ClassNotes5

# ClassNotes5 - ECE595 Wireless Communications Lecture 5...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE595 Wireless Communications Lecture 5 Spring 2009 [Review of log-normal fading] ψ dB = 10log | ch | 2 , P ψ dB = 1 √ 2 πσ dB e- ( ψ dB- m dB ) 2 2 σ 2 dB , m dB = α dB- 10 n log d [Example] Best path loss fit is- 31 . 54 dB- 37 . 1log d for the data below. Find σ 2 dB , variance of log-normal fading d P r /P t 10 m-70 dB 20 m-75 dB 50 m-90 dB 100 m-110 dB 300 m-125 dB [Solution] Take the average out and average the rest for variance: σ 2 dB = 1 5 [(- 70 + 31 . 54 + 37 . 1log 10) 2 + + (- 75 + 31 . 54 + 37 . 1log 20) 2 + ··· = 13 . 29 σ dB = 3 . 65 2 Lecture 5 [Outage Probability] prob { P r dB ≤ P min } P r dB = P T dB + ψ dB (Normal distribution) prob { P r dB ≤ P min } = 1- Q P min- α dB- 10 n log d z }| { ( P T dB + K dB- 10 γ log d ) σ ψ dB Q ( z ) = 1 √ 2 π Z ∞ z e- x 2 / 2 dx [Example 2.5 of Goldsmith’s] Find outage probability at 150 m, P min =- 110 . 5 dBm ,P t = 10mW for previous example....
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

ClassNotes5 - ECE595 Wireless Communications Lecture 5...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online