# Frequency Response - Gain & Network Function

Discussion in 'Homework Help' started by jegues, Nov 28, 2010.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Everything given in the problem is contained in my figure. The input voltage is Vi(t) and the response is the voltage across RL, Vo(t).

R1 is given to 10k and C to be 0.1uF.

They want is to design this circuit to satisfy these 2 conditions:

a) The gain at low frequencies is 5.

b) The gain at high frequencies is 2.

I've got my transfer function H(w) and I've managed to get the equation for condition a, but I don't know what to do for condition b.

How do I handle this? It looks like if I let omega approach infinity everything it's going to give me something similar to infinity over infinity, and I can't work with that.

How should I proceed?

Thanks again!

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2. ### hgmjr Moderator

Jan 28, 2005
9,030
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How about the value for R2?

hgmjr

3. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
We are suppose to choose values for R2 and R3 to satisfy the conditions mentioned above.

4. ### Fraser_Integration Member

Nov 28, 2009
142
5
don't know if I'm missing a step, or if you must solve it quantatively but qualatitively:

C1 will have a very low impedance at high frequencies, so will effectively short out R3. This means that only R2 will be in the feedback path, so it must be 20kohm to give you the -2 gain.

At low frequencies, C1 is high impedance, so the parallel combination of it and R3 is almost exactly R3. So both R2 and R3 are in feedback path. Then R3 must equal 30k to give a total Rf of 50k, which gives you your minus 5 gain

5. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Wow that's actually really simple after noticing what I should have been doing. I never thought of actually looking back at the original circuit and thinking, my attention was focused on my nasty expression for H(w).

There's only one part of your explanation that confuses me.

Agreed.

How do you draw this conclusion? The way I see it is:

There is a really big number, call it, $z_{c}$ in parallel with R3 like so,

$\frac{z_{c} \cdot R_{3}}{z_{c} + R_{3}}$

How does this become $R_{3}$?

6. ### hgmjr Moderator

Jan 28, 2005
9,030
214
I understand now.

hgmjr

7. ### Fraser_Integration Member

Nov 28, 2009
142
5
It's your lucky day I have not started drinking yet so can offer you a decent quantative answer!

Your transfer function is correct. Put w as 0 onto it and you end up with R2 + R3 / R1

great.

if you try and put infinity in there though, you end up with infinity/infinity which is meaningless. what you must do is apply l'hopitals rule to the expression. you can look it up but essentially take the derivative of top and bottom with respect to w and you will be left with jR2R3C / jR1R3C which helpfully reduces to R2 over R1 and hey presto!

8. ### Fraser_Integration Member

Nov 28, 2009
142
5
Erm I'm not sure how to put in in general terms, it's just one of those rules of thumb. Like two equal resistors will give you exactly half in parallel. Put it this way, as R becomes low compared with Zc, then Zc+R in the denominator roughly equals Zc, so you are left with just (Zc*R)/Zc which equals R

Try it with a 1k resistor and a 100k resistor and you will see what I mean.

Last edited: Nov 28, 2010
9. ### Fraser_Integration Member

Nov 28, 2009
142
5
here is l'hopitals rule

10. ### hgmjr Moderator

Jan 28, 2005
9,030
214
The expression for R3 in parallel with infinity need not force you to resort to L'hopital's since the identity for the inverse of infinity is 0.

R3||∞ = $\frac{1}{\frac{1}{\infty}+\frac{1}{R3}}$

R3||∞ = $\frac{1}{0+\frac{1}{R3}}$

R3||∞ = ${R3}$

hgmjr

11. ### Fraser_Integration Member

Nov 28, 2009
142
5
L'Hopital was for the transfer function query, not the parallel resistors. I would agree that would be overkill for just that!

12. ### hgmjr Moderator

Jan 28, 2005
9,030
214
I think for the gain it is feasible to resort to "Initial Value Theorem" and "Final Value Theorem"

13. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Thanks to both of you!

All my confusions have been cleared up!

14. ### Fraser_Integration Member

Nov 28, 2009
142
5
no problem.

hgmjr - funny you should say that, I think those two are coming up in my course after Christmas.