# help with transfer function

Discussion in 'Homework Help' started by kanin, Aug 29, 2010.

1. ### kanin Thread Starter New Member

Aug 29, 2010
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0
Hi!

i need help determining the transfer function, or at least the currents in the circuit,
will the current be the same thru R1 and R2?
will Vin = i2*R1?

grateful for any help or feedback!

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2. ### hgmjr Retired Moderator

Jan 28, 2005
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Go ahead and post what you have done so far toward finding a solution so that our members can assist you in getting past your snag.

hgmjr

3. ### kanin Thread Starter New Member

Aug 29, 2010
6
0
ok, this is what i have done,

H(w) = Vout/Vin

Vout = i2 * R4

the part i have problem with is:

is i1 = i2 + i3
or
is i1 = i2 ?

this i need to determine the total impedance Ztot, if statement 1 is correct, i just use the paralell impedance, if statement 2 is correct i just add the impedances affecting i2

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Jan 28, 2005
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5. ### Georacer Moderator

Nov 25, 2009
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When working with OpAmps we need to keep two things in mind.
1. The voltages at the input terminals are equal: V+=V-
2. The input terminals don't draw current: I+=I-=0

That said, you get I1=I2 in your schematic.

All the above, of course, stand for ideal OpAmps, but you can use this approach in a circuit of small demands like this.

6. ### kanin Thread Starter New Member

Aug 29, 2010
6
0
ok, thanks!

a supplementary question then

if i1 = i2 then i get Ztot = R2 + R3 +1/(jwc) + R4

this results in

Vin = i1 *(R2 + R3 + R4 + 1/(jwc) )
and
Vout = i1 * R4

and the transfer function H(w)=(Vout Vin)

H(w) = R4/(R2 + R3 + R4 + 1/(jwc))

is this correct and will the plot look like attachment?

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7. ### hgmjr Retired Moderator

Jan 28, 2005
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You need to revisit your solution.

One way to make this circuit more analysis-friendly is to take advantage of the opamp's features to permit you to break the overall circuit up into sub-circuits. You can actually analyze the opamp circuit on its own.

Next you analyze the circuit composed of R3, C, and R4.

Then you can obtain the overall transfer function by multiplying the transfer function of the opamp by the transfer function of the circuit made up of R3, C, and R4.

hgmjr

8. ### kanin Thread Starter New Member

Aug 29, 2010
6
0

didn't realy understand that, how do i get the transfer function of the opamp? by replace it with its components?

9. ### Georacer Moderator

Nov 25, 2009
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1,271
When I said above that I1=I2, I was referring to the currents flowing trough resistors R1 and R2. You hadn't marked the current on the sketch yet. From now on I' use your sketch's notation.

With the new notation, we have I2=-I3 and I1=0
Vout is indeed R4*I4 but
Vin=R1*I3

A way to solve the problem is the followin:
Since I3 is regulated by Vin, find it, and you automatically have I2 too.
If you know I2, you know the voltage drop across R2. That allows you to find the voltage at the output terminal of the OpAmp. From there on, you have a simple voltage divider, where you need to calculate the voltate after the capacitor (Vout).

Does your problem ask for Zout too, or you assumed you needed it?

10. ### kanin Thread Starter New Member

Aug 29, 2010
6
0

no, i was asked for the transfer function H(w) =Vout/Vin and the plot of |H(w)| as a function of w

what i have come up with so far is

i splitted the circuit to the left of R3 and got the voltage V2 and current I through R3 C and R4

V2=(Vin/R1)*(R1+R2)

I=V2/(R3+R4+1/(jwc))
and
Vout = R4 *I

this i put together to H(w)

is this correct?, how do i get the plot from this?

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Looks like you are close. After collecting and re-arranging all the algebraic terms what do you have for the final form of H(ω)?

Last edited: Aug 30, 2010
12. ### kanin Thread Starter New Member

Aug 29, 2010
6
0
i got

|H(w)| = (R4(R1+R2))/(R1*sqrt(1+(w/w0)^2))

13. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Presumably you have

$\omega_0=\frac{1}{(R_3+R_4)C}$

....?

Your |H(ω)| relationship seems to be missing an ωC (?) term in the numerator.

As a point of reference, keep in mind that as ω->∞ the value of |H(ω)| should tend to

$|H(\infty)|=\frac{R_4(R_1+R_2)}{R_1(R_3+R_4)}$

and at ω=0

$|H(0)|=0$

because the capacitor cannot pass DC

14. ### Georacer Moderator

Nov 25, 2009
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1,271
The expression for the transfer function is:

$H(w)=\frac{R_4}{R_1} \cdot \frac{R_1+R_2}{R_3+R_4+\frac{1}{jwC}}$

The amplitude of the transfer function is thus:

$|H(w)|=\frac{R_4}{R_1} \cdot \frac{R_1+R_2}{\sqrt{(R_3+R_4)^2+(\frac{1}{wC})^2}}$

You can input this expression in mathematica or excel to get an idea of its plot.
As the expression is in the form $|H(w)|=a \cdot (\beta + \gamma \cdot w^{-2})^{-\frac{1}{2}}$ I guess (I haven't actually tested it) it will look like a logarithmic curve, starting from zero and reaching a maximum amplitude after a certain frequency.

I suggest you pay more attention while replacing terms and performing operations while solving an equation. It costs only a few seconds of attention and you don't have to lose face for such a trivial set of steps.

Last edited: Aug 31, 2010
15. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I would like to stand to the OP's defence. If making mistakes is tantamount to losing face then count me in - happy to look foolish if it means I actually learn something.

I've made more than my share of mistakes in my professional life - let alone as a 'green' student struggling to learn what others seemed to find trivially simple.

16. ### t_n_k AAC Fanatic!

Mar 6, 2009
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789

$H(w)=\frac{(R_1+R_2)}{R_1}\frac{R_4}{(R_3+R_4+ \frac{1}{j \omega C})}$

$H(w)=\frac{(R_1+R_2)}{R_1}\frac{j \omega C R_4}{((R_3+R_4) j \omega C+1)}$

$H(w)=\frac{(R_1+R_2)}{R_1}\frac{j \omega C R_4}{(j \frac{\omega} {\omega_0}+1)}$

where

$\omega_0=\frac{1}{(R_3+R_4)C}$

I'll leave it to you to formulate the |H(ω)|

17. ### Georacer Moderator

Nov 25, 2009
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Sure, we all make mistakes everyday. But since a forum is a written form of expression and in a public environment, it would be best if we reviewed our post for mistakes, once we write it. It helps others under it better and saves us the time to come back later and try to remember what went wrong.

18. ### hgmjr Retired Moderator

Jan 28, 2005
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I have always considered mistakes to be the inevitable and unavoidable by-product of the learning process. If there are no mistakes then the student is probably not fully engaged in the learning process.

hgmjr

19. ### Georacer Moderator

Nov 25, 2009
5,150
1,271
Oh, well! Maybe I 'm too young not to think in the most competitive way I can.
Maybe I should let everyone have their own learning curves.

20. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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From my perspective it's probably a case of cutting the struggling student some slack on the homework forum. As everyone tells me, it takes ten positive statements to counteract a single negative statement directed at an individual.

In the real world it's usually a case of victory to the strongest & smartest. Work hard at training body & mind.

Who said the meek shall inherit the Earth?

BTW - You have a reasonable right to state your opinion - it's as valid as anyone else's. I'm not offended by your comments.