ∧ 𝐴𝐴𝐴𝐴

𝑃𝑃 𝑀𝑀𝑀𝑀𝑀𝑀𝑐𝑐𝑟𝑟𝑤𝑐𝑐𝑀𝑀 = 𝑂𝑂𝑂𝑂 = 𝑃𝑃

+ 3 � 𝑃𝑃

⋅ 𝑃𝑃

𝑃𝑃 𝑀𝑀𝑀𝑀𝑀𝑀𝑐𝑐𝑟𝑟𝑤𝑐𝑐𝑀𝑀 = 𝑂𝑂𝑂𝑂 = 0,97725

+ 3 ⋅ 0,97725

� 0,02275

𝑃𝑃 𝑀𝑀𝑀𝑀𝑀𝑀𝑐𝑐𝑟𝑟𝑤𝑐𝑐𝑀𝑀 = 𝑂𝑂𝑂𝑂 = 0,9985

Exercise 3

𝑃𝑃 = 0,99 ⇒ Φ 𝑦𝑦

5 ⇒ 𝑦𝑦

5 = 2,326 ⇒ 𝑦𝑦 = 11,632 𝑃𝑃_{𝑅𝑅𝑅𝑅} 𝑑𝑑 = 𝑃𝑃_{𝑇𝑇𝑅𝑅} − (𝑃𝑃𝑃𝑃₀(𝑑𝑑₀) + 10 �2,5 � log 𝑑𝑑

𝑑𝑑₀ + 𝑋𝑋_𝛿𝛿) 𝑃𝑃_{𝑅𝑅𝑅𝑅} 𝑑𝑑 ≥ −80 𝑑𝑑𝑑𝑑𝑐𝑐

−10 − 10 − 25 �log 𝑑𝑑

𝑑𝑑₀ − 11,632 ≥ −80 log 𝑑𝑑

𝑑𝑑₀ ≤ 1,9347 ⇒ 𝑑𝑑 ≤ 10^1,9347 ⇒ 𝑑𝑑 = 86 𝑐𝑐

Exercise 4

• Multiple digital signals (or analog signals carrying digital data) carried on a single transmission path by interleaving portions of each signal in time.

• GSM: 8 timeslots in 120/26 ms frame (many auxiliary channels) DECT: 24 time slots in 10 ms frames

ISDN: 32 time slots in 125 microseconds frame (PRI, E1)

• SDMA transmits different information in different physical areas

• Examples: simple cellular radio systems and advanced cellular systems which use directional antennas and power modulation

• Mobile phone network cell structure - same frequencies reused in areas that are far enough apart.

• FM Radio

Exercise 4

• FDMA – multiple signals are carried simultaneously on the same medium allocating each signal to a different frequency band.

• Modulation and multiplexing equipment are needed to move each signal to the required frequency band and to combine modulated signals.

• TV/Radio

• CDMA – every signal has its own code (chip sequence), receiver knows these codes and uses them to decode the received information

• UMTS, LTE

Exercise 5

Phase converges to 180 degrees

Explanation:

Try to calculate phase shift (using LOS distance and reflected distance) 𝑑𝑑_{𝐿𝐿𝐿𝐿𝑆𝑆} = 2 ⋅ 𝑑𝑑

𝑑𝑑_{𝑅𝑅𝑅𝑅𝑓𝑓} = 2 � 𝑑𝑑² + 𝑤² 𝜑𝜑 = 2𝜋𝜋 � 𝑑𝑑_{𝑅𝑅𝑅𝑅𝑓𝑓} − 𝑑𝑑_{𝐿𝐿𝐿𝐿𝑆𝑆}

𝜆𝜆 + 𝜋𝜋

Exercise 5

Rice, Rayleigh

Log normal shadowing

Rice, Rayleigh

Exercise 6

Number of codes, interferences

Number of channels, bandwidth

Number of time slots

Exercise 6

Decreasing – if cell is overloaded neighboring cells take over the part of the load

Increasing – to help neighboring nodes when they are overloaded

Exercise 7

Exercise 7

Exercise 7

Exercise 8

Exercise 8

1 8

2

Exercise 8

8 = 2

⇒ 3 𝑏𝑏𝑟𝑟𝑐𝑐𝑠𝑠

Distance is not evenly distributed => non optimal number of phases

𝐴𝐴𝑐𝑐𝑀𝑀𝑀𝑀𝑟𝑟𝑐𝑐𝑀𝑀𝑑𝑑𝑐𝑐 = 𝑄𝑄_𝑋𝑋² + 𝐼𝐼_𝑅𝑅² = 4² + 1² = 17 𝑃𝑃𝑤𝑃𝑃𝑠𝑠𝑐𝑐 = tan⁻¹𝑄𝑄_𝑋𝑋

𝐼𝐼_𝑅𝑅 = tan⁻¹1

4 = 14,04°

Exercise sheet 3

• Will be published soon

• Next tutorial:

Tuesday 11.12.2018 from 16:00 – 18:00 in B016