ECE 101 problem set solution3A

ECE 101 problem set solution3A - ECE 101 Linear Systems...

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ECE 101 – Linear Systems Fundamentals Problem Set # 3A Solutions February 5, 2008 (send questions/comments to psiegel@ucsd.edu) 1. SYNTHESIS 3.21 The Fourier coefficients are given, so we can plug them into the synthesis equation (3.38) to obtain the signal x(t): x(t) = = a −∞ = k t jk k o e a ω 1 e j ω o t + a -1 e -j ω o t + a 5 e j5 ω o t + a -5 e -j5 ω o t = je j( π /4)t + -je -j( π /4)t + 2e j5( π /4)t + 2e -j5( π /4)t , noting that ω o = 2 π /T = 2 π /8 = π /4. Now use Euler’s relation sin( ω o t) = (1/2j)[e j ω o t – e j ω o t ] on the first 2 terms, and cos( ω o t) = (1/2)[e j ω o t + e j ω o t ] on the second 2 terms, to obtain: x(t) = –2sin( π t/4) + 4cos(5 π t/4) To get x(t) into the desired form of purely cosine functions, note that the sin function is just the cos function shifted π /2 to the right. Thus, we have: x(t) = –2cos( π t/4 – π /2) + 4cos(5 π 3.23(c) Here the only non-zero Fourier coefficients are a ±1 and a ±2 . Note that the non-zero coefficients are also purely imaginary numbers, so we should expect x(t) to be an odd function (i.e., a sum of sines, no cosines). To find x(t), we just plug the a k ’s into the synthesis equation, as in the previous problem: x(t) = a 1 e j ω o t + a -1 e -j ω o t + a 2 e j2 ω o t + a -2 e -j2 ω o t = je j( π /2)t + -je -j( π /2)t + 2je j2( π /2)t + -2je -j2( π /2)t noting that ω o = 2 π /T = 2 π /4 = π /2. Again, using Euler’s relation, we can write the complex exponentials as a sum of sines: x(t) = –2sin( π t/2) – 4sin( π t) 2. ANALYSIS 3.22(a), fig (b) Examining the graph, T = 6 ⇒ ω o = π /3. Note that the average value, over one period, of this function is 0.5. Thus, a 0 = 0.5. To compute the other a k ’s, we need to use the analysis equation in (3.39) and do out the integral: a k = dt e t x t jk o 3 3 ) ( 6 1 = ] ) 2 ( 1 ) 2 ( [ 6 1 2 1 ) 3 / ( 1 1 ) 3 / ( 1 2 ) 3 / ( dt e t dt e dt e t t jk t jk t jk + + + + π The first integral is solved by a substitution and by-parts integration:
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du ue u t dt du t u dt e t u jk t jk = = + = + 1 0 ) 2 )( 3 / ( 1 2 ) 3 / ( 2 , 2 ) 2 ( π For convenience, let A = –jk π /3. Then the integral becomes: A A A A A Au A u A e A e A e A A e A A e e A A u e du ue 2 2 2 2 2 2 1 0 2 2 1 0 ) 2 ( 1 1 1 ] 1 ) 1 1 [( ) 1 ( + = + = = The second integral is simply integrating e At
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This note was uploaded on 03/16/2009 for the course ECE 101 taught by Professor Siegel during the Spring '08 term at UCSD.

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ECE 101 problem set solution3A - ECE 101 Linear Systems...

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