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# The present problem statement only requires that we

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the present problem statement only requires that we evaluate equation ( c ) for the one value of time

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ECE 303 - Sum2018 Turn-In Assignment 1: 25 May 2018 2 Problem 1. (cont.) given by t = 10 ns = 10 8 sec. Therefore, substitute t = 10 8 into ( c ). Using v p = 2(10) 8 , the argument of the exponential in ( c ) thus becomes 2(10) 8 parenleftBig t - ( x / v p ) parenrightBig = 2(10) 8 parenleftbigg 10 8 - x 2(10) 8 parenrightbigg = 2 - x ( d ) Using these same values, the argument of the cosine in ( c ) thus becomes 10 9 π parenleftBigg t - x v p parenrightBigg = 10 9 π parenleftbigg 10 8 - x 2(10) 8 parenrightbigg = 10 π - 5 π x ( e ) Using the same values, the bounded inequality on the right side of ( c ) becomes v p ( t - T ) x v p t 2(10) 8 parenleftBig 10 8 - 4(10) 9 parenrightBig x 2(10) 8 × 10 8 ( f ) Completing the computations in the inequality on the right side of ( f ) then gives the following simpler form: 2(10) 8 parenleftBig 10 8 - 4(10) 9 parenrightBig x 2(10) 8 × 10 8 1 . 2 x 2 ( g ) Substituting ( d ), ( e ), and ( g ) into equation ( c ) then gives g ( x ) = f parenleftBig t 1 - x v p parenrightBig = braceleftBigg 5 e x 2 cos parenleftBig 10 π - 5 π x parenrightBig , 1 . 2 x 2 0 , else ( h ) We can simplify ( h ) somewhat by rewriting the exponent as 5 e x 2 = 5 e 2 e x = 0 . 677 e x ( i ) Now recall that cos( α ) = cos( - α ), and cos( α ) = cos( α + 2 ) for any integer n . Using these properties then gives the following: cos parenleftBig 10 π - 5 π x parenrightBig = cos parenleftBig - 5 π x parenrightBig = cos parenleftBig 5 π x parenrightBig ( j ) Substituting ( i ) and ( j ) into ( h ) then gives the desired form for the solution: g ( x ) = braceleftBigg 0 . 677 e x cos parenleftBig 5 π x parenrightBig , 1 . 2 x 2 0 , else ( k ) Comparing ( k
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• Spring '08
• ALEXANDER
• Following, Trigraph, Transmission line, V1, Chebyshev's inequality

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