ECE 101 – Linear Systems Fundamentals Problem Set # 3A Solutions February 5, 2008 (send questions/comments to [email protected]) 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∑∞−∞=ktjkkoeaω1ejωot + a-1e-jωot+ a5ej5ωot + a-5e-j5ωot = jej(π/4)t + -je-j(π/4)t + 2ej5(π/4)t + 2e-j5(π/4)t, noting that ωo= 2π/T = 2π/8 = π/4. Now use Euler’s relation sin(ωot) = (1/2j)[ejωot– ejωot] on the first 2 terms, and cos(ωot) = (1/2)[ejωot+ ejωot] on the second 2 terms, to obtain: ⇒x(t) = –2sin(πt/4) + 4cos(5πt/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(πt/4 – π/2) + 4cos(5πt/4) 3.23(c)Here the only non-zero Fourier coefficients are a±1and 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 ak’s into the synthesis equation, as in the previous problem: x(t) = a1ejωot + a-1e-jωot+ a2ej2ωot + a-2e-j2ωot= jej(π/2)t + -je-j(π/2)t+ 2jej2(π/2)t + -2je-j2(π/2)tnoting that ωo= 2π/T = 2π/4 = π/2. Again, using Euler’s relation, we can write the complex exponentials as a sum of sines: ⇒x(t) = –2sin(πt/2) – 4sin(πt) 2. ANALYSIS 3.22(a), fig (b) Examining the graph, T = 6 ⇒ ωo= π/3. Note that the average value, over one period, of this function is 0.5. Thus, a0= 0.5. To compute the other ak’s, we need to use the analysis equation in (3.39) and do out the integral: ak= dtetxtjko∫−−33)(61ω= ])2(1)2([6121)3/(11)3/(12)3/(dtetdtedtettjktjktjk∫∫∫−−−−−−+−+++πππThe first integral is solved by a substitution and by-parts integration:
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