# Mathematical question? (Impedance)

Discussion in 'Homework Help' started by ihaveaquestion, May 8, 2009.

1. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
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http://img253.imageshack.us/img253/9572/23525323.jpg

Looking at the example I did for the RC circuit at the top:

Vi = e^jt
H(jw) = Vo/Vi

If I divide over Vi that's essentially dividing over e^jt
In the answer though, for all four of these types of problems for that matter, in the book's answer, there is still a e^(phi)t left on the right hand side even though Vi is already divided over (which is the only reason I wrote them in my work)..

Why is that?

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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The transfer function H(jω) will not have any reference to either the input or output voltages - the exponential [=Vi] term should not be there. If the book has it there, it is wrong.

3. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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I'm not quite sure what you mean tnk...

H(jw) is not equal to Vo/Vi?

Or are you just saying the e^(phi)t that I'm asking about should not be there and the book is wrong for having it there?

4. ### steveb Senior Member

Jul 3, 2008
2,433
469
Note that the phase term is exp(j*phi) and not exp(phi*t)

You don't just put a exp(j*phi) into the equations like that. The phase angle is just that angle that shows up in the polar representation of the complex number which is the frequency response.

To clarify, the concept goes as follows. Even though complex exponentials do not exist in the real world, we can use them as mathematical tools. So, you think about putting a complex exponential as an input to the system (for example Vi=A*exp(jwt)). A linear system will always generate an output signal at the same frequency, but the amplitude and phase may change. (for example Vo=Ao*exp(jwt+j*phi).

You can now do very simple math to get the ratio of the output divided by the input. The answer is T=Vo/Vi=(A/A0)*exp(j*phi) which is just a complex number written in polar form. This can be converted to a complex number in rectangular form (T=x+jy). Note that in your problem you should not have just thrown in the exp(j*phi). Instead you calculate it by taking your answer in rectangular form and converting it to polar form.

You actually started correctly, but the artificial insertion of the exp(j*phi) was not mathematically correct. You eventually obtained the transfer function in rectangular form (ignoring the phase factor). At that point you need to just convert to polar form to get magnitude and phase.

5. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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Ok I'm pretty sure I see what you're saying steveb..

So basically my answer is correct w/out the e^j(phi)?

6. ### steveb Senior Member

Jul 3, 2008
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469
Just looking quickly, I'd say you did it correctly.

May 1, 2009
314
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cool thanks

8. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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I don't need to convert to polar form or anything though right?

Is the way I have it now (I erased the e^(phi)j stuff on them) correct?

http://img141.imageshack.us/img141/9572/23525323.jpg

Because it asked for the magnitude |H(jw)| and the phase angle and I found those

*sorry if I seem so redundant in asking silly questions over and over but my test is monday and I just want to make sure I'm ready and doing it right

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Sure, H(jw) is defined by the relationship Vo/Vi - but of itself the function H(jw) will not contain any reference to Vo or Vi.

Your original derivation incorrectly showed the input voltage Vi = exp(j phi) as being part of the function H(jw).

10. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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Can someone please confirm that what I have circled doing my way in the blue is equivalent to the teacher's way circled in the red, i.e. our two phase angles are the same.

http://img172.imageshack.us/img172/7593/555555yt.jpg

http://img139.imageshack.us/img139/4520/656766y.jpg

Letting A be whatever she has in her inverse tangent parenthesis

Her phase is

PI/2 + A

Whereas my phase is simply

1/A

I'm pretty sure the inverse tangent of PI/2 + A is equal to the inverse tangent of 1/A

More importantly is this part:

To find the magnitude, I would just go from where I have circled in green... I can see that our denominators would match when I take the magnitude of that... but when I look at taking the magnitude of the numerator I get a little confused because I have C*e^jt... If I just took the magnitude of C (which is itself C) I would be fine, but I'm just confused about that e^jt being there.

11. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
Hmm..

Thinking about it now, I suppose I don't even need to include the e^jt when I'm finding the magnitude:

The whole reason for using the e^jt was to take the real part (cos) of it in the end, so I don't think I need to include it in my calculation for magnitude.

So:
From the 'my answer' link, starting at the part circled in green

I would just move the e^jt outside of the green circle, take the magnitude and obtain
C/sqrt((1-LC)^2 + (RC)^2)

and my magnitude and my phase angle would be what is circled in blue
inverse tangent of (1-LC)/RC

Then to take the real part in the end just take the cosine

This would give me the same answer as her I believe

Any thoughts on whether what I'm saying makes sense/is accurate?

12. ### steveb Senior Member

Jul 3, 2008
2,433
469
Yes. The idea of the Fourier Transform (or Laplace Transform) is to eliminate the time dependence from the equations. The e^jwt is implied in all signals. If you write i(t)= some expression, then it is correct to keep the e^jwt, but usually we want to write I(w).

Another confusing thing is why does your teacher set w=1 and talk about e^jt rather than e^jwt. What is so special about w=1? Is she trying to implement some type of normalization?

Another thing that seems unusual is the use of Fourier (jw) rather than Laplace (s) transforms. It's much better to work in the s domain. It's easier to carry the "s" around rather than jw. Later you can substitute s=jw if you want to get back to the Fourier transform for frequency plots of magnitude and phase.

Last edited: May 9, 2009
13. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
Part of the problem tells you to let w=1, perhaps to make calculations easier I don't know.

And I don't know what Fourier/Laplace transforms are yet, that's next semester.

I've done every single homework problem over.. now just have to do the old test problems... test is Monday

14. ### steveb Senior Member

Jul 3, 2008
2,433
469
OK, the w=1 is just a special case for the particular problems you are solving. You should think of this w as a variable and realize that you are characterizing the response of the system as a function of frequency w.

Actually, you are doing Fourier Transforms now without realizing it. The magnitude and phase functions you are calculating are a function of frequency (w). This is the magnitude and phase of the Fourier Transform. Later they will make a big deal about it with integral transforms, but as you can see it is really very simple. It's all about how linear systems respond to sinusoidal signals.

Don't worry about my comment now. Later you will find that using the complex variable $s=\sigma+j\;\omega$ is more general and more flexible.