6. 2 ray propagation model part 2
•
1 like•321 views
The document summarizes the two-ray propagation model for wireless communication. It defines key terms like path difference, phase difference, and time difference. The total electric field is calculated as the sum of the direct and ground reflected electric fields. The power received is proportional to the transmitted power divided by the distance to the fourth power. The path loss is calculated as 40 log(d) minus gains and heights.
![WIRELESS COMMUNICATION
SERIES
2 Ray Propagation Model
[Part – 2]](https://image.slidesharecdn.com/6-200317150450/85/6-2-ray-propagation-model-part-2-1-320.jpg)
Recommended
More Related Content
What's hot
What's hot (20)
Similar to 6. 2 ray propagation model part 2
Similar to 6. 2 ray propagation model part 2 (20)
More from JAIGANESH SEKAR
More from JAIGANESH SEKAR (20)
Recently uploaded
Recently uploaded (20)
6. 2 ray propagation model part 2
- 1. WIRELESS COMMUNICATION SERIES 2 Ray Propagation Model [Part – 2]
- 2. Lecture Videos are available for this slides at www.youtube.com/gurukula Support by Subscribing to my Channel
- 3. 𝑬 𝒕𝒐𝒕 = 𝑬 𝒐 𝒅 𝒐 𝒅′ . 𝑪𝒐𝒔 𝝎 𝒄(𝒕 − 𝒅′ 𝒄 ) − 𝑬 𝒐 𝒅 𝒐 𝒅′′ . 𝑪𝒐𝒔 𝝎 𝒄(𝒕 − 𝒅′′ 𝒄 ) Path Difference (Δ) ≈ 𝟐. 𝒉 𝒕 𝒉 𝒓 𝒅 𝑷𝒉𝒂𝒔𝒆 𝑫𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 (𝜽∆) = 𝝎 𝒄.∆ 𝒄 𝑻𝒊𝒎𝒆 𝑫𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 (𝝉 𝒅) = 𝜽∆ 𝟐𝝅𝒇 It should be noted that when d becomes large, the difference between the distances d’ and d’’ becomes very small, and the amplitudes Elos and Eg are virtually identical and differ only in phase 𝐸 𝑜 𝑑 𝑜 𝑑 = 𝐸 𝑜 𝑑 𝑜 𝑑′ = 𝐸 𝑜 𝑑 𝑜 𝑑′′ If the received E – Field is evaluated at some TIME , say t = d’’ / c, then E Field Eqn. can be written as 𝐸𝑡𝑜𝑡(𝑑) = 𝐸 𝑜 𝑑 𝑜 𝑑 cos 𝜔𝑐. ∆ 𝑐 − cos(0) 𝐸𝑡𝑜𝑡(𝑑, 𝑡 = 𝑑′′ 𝑐 ) = 𝐸 𝑜 𝑑 𝑜 𝑑′ . 𝐶𝑜𝑠 𝜔𝑐( 𝑑′′ 𝑐 − 𝑑′ 𝑐 ) - 𝐸 𝑜 𝑑 𝑜 𝑑′′ . 𝐶𝑜𝑠 𝜔𝑐( 𝑑′′ 𝑐 − 𝑑′′ 𝑐 ) 𝐸𝑡𝑜𝑡(𝑑) = 𝐸 𝑜 𝑑 𝑜 𝑑 cos 𝜃∆ − cos(0)
- 4. 𝑐𝑜𝑠 𝜔𝑐 𝑡 − 𝑐𝑜𝑠 𝜔𝑐 𝑡 + 𝜃∆ = 2 𝑠𝑖𝑛 𝜃∆ 2 𝑠𝑖𝑛 𝜔𝑐 𝑡 + 𝜃∆ 2 𝐸𝑡𝑜𝑡(𝑑) = 𝐸 𝑜 𝑑 𝑜 𝑑 cos 𝜃∆ − cos(0) 𝐸𝑡𝑜𝑡(𝑑) = 2. 𝐸 𝑜 𝑑 𝑜 𝑑 𝑠𝑖𝑛 1 2 𝜔𝑐∆ 𝑐 𝐸𝑡𝑜𝑡(𝑑) = 2. 𝐸 𝑜 𝑑 𝑜 𝑑 𝑠𝑖𝑛 1 2 . 1 𝑐 . 2𝜋𝑓. 2ℎ 𝑡ℎ 𝑟 𝑑 𝑬 𝒕𝒐𝒕(𝒅) = 𝟐. 𝑬 𝒐 𝒅 𝒐 𝒅 𝒔𝒊𝒏 𝟐𝝅𝒉 𝒕 𝒉 𝒓 𝝀𝒅 𝑠𝑖𝑛 𝜃∆ 2 = 𝜃∆ 2 for θ < 0.3 𝑟𝑎𝑑 𝐸𝑡𝑜𝑡(𝑑) = 2. 𝐸 𝑜 𝑑 𝑜 𝑑 2𝜋ℎ 𝑡ℎ 𝑟 λ𝑑 𝑬 𝒕𝒐𝒕(𝒅) = 𝑬 𝒐 𝒅 𝒐 𝟒𝝅𝒉 𝒕 𝒉 𝒓 𝝀𝒅 𝟐 By using the identity, By using the Trigonometric identity, Relating Field to Power by using the simple relation (Field)2 = Power 𝑷 𝒓 = 𝑬 𝒕𝒐𝒕 𝟐 . 𝑨 𝒆 = 𝑬 𝒐 𝒅 𝒐 𝟐 . (𝟒𝝅) 𝟐 𝒉 𝒕 𝒉 𝒓 𝝀𝒅 𝟐 𝟐 . 𝑮 𝒓 𝝀 𝟐 𝟒𝝅 𝑷 𝒓 = 𝑬 𝒕𝒐𝒕 𝟐 . 𝑨 𝒆 = 𝑷 𝒕 𝑮 𝒕 . 𝒉 𝒕 𝒉 𝒓 𝝀𝒅 𝟐 𝟐 . 𝑮 𝒓 𝝀 𝟐 𝑷 𝒓 = 𝑷 𝒕 𝑮 𝒕 𝑮 𝒓 𝒉 𝒕 𝒉 𝒓 𝟐 𝒅 𝟒
- 5. 𝑷 𝒓 = 𝑷 𝒕 𝑮 𝒕 𝑮 𝒓 𝒉 𝒕 𝒉 𝒓 𝟐 𝒅 𝟒 𝑃𝑟 𝑃𝑡 = 𝑮 𝒕 𝑮 𝒓 𝒉 𝒕 𝒉 𝒓 𝟐 𝑑4 Calculating the Path Loss (PL) = Pt / Pr 𝑃𝑡 𝑃𝑟 = 𝑑4 𝑮 𝒕 𝑮 𝒓 𝒉 𝒕 𝒉 𝒓 𝟐 𝑷𝑳 𝒅𝑩 = 𝟒𝟎 𝐥𝐨𝐠 𝒅 − (𝟏𝟎 𝒍𝒐𝒈 𝑮 𝒕 + (𝟏𝟎 𝒍𝒐𝒈 𝑮 𝒓 + (𝟐𝟎 𝒍𝒐𝒈 𝒉 𝒕 + 𝟐𝟎𝒍𝒐𝒈(𝒉 𝒓) Path Loss for 2 Ray model is given by:
- 6. 𝑬 𝒕𝒐𝒕 = 𝑬 𝒐 𝒅 𝒐 𝒅′ . 𝑪𝒐𝒔 𝝎 𝒄(𝒕 − 𝒅′ 𝒄 ) − 𝑬 𝒐 𝒅 𝒐 𝒅′′ . 𝑪𝒐𝒔 𝝎 𝒄(𝒕 − 𝒅′′ 𝒄 ) Path Difference (Δ) ≈ 𝟐. 𝒉 𝒕 𝒉 𝒓 𝒅 𝑷𝒉𝒂𝒔𝒆 𝑫𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 (𝜽∆) = 𝝎 𝒄.∆ 𝒄 𝑻𝒊𝒎𝒆 𝑫𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 (𝝉 𝒅) = 𝜽∆ 𝟐𝝅𝒇 𝐸 𝑑, 𝑡 = 𝐸 𝑜 𝑑 𝑜 𝑑 . 𝐶𝑜𝑠 𝜔𝑐(𝑡 − 𝑑 𝑐 ) 𝐸𝑙𝑜𝑠 𝑑′, 𝑡 = 𝐸 𝑜 𝑑 𝑜 𝑑′ . 𝐶𝑜𝑠 𝜔𝑐(𝑡 − 𝑑′ 𝑐 ) 𝐸𝑔 𝑑′′, 𝑡 = 𝜞 𝐸 𝑜 𝑑 𝑜 𝑑′′ . 𝐶𝑜𝑠 𝜔𝑐(𝑡 − 𝑑′′ 𝑐 ) 𝑬 𝒕𝒐𝒕 = 𝑬𝒍𝒐𝒔 + 𝑬 𝒈 Δ = d’’ – d’ 𝒅′ = (𝒉 𝒕 − 𝒉 𝒓) 𝟐 + 𝒅 𝟐 𝒅′′ = (𝒉 𝒕 + 𝒉 𝒓) 𝟐 + 𝒅 𝟐 𝐸 𝑜 𝑑 𝑜 𝑑 = 𝐸 𝑜 𝑑 𝑜 𝑑′ = 𝐸 𝑜 𝑑 𝑜 𝑑′′ 𝑬 𝒕𝒐𝒕(𝒅) = 𝑬 𝒐 𝒅 𝒐 𝟒𝝅𝒉 𝒕 𝒉 𝒓 𝝀𝒅 𝟐 Relating Field to Power by using the simple relation (Field)2 = Power 𝑷 𝒓 = 𝑷 𝒕 𝑮 𝒕 𝑮 𝒓 𝒉 𝒕 𝒉 𝒓 𝟐 𝒅 𝟒 𝑷𝑳 𝒅𝑩 = 𝟒𝟎 𝐥𝐨𝐠 𝒅 − (𝟏𝟎 𝒍𝒐𝒈 𝑮 𝒕 + (𝟏𝟎 𝒍𝒐𝒈 𝑮 𝒓 + (𝟐𝟎 𝒍𝒐𝒈 𝒉 𝒕 + 𝟐𝟎𝒍𝒐𝒈(𝒉 𝒓) Path Loss for 2 Ray model is given by: