How Do I Solve This Complex Fourier Series Problem?

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Hi there,

I have been having problems with the question that I have attached, I have given it a go but unfortunately don't have a scanner to scan any of my work in but I will post it as soon as I can. Until then I was hoping that someone would be willing to give me a worked example of a similar question or just point me in the right direction as I am almost completely lost and the only part that I can answer confidently is part (a).


Many thanks in advance


Ram
 

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I have gone as far as doing the Fourier transform but again I'm not entirely sure that this is correct, I got:


[tex]G(f) =[/tex][tex]\int^{\pi}_{-\pi}Sin(t)e^{-j}^{\omega}^{t}dt[/tex]

I am not sure if this is as far as I need to go or not, I tried to complete the integration by using the integration by parts technique but got a very jumbled answer. The last part of the question (sketching the imaginary part of the transform) I haven't been able to do at all but I get the feeling it's something I could do if pointed in the right direction and if I manage to get the first part of the question right.
 
Thank you very much for the welcome.

The explisit expression I think is:

[tex]g(t)=Sin(\omega t)[/tex]

Would I be right in saying that [tex]e^{-j \omega t}[/tex] would be written as:

[tex]e^{-j \omega t}=Cos(\omega t)-j Sin(\omega t)[/tex]?

and yes we do have a limited table of integrals and derivatives.
 
Ram_1 said:
Thank you very much for the welcome.

The explisit expression I think is:

[tex]g(t)=Sin(\omega t)[/tex]

Would I be right in saying that [tex]e^{-j \omega t}[/tex] would be written as:

[tex]e^{-j \omega t}=Cos(\omega t)-j Sin(\omega t)[/tex]?

and yes we do have a limited table of integrals and derivatives.

Yes, [itex]e^{-i\omega t}= cos(-\omega t)+ i sin(-\omega t)= cos(\omega t)- i sin(\omega t)[/itex] because cosine is an even function and sine is an odd function.

You could also do this the other way around:
[tex]sin(\omega t)= \frac{e^{\omega t}- e^{-\omega t}}{2i}[/tex]

(Sorry, I just can't bring myself to write "j" instead of "i"!)
 
Ram_1 said:
Thank you very much for the welcome.

The explisit expression I think is:

[tex]g(t)=Sin(\omega t)[/tex]
I see a problem with that expression. For example, it gives g(3pi/2) is -1, whereas the graph you provided clearly shows g(3pi/2) is zero.

Would I be right in saying that [tex]e^{-j \omega t}[/tex] would be written as:

[tex]e^{-j \omega t}=Cos(\omega t)-j Sin(\omega t)[/tex]?

Yes, that's right. But HallsofIvy gives an even better way to evaluate the integral, using the substitution

[tex] \sin (t) = \frac{e^{jt}-e^{-jt}}{2j}[/tex]

and yes we do have a limited table of integrals and derivatives.

Just as a general suggestion, it could help in your studies to have a full table of integrals handy. You can tell it's a full table if it has something like 700 or 800 integrals (or possibly more) included.
 
Ok, I think I understand all that now but the next part of the question

"sketch a labelled graph of the imaginary parts of the transform"

is where I really start to struggle as I don't really know what it's asking for, do you think you could point me in the right direction as I haven't even been able to start this part?
 
Ok I have tried to get an expression using that substitution but I still can't figure it out.