Übung zu Drahtlose Kommunikation

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Übung zu Drahtlose Kommunikation

1. Übung

06.11.2018

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Dr. Jovan Radak

– Room: B 229

– Mail: radak@uni-koblenz.de

– Exercises in English

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Exercise 1

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Exercise 1

k=1 sin(1*t) k=3 sin(3*t) k=5 sin(5*t)

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Exercise 1

Odd numbered multiplications of the base frequency 𝜔𝜔 = 2𝜋𝜋𝜋𝜋

k=1 sin(1*t) => 𝜋𝜋 = 1 � _2𝜋𝜋¹ k=3 sin(3*t) => 𝜋𝜋 = 3 � _2𝜋𝜋¹

k=5 sin(5*t) => 𝜋𝜋 = 5 � _2𝜋𝜋¹

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Exercise 1

We would need an infinite number of elements in the sum (which is practically impossible)

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Exercise 1

Not perfect, edges are not steep and it is not completely flat

A bit smaller than the original (on average)

In theory bandwidth does not change, however we have attenuation which affects signal

Not changed

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Exercise 2

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Exercise 2

20𝑑𝑑𝑑𝑑 = 10 log

₁₀

𝐺𝐺 ⇒ 𝐺𝐺 = 10

²⁰¹⁰

= 100 ⇒ 1%

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Exercise 2

𝑃𝑃

_{𝑅𝑅𝑅𝑅}

𝑑𝑑𝑑𝑑𝑑𝑑 = 𝑃𝑃

_{𝑇𝑇𝑅𝑅}

𝑑𝑑𝑑𝑑𝑑𝑑 + 𝐺𝐺

_{𝑇𝑇𝑅𝑅}

𝑑𝑑𝑑𝑑 − 𝑃𝑃𝑃𝑃 𝑑𝑑𝑑𝑑 + 𝐺𝐺

_{𝑅𝑅𝑅𝑅}

𝑑𝑑𝑑𝑑

𝑃𝑃

_{𝑅𝑅𝑅𝑅}

𝑑𝑑𝑑𝑑𝑑𝑑 = 10 + 20 − 60 = −30𝑑𝑑𝑑𝑑𝑑𝑑

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Exercise 2

Yes!!!

Direct consequence of the result from the previous task.

Remember: Friis free space model also takes into

account antenna gain.

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Exercise 2

dBm – refers to the real power (power is compared against reference power of 1mW)

dB – refers to the factor by which values differ from

each other (simple division of two values)

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Exercise 3

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Exercise 3

𝜆𝜆 = 𝑐𝑐

𝜋𝜋 = 3 � 10⁸ 𝑑𝑑 𝑠𝑠 868 � 10⁶ 1

𝑠𝑠

= 300

860 𝑑𝑑 = 34,6 𝑐𝑐𝑑𝑑

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Exercise 3

𝜆𝜆 = 𝑐𝑐

𝜋𝜋 = 3 � 10⁸ 𝑑𝑑 𝑠𝑠 868 � 10⁶ 1

𝑠𝑠

= 300

860 𝑑𝑑 = 34,6 𝑐𝑐𝑑𝑑

𝜋𝜋 = 𝑐𝑐

𝜆𝜆 = 3 � 10⁸ 𝑑𝑑

6 � 10⁻²𝑠𝑠𝑑𝑑 = 5 � 10⁹1

𝑠𝑠 𝑑𝑑 = 5 𝐺𝐺𝐺𝐺𝐺𝐺

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Exercise 3

𝜆𝜆 = 𝑐𝑐

𝜋𝜋 = 3 � 10⁸ 𝑑𝑑 𝑠𝑠 868 � 10⁶ 1

𝑠𝑠

= 300

860 𝑑𝑑 = 34,6 𝑐𝑐𝑑𝑑

𝜋𝜋 = 𝑐𝑐

𝜆𝜆 = 3 � 10⁸ 𝑑𝑑

6 � 10⁻²𝑠𝑠𝑑𝑑 = 5 � 10⁹1

𝑠𝑠 𝑑𝑑 = 5 𝐺𝐺𝐺𝐺𝐺𝐺

𝜋𝜋 = 𝑐𝑐

𝜆𝜆 ⇒ 𝜋𝜋 = 𝑐𝑐 4𝑥𝑥

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Exercise 4

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Exercise 4

𝐺𝐺 = 𝑒𝑒_𝑎𝑎 � 𝜋𝜋 � 𝜋𝜋 � 𝑑𝑑 𝑐𝑐

2

= 0,6 � 𝜋𝜋 � 4 𝐺𝐺𝐺𝐺𝐺𝐺 � 0,6 𝑑𝑑 3 � 10⁸ 𝑑𝑑

𝑠𝑠

2 = 0,6 � 𝜋𝜋 � 24 3

2

= 379

⇒ 𝐺𝐺 𝑑𝑑𝑑𝑑𝑑𝑑 = 10 � log₁₀𝐺𝐺 = 25,8 𝑑𝑑𝑑𝑑𝑑𝑑

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Exercise 4

Diameter increased:

0,9 𝑑𝑑 = 1,5 � 0,6 𝑑𝑑 ⇒ 50% ⇒ 𝑑𝑑₁ = 1,5 � 𝑑𝑑 That means:

⇒ 𝐺𝐺₁ 𝑑𝑑𝑑𝑑 = 10� log𝑒𝑒_𝑎𝑎 � 𝜋𝜋 � 𝜋𝜋 � 𝑑𝑑₁ 𝑐𝑐

2

= 10 log𝑒𝑒_𝑎𝑎 + 20 log𝜋𝜋 � 𝜋𝜋

𝑐𝑐 + 20 log 1,5 � 𝑑𝑑

𝐺𝐺₁ 𝑑𝑑𝑑𝑑 = 10 �log𝑒𝑒_𝑎𝑎 � 𝜋𝜋 � 𝜋𝜋 � 𝑑𝑑₁ 𝑐𝑐

2

= 10 log𝑒𝑒_𝑎𝑎 + 20 log𝜋𝜋 � 𝜋𝜋

𝑐𝑐 + 20 log𝑑𝑑 + 20 log 1,5 𝐺𝐺₁ 𝑑𝑑𝑑𝑑 = 𝐺𝐺 + 20 log 1,5 = 𝐺𝐺 + 3,52 𝑑𝑑𝑑𝑑

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Exercise 5

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Exercise 5

Wired transmission attenuates 3dB per km

⇒ We have linear attenuation that depends on the distance

⇒ attenuation = distance * attenuation (1km)

Wireless transmission is modeled according to Friis free space model

⇒ 𝑃𝑃_{𝑅𝑅𝑅𝑅}

𝑃𝑃_{𝑇𝑇𝑅𝑅} = 𝐺𝐺_𝑡𝑡𝐺𝐺_𝑟𝑟 𝜆𝜆 4𝜋𝜋𝜋𝜋

2

⇒ 10 log𝑃𝑃_{𝑅𝑅𝑅𝑅}

𝑃𝑃_{𝑇𝑇𝑅𝑅} = 10 log𝐺𝐺_𝑡𝑡 + 10 log𝐺𝐺_𝑟𝑟 + 20 log 𝜆𝜆

4𝜋𝜋 − 20 log𝜋𝜋

⇒ The only part that changes is:

⇒ 20 log𝜋𝜋

⇒ And the distances are doubled

⇒ 20 log 2𝜋𝜋 = 20 log𝜋𝜋 + 20 log 2

⇒ 20 log𝜋𝜋 + 20 log 2

⇒ 20 log𝜋𝜋 + 20 log 4

⇒ 20 log𝜋𝜋 + 20 log 8

⇒ 20 log𝜋𝜋 + 20 log 16

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Exercise 5

distance (km) wireless (dB) wired (dB)

1 -6 -3

2 -12 -6

4 -18 -12

8 -24 -24

16 -30 -48

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Exercise 6

A base station (BS) is located at position (7,0). A mobile terminal (MT) is located at position (3,0). A wall is

located at y=3. The radio link between BS and MT consists of two paths, the line of sight (LOS) and the reflection at the wall.

The MT transmits to BS. The data is modulated on a carrier wave with a wavelength of 40 cm.

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Exercise 6

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Exercise 6

Inverted, we have a phase shift of 180°

Distances:

𝑑𝑑₁ = 4𝑑𝑑

𝑑𝑑₂ = 2 � 2² + 3² = 2� 3,61 = 7,22 𝑑𝑑 Phases:

𝑃𝑃₁ = 4

0,4 = 10 ⇒ 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑒𝑒 𝑛𝑛𝑛𝑛𝑑𝑑𝑛𝑛𝑒𝑒𝑛𝑛 ⇒ 0° 𝑝𝑝𝑤𝑝𝑝𝑠𝑠𝑒𝑒 𝑤𝑤𝜋𝜋𝜋𝜋𝑠𝑠𝑒𝑒𝑜𝑜 𝑃𝑃₂ = 7,22

0,4 = 18,025 ⇒ 0,025 �360° + 180° = 9° + 180° = 189°

𝑃𝑃₂ − 𝑃𝑃₁ = 189°

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Exercise 6

𝑃𝑃₂ − 𝑃𝑃₁ = 189°

Destructive interference! Because the phase is bigger than 180 degrees

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Exercise 6

Received power for Friis free space model:

𝑃𝑃_{𝑅𝑅𝑅𝑅} = 𝑃𝑃_{𝑇𝑇𝑅𝑅} � 𝐺𝐺_{𝑇𝑇𝑅𝑅} � 𝐺𝐺_{𝑅𝑅𝑅𝑅} � 𝜆𝜆 4𝜋𝜋𝑑𝑑𝑃𝑃

2

𝑃𝑃_{𝑇𝑇𝑅𝑅}

𝑃𝑃_{𝑅𝑅𝑅𝑅} = 1

𝐺𝐺_{𝑇𝑇𝑅𝑅} � 𝐺𝐺_{𝑅𝑅𝑅𝑅} � 4𝜋𝜋𝑑𝑑𝑃𝑃 𝜆𝜆

2

Attenuation for the direct path: ^{4𝜋𝜋𝑑𝑑}_𝜆𝜆¹^𝐿𝐿 ²= 15971 ⇒ 41,98 𝑑𝑑𝑑𝑑

Attenuation for the reflected path: ^{4𝜋𝜋𝑑𝑑}_𝜆𝜆²^𝐿𝐿 ² = 51306 ⇒ 47,1 + 10 = 57,1𝑑𝑑𝑑𝑑

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Exercise 6

New point P(1,0):

𝑑𝑑₁ = 6 𝑑𝑑

𝑑𝑑₂ = 2 � 3² + 3² = 8,48 𝑑𝑑

For P(3,0): ∆𝑑𝑑 = 7,22 − 4 = 3,22 𝑑𝑑 For P(1,0): ∆𝑑𝑑 = 8,48 − 6 = 2,48 𝑑𝑑

⇒ Destructive maximum: 3,2m ; 2,8m ; 2,4m

⇒ Constructive maximum: 3m ; 2,6m

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Exercise 7

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Exercise 7

𝜆𝜆 = 𝑐𝑐

𝜋𝜋 = 3 � 10⁸

900 � 10⁶ = 0,333 𝑑𝑑 = 33,33 𝑐𝑐𝑑𝑑

⇒ ^𝜆𝜆₄ = 8,33 𝑐𝑐𝑑𝑑

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Exercise 7

𝜆𝜆 = 𝑐𝑐

𝜋𝜋 = 3 � 10⁸

900 � 10⁶ = 0,333 𝑑𝑑 = 33,33 𝑐𝑐𝑑𝑑

⇒ ^𝜆𝜆₄ = 8,33 𝑐𝑐𝑑𝑑

𝐺𝐺 = 4 � 𝜋𝜋 � 𝐴𝐴

𝜆𝜆² ⇒ 𝐴𝐴 = 𝜆𝜆² � 𝐺𝐺

4 � 𝜋𝜋 = 10¹⁰³ � 0,333²

4 � 𝜋𝜋 = 0,0176 𝑑𝑑²

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Exercise 7

𝜆𝜆 = 𝑐𝑐

𝜋𝜋 = 3 � 10⁸

900 � 10⁶ = 0,333 𝑑𝑑 = 33,33 𝑐𝑐𝑑𝑑

⇒ ^𝜆𝜆₄ = 8,33 𝑐𝑐𝑑𝑑

𝐺𝐺 = 4 � 𝜋𝜋 � 𝐴𝐴

𝜆𝜆² ⇒ 𝐴𝐴 = 𝜆𝜆² � 𝐺𝐺

4 � 𝜋𝜋 = 10¹⁰³ � 0,333²

4 � 𝜋𝜋 = 0,0176 𝑑𝑑²

𝑃𝑃_{𝑅𝑅𝑅𝑅} = 𝑃𝑃_{𝑇𝑇𝑅𝑅} � 𝐺𝐺_{𝑇𝑇𝑅𝑅} � 𝐺𝐺_{𝑅𝑅𝑅𝑅} � 𝜆𝜆 4𝜋𝜋𝑑𝑑𝑃𝑃

2

= 10𝑊𝑊 � 10¹⁰³ � 10¹⁰³ � 0,333

4 � 𝜋𝜋 � 1 �2 � 10³

2

= 7,0 � 10⁻⁹𝑊𝑊

⇒ 𝑃𝑃_{𝑅𝑅𝑅𝑅} 𝑑𝑑𝑑𝑑 = −81,5𝑑𝑑𝑑𝑑

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Exercise 8

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Exercise 8

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Exercise 8

𝑃𝑃𝑃𝑃 = 𝑃𝑃

_{𝑇𝑇𝑅𝑅}

𝑑𝑑𝑑𝑑𝑑𝑑 − 𝑃𝑃

_{𝑅𝑅𝑅𝑅}

𝑑𝑑𝑑𝑑𝑑𝑑 = 𝑃𝑃𝑃𝑃

₀

+ 10 � 𝛾𝛾 � log 𝑑𝑑

𝑑𝑑

₀

+ 𝑋𝑋

_𝛿𝛿

𝑋𝑋

_𝛿𝛿 – Gaussian (normal) random variable expressing attenuation (in dB) caused by fading

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Übung zu Drahtlose Kommunikation