Transfer function of this circuit

[cocopa]

Hi! What is the transfer function of this circuit when the potentiometer is at 10%,50% and 90% respectivelly? Can anyone help me?
Thank you.

 

[FvM]

The circuit doesn't work at any potentiometer position due to missing DC feedback.

Apart from this problem, the potentiometer divider factor will be multiplied with the feedback factor of the basic circuit, so you should be able to calculate total feedback factor and resulting gain.

 

[cocopa]

Thank for your reply, but what do you mean by DC feedback? How should it be connected to work?

 

[FvM]

You need to remove C2 to make the circuit work.

 

[_Eduardo_]

Also you can add a high value resistor in parallel with C2.
Anyway, will continue in problems when the pot is near closed.

 

[Ratch]

cocopa,

See the attachment for the transfer function.

Ratch

 

[FvM]

My comment about missing feedback didn't consider a possible outer feedback loop. If the circuit is used e.g. as PI controller, there's no problem. As Eduardo stated, the potentiometer ratio shouldn't approach zero.

The potentiometer simply introduces a gain multiplier for the circuit (assuming sufficient OP loop gain and neglecting potentiometer resistance). If k is the potentiometer ratio, e.g. 0.1, than the gain is multiplied by 1/k, e.g. 10.

Eo/Ei = -1/k * (1 + jωR3C2)/jωR4C2

 

[surz90]

yes, i agree with FvM,
the transfer function is Eo/Ei = -1/k*(R3/R4 + 1/jωR4C2).
but i think there is an effect because of resistance of potensiometer which is parallel with C2 and R3.
the output resistance of opamp is small,but in that circuit the output resistance will be Rp1 parallel with Rp2..see my attachment..
So i think the transfer function will be
Eo/Ei = -Rp1/(Rp1+Rp2)*((R3+Rp1||Rp2)/R4 + 1/jωR4C2).

 

[Ratch]

FvM,

Your calculation agrees with mine, IF the resistance of the pot is low enough so as not to compete in value with R3 and C2. I did calculate the transfer function in post #6 assuming a general value for the pot, and it becomes a much more complicated expression.

surz90,

I don't get the same transfer function as you posted. Perhaps you can look at the equations in post #6 and see if they agree with what you came up with.

Ratch

 

[FvM]

Your result is identical for Rpot = 0, suggesting that the derivation is correct as well. But it's in a rather confusing form. Numerator and denominator should be preferably written as products of zeros and poles.

 

[surz90]

Hi Ratch, i have see your attachment, actually i rather confuse with how you solve that equation..in your result i see an j is alone,but i consider in equation j must follow with jωC2..
I solve your equation using my way...you can see in my attachment..
I solve the transfer function of the circuit in 2 ways,and all of that give me the same transfer function..i directly assume that OP is ideal..because if it is non ideal the analysis become very confusing..
Hi FvM,,the complete analysis see in my attachment...i have written numerator and denominator in zero pole form..

if you have see my attachment, if the potensiometer don't burden the circuit, the transfer function will be same with transfer function which FvM did..
what do you think?

 

[surz90]

i think that the potensiometer doesn't give effect to the circuit, it must require that R3 >> Rpot..

 

[Ratch]

surz90,

Hi Ratch, i have see your attachment, actually i rather confuse with how you solve that equation..in your result i see an j is alone,but i consider in equation j must follow with jωC2..
I solve your equation using my way...you can see in my attachment..
I solve the transfer function of the circuit in 2 ways,and all of that give me the same transfer function..i directly assume that OP is ideal..because if it is non ideal the analysis become very confusing..
Hi FvM,,the complete analysis see in my attachment...i have written numerator and denominator in zero pole form..

I followed your solution, and you came up with the correct solution both times. You should multiply both the numerator and denominator of your solution by j to get rid of j in the denominator. Then your solution will have a lone j in the numerator. Othewise, your solution is identical to mine. Look at equation #6 in my solution. It is identical to yours. I did solve the problem with the amplification not equal to infinity (see equation #5) and there many more terms.

if you have see my attachment, if the potensiometer don't burden the circuit, the transfer function will be same with transfer function which FvM did..
what do you think? .

Yes, that is what I posted earlier.

Ratch

 

[surz90]

thank you Ratch,
i see now..now,i understand your solution..
i didn't aware that if i multiply both numerator and denominator with j,i got same solution with you in your equation #6..

 

[FvM]

Hi Ratch, i have see your attachment, actually i rather confuse with how you solve that equation..in your result i see an j is alone,but i consider in equation j must follow with jωC2..

As already mentioned, the form with the "lonely j" is confusing. Writing transfer functions in a standard form serves a purpose, I assume that Ratch will learn it as well, the latest when presenting similar calculations in an examination.

 

[Ratch]

FvM,

What is the "standard form"? Who designates it or what is its rationale. I usually rationalize my equations, i.e. remove "j"s from the denominator. I will never sit for an examination again.

Ratch

 

[FvM]

I will never sit for an examination again.

Me neither. But others contributing to this forum are required to use an understandable form, because they are teaching the stuff, writing reports for coworkers or customers, or passing examinations.

What is the "standard form"?

I think, writing numerator and denominator as products of zeros and poles. It gives an intuitive view on the transfer function's properties. Or other forms found in text books. Surely no detached "j".

 

[Ratch]

Fvm,

Do you mean in the frequency domain like they do in tables of LaPlace transforms? If not, then perhaps you could show me an example.

Ratch

 

[surz90]

Hi Ratch, i usually see the transfer function represent as a function of jw..F(jω) or F(s) in Laplace..
i never see tf write like yours, so at first it makes me confuse..
if tf present as a product of zeros and poles form, i think it is more easy to consider the frequency response of the circuit
for example(consider the magnitude response),

at 1st sight, i can see that the circuit have pole at ω = 0 and zero at -1/R3C2,
so i know that magnitude response will decrease -20dB each decade (start at ω = 0) until ω = 1/R3C2 after that the magnitude response will be constant.

and if the tf represent like you did,

Quote from your attachment,
Eo/Ei = (-R3ωC2 + j)/ωC2R4.

i can't make analysis like that at first sight.

How do you think?
i'm an undergraduate student.i'm learning about this and i'm not yet understanding it well.so,correct me if i'm wrong,

 

[Ratch]

surz90,

How do you think?
i'm an undergraduate student.i'm learning about this and i'm not yet understanding it well.so,correct me if i'm wrong,

I think that once you have the correct equation, you can convert it into any form you like.

Ratch