From awerxrm at ece.msstate.edu Thu Feb 1 02:19:57 2007 From: awerxrm at ece.msstate.edu (Kerry) Date: Thu Feb 1 02:20:22 2007 Subject: [ece3163] [SPAM]heir with st Message-ID: <092b01c74625$01c74625$e0e4f1db@ece.msstate.edu> Skipped content of type multipart/alternative-------------- next part -------------- A non-text attachment was scrubbed... Name: not available Type: image/gif Size: 4920 bytes Desc: image475.gif Url : http://www.ece.msstate.edu/pipermail/ece3163/attachments/20070201/11a2896a/attachment.gif From uryb at ece.msstate.edu Mon Feb 5 05:57:33 2007 From: uryb at ece.msstate.edu (Connie) Date: Mon Feb 5 07:58:02 2007 Subject: [ece3163] [SPAM]Hot Stock Report pituitary Message-ID: <79d601c748ea$01c748ea$62e0d2d8@ece.msstate.edu> Tammy -------------- next part -------------- An HTML attachment was scrubbed... URL: http://www.ece.msstate.edu/pipermail/ece3163/attachments/20070205/85e1934c/attachment.html From np1 at ece.msstate.edu Thu Feb 8 21:18:21 2007 From: np1 at ece.msstate.edu (Naveen Parihar) Date: Thu Feb 8 21:18:27 2007 Subject: [ece3163] Re: Homework!! In-Reply-To: <932242.1170950934502.JavaMail.dmj52@msstate.edu> References: <932242.1170950934502.JavaMail.dmj52@msstate.edu> Message-ID: > I ran into some trouble on > the first problem c-f. I am not quite sure how to go from the step of > substituting. I have: > > c). x(t) = cos(t-1) + sin(t-1/2) > = 1/2[e^(j(t-1)) + e^(-j(t-1))] + 1/2j[e^(j(t-1/2)) - > e^(-j(t-1/2))] > This is what I went to next: > = 1/2[e^(jt)e^(-j) +e^(-jt)e^(j)] + 1/2j[e^(jt)e^(-j/2) - > e^(-jt)e^(j/2)] > Now I'm not sure as to what to do. > I said when k=1 = e^(-j) and e^(-j/2) ; k=-1 = e^(j) and > e^(j/2) . Not exactly. You need to group the terms so that you are left with two terms - one will have some factor multiplied by e^(jt) and another one will have some other factor mutlplied by e^(-jt). We are trying to arrange the terms so that we can get to the exponential fourier series similar to what we did for the first two examples in the class. So, in this example, we get, x(t) = ((1/2)e^(-j)) - (j/2)e^(-j/2)) e^jt + ((1/2)e^j + (j/2)e^(j/2)) e^(-jt) So, the coefficient of e^jt is Ck when k=1, and the coefficinet of e^(-jt) is Ck when k=-1. > d). x(t) = cos2t sin3t > = 1/2[e^(2jt) +e^(-2jt)] * 1/2[e^(3jt)-e^(-3jt)] You have missed the j in the denomintor while expanding sin as exponential. Remmeber, sin(x) = (1/2j)(e^(jx) - e^(-jx)) > wo = 1 and k = 2, k=-2, k= 3 and k=-3. Then I plotted on -2, -3, 2, > and 3 to ' 1/4' I could get a phase graph from those numbers. > > e). x(t) = cos(5t)cos(5t) > = 1/2[e^(5jt) + e^(-5jt)] * 1/2[e^(5jt) + e^(-5jt)] > = 1/2[(5-j)e^(5jt)] * 1/2[(5-j)e^(5jt)] Not sure how you got this from the previous step. You need to multiple using the following formulae: (a+b)(c+d) = ac + ad + bc + bd > = 1/4(5-j)^2e^(10jt)] > Plot on 10 and -10 to 'sqrt(26)/4 or 1.27' > phase graph at 78.7deg > > > f). x(t) = cos(3t) + cos(5t) > = 1/2[e^(3jt) + e^(-3jt)] + 1/2[e^(5jt) + e^(-5jt)] > = 1/2[(3-j)e^(3jt)] + 1/2[(5-j)e^(5jt)] This step is not correct - two exponents with different powers cannot be combined in addition. For example, e^x + e^y cannot be combined. For this specific problem, you will be left with sum of four terms. Hence, there are four values of k where Ck is not zero. These vour values of k's are k=3, -3, 5, -5. > w = 1; k= 3 and k= 5 @ k=3 = 1/2(3-j) and k=5 = 1/2(5-j) > Ploted at 3 and 5 @ sqrt(26)/2 and phase angle at (-78.7deg) > > These are the steps that where taking to solve these problems and I am > not sure if I did them right, so if you would look at what I have. I > would really appreciate it. > > Also on 3.13d. I got the same results as we did for c. is that correct. No. Similar to the problem 3.13c that we solved in class, you can solve 3d. It is just that it has two components. For 3.13c, we substituted, k' = k+1. One the term in 3.13d will have to dealt with in same manner. However, for the second tern, you will need to substitute k'=k-1. Once, these substitutions are done, you will get sum of two complex exponential fourier serier. This sum can next be combined into one by taking e^(jkw0t) ommon. -Naveen From ggdz3dz2 at ece.msstate.edu Thu Feb 15 07:04:32 2007 From: ggdz3dz2 at ece.msstate.edu (Blanche Jacobsen) Date: Thu Feb 15 07:08:22 2007 Subject: [ece3163] cash flow slow? get a diploma Message-ID: <5009D187.467812.42851@LTGE> dz2@ece.msstate.edu guess what! Go to col!llege online: !Accounting !Business !Criminal! justice !Educ!ation !Health !Technology No study required 100% verifiable Call Today 206-333-1249 (7 days a week, 24 hoursw a day) Take Care Juliana Roark verve noble frivolity gmt monaco project slog broom pragmatist truly hornbeam constraint pest beltsville aquarius sepulchral cuff bethought lingo antiquated norm penance offensive blackmail From rj at ece.msstate.edu Mon Feb 26 06:26:52 2007 From: rj at ece.msstate.edu (Marion) Date: Mon Feb 26 01:03:30 2007 Subject: [ece3163] [SPAM]Personal Finance Message-ID: <00a401c759a1$01c759a1$1295becb@ece.msstate.edu> transmitting Ronny basket -------------- next part -------------- An HTML attachment was scrubbed... URL: http://www.ece.msstate.edu/pipermail/ece3163/attachments/20070226/d7cb13a1/attachment.html