# Finding the gain of a circuit (Part Two)

1. Jul 11, 2011

### NHLspl09

As previously stated in another thread, I have started to work on a summer assignment for my Fall semester EE course. I was given a circuit with an op amp and asked to determine the transfer function and gain. This problem I have completed (not sure if I'm correct) based off of notes from an EE course I took last semester, and am just asking for a simple check as to whether I'm horribly wrong or I did well. Any input is greatly appreciated!!

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

(Attatchment below - Page 1)
Determine the transfer function of the circuit given.

2. Relevant equations

(Attatchment below - Page 2)
Node equations

3. The attempt at a solution

(Attatchment below - Page 2)

#### Attached Files:

• ###### EE P4.pdf
File size:
73.7 KB
Views:
78
2. Jul 11, 2011

### Staff: Mentor

I don't think the mathematics you committed in the last two lines of the v1 node calculation is kosher, so the line you've designated (1) is not correct.

3. Jul 11, 2011

### The Electrician

gneill is referring to the fact that if you have a set of parentheses with 4 terms inside, the reciprocal of the content of the parentheses is not just the sum of the reciprocals of the individual terms.

Furthermore, earlier you have the left side of the equation equal to zero, then you set that same left side equal to Vin/R1, which implies that Vin/R1 is equal to zero. This is an error; Vin/R1 is only equal to zero if Vin is zero.

Just as a hint, I get ALF = -R3/R1

4. Jul 12, 2011

### NHLspl09

I think I forgot the -1 outside of the parenthases, making the second to last line equivilent to the last line. Is that not right?

What I did was bring the -$\frac{Vin}{R_1}$ to the right side of the equation making it $\frac{Vin}{R_1}$

5. Jul 12, 2011

### Staff: Mentor

$\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \neq \left( a + b + c\right)^{-1}$

6. Jul 12, 2011

### NHLspl09

You're right, I just went back and checked with you. So my best bet would be to just keep it as is in line three? I just feel like that'd get really disgusting with $\frac{1}{1/sC_1}$

7. Jul 12, 2011

### Staff: Mentor

The individual term can become s*C1.

What I might suggest is putting everything over a common denominator to start with. Since the expression equals zero you can then discard the denominator. Then collect the voltage terms, make your substitution for v1, and proceed. It ain't pretty, but it'll get ya there.

8. Jul 12, 2011

### NHLspl09

That doesn't sound pretty at all - when you say putting everything over a common denominator do you mean what I have in the first step? Or go another route

9. Jul 12, 2011

### Staff: Mentor

Ah. I see that you've essentially done this by the time you first write vo/vi on the LHS of the expression. (Sorry about missing that...).

From that point, instead of trying to get the coefficient of the s2 term to be one, look to make the constant term 1. That will put the expression in the form desired:
$$A_v(s) = A_{LF}\left(\frac{1}{1 + b_1 s + b_2 s^2}\right)$$

10. Jul 12, 2011

### NHLspl09

Haha it's no problem at all, I was really confused for a moment :tongue:

Because of the way I have it in my final answer, 1/((something) + s*(something) + s2), is there an easy way to do what you're saying by making the constant term 1?

11. Jul 12, 2011

### Staff: Mentor

I would just go back to your first vo/vin line and divide numerator and denominator by R1R2.

12. Jul 12, 2011

### NHLspl09

Right I'm going to give that a shot and see how it goes. Updates soon.

Last edited: Jul 12, 2011
13. Jul 12, 2011

### NHLspl09

Took your advice by dividing the numerator and denomenator by R1R2 and it was much cleaner it seemed and also I now have an answer for b2.

#### Attached Files:

• ###### EE P4.1.pdf
File size:
28 KB
Views:
50
14. Jul 12, 2011

### Staff: Mentor

Looks fine to me. Well done!

15. Jul 12, 2011

### The Electrician

I see that now. It was late when I first looked at your work and I missed that point. :-(

You have now got the correct answer. The attachment shows how I got the result using another method.

#### Attached Files:

• ###### P4EE.png
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6.9 KB
Views:
50
16. Jul 12, 2011

### NHLspl09

Thank you for all the help :D

Haha it's no worries at all! Thank you very much for your input and help, and that attatchment makes for a nice check!