# Transfer function of Op-Amp circuit

1. May 6, 2014

### Maylis

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
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

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2. May 6, 2014

### Maylis

Here is my attempt on the top circuit

3. May 6, 2014

### rude man

Assuming you made no math errors (your first 3 equations are correct), what's the problem? You have w in your gain expression.

4. May 6, 2014

### Maylis

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: May 6, 2014
5. May 6, 2014

### rude man

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.

6. May 6, 2014

### Maylis

I haven't learned Laplace transforms, we always made the plots in the frequency domain

7. May 6, 2014

### rude man

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.

8. May 6, 2014

### rude man

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.

9. May 10, 2014

### Maylis

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. May 10, 2014

### Maylis

11. May 10, 2014

### rude man

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.

12. May 10, 2014

### rude man

You don't want the 1/jwC2 in the numerator. Multiply num & denom by jwC2 to get rid of it.

13. May 10, 2014

### Maylis

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. May 10, 2014

### Maylis

I just put bode plots on my regular scratch paper. I'm sure the spacing is way off

Magnitude

Phase

15. May 10, 2014

### Maylis

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??

16. May 11, 2014

### rude man

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. May 11, 2014

### rude man

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. May 11, 2014

### Staff: Mentor

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. May 11, 2014

### Maylis

Yes those are femtofarads. We were given values to make the bode plot

20. May 11, 2014

### rude man

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.