OP amp analysis help-exam paper question

[aruna1]

Homework Statement

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

2. The attempt at a solution

i derived

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

for second question
magnitude = |Vo/Vi|=|2jwRC-1|
and don't 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

 

[yungman]

Homework Statement

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

2. The attempt at a solution

i derived

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

for second question
magnitude = |Vo/Vi|=|2jwRC-1|
and don't 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

I assume it is an ideal voltage source which is 0 impedance. You basically divide the circuit into 2 half.

1) the C and R that form the high pass into the +ve input. The output is just gain of 2 of the voltage at the junction of the C and R.
2) the Inverting configuration form by the two resistors into the -ve input and feedback from the output. This has an inverted gain of 1 at the output.
3) Using super position to sum the two transfer functions as the final output. Basically, at low freq, no signal input into the +ve input so the circuit has a gain of -1. At break frequency of C and R, signal start to drive into +ve input and start canceling the other side so output roll off. At break frequency, output is 0. Beyond break freq, the circuit has a gain of +1.

I post a wrong answer 2 hours ago, sorry. I am also experimenting puting a jpeg on this also, apparantely it still won't display on the post. Still come out as an attached file. Please double check my work. I have been making mistake lately at my old age! But I am sure the idea of super position stand. Just calculate the result in polar form to get the phase angle.

 

[aruna1]

thank you sir for your help.but i can't download your attachment.no download link appears.can you re upload it or send it to my email
aruna.rubasinghe@gmail.com

 

[yungman]

I have send it to your email.

I can down load from here. Must be still waiting for approval. I still have not find a way to put the drawing directly onto the post, any attachment need a day for the moderator for approval. Bummer!

Almost forgot! Break freq. is 1/(2 \pi RC).

Let me know if you receive it and we can talk more. Just remember I am not a teacher or tutor, double check my stuff.

 

[aruna1]

3) Using super position to sum the two transfer functions as the final output. Basically, at low freq, no signal input into the +ve input so the circuit has a gain of -1. At break frequency of C and R, signal start to drive into +ve input and start canceling the other side so output roll off. At break frequency, output is 0. Beyond break freq, the circuit has a gain of +1.
.

it seems your calculations are right but i have a question.

well when at break frequency you say out put is zero but i was wandering doesn't capacitors lag voltage by 90 degrees? so two inputs reach volt V at two different moments there fore there will be non zero output?
i don't know whethere I'm right or wrong so correct me.
thanks

 

[yungman]

it seems your calculations are right but i have a question.

well when at break frequency you say out put is zero but i was wandering doesn't capacitors lag voltage by 90 degrees? so two inputs reach volt V at two different moments there fore there will be non zero output?
i don't know whethere I'm right or wrong so correct me.
thanks

Good question! One thing is at break frequency, it is only 45 deg. lag. At low freq., Point A is 90 deg lag but the amplitude is low, don't matter. At freq>>break freq. there is very little lag so it is pretty much no lag.

I think if you plot the equaion out, you should see the respond. Seems like the amplitude will go very low at break frequency. The signal actually go through 180 deg shift( from -Vin to +Vin).

I am an engineer and I have been designing opamp stuff for many years. In real life, we have a requirement and we provide a solution! This will not be a circuit I would use for a notch filter.

 

[aruna1]

Good question! One thing is at break frequency, it is only 45 deg. lag. At low freq., Point A is 90 deg lag but the amplitude is low, don't matter. At freq>>break freq. there is very little lag so it is pretty much no lag.

I think if you plot the equaion out, you should see the respond. Seems like the amplitude will go very low at break frequency. The signal actually go through 180 deg shift( from -Vin to +Vin).

I am an engineer and I have been designing opamp stuff for many years. In real life, we have a requirement and we provide a solution! This will not be a circuit I would use for a notch filter.

I'm a electronic engineering student in sri lanka

 

[yungman]

I'm a electronic engineering student in sri lanka

Good for you. You are happy with my answers?

 

[aruna1]

Good for you. You are happy with my answers?

indeed.I'm currently studying electrical measurement subject and this question is related to Principles of electronics subject.when i finished studying measurements i'll go back to electronic and analyze your answer.thank you for helping me.

 

[yungman]

indeed.I'm currently studying electrical measurement subject and this question is related to Principles of electronics subject.when i finished studying measurements i'll go back to electronic and analyze your answer.thank you for helping me.

You are very welcome.

 

[The Electrician]

I think if you plot the equaion out, you should see the respond. Seems like the amplitude will go very low at break frequency.

I don't think this is right, Yungman. This circuit is an all-pass network, and the amplitude response is constant with frequency; only the phase shift changes with frequency.

Your last expression for Vout in post #2 can be re-written as:

jwRC-1
------- * Vin
jwRC+1

which is a classic all-pass response.

 

[yungman]

I don't think this is right, Yungman. This circuit is an all-pass network, and the amplitude response is constant with frequency; only the phase shift changes with frequency.

Your last expression for Vout in post #2 can be re-written as:

jwRC-1
------- * Vin
jwRC+1

which is a classic all-pass response.

Maybe you are right. I got the transfer function and I just guess the output will get lower at \omega = 1/(RC). because it is in opposite direction. I never really sit down and go through it. But since you mentioned, I calculate at break freq. The output is -j and amplitude still is 1 at -90 deg.

Just a very uncommon circuit. Been designing opamp type of circuits for years, never have any use of this type yet. More the fancy type of circuits puting in the test.