Check my working for an integral please?

[dariel]

Hiya,

I'm going through a worked solution, but my answer isn't matching my lecturers, if someone could point out the disparity I'd really appreciate it.
It's Elec Eng so j=i

First, the solution states:

[MATH] \alpha_n = \frac{1}{T_0}\left[ \int _0^\frac{T_0}{4}{4Ae^{-2jn \pi \frac{t}{T0}}}dt+\int _\frac{T_0}{4}^\frac{T_0}{2}{-4Ae^{-2jn \pi \frac{t}{T0}}}dt\right] [/MATH]
[MATH]\quad\; = \frac{-2A}{jn \pi T_0}\left[e^{-jn \frac{\pi}{2}}-1-(-1)^n\right][/MATH]
I understand [MATH]e^{-j n \pi} = \cos {(n \pi)} = (-1)^n [/MATH]but I don't see how the [MATH]T_0[/MATH] ends up in the denominator in the solution. When I work it out, it is in the numerator and cancels with the original [MATH]\frac{1}{T_0}[/MATH]I also end up with [MATH]e^{-jn \frac{\pi}{2}}[/MATH] occurring in the solution to both integrals, giving my final answer as
[MATH] \frac{-2A}{jn \pi}\left[2e^{-jn \frac{\pi}{2}}-1-(-1)^n\right][/MATH]
Can anyone see where I have gone wrong?

Many thanks,

D

 

[Otis]

Hello dariel. I used a computer algebra system (Maple). It has T_0 cancelled out, but it has j in the numerator, multiplied by positive 2.

\(\displaystyle \frac{2~j~A}{\pi~n} \bigg( 2~e^{-j~n~\pi/2} - 1 - (-1)^{n} \bigg)\)

By the way, we can't really help you to find disparities in your work without seeing it. There's more info about how to submit questions in the Read Before Posting notice. Cheers

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[dariel]

That's great! Thank you for your efforts.
It confirms what I thought, my answer matches the one from Maple, as if I multiply the initial term by [MATH]\frac{j}{j}[/MATH] the [MATH]j\cdot j[/MATH] in the denominator becomes [MATH]-1[/MATH] (being the property of the imaginary number).

I'm so sorry about the lack of working! I think I was so wrapped up in my question that I didn't think to look at it from an outsider's eyes.

I'll check out doing this on Maple too, have a student copy but not used it much.

 

[Deleted member 4993]

That's great! Thank you for your efforts.
It confirms what I thought, my answer matches the one from Maple, as if I multiply the initial term by [MATH]\frac{j}{j}[/MATH] the [MATH]j\cdot j[/MATH] in the denominator becomes [MATH]-1[/MATH] (being the property of the imaginary number).

I'm so sorry about the lack of working! I think I was so wrapped up in my question that I didn't think to look at it from an outsider's eyes.

I'll check out doing this on Maple too, have a student copy but not used it much.

This is a pretty simple integral - you don't have to use Maple to "solve" it. You can also use Wolframalfa.com to check your answer.

 

[Otis]

… you don't have to use Maple to "solve" it.

I'm thinking the OP had solved it before posting, but they couldn't match their answer to the instructor's. Seems like a typo in the material.

… You can also use Wolframalfa.com to check …

I tried that first (to post a link), but I couldn't get wolfram to recognize the entire expression. (It worked in parts, so not sure what's up with entering them all at once.) So, I used Maple, to check the answer.

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[dariel]

Yes, this is correct, Otis.

I'm coming to this level of engineering (Fourier series in this case) without any previous maths experience, so finding a new way (Maple) of checking my work is correct is really useful, though now I need to figure out how to use it accurately!
I'd actually spent a few hours trying to find why my answer didn't match my lecturers (usually it is my error )