Understanding Fourier Transform: Solving a Homework Problem Step by Step

Aows

Homework Statement


Hello everyone,
am trying to solve this Fourier Trans. problem,
here is the original solution >> https://i.imgur.com/eJJ5FLF.pngQ/ How did he come up with this result and where is my mistake?

Homework Equations


All equation are in the above attached picture

The Attempt at a Solution


here is my attempt,

part 1>> https://i.imgur.com/DT2tJ0y.jpg

part 2>> https://i.imgur.com/jopEoQd.jpg

part 3>> https://i.imgur.com/cKXoekT.jpg
 
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Aows said:

Homework Statement


Hello everyone,
am trying to solve this Fourier Trans. problem,
here is the original solution >> https://i.imgur.com/eJJ5FLF.pngQ/ How did he come up with this result and where is my mistake?

Homework Equations


All equation are in the above attached picture

The Attempt at a Solution


here is my attempt,

part 1>> https://i.imgur.com/DT2tJ0y.jpg

part 2>> https://i.imgur.com/jopEoQd.jpg

part 3>> https://i.imgur.com/cKXoekT.jpg

The answer obtained in the "original solution" you posted happens to be correct, despite several blunders the author made along the way (for example, there is no ##\theta## in the exponential expression for ##\sin(t)##, and the limits on the indefinite integration are ##|_0^{\pi}##, not the ##|_{\pi}^0## written by the author). I could follow your work until the middle of part 2, then it got too messy and I gave up. Please try to follow PF standards and type out your work, or else at least write it out clearly and neatly before photographing it. Your final result has an extra ##\omega## in front of the ##\sin (\omega \pi)##.

BTW: many helpers will not look at posted images (rather than typed work); usually I will not look at them either, but today I made an exception, and then had to give up part way through.
 
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In part 2, going from line 3 to line 4, you forget to distribute accordingly: e^{j\pi(1-\omega)}=e^{j\pi}e^{-j\pi\omega} Instead, you distribute as if the negative sign is not there.
 
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Daniel Gallimore said:
In part 2, going from line 3 to line 4, you forget to distribute accordingly: e^{j\pi(1-\omega)}=e^{j\pi}e^{-j\pi\omega} Instead, you distribute as if the negative sign is not there.
thanks indeed,
is this the only mistake ?
 
Ray Vickson said:
The answer obtained in the "original solution" you posted happens to be correct, despite several blunders the author made along the way (for example, there is no ##\theta## in the exponential expression for ##\sin(t)##, and the limits on the indefinite integration are ##|_0^{\pi}##, not the ##|_{\pi}^0## written by the author). I could follow your work until the middle of part 2, then it got too messy and I gave up. Please try to follow PF standards and type out your work, or else at least write it out clearly and neatly before photographing it. Your final result has an extra ##\omega## in front of the ##\sin (\omega \pi)##.

BTW: many helpers will not look at posted images (rather than typed work); usually I will not look at them either, but today I made an exception, and then had to give up part way through.
thanks a lot for your advices and comment, can you spot more mistake if any available ? that would help me a lot.
 
Aows said:
thanks a lot for your advices and comment, can you spot more mistake if any available ? that would help me a lot.

I stated already that I would not attempt to read what you did past the middle of part 2, because it is too messy.
Aows said:
thanks a lot for your advices and comment, can you spot more mistake if any available ? that would help me a lot.

You are doing it the hard way. It is much easier to evaluate ##F(b) = \int_0^{\pi} \sin(t) e^{bt} \, dt## using integration by parts (twice)---anyway, a standard calculus 101 homework problem---simplifying as much as possible. THEN put ##b = -j \omega.##
 
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Ray Vickson said:
I stated already that I would not attempt to read what you did past the middle of part 2, because it is too messy.You are doing it the hard way. It is much easier to evaluate ##F(b) = \int_0^{\pi} \sin(t) e^{bt} \, dt## using integration by parts (twice)---anyway, a standard calculus 101 homework problem---simplifying as much as possible. THEN put ##b = -j \omega.##
it is necessary to solve it like this but i don't think it is messy (dont pay attention to the RHS of the paper) the solution is on the LHS.
 
Aows said:
Hello Dr. Ray and Daniel,
I've made some corrections which made my solutions very close to the original solution but still some mistake, can you identify them ??
here is my attempt:
https://i.imgur.com/jkVdK7z.jpg

and here is the original solution:
https://i.imgur.com/eJJ5FLF.png
You should really write your math out in your post instead of attaching a file. There are tutorials for typesetting math using LaTeX commands on the site (highly recommend). Additionally, people on the forum will likely get tired of pointing out individual mistakes to you. Nevertheless, from line 2 to 3 on the document where you provide your most recent attempt, you pull the (-1)s out, but not the -1s beside the exponentials. If you factor out the minus signs correctly, the -1s will turn into +1s. Carry it forward, and you'll get the correct answer.
 
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Mr. Daniel where is the mistake in my pulling the (-1), kindly, can you explain more ? that would help me a lot?
 

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