Transfer function of Op-Amp circuit

Click For Summary

Discussion Overview

The discussion revolves around the transfer function of an operational amplifier (op-amp) circuit, focusing on the challenges of deriving the transfer function in terms of frequency (ω) and preparing Bode plots. Participants explore mathematical manipulations, the use of Laplace transforms, and the implications of different forms of the transfer function for Bode plotting.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in converting impedance to frequency (ω) for their transfer function.
  • Another participant suggests that the gain expression already contains ω and questions the original poster's approach.
  • There is a discussion about the necessity of using Laplace transforms, with some arguing that it is not required for Bode plots.
  • Participants debate the correct form of the transfer function for Bode plotting, emphasizing that the numerator should be in terms of jw and the denominator should be real.
  • Concerns are raised about the range of frequencies on the semilog plot, with one participant noting that critical values exceed the available cycles.
  • Another participant mentions the importance of maintaining the correct form of the transfer function and avoiding unnecessary substitutions.
  • There is confusion about the significance of a pole at ω=1 and its implications for the slope of the Bode plot.
  • Participants discuss the need to sketch Bode plots based on given values and the importance of accurately representing the gain and phase characteristics.

Areas of Agreement / Disagreement

Participants do not reach consensus on the necessity of using Laplace transforms, the interpretation of the pole at the origin, and the correct approach to preparing Bode plots. Multiple competing views remain regarding the mathematical manipulations required for the transfer function.

Contextual Notes

Limitations include potential missing assumptions about the familiarity with Laplace transforms and Bode plot conventions. There are unresolved mathematical steps in the derivation of the transfer function and its representation in the frequency domain.

gfd43tg
Gold Member
Messages
949
Reaction score
48

Hello,

I am working on this previous final exam question. I am totally stumped with getting my answer with frequency. I believe my answer is correct with impedance, but I am having a hard time changing this to ω
 

Attachments

Physics news on Phys.org
Here is my attempt on the top circuit
ImageUploadedByPhysics Forums1399359300.379890.jpg


ImageUploadedByPhysics Forums1399359315.616875.jpg
 
Assuming you made no math errors (your first 3 equations are correct), what's the problem? You have w in your gain expression.
 
I have to clean up that expression and it really sucks to do that!

I need to put it into a standard form incase I need to make a bode plot
 
Last edited:
Bode plot? OK, have you been introduced to the Laplace transform? If so, replace jw by s and therefore -w^2 by s^2. Do that with your last expression (I assume you did the math correctly). Then I can help you with your bode plot.
 
I haven't learned Laplace transforms, we always made the plots in the frequency domain
 
Just FYI, I think teaching Bode plots without the s variable is cuckoo.

BUT - you're stuck, so:

1. don't multiply by comlex-conjugates as you did in the last line of your sheet 1.
2. (1/jwC1)/R1 + 1 = 1/jwR1C1 + 1 = (jwR1C1 + 1)/jwR1C1
Then the remaining terms are suitable for Bode plotting. Do not go (jw)^2 = -w^2. Leave everything in terms of jw. Your transfer function will always be a function of jw.

EDIT: wait, you have to multiply numerators and denominators by jw until you don't have jw in any denominator of any fraction.
 
Maylis said:
I haven't learned Laplace transforms, we always made the plots in the frequency domain

Since your transfer function is always a function of jw you can substitute s=jw and get rid of all complex numbers. You don't have to have "had" the Laplace transform. It's just a simple substitution that they should have taught you.

And BTW s is a frequency variable. A transfer function is always in terms of a frequency variable unless it's just a constant gain.
 
Here is where I am so far. I don't know how this is suitable for a bode plot yet.

I am confused, to make the bode plots, is the idea that the numerator is in terms of jw, and the denominator is all real?
 
  • #10
ImageUploadedByPhysics Forums1399708150.612172.jpg
 
  • #11
Maylis said:
Here is where I am so far. I don't know how this is suitable for a bode plot yet.

I am confused, to make the bode plots, is the idea that the numerator is in terms of jw, and the denominator is all real?

No.
Example: a simple R-C low-pass network has transfer function 1/(1 + jwRC). That surely is easy to Bode-plot. But if you wrote (1 - jwRC)/[1 + (wRC)^2] you'd be all messed up.

Leave everything in jw as much as possible.

Again, what you should really do is change all jw to s but if they didn't teach you that I guess you'd better hold off. Personally I can't imagine doing Bode plots without that transformation. You don't even need to have had the Laplace transform.
 
  • #13
ImageUploadedByPhysics Forums1399777817.270796.jpg


here is my latest work. Now I am running into the problem that I don't have enough cycles on my semilog plot (given on the exam) to put everything in there! that jω pole should go through ω=1. However, my critical values for ω are at 10^10, and 5x10^8. There are not enough cycles to fit that range of frequencies!
 
  • #14
I just put bode plots on my regular scratch paper. I'm sure the spacing is way off

Magnitude
ImageUploadedByPhysics Forums1399779238.600629.jpg
Phase
ImageUploadedByPhysics Forums1399779359.578791.jpg
 
  • #15
I did the bottom circuit, this one seems to be a lot easier. That one switch of a capacitor and resistor makes all the difference huh??

ImageUploadedByPhysics Forums1399780295.562625.jpg


ImageUploadedByPhysics Forums1399780316.833029.jpg
 
  • #16
Maylis said:
View attachment 69654

here is my latest work. Now I am running into the problem that I don't have enough cycles on my semilog plot (given on the exam) to put everything in there! that jω pole should go through ω=1. However, my critical values for ω are at 10^10, and 5x10^8. There are not enough cycles to fit that range of frequencies!

going thru w=1 is not necessary. in fact, it makes no sense per se. you have plenty of cycles along the x axis. You now nearly have the right form for doing the Bode plots, assuming no math slipups along the way. Just one more change: multiply num. & denom. by w1*w2 to get H(jw) = (1e10*w1/w2)(jw+w1)/[(jw)(jw+w2)].

Now, pick an w << 5e8, say w=1e7. Calculate |H(jw)| at that freq. & plot it. Note that at w=1e7 your |H(jw)| simplifies to ~ 1e10/1e7 = 1e3. Let w=1e7 be the origin along the x axis, then mark off w=1e8, 1e9, 1e10, 1e11, 1e12. That's a total of just 5 decades. Now you're ready to finish the gain Bode plot, right?

Remember, I did not check your math. Make sure w1 and w2 are what you computed, and that they're not swapped in value.
 
  • #17
Maylis said:
View attachment 69654

here is my latest work. Now I am running into the problem that I don't have enough cycles on my semilog plot (given on the exam) to put everything in there! that jω pole should go through ω=1. However, my critical values for ω are at 10^10, and 5x10^8. There are not enough cycles to fit that range of frequencies!

This actually looks good for the gain plot except there is no flat section on the left. There is nothing magic about w=1!

The phase plot looks bad though. The phase angle is -90 solid until w=5e8, then goes even more negative before returning back to eventually settle at -90 for very high frequencies (say w=1e11).
I'll let you determine the break angles. Note the correspondence between the gain slopes and the phase angles.
 
  • #18
Maylis said:
Here is where I am so far. I don't know how this is suitable for a bode plot yet.

I am confused, to make the bode plots, is the idea that the numerator is in terms of jw, and the denominator is all real?
A lot of algebra is involved. Being a sketch, perhaps the intention is that you separately sketch each first-order low-pass Bode plot, then add them graphically? Are you given the element values before being asked to sketch the Bode plot?

Those capacitor values are in femto-Farads?
 
  • #19
Yes those are femtofarads. We were given values to make the bode plot
 
  • #20
Since youi're only allowed 4 decades on your log paper, mark the x-axis as 1e7 (origin), 1e8, 1e9, 1e10 so 1e11 is the extreme right-hand frequency.
 
  • #21
ImageUploadedByPhysics Forums1399843707.997897.jpg


Rude man, I want to show you the table that are given in order to construct bode plots. Basically, the way we do it is that we have to get our transfer function to look like one of the factors here, and then use that to construct the plot.

I want to take note in particular of the pole at the origin. Notice how it has a slope of -20N dB/decade at exactly 1 rad/s. That is what I am talking about in regards to that 1 rad/s that I have included in my bode plot, because I have a pole at the origin in my transfer function.
 
  • #22
Maylis said:
View attachment 69684

Rude man, I want to show you the table that are given in order to construct bode plots. Basically, the way we do it is that we have to get our transfer function to look like one of the factors here, and then use that to construct the plot.

I see no problem with your table. The desired form is just how I've tried to get you to arrange your transfer function - i.e. as functions of jw.
I want to take note in particular of the pole at the origin. Notice how it has a slope of -20N dB/decade at exactly 1 rad/s. That is what I am talking about in regards to that 1 rad/s that I have included in my bode plot, because I have a pole at the origin in my transfer function.

Your post 14 clearly shows a flat segment from the left-hand end to w=1. This is wrong. The slope runs -20dB/decade all the way from the y-axis to your first break point at w = 5e8 rad/s.

Again - there is nothing sacred about w=1. Do not use it in any way unless your break frequencies are so low as to make that necessary.

A pole at the origin means infinite gain at dc so no matter how low w goes, the slope goesd as -20 dB/decade all the way from the y-axis to wherever the next break frequency is. It does not start at w=1. Even if you could plot for w=1 which you can't, as you yourself pointed out (not enough decades on the x axis!).
 
  • #23
I don't get it, my transfer function is a function of jw, no??
 
  • #24
Maylis said:
I don't get it, my transfer function is a function of jw, no??

Yes, and a "jw" by itself (no "+1" added) IS a pole at the origin, i.e represents an integrator with a -20 dB/decade rolloff all the way to zero rad/s. You showed that correctly in an earlier post (post 14) except you straightened out the slope from the y-axis to w=1. In fact, that magnitude plot was otherwise spot-on.
 
  • #25
ImageUploadedByPhysics Forums1399928456.519430.jpg


But look now, if I have 10^7 as my starting point, how will I know what magnitude to start its slope at?
 
  • #26
Ok I think I am getting closer. I know at w=0, the value should be 200. It is sloping at -20 dB/decade, so at 10^7, it should be at 200 - 160 = 40 dB, no? I put 60 dB because I am trying to think in my head, but is 40 dB supposed to be the starting point?

ImageUploadedByPhysics Forums1399929047.894460.jpg
 
  • #28
Here is my phase plot

By the way was I right about 40 db?

ImageUploadedByPhysics Forums1399930235.341957.jpg
 
  • #29
Maylis said:
Here is my phase plot

By the way was I right about 40 db?

View attachment 69727

No, you had 60 dB gain at w=1e7 in your post and I showed you how I got 60 dB in my post 14. It's important that you understand how we got 60 dB. Please study my post 14 carefully. I wonder what put 40 dB in your head.

Don't have time to check your phase plot now. Will try later.
But follow your table and you should be OK. Just thank your lucky stars you haven't encountered any quadratic poles or zeros yet ... :bugeye:
 
  • #30
ImageUploadedByPhysics Forums1399937260.208876.jpg


Just to show you what was going on in my head.

Some great news as a result of telling me to find the magnitude at w=10^7 rad/s. I just imputed the function into my calculator, and I discovered that my calculator is able to find the magnitude of complex expressions. Then I just did 20 log [magnitude] and got 60 dB! I never knew.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
Replies
21
Views
3K
Replies
15
Views
3K
Replies
34
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 44 ·
2
Replies
44
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 13 ·
Replies
13
Views
3K