How to derive equation from op amp circuit?

Click For Summary

Discussion Overview

The discussion revolves around deriving a differential equation from a specific operational amplifier (op amp) circuit. Participants are focused on the mathematical formulation necessary to express the circuit's behavior in standard differential equation form, which is essential for analyzing its characteristic modes and responses.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • The initial attempt at deriving the differential equation involves using the relationships between input and output voltages across capacitors and resistors in the op amp circuit.
  • One participant suggests solving for the output of the first op amp, y1(t), and using it as an input for the second stage to derive the equation for y2(t).
  • A later reply presents a derived equation involving second derivatives and suggests simplifying terms by introducing new variables for resistances and capacitances.
  • There is a correction regarding the sign of the right-hand side of the equation, indicating that the output should be positive due to the nature of the inversions in the circuit.

Areas of Agreement / Disagreement

Participants generally agree on the approach to derive the differential equation, but there is a disagreement regarding the sign of the right-hand side of the final equation. The discussion remains unresolved on this specific point.

Contextual Notes

Some participants note that the lack of familiarity with the Laplace transform may limit the ability to fully solve the problem. There are also suggestions to simplify the mathematical expressions for clarity.

Nat3
Messages
69
Reaction score
0

Homework Statement



Derive the differential equation from the following circuit

m10a.png

Homework Equations



The solution needs to be a differential equation in standard form, so that I can solve the equation for the characteristic modes, zero-input response, zero-state response, etc.

The Attempt at a Solution



For the first op amp, I did this:

\frac{x(t)-0}{R_1} = C_1\frac{dV_{C_1}}{dt}

V_{C1} = 0 - y_1(t) = -y_1(t)

\frac{x(t)}{R_1} = C_1\frac{dy_1(t)}{dt}

\frac{dy_1(t)}{dt} = \frac{x(t)}{R_1C_1}

Then, for the second op amp:

\frac{y_1(t)}{R_2} = C_2\frac{dV_{C_2}}{dt} + \frac{0-y(t)}{R_3}

V_{C_2}=0-y(t)

\frac{y_1(t)}{R_2} = -C_2\frac{dy(t)}{dt} - \frac{y(t)}{R_3}

At this point, I'm stuck.

Any help would be greatly appreciated!
 
Physics news on Phys.org
Have you had the Laplace transform?

If not, solve for y1(t) and then write the d.e. for the second stage with y1(t) the input for the second stage.

So y2(t) is a d.e with y1(t) as the excitation function.

You did OK but insert y1(t) as part of your d.e. for y2(t) after you solved for y1(t).
 
  • Like
Likes   Reactions: 1 person
Thanks for your help!

No Laplace transform yet..

I plugged the first equation into the second and got this:

\frac{\int \frac{x(t)}{R_1C_1}}{R_2} = -C_2\frac{dy(t)}{dt}-\frac{y(t)}{R_3}

Differentiating both sides results in:

\frac{\frac{x(t)}{R_1C_1}}{R_2} = -C_2\frac{d^2y(t)}{dt^2}-\frac{1}{R_3}\frac{dy(t)}{dt}

or

\frac{x(t)}{R_1R_2C_1} = -C_2D^2y(t)-\frac{Dy(t)}{R_3}

or

D^2y(t)+\frac{Dy(t)}{R_3C_2} = -\frac{x(t)}{R_1R_2C_1C_2}

Is that correct?
 
Nat3 said:
D^2y(t)+\frac{Dy(t)}{R_3C_2} = -\frac{x(t)}{R_1R_2C_1C_2}

Is that correct?

Yes, except for the - sign on the rhs.

There arer two inversions so the output is + if the input is +.

Suggestion: if you let R3/R2 = k , R1C1 = T1 and R3C2 = T2 the math is cleaner. The equation then is
y'' + y'/T2 = kx/T1T2.

If you look at the 2nd stage the dc gain is k.

When you get the Laplace transform all this becomes much easier.
 
  • Like
Likes   Reactions: 1 person

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
Replies
34
Views
4K
Replies
15
Views
3K
  • · Replies 28 ·
Replies
28
Views
4K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K