How Do I Solve This Complex Fourier Series Problem?

[Ram_1]

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

 

[Redbelly98]

What have you been able to do or try so far for part (b), "write a mathematical specification for the function"?

 

[Ram_1]

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


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

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.

 

[Redbelly98]

Looks good so far. From the question wording, it sounds like they also want an explicit expression for g(t).

For the integral, it might be easier if you express e^(-jwt) in terms of sines and cosines.

Are you permitted to use a table of integrals?

edit:
p.s Welcome to Physics Forums!

 

[Ram_1]

Thank you very much for the welcome.

The explisit expression I think is:

g(t)=Sin(\omega t)

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

e^{-j \omega t}=Cos(\omega t)-j Sin(\omega t)?

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

 

[HallsofIvy]

Thank you very much for the welcome.

The explisit expression I think is:

g(t)=Sin(\omega t)

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

e^{-j \omega t}=Cos(\omega t)-j Sin(\omega t)?

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

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

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

(Sorry, I just can't bring myself to write "j" instead of "i"!)

 

[Redbelly98]

Thank you very much for the welcome.

The explisit expression I think is:

g(t)=Sin(\omega t)

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 e^{-j \omega t} would be written as:

e^{-j \omega t}=Cos(\omega t)-j Sin(\omega t)?

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

<br /> \sin (t) = \frac{e^{jt}-e^{-jt}}{2j}<br />

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.

 

[Ram_1]

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?

 

[Redbelly98]

Did you get an expression for the transform? You need to do that first. Then find the imaginary part of the expression.

 

[Ram_1]

Ok I have tried to get an expression using that substitution but I still can't figure it out.

 

[Defennder]

What have you got so far?

 

[Redbelly98]

Yes, please write out the integral you have after making the substitution -- the one for sin(t) given in Post #7.