# Finding the gain of a circuit (Part Two)

• Engineering
• NHLspl09
In summary, the student attempted to solve a problem in an EE class from last semester, but was unable to complete it. They were given help by another student and the help was accepted, but the student was not able to complete the solution. The student was able to provide a summary of their work and explain why they believe their work is correct.
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!

## Homework Statement

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

## Homework Equations

(Attatchment below - Page 2)
Node equations

## The Attempt at a Solution

(Attatchment below - Page 2)

#### Attachments

• EE P4.pdf
73.7 KB · Views: 277
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.

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

gneill said:
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.

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?

The Electrician said:
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

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

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

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

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 $\frac{1}{1/sC_1}$

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 you there.

gneill said:
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 you there.

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

NHLspl09 said:
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

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:
$$A_v(s) = A_{LF}\left(\frac{1}{1 + b_1 s + b_2 s^2}\right)$$

gneill said:
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:
$$A_v(s) = A_{LF}\left(\frac{1}{1 + b_1 s + b_2 s^2}\right)$$

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

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?

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

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

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

Last edited:
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.

#### Attachments

• EE P4.1.pdf
28 KB · Views: 254
Looks fine to me. Well done!

NHLspl09 said:
What I did was bring the -$\frac{Vin}{R_1}$ to the right side of the equation making it $\frac{Vin}{R_1}$

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.

#### Attachments

• P4EE.png
7.2 KB · Views: 472
gneill said:
Looks fine to me. Well done!

Thank you for all the help :D

The Electrician said:
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.

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

## What is the gain of a circuit?

The gain of a circuit is a measure of how much the output signal of the circuit is amplified compared to the input signal. It is usually expressed in decibels (dB) and represents the ratio of the output voltage or current to the input voltage or current.

## How do you calculate the gain of a circuit?

The gain of a circuit can be calculated by dividing the output signal by the input signal and taking the logarithm of the result. This can be represented by the formula: Gain (dB) = 20 log (Vout/Vin) or Gain (dB) = 10 log (Iout/Iin), where V represents voltage and I represents current.

## What factors can affect the gain of a circuit?

The gain of a circuit can be affected by various factors such as the components used in the circuit, the design and layout of the circuit, the operating frequency, and external factors such as temperature and noise. Any changes to these factors can result in a change in the gain of the circuit.

## Why is finding the gain of a circuit important?

Determining the gain of a circuit is important as it helps in understanding the behavior and performance of the circuit. It also allows for predicting the output signal based on the input signal and helps in designing and optimizing circuits for specific applications.

## What are some common methods for measuring the gain of a circuit?

Some common methods for measuring the gain of a circuit include using an oscilloscope to measure the input and output signals, using a multimeter to measure the voltage or current, and using a signal generator to provide a known input signal and measuring the resulting output signal. There are also specialized instruments such as a spectrum analyzer that can be used for more precise measurements of gain at different frequencies.

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