{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Turbulence lecture 6

# Turbulence lecture 6 - Turbulence Lecture 6 Fix errors on...

This preview shows pages 1–4. Sign up to view the full content.

Turbulence Lecture 6 Fix errors on page 4 & 7. If U is uniformly distributed on the interval 0< U <2, zero elsewhere Show that average of ( ) ( ) ( ) u f P u f u du −∞ = ( ) ( ) 2 0 2 2 2 0 0 2 4 2 3 3 0 0 1 1 2 0 1 2 2 1 2 1 1 u 4 u 1 2 2 2 4 1 1 u 16 u u 2 2 4 8 du P u u du du = = = = = = = = = = 2 = Area under curve = 2 3 0 u u 16 d = Average value = 3 0 1 u u L d L = 2 3 0 1 2 u du 1

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

View Full Document
But incorrect to use. ( ) 2 3 3 0 P u u u What is anyhow? Not measured, or possibly if measured must be ( 3 P u ) 3 1 u . You could also show simply with a line or other function (*good test question* ) Joint Probability Define { } ( ) ( 1 if u , and , , , , ; , 0 otherwise u x t u u V v x t V u u V V x t χ < + ∆ < = ± ± ) V + ∆ [ ] [ ] ( ) ( ) 1 u , 1 lim , , , ; , , , , ; , prob V v , N N i U x t U u u V V x t u u V V x t U x t V V N χ χ →∞ = < + ∆ = = < + ∆ ± ± ± ± and if are small and U V ( ) , ; , P U V x t U V = ± 2
Check say and u , 1000 N = ( ) x t ± n 10 hit is uniformly distributed over interval . 0 10 and (the s per interval) < 0.1 u u

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.

{[ snackBarMessage ]}

### Page1 / 5

Turbulence lecture 6 - Turbulence Lecture 6 Fix errors on...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online