# Fourier Question

1. Aug 20, 2008

### 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

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2. Aug 20, 2008

### Redbelly98

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

3. Aug 21, 2008

### 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.

4. Aug 21, 2008

### Redbelly98

Staff Emeritus
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!

5. Aug 21, 2008

### 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.

6. Aug 21, 2008

### HallsofIvy

Staff Emeritus
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"!)

7. Aug 21, 2008

### Redbelly98

Staff Emeritus
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.

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

$$\sin (t) = \frac{e^{jt}-e^{-jt}}{2j}$$

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.

8. Aug 25, 2008

### 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?

9. Aug 25, 2008

### Redbelly98

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

10. Aug 26, 2008

### Ram_1

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

11. Aug 26, 2008

### Defennder

What have you got so far?

12. Aug 26, 2008

### Redbelly98

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