How to derive equation from op amp circuit?

[Nat3]

Homework Statement

Derive the differential equation from the following circuit

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!

 

[rude man]

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).

 

[Nat3]

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?

 

[rude man]

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

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.

 

[rezeew]

To derive the differential equation from an op amp circuit, we can use Kirchhoff's current law and voltage law to analyze the circuit. First, we can assume that the op amps are ideal, meaning that the input impedance is infinite and the output impedance is zero. This allows us to simplify the circuit and treat the op amps as virtual short circuits, meaning that the voltage at the input terminals of the op amps are equal.

Next, we can use Kirchhoff's current law to write the following equation for the first op amp:

\frac{x(t) - y_1(t)}{R_1} = C_1\frac{dV_{C_1}}{dt}

Since the input voltage at the op amp is equal to the output voltage, we can substitute y_1(t) for x(t) to get:

\frac{y_1(t) - y_1(t)}{R_1} = C_1\frac{dV_{C_1}}{dt}

This simplifies to:

0 = C_1\frac{dV_{C_1}}{dt}

Using the relationship between voltage and capacitor current, we can rewrite this as:

0 = C_1\frac{dy_1(t)}{dt}

Now, we can use Kirchhoff's voltage law to write the following equation for the second op amp:

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

Again, we can substitute y_1(t) for x(t) to get:

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

Simplifying this equation gives us:

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

From here, we can use the relationship between voltage and capacitor current to rewrite this as:

\frac{y_1(t)}{R_2} = C_2\frac{dy_2(t)}{dt}

where y_2(t) represents the voltage across