Question 12.8: Deriving a Differential Equation from an Analog Computer Cir...

Known Quantities: Resistor and capacitor values.

Find: Differential equation in x(t).

Schematics, Diagrams, Circuits, and Given Data: R_{1}=0.4  \mathrm{M} \Omega ; R_{2}=R_{3}=R_{5}=1  \mathrm{M} \Omega ; R_{4}=2.5  \mathrm{k} \Omega ; C_{1}=C_{2}=1  \mu \mathrm{F}.

Assumptions: Assume ideal op-amps.

Analysis: We start the analysis from the right-hand side of the circuit, to determine the intermediate variable z as a function of x :

x=-\frac{R_{5}}{R_{4}} z=-400 z

Next, we move to the left to determine the relationship between y and z :

z=-\frac{1}{R_{3} C_{2}} \int y\left(t^{\prime}\right) d t^{\prime} \quad \text { or } \quad y=-\frac{d z}{d t}

Finally, we determine y as a function of x and f :

y=-\frac{1}{R_{2} C_{1}} \int x\left(t^{\prime}\right) d t^{\prime}-\frac{1}{R_{1} C_{1}} \int f\left(t^{\prime}\right) d t^{\prime}=-\int\left[x\left(t^{\prime}\right)+2.5 f\left(t^{\prime}\right)\right] d t^{\prime}

or

\frac{d y}{d t}=-x-2.5 f

Substituting the expressions into one another and eliminating the variables y and z, we obtain the differential equation in x :

\begin{aligned}&x=-400 z \\&\frac{d x}{d t}=-400 \frac{d z}{d t}=400 y \\&\frac{d^{2} x}{d t^{2}}=400 \frac{d y}{d t}=400(x-2.5 f)\end{aligned}

and

\frac{d^{2} x}{d t^{2}}+400 x=-1,000 f

Comments: Note that the summing and integrating functions have been combined into a single block in the first amplifier.