OP amp analysis help

[aruna1]

i have this exam paper question.

1. derive an expression for transfer function
2. find expression for the magnitude and phase of the response

i have attached circuit diagram below

i derived

for first question
Vo= 2RC(dVi/dt) - Vi
(Vo/Vi)= 2sRC - 1

for second question
magnitude = |Vo/Vi|=|2jwRC-1|
and dont know how to find phase.

I'm not sure about my answers as i learned them from book that has no this kind a examples. so I'm glad if someone can show me how to solve this

Thanks

 

[Hayato]

Well, I'll try to show you the path. ;D

In the frequency dominium, a transfer function is Vo(w)/Vi(w), isn't it?

The first thing to do, is to find Vo(w). Apply a simple nodal/mesh analysis to do that:
->You know what are the resistor impedances (r) and capacitor impedances (1/jwc).
->You know that V- = V+, for an ideal OPAMP, and that there is no current flowing through V+ and V- terminals.

Well, to find the magnitude, is simple, you were able to do find.

To find the phase, is simple as well. Let the H(w) be the function that you want to find the phase.
So the phase will be arctan(|Im{H(w)}|/|Re{H(w)}|).

 

[Roff]

i have this exam paper question.

1. derive an expression for transfer function
2. find expression for the magnitude and phase of the response

i have attached circuit diagram below

i derived

for first question
Vo= 2RC(dVi/dt) - Vi
(Vo/Vi)= 2sRC - 1

for second question
magnitude = |Vo/Vi|=|2jwRC-1|
and dont know how to find phase.

I'm not sure about my answers as i learned them from book that has no this kind a examples. so I'm glad if someone can show me how to solve this

Thanks

Your answer to the first question is wrong. Think about it some more.

 

[Roff]

Well, I'll try to show you the path. ;D

In the frequency dominium, a transfer function is Vo(w)/Vi(w), isn't it?

The first thing to do, is to find Vo(w). Apply a simple nodal/mesh analysis to do that:
->You know what are the resistor impedances (r) and capacitor impedances (1/jwc).
->You know that V- = V+, for an ideal OPAMP, and that there is no current flowing through V+ and V- terminals.

Well, to find the magnitude, is simple, you were able to do find.

To find the phase, is simple as well. Let the H(w) be the function that you want to find the phase.
So the phase will be arctan(|Im{H(w)}|/|Re{H(w)}|).

Hayato, this is not meant as criticism. I'm just trying to help you make your posts more readable.
If you click on "Go Advanced" below the "Quick Reply" window, you should see a table of symbols next to the window. One which would have been useful in your previous reply is ω, which is more readable than w.

 

[Hayato]

Hayato, this is not meant as criticism. I'm just trying to help you make your posts more readable.
If you click on "Go Advanced" below the "Quick Reply" window, you should see a table of symbols next to the window. One which would have been useful in your previous reply is ω, which is more readable than w.

Thanks.
Since the advent of the "quick reply" button, I've been turning lazy.

 

[Roff]

Thanks.
Since the advent of the "quick reply" button, I've been turning lazy.

I understand completely. I sometimes type "ohms" because it is quicker than getting "Ω" from the symbol table.

 

[Hayato]

I understand completely. I sometimes type "ohms" because it is quicker than getting "Ω" from the symbol table.

Yes. And sometimes I'm afraid that the "Ω" would show as an unknown symbol like "?" or some strange character.

Sometimes I miss an equation editor like Mathtype.

 

[MrAl]

Hi,


I usually type 'ohms' too because it is more universal than typing
the actual symbol Ω, so some text editors may not recognize it.
Notepad does though.

A couple hints for the OP:

1. The time domain solution is an exponential.
2. The frequency response is the simplest possible.

Just curious, did anyone try to simulate this circuit yet?

 

[Hayato]

Nope but seems some sort of active filter to me.

 

[Roff]

Hi,


I usually type 'ohms' too because it is more universal than typing
the actual symbol Ω, so some text editors may not recognize it.
Notepad does though.

A couple hints for the OP:

1. The time domain solution is an exponential.
2. The frequency response is the simplest possible.

Just curious, did anyone try to simulate this circuit yet?

I have used this circuit in the past. It's an allpass network. With an ideal op amp, it has a gain magnitude of 1 from DC to blue light. Phase shift goes from 180 degrees at DC to zero degrees at high frequency. Even with a real op amp it is pretty darned flat if you choose one with decent gain at f=1/(2piRC). If you swap the R and the C, the phase shift is zero at DC and 180 degrees at high frequency. Gain is still flat.

 

[Roff]

Nope but seems some sort of active filter to me.

Yes. An allpass network is sort of a filter in the phase domain. The are very useful in correcting group delay problems in filters.

 

[xanadunow]

aruna1.. there will, probably be some challenges to my statent, that your "integral" insertion of a capacitor within the positive feedback will lead to some oscillation or just plain saturation of the output. No, I did not look in the books - it is just a feeling about it. Do not concern yourself with the phase considerations until you have this circuit working. Another approach taking dV/dt indicate the strong willingness to add a "kick" to the response in the PHASE opposite to the one you are probably after.

I do feel;you will achieve the saturation of the output within the capabilitiees of the +/- Vcc applied to your opamp. There is just something that dioes not sund right about this circiut?

Regards,
xanadunow

 

[xanadunow]

I would also like to see the resistor (half of the value of R1) going to the GND of the +/- Vcc source, attached at the + input of the opamp. Othrewise your + input is not very much referenced even thought it is preetty much - sitting on "0" V (thanks to the OpAmp)

Regards,
xanadunow

 

[xanadunow]

aruna1.. there will, probably be some challenges to my statement that your "integral" insertion of a capacitor within the positive feedback will lead to some oscillation or just plain saturation of the output. No, I did not look in the books - it is just a feeling about it. Do not concern yourself with the phase considerations until you have this circuit working. Another approach taking dV/dt indicates your strong willingness to add a "kick" in the response (more like a "D" action) in the PHASE, quite opposite to the one you are probably after.

I do feel;you will acheve the saturation of the output within the capabilitiees of the +/- Vcc app aligned/choosen to your opamp. There is just something that does not sound right about this circiut?

Regards,
xanadunow

 

[xanadunow]

Nope. there is no positive feedback from the output, just the integral action of the positive input.. the result is the "runaway" - saturation

 

[xanadunow]

did I say integral?

 

[xanadunow]

The "R" wasn't determined. I do feel the "D" action being to strong for the sme result.

 

[Roff]

The "R" wasn't determined. I do feel the "D" action being to strong for the sme result.

I don't really understand what you are saying, but I think it's good you defend the right to be wrong, since I think you are.
The circuit is perfectly stable. As I said, it's an allpass filter. Look it up.

The transfer function is

H(s) = (sRC-1)/(sRC+1).

 

[xanadunow]

I do fell it will run away..(in the common envirovement) Please give me the envirovement.. it was not supplied.. in this such basic circuit.

Regards,
xanadunow

 

[Roff]

I do fell it will run away..(in the common envirovement) Please give me the envirovement.. it was not supplied.. in this such basic circuit.

Regards,
xanadunow

I suspect you are not including a voltage source from Vi to GND in your "analysis".