Turbulence lecture 6

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

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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 Pu f udu −∞ = 2 0 2 2 2 0 0 2 4 2 33 0 0 11 20 1 22 1 2 u4 u1 24 6 uu 48 du Pu ud u du =− = = === = == = 2 = Area under curve = 2 3 0 uu1 6 d = Average value = 3 0 1 L d L = 2 3 0 1 2 udu 1
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But incorrect to use. () 2 33 0 Pu u u What is anyhow? Not measured, or possibly if measured must be ( 3 Pu ) 3 1 u . You could also show simply with a line or other function (*good test question* ) Joint Probability Define {} ( )( 1 if u , and , ,, , ; , 0o t h e r w i s e ux t u u V v x t V uu VV x t χ ≤ < + ∆≤ < ∆∆ = ± ± ) V + [] ( ) 1 u, 1 lim , , , ; , , , , ; , prob Vv, N N i Ux t U x t x t U x tV V N χχ →∞ = <+ = ∆∆ = ± ±± ± and if are small and U V ,;, PUVxt U V =∆ ± 2
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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 uu ∆= [] 1000 1 11 ,; , 0 . 0
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This note was uploaded on 06/07/2011 for the course EGM 6341 taught by Professor Mei during the Spring '09 term at University of Florida.

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

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