Questions

Design an analog circuit to solve this equation and determine \(v_{o}(t)\) for the equation \[\frac{dv_{o}}{dt} + 2v_{o} = 0\] with Initial condition: \(v_{o}(0) = 5\text{ V}\) (First-Order decay problem).
Build the analog computer circuit \[\frac{dv_{o}}{dt} + 4v_{o} = 8\] and find the steady-state voltage considering the initial condition as \(v_{o}(0) = 0\text{ V}\) (First-Order with constant input problem).
Design a two-integrator loop and predict the oscillation frequency as per the differential equation \[\frac{d^2v_{o}}{dt^2} + 9v_{o} = 0\] with initial conditions taken as \(v_{o}(0) = 3\text{ V}, \frac{dv_{o}}{dt}(0) = 0\) (Second-Order undamped oscillator problem).
Implement the analog solution of the differential equation, \[\frac{d^2v_{o}}{dt^2} + 6\frac{dv_{o}}{dt} + 8v_{o} = 0\] with initial conditions taken as \(v_{o}(0) = 4\text{ V}\), \(\frac{dv_{o}}{dt}(0) = 0\) and classify the damping type (Second-Order damped oscillator problem).
Build the complete analog computer to compute the solution of \[\frac{d^2v_{o}}{dt^2} + 2\frac{dv_{o}}{dt} + 5v_{o} = 10\] having initial conditions: \(v_{o}(0) = 0\text{ V}\), \(\frac{dv_{o}}{dt}(0) = 0\) and find both the transient and steady-state response (Second-Order damped oscillation problem with forcing function).

Answers

Rearrange for the highest derivative as \[\frac{dv_{o}}{dt} = -2v_{o}\] The circuit design will be consisting of one inverting integrator with output \(v_{o}\), which is fed back through a resistor network that scales by \(2\). Since the integrator inverts, and \(-2v_{o}\) is at the input, which gives feed \(v_{o}\) directly with gain \(2\). Pre-charge capacitor to \(-5\text{ V}\) (inverted) to achieve initial condition. The integrator should have \(\text{R} = 500\text{ k}\Omega\), \(\text{C} = 1\text{ }\mu\text{F}\) (time constant = \(0.5 \,\text{s}\), effective gain = \(2\)). The analytical solution is \(v_{o}(t) = 5e^{-2t}\text{ V}\).
Rearrange the equation as \[\frac{dv_{o}}{dt} = 8 - 4v_{o}\] The circuit design will be consisting of one summing amplifier followed by an integrator. The inputs of summing amplifiers are \(+8\text{ V}\) reference through \(\text{R}_{1} = 100\text{ k}\Omega\), \(v_{o}\) fed back through \(\text{R}_{2} = 25\text{ k}\Omega\) (gain of \(4\)) and Feedback resistor: \(\text{R}_{f} = 100\text{ k}\Omega\) to obtain the output of summing amplifier as \(-(8 - 4v_{o})\). The integrator is constructed with \(\text{R} = 1\text{ M}\Omega\), \(\text{C} = 1\text{ }\mu\text{F}\) keeping the Time constant = \(1 \,\text{s}\). The analytical solution is given as \(v_{o}(t) = 2(1 - e^{-4t})\text{ V}\).
Rearrange the equation as \[\frac{d^2v_{o}}{dt^2} = -9v_{o}\] Let \(V_{1} = \frac{dv_{o}}{dt}\), then \(\frac{dV_{1}}{dt} = -9v_{o}\) and \(\frac{dv_{o}}{dt} = V_{1}\). The The circuit design will be consisting of two integrators and an inverter circuit. The first Integrator should have an input \(v_{o}\) through \(\text{R}_{1} = 111\text{ k}\Omega\), feedback capacitor \(\text{C}_{1} = 1\text{ }\mu\text{F}\) having an effective gain \(= 9\) (since \(1/(\text{R}_{1} \text{C}_{1}) = 9\)) and initial condition: \(V_{1}(0) = 0\text{ V}\) to produces \(\text{V}_{1}\) from \(-9v_{o}\). The second integrator should have \(\text{R}_{2} = 1\text{ M}\Omega\), \(\text{C}_{2} = 1\text{ }\mu\text{F}\), where the capacitor is pre-charge to \(-3\text{ V}\) (inverts to \(+3\text{ V}\)) to produce \(v_{o}\) from \(\text{V}_{1}\). An inverter is required between integrators to correct sign (unity gain, \(\text{R} = 100\text{ k}\Omega\)). The analyical solution is given as \(v_{o}(t) = 3\cos(3t)\text{ V}\). Hence the oscillation frequency is obtained as \(\omega = 3\text{ rad/s}\), \(f = 0.477\text{ Hz}\), Time period\( = 2.09 \,\text{s}\).
Rearrange the equation as \[\frac{d^2v_{o}}{dt^2} = -6\frac{dv_{o}}{dt} - 8v_{o}\]. The circuit design will be consisting of two integrators in cascade with a summing amplifier at the input. The summing amplifier combines \(-8v_{o}\) and \(-6V_{1}\) (where \(V_{1} = dv_{o}/dt\)) using the two inputs, namely, first input from \(v_{o}: \text{R}_{1} = 125\text{ k}\Omega\) (gain \(8\) with \(\text{R}_{f} = 1\text{ M}\Omega)\) and second input from \(\text{V}_{1}\) through \(\text{R}_{2} = 167\text{ k}\Omega\) (gain \(6\)) to produce the output of \(\frac{d^2v_{o}}{dt^2}\). The first and second integrator has \(\text{R} = 1\text{ M}\Omega\), \(\text{C} = 1\text{ }\mu\text{F}\). The first integrator is set with the initial condition: \(\text{V}_{1}(0) = 0\) to produce \(\text{V}_{1}\) from \(d^2v_{o}/dt^2\) and the second integrator is set with initial condition of \(v_{o}(0) = 4\text{ V}\) to produce \(v_{o}\) from \(\text{V}_{1}\). The system is a Overdamped oscillator as the characteristic equation is obtained as \(s^2 + 6s + 8 = 0\), whose roots are \(s = -2, -4\) (both real and negative). The analytical solution of the differential equation is obtained as \(v_{o}(t) = 8e^{-2t} - 4e^{-4t}\text{ V}\).
Rearrange the equation as \[\frac{d^2v_{o}}{dt^2} = 10 - 2\frac{dv_{o}}{dt} - 5v_{o}\] The circuit design will be consisting of two integrator, a summing amplifier, and an inverter. Summing amplifier has \(3\) inputs, namely, \(+10\text{ V}\) DC reference through \(\text{R}_{1} = 100\text{ k}\Omega\), \(V_{1}\) (derivative) through \(\text{R}_{2} = 500\text{ k}\Omega\) (gain \(2\)) and \(v_{o}\) through \(\text{R}_{3} = 200\text{ k}\Omega\) (gain \(5\)) with \(\text{R}_{f} = 1\text{ M}\Omega\). The first and second integrator has \(\text{R} = 1\text{ M}\Omega\), \(\text{C} = 1\text{ }\mu\text{F}\). The first integrator produces \(-V_{1}\) as output. The inverter has the unity gain gives \(+\text{V}_{1}\). The second integrator produces the \(v_{o}\) as output. The damping type is underdamped which expects oscillatory decay, because the characteristic equation of the differential equation is obtained as \(s^2 + 2s + 5 = 0\) having the roots as \(s = -1 \pm 2j\) which are complex conjugates. The complete analytical solution is given as \(v_{o}(t) = 2 - e^{-t}\left(2\cos(2t) + \sin(2t)\right)\text{ V}\) or \(v_{o}(t) = 2 - \sqrt{5}e^{-t}\cos(2t - 0.464)\text{ V}\) with the steady-state as \(V_{\text{ss}} = \frac{10}{5} = 2\text{ V}\). The frequency of the damped oscillation is given as : \(\omega = 2\text{ rad/s}\), Time period \(\approx 3.14 \,\text{s}\).