Hi there

I need help analysing this circuit.

I have taken the op amp on the left to be amp A and the one on the right to be amp B.
I calculated the following:
The gain for A is (1 + 10k/1k = 11).
The gain for B is (1 + 56k/1k = 57).
3n3 capacitor : Xc = 1/2pi f c = 3215.25 ohms.

It says the input signal is a 15kHz sine wave with a voltage of 0.1 Vp-p.
I know it is an AC circuit and need to know what else to do.

I also need to find the resonant frequency of the LCR circuit.

http://i1362.photobucket.com/albums/r687/JasonMcG1/OpAmp_zps70d48f9a.jpg

Regards
Jason McG

 

Jason, are you sure about the values?
For IDEAL components (L, C) the resonant peak is at 27.7 kHz and for 15kHz the damping is app. 45 dB. Thus, the real properties of the components (parasitic resistances) play a major role.

 

Hi there

I need help analysing this circuit.

I have taken the op amp on the left to be amp A and the one on the right to be amp B.
I calculated the following:
The gain for A is (1 + 10k/1k = 11).
The gain for B is (1 + 56k/1k = 57).
3n3 capacitor : Xc = 1/2pi f c = 3215.25 ohms.

It says the input signal is a 15kHz sine wave with a voltage of 0.1 Vp-p.
I know it is an AC circuit and need to know what else to do.

I also need to find the resonant frequency of the LCR circuit.

http://i1362.photobucket.com/albums/r687/JasonMcG1/OpAmp_zps70d48f9a.jpg

Regards
Jason McG

You need to show YOUR work that you have done in an effort to solve YOUR problem. This is Homework Help, not Homework Done For you.

What is the transfer function for LCR circuit in the dotted box, at least at 15kHz?

 

I also need to find the resonant frequency of the LCR circuit.

For IDEAL components (L, C) the resonant peak is at 27.7 kHz

Well, it didn't USE to be the Homework Done For You forum.

 

I'll help you to get started. The L, C and R in the middle of the circuit for a voltage divider. Do you know how to formulate the equation for a voltage divider? So, you have 3 impeadances. Lable them Z1, Z2,and Z3. Now, use these Z's to formulate the divider equation. Then substitute the expressions for each Z (hint, you formula for the capacitor will be 1/jωC) Now, you have an equation in terms of the frequency dependent impeadances. Combine any terms and use any algebraic simplification to get the simplified equation. Now, you can find a value for ω that makes the expression zero, or makes the denominator zero.

That's all you get. I haven't actually solved one of these in years, and so I'm not going to try and embarass myself

 

The OP never bothered to come back. I'd be surprised if they actually see your work.
Anyway - good to get some practice in helping with the theory.

I'd suggest reviewing your comments for accuracy.

 

I have since been able to calculate the answer.
I calculated the gain for the first op amp using the 0.1V and got 1.1 V coming out the first amp. Using the inductive reactance and capacitive reactance I found the impedance to be 2.27 k ohms. Using this and (I=V/Z) the current is 0.484mA. V10ohm = Vamp2= (0.484 * 10) = 4.84mV. Using the gain on the second amp i got the output to be 0.28 Vp-p.

 

You need to show YOUR work that you have done in an effort to solve YOUR problem. This is Homework Help, not Homework Done For you.

What is the transfer function for LCR circuit in the dotted box, at least at 15kHz?[/QU

I dont think a few answers to questions are going to make or brake someones education. When I get the answers I go back and try the calculations to see what I did wrong or until I come up with the correct answer. Everyone learns differently and that is why some books have the answer key in the back. Unless you are saying there is only one way to learn??

 

Why should we think that you do any checking at all, since you pretty firmly refuse to show any of it here? Anyone can say, "Give the the answer. Trust me, I always go back and work it out myself." It's a lot more believable when you make the effort to show your work up front. In this case, the OP wasn't being asked to even make an attempt at a solution to the whole problem, but just one tiny portion of it. Why is that too much to expect? Remember, this is THEIR homework problem.

 

Why should we think that you do any checking at all, since you pretty firmly refuse to show any of it here? Anyone can say, "Give the the answer. Trust me, I always go back and work it out myself." It's a lot more believable when you make the effort to show your work up front. In this case, the OP wasn't being asked to even make an attempt at a solution to the whole problem, but just one tiny portion of it. Why is that too much to expect? Remember, this is THEIR homework problem.

If I knew how to post all the symbols yes I would show all my work. I think if a person is going to take the time to reach out to ask for help they are serious about learning

 

If I knew how to post all the symbols yes I would show all my work.

You could scan your work to a photo and post the photo, you could do your work in word and print it to a pdf and attach that. Knowing the symbols here is minor.

The point of you showing your work is for people to identify where you went astray, if you did, to help you.

Identifying that point allows you to have that "ah ha" moment sooner.

 

If I knew how to post all the symbols yes I would show all my work.

Excuses, excuses.

You can attach a picture. You can use use the symbol palette. You can use the TEX tags. There are stickies and FAQs telling you how to do each of these. You've replied to threads that have used all of these. You could have ASKED how to do any of these. But you didn't, because you had no use for them because you had no intention of actually using any of them to post any effort.

You are always going to have some excuse as to why you shouldn't be expected to show any effort but that everyone else should just spoon feed you the complete answers.

I think if a person is going to take the time to reach out to ask for help they are serious about learning

Not so. How serious does someone have to be about learning to ask someone else to do their work for them?

 

One who learns by finding out, has seven fold the skills of him who learns by being told.
-- Arthur Gutterman

 

I have since been able to calculate the answer.
I calculated the gain for the first op amp using the 0.1V and got 1.1 V coming out the first amp. Using the inductive reactance and capacitive reactance I found the impedance to be 2.27 k ohms. Using this and (I=V/Z) the current is 0.484mA. V10ohm = Vamp2= (0.484 * 10) = 4.84mV. Using the gain on the second amp i got the output to be 0.28 Vp-p.

Your second and third calculation is incorrect. I'll wait till you post your work to illustrate where the error originates.

 

One who learns by finding out, has seven fold the skills of him who learns by being told.
-- Arthur Gutterman

I wonder how he determined that quantitatively? What are the error bars? I wish he would show his work!

Seriously, though, I couldn't agree more. I like to think that I've got very good skills at learning material from reading it or having someone present it and work some examples while I take notes. I also know that I can learn very effectively by seeing the solution to a problem I couldn't solve on my own, but this is primarily because I am not satisfied with just seeing the answer, but in wanting to know not only what each step in the solution process was, but in understanding why that step was taken given only the situation prior to that step being taken. Even so, by far my most valuable and extensive learning has come from when I have struggled with a problem and had to fight it for hours and hours and come at it from several different directions and see a little flicker of insight here and there until I was finally able to beat it into submission. Then I will usually go back and work it again, from scratch, utilizing all the hard-won knowledge I've scraped out of it, usually to discover that the problem was solvable in a much shorter fashion had I just looked at it properly -- and knowing that now I CAN look at it properly brings a huge sense of accomplishment making all the pain and effort worth it.

 

Even so, by far my most valuable and extensive learning has come from when I have struggled with a problem and had to fight it for hours and hours and come at it from several different directions and see a little flicker of insight here and there until I was finally able to beat it into submission. Then I will usually go back and work it again, from scratch, utilizing all the hard-won knowledge I've scraped out of it, usually to discover that the problem was solvable in a much shorter fashion had I just looked at it properly -- and knowing that now I CAN look at it properly brings a huge sense of accomplishment making all the pain and effort worth it.

The most valuable: firsthand experience.

 

Well, since the OP received a positive response to their calculations elsewhere (ETO) I doubt they will return here.

Anyway ... Here were my results ... and no, I didn't show my work.