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Basic Trig Integral Question. |
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| Apr9-12, 02:07 AM | #1 |
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Basic Trig Integral Question.
1. The problem statement, all variables and given/known data
This question is part of Fourier Series in Circuit Analysis. There were fairly straightforward integrals which I calculated and confirmed using MAPLE to be correct, however the book gives somewhat different answers. I would presume that what I did was correct and the solutions manual made an error, however since it's a fairly large question with answers being carried forward I want to make doubly sure. Sorry about the size of the images, I will remove them after the problem This is the integral essentially, the definite integral from 2 to 4 is left out because it's zero, f(t) = 5 for 0 < t < 1 f(t) = 10 for 1 < t < 2 ![]() 2. Relevant equations cos (Pi/2) = (-1)[itex]^{\frac{n-1}{2}}[/itex] cos (Pi) = (-1)[itex]^{n}[/itex] 3. The attempt at a solution My answer came to this: ![]() EDIT: cos(nPi/2) goes to (-1)^n/2 - still doesn't reconcile my answers with the book though. The MAPLE output was: [itex]5\,{\frac {1+\cos \left( 1/2\,n\pi \right) -2\,\cos \left( n\pi \right) }{n\pi }}[/itex] The answer in the book was (last line before the table): ![]() As you can imagine, because the answers are different, the values in the table are going to be different and hence whatever I have to plot afterwards will be different. |
| Apr9-12, 09:45 AM | #2 |
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Recognitions:
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##(-1)^{integer}## cannot produce any zeros, it can only produce +1 or -1. So it doesn't replace ##cos(n \pi / 2)##.
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| Apr9-12, 10:29 AM | #3 |
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I should have been clearer, cos (n*Pi/2) is replaced by (-1)^n/2
So if n = 3, I'm guessing that term is ignored because you can't compute that. At least that's the identity they gave in the book. |
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