Electronics Guide

Solutions and Initial-Value Problems

A solution satisfies the equation identically on an interval of t. Substitution shows that e^−t solves the second-order equation above, since e^−t − 3e^−t + 2e^−t = 0, and so does e^−2t. An equation of order n normally has a general solution with n arbitrary constants, here K₁e^−t + K₂e^−2t, and n conditions select one member of the family. Conditions on y and its first n − 1 derivatives at one instant make an initial-value problem; conditions at two instants make a boundary-value problem, such as finding a periodic steady state.

Why Circuits Produce Differential Equations

Kirchhoff's laws are algebraic, but the element laws of capacitors and inductors contain derivatives. A current source i_S(t) driving a resistor R and a capacitor C in parallel gives, by Kirchhoff's current law,

C dv/dt + v/R = i_S(t)

a first-order equation in the capacitor voltage. In a circuit of resistors, capacitors, inductors, and independent sources, the order, meaning the number of independent initial conditions, equals the number of capacitors and inductors, less one for each independent loop made only of capacitors and voltage sources and one for each independent cutset made only of inductors and current sources. A circuit of order n gives n coupled first-order equations, or a single equation of order n or less in any one variable. Thermal networks and rotating machines obey equations of the same form.

Existence and Uniqueness

Write a first-order equation as dy/dt = f(t, y) with y(t₀) = y₀. The Picard–Lindelöf theorem guarantees exactly one solution on some interval around t₀ when f is continuous near (t₀, y₀) and Lipschitz in y there: a constant M exists with |f(t, y₁) − f(t, y₂)| ≤ M|y₁ − y₂| for nearby y₁ and y₂. A continuous partial derivative ∂f/∂y near the point implies the Lipschitz condition, and the theorem extends to systems and so to equations of any order.

Continuity alone guarantees a solution but not a unique one: dy/dt = 3y^2/3 with y(0) = 0 is solved by both y = 0 and y = t³, because the slope of y^2/3 is unbounded at zero. The guarantee is also local: dy/dt = y² with y(0) = 1 gives y = 1/(1 − t), which escapes to infinity as t approaches 1. Linear equations escape both limits: when the coefficients and input are continuous on an interval and the leading coefficient never vanishes there, the solution exists and is unique on that whole interval.

Separation of Variables

An equation is separable when it can be written dy/dt = g(t)h(y). Dividing by h(y) and integrating each side with respect to its own variable gives

∫ dy/h(y) = ∫ g(t) dt + K

The source-free RC circuit, dv/dt = −v/τ with τ = RC, separates into dv/v = −dt/τ, so ln|v| = −t/τ + K and v(t) = V₀e^−t/τ. Dividing by h(y) discards constant solutions where h(y) = 0, here v = 0, so they must be checked separately.

The method also works when h(y) is nonlinear. Model a capacitor C discharging through a diode-connected MOSFET by the square law, i = (β/2)(v − V_th)² above the threshold voltage V_th, where β is the transconductance parameter. With u = v − V_th, C du/dt = −(β/2)u² separates into du/u² = −[β/(2C)] dt, and integrating from u₀ gives

u(t) = u₀/[1 + βu₀t/(2C)]

With C = 100 nF, β = 10 mA/V², and u₀ = 2 V, the discharge starts at 20 mA, and the voltage above threshold halves after 10 µs, again 20 µs later, and again 40 µs after that. The decay is hyperbolic and has no time constant. A real transistor keeps conducting below threshold, so the model covers only the discharge above it.

The Integrating Factor

Every first-order linear equation can be written in the standard form

dy/dt + p(t)y = q(t)

Multiplying by the integrating factor μ(t) = exp[∫p(t) dt] makes the left side an exact derivative, because d(μy)/dt = μ dy/dt + μpy. The equation becomes d(μy)/dt = μq, and one integration gives

y(t) = [∫μ(t)q(t) dt + K]/μ(t)

The formula holds when p varies with time, provided the integrals can be evaluated. With constant coefficients, as in an RC or RL circuit, write dy/dt + y/τ = x(t)/τ, with the input x(t) scaled so that a constant input X gives the final value X; then μ = e^t/τ, and integrating from 0 to t gives

y(t) = y(0)e^−t/τ + (1/τ) ∫₀^t e^{−(t − σ)/τ} x(σ) dσ

The first term is the zero-input response, set by the initial state, and the second is the zero-state response, the convolution of the input with the impulse response (1/τ)e^−t/τ. A constant input X gives the familiar y(t) = X + [y(0) − X]e^−t/τ.

The Characteristic Equation

Because the derivatives of an exponential are multiples of it, the trial solution y = e^λt reduces the homogeneous equation to (aλ² + bλ + c)e^λt = 0, so λ must satisfy the characteristic equation

aλ² + bλ + c = 0, with roots λ = [−b ± √(b² − 4ac)]/(2a)

The roots are the natural frequencies of the system and the poles of its transfer function, and the sign of the discriminant, b² − 4ac, sets the form of the homogeneous solution.

In the complex case σ = −b/(2a) and ω_d = √(4ac − b²)/(2a). Every root has a negative real part, and every homogeneous solution decays, exactly when b and c are also positive. For c > 0 the equation can be written d²y/dt² + 2ζω₀ dy/dt + ω₀²y = f(t)/a, with ω₀ = √(c/a) and ζ = b/[2√(ac)], and for b ≥ 0 the three forms become the overdamped (ζ > 1), critically damped (ζ = 1), and underdamped (0 ≤ ζ < 1) responses of the transient-response article.

Why the Three Forms Work

Distinct roots each supply an exponential: for d²y/dt² + 5 dy/dt + 4y = 0, (λ + 1)(λ + 4) = 0 gives y_h = K₁e^−t + K₂e^−4t. A repeated root supplies one exponential, and te^λt is the second solution: for distinct roots, (e^λ₁t − e^λ₂t)/(λ₁ − λ₂) is a solution, and as λ₂ approaches λ₁ it becomes the derivative of e^λt with respect to λ, which is te^λt.

Complex roots arrive in conjugate pairs because the coefficients are real, and by Euler's formula, e^jθ = cos θ + j sin θ, the complex solutions e^{(σ ± jω_d)t} combine into the real solutions e^σt cos ω_dt and e^σt sin ω_dt. The combination e^σt(K₁ cos ω_dt + K₂ sin ω_dt) equals Ae^σt cos(ω_dt − φ), with A = √(K₁² + K₂²), cos φ = K₁/A, and sin φ = K₂/A.

Higher Orders

At order n, each distinct real root λ of the characteristic polynomial contributes e^λt, a root of multiplicity m contributes e^λt, te^λt, …, t^m−1e^λt, and each complex pair contributes cosine and sine terms with the same powers of t. The denominator of a third-order Butterworth low-pass filter normalized to 1 rad/s gives λ³ + 2λ² + 2λ + 1 = (λ + 1)(λ² + λ + 1), with roots −1 and −½ ± j√3/2, so that

y_h = K₁e^−t + e^−t/2[K₂ cos(√3 t/2) + K₃ sin(√3 t/2)]

Every homogeneous solution decays exactly when every root has a negative real part, the stability condition for poles, and the Routh–Hurwitz test settles that question without finding the roots.

Exponential and Constant Inputs

Write the characteristic polynomial as P(λ), so that the equation reads P(d/dt)y = f(t). For the input Ke^at, the trial solution Ae^at gives P(a)A = K, so

y_p = Ke^at/P(a), provided that P(a) is not zero

A constant input F is the case a = 0, which gives y_p = F/c for the second-order standard form. The factor 1/P(a) is the transfer function evaluated at s = a. When a is a simple root of P, the rule becomes y_p = Kte^at/P′(a), and for a root of multiplicity m, y_p = Kt^me^at/P^(m)(a), with P^(m) the mth derivative of P. The input then matches a natural mode, a form of resonance that the first-order RC example below shows.

Sinusoidal Inputs

A sinusoid is the real part of a complex exponential, cos ωt = Re{e^jωt}, so for real coefficients the exponential rule gives the particular solution for K cos ωt as the real part of Ke^jωt/P(jω):

y_p = [K/|P(jω)|] cos(ωt − θ), where θ is the angle of P(jω)

This is the phasor method of AC Circuit Analysis in another notation. For dy/dt + 1000y = 1000 cos 1000t, an RC low-pass filter with a 1 ms time constant driven at 1000 rad/s, P(j1000) = 1000 + j1000 has the magnitude 1414 and the angle 45°, so y_p = 0.707 cos(1000t − 45°), which equals the 0.5 cos 1000t + 0.5 sin 1000t that the trial solution gives.

Other Inputs and the Steady State

Inputs outside the table need a general method: variation of parameters, which replaces the constants of y_h with functions of time, or, for a system starting at rest, the convolution of the input with the impulse response. A pulse or other piecewise input is solved interval by interval, each final state becoming the next initial state.

When every characteristic root has a negative real part, y_h decays and y_p remains, so for a constant or sinusoidal input the table's particular solution, which contains no natural modes, is the steady-state response.

Undamped Resonance and Beats

For ω different from ω₀, the trial solution A cos ωt gives y_p = F cos ωt/(ω₀² − ω²), whose amplitude grows without limit as ω approaches ω₀. From rest, with y(0) = 0 and y′(0) = 0, the complete solution is

y(t) = F(cos ωt − cos ω₀t)/(ω₀² − ω²) = [2F/(ω₀² − ω²)] sin[(ω₀ − ω)t/2] sin[(ω₀ + ω)t/2]

an oscillation at the mean frequency inside an envelope that varies at half the difference frequency, so that the amplitude swells and fades at the difference frequency itself: the beats. At ω = ω₀, cos ω₀t solves the homogeneous equation, so the trial solution must be multiplied by t, and substitution gives

y(t) = [F/(2ω₀)] t sin ω₀t

which already satisfies both initial conditions. The amplitude grows in proportion to t without bound, and the same function is the limit of the beat solution as ω approaches ω₀.

Damped Resonance

Damping bounds the growth. For d²y/dt² + 2ζω₀ dy/dt + ω₀²y = F cos ωt with ζ > 0, the natural modes decay and leave the steady-state solution

y_p = F cos(ωt − θ)/√[(ω₀² − ω²)² + (2ζω₀ω)²]

where θ, between 0 and 180°, is the angle of (ω₀² − ω²) + j2ζω₀ω. Dividing the amplitude by F/ω₀², the response to a constant input of the same size, gives the magnification, which equals the quality factor Q = 1/(2ζ) at ω = ω₀, where the response lags by 90°. For ζ < 1/√2 the magnification peaks at ω₀√(1 − 2ζ²) with the value 1/[2ζ√(1 − ζ²)]; with more damping it has no peak. The amplitude of dy/dt, which is proportional to the current when y is the capacitor voltage of a series RLC circuit, peaks exactly at ω₀.

Equilibria and the Jacobian

An autonomous system dx/dt = f(x) is at equilibrium where f(x^*) = 0. A small deviation δx = x − x^* then obeys, to first order,

d(δx)/dt = J δx

where J is the Jacobian matrix of partial derivatives ∂f_i/∂x_k evaluated at x^*. By Lyapunov's indirect method, the eigenvalues of J decide local stability: if every eigenvalue has a negative real part, the equilibrium is locally asymptotically stable; if any has a positive real part, it is unstable; and if the largest real part is exactly zero, the linearization cannot decide.

A diode fed by a constant current I, with a capacitor C across it, obeys C dv/dt = I − I_S[e^v/(nV_T) − 1]. At equilibrium the diode current equals I, and the slope of the diode law there, the small-signal conductance g_d = (I + I_S)/(nV_T), gives small deviations the time constant C/g_d. With I = 1 mA, n = 1, V_T = 25.85 mV at 300 K, and C = 10 µF, g_d = 38.7 mS and the time constant is 258.5 µs. In the nonlinear equation, a 2 mV rise decays to 1/e in 252 µs and a 2 mV fall in 265 µs, near the linear value, but a 26 mV rise does so in 186 µs and a 26 mV fall in 350 µs, because the diode's conductance changes exponentially with voltage.

A Self-Sustained Oscillator

The van der Pol equation, d²x/dt² − μ(1 − x²) dx/dt + x = 0 with μ > 0, models an oscillator whose amplifier supplies energy at small amplitudes and absorbs it at large ones; Balthasar van der Pol published "On 'Relaxation-Oscillations'" in the Philosophical Magazine in 1926. With x₁ = x and x₂ = dx/dt, the Jacobian at the origin is [0, 1; −1, μ], whose eigenvalues, [μ ± √(μ² − 4)]/2, have positive real parts, so any disturbance grows. Only the nonlinear equation says where the growth stops: numerical solution shows a limit cycle with a peak amplitude of 2.000 and a period of 6.287 for μ = 0.1, and 2.009 and 6.663 for μ = 1, in the equation's normalized time.

Implicit Methods

Implicit methods evaluate the slope at the new point, so each step solves an equation for y_n+1:

Backward Euler: y_n+1 = y_n + h f(t_n+1, y_n+1)

Trapezoidal rule: y_n+1 = y_n + (h/2)[f(t_n, y_n) + f(t_n+1, y_n+1)]

On the test equation they multiply y by 1/(1 + h/τ) and by [1 − h/(2τ)]/[1 + h/(2τ)] per step, factors smaller than 1 in magnitude for every positive h. Both are A-stable: every decaying linear mode also decays numerically, whatever the step.

At long steps, backward Euler damps fast modes strongly, at the price of first-order accuracy. The trapezoidal rule is second order and preserves the amplitude of an undamped oscillation, but for a fast mode and a long step its factor approaches −1, and the mode lingers as a slowly decaying step-to-step alternation: the trap ringing that circuit simulators can show after an abrupt edge.

By Germund Dahlquist's second barrier, from his 1963 paper in BIT, no A-stable linear multistep method exceeds order 2, which the trapezoidal rule attains. The backward differentiation formulas, which circuit simulators call Gear methods, exceed order 2 by giving up full A-stability above that order. The second-order formula, y_n+1 = (4/3)y_n − (1/3)y_n−1 + (2/3)h f(t_n+1, y_n+1), is A-stable and damps fast modes as backward Euler does; orders 3 through 6 are stable over a wedge of the left half-plane containing the negative real axis; and higher orders are unstable.

How SPICE Integrates

SPICE applies these methods to modified nodal analysis, whose equations mix differential and algebraic ones, because a node without capacitance contributes no derivative. At each time point the integration formula turns each capacitor and inductor into a companion model. Under backward Euler, i = C dv/dt becomes i_n+1 = (C/h)v_n+1 − (C/h)v_n, a conductance C/h with a current source set by the previous voltage; under the trapezoidal rule it becomes i_n+1 = (2C/h)v_n+1 − [(2C/h)v_n + i_n]. Each time point is then a resistive circuit, solved by Newton–Raphson iteration when devices are nonlinear, and an estimate of the local truncation error decides whether to accept the point and how long the next step may be.

In ngspice, trapezoidal integration is the default; .options method=gear selects Gear integration, with a maximum order, MAXORD, from 2, the default, to 6; and MAXORD = 1 with the trapezoidal method gives backward Euler. The manual notes that reducing XMU, the trapezoidal damping factor, slightly below its default of 0.5, for example to 0.495, may suppress trap ringing, though too much damping can make a circuit prone to ringing look stable. Transient Response of RC, RL, and RLC Circuits shows how to recognize trap ringing, and Circuit Simulation (SPICE) covers convergence settings.

An RC Stage Driven by an Exponential

A unity-gain buffer applies the decaying voltage Ve^−t/τ₁, with V = 5 V, to an RC low-pass filter with R = 10 kΩ and C = 100 nF, so that τ = RC = 1 ms. The capacitor starts discharged. Find the output v(t) for input time constants of 2 ms, 0.5 ms, and 1 ms.

1. Write the equation. Kirchhoff's current law at the output gives C dv/dt = (Ve^−t/τ₁ − v)/R, or τ dv/dt + v = Ve^−t/τ₁, with v(0) = 0.
2. Apply the integrating factor. Dividing by τ and multiplying by e^t/τ gives d(ve^t/τ)/dt = (V/τ) exp[t(1/τ − 1/τ₁)].
3. Integrate from zero. For τ₁ different from τ, v(t) = [Vτ₁/(τ₁ − τ)](e^−t/τ₁ − e^−t/τ), which peaks when t = [ττ₁/(τ₁ − τ)] ln(τ₁/τ).
4. Evaluate. For τ₁ = 2 ms, v(t) = 10(e^{−t/2 ms} − e^{−t/1 ms}) V, which peaks at 2.50 V at t = 1.39 ms. For τ₁ = 0.5 ms, v(t) = 5(e^{−t/1 ms} − e^{−t/0.5 ms}) V, which peaks at 1.25 V at t = 0.693 ms.
5. Treat the matching case. At τ₁ = τ the input is the filter's own natural mode, the integrand is the constant V/τ, and v(t) = V(t/τ)e^−t/τ, which peaks at V/e = 1.84 V at t = 1 ms.

Every solution starts at zero with the slope V/τ = 5,000 V/s. The matching case is the limit of the general solution as τ₁ approaches τ, and the exponential rule gives it directly, since P(λ) = τλ + 1 has the simple root −1/τ and P′ = τ.

A Series RLC Circuit Driven at Resonance

A source v_S(t) = sin 10,000t volts, switched on at t = 0, drives a series loop of R = 20 Ω, L = 10 mH, and C = 1 µF that starts at rest. Find the capacitor voltage v(t).

1. Write the equation. From LC d²v/dt² + RC dv/dt + v = v_S, dividing by LC = 10⁻⁸ s² gives d²v/dt² + 2000 dv/dt + 10⁸v = 10⁸ sin 10,000t, with v(0) = 0 and dv/dt(0) = i(0)/C = 0.
2. Find the roots. λ² + 2000λ + 10⁸ = 0 gives λ = −1000 ± j9950, so ω₀ = 10,000 rad/s, ζ = 0.1, and Q = 5. The source runs at ω₀, but ±j10,000 are not roots, so the trial solution needs no factor of t.
3. Find the particular solution. Substituting A cos 10,000t + B sin 10,000t, the terms in ω₀² cancel, and the damping term must supply the input: 2 × 10⁷(B cos 10,000t − A sin 10,000t) = 10⁸ sin 10,000t, so B = 0, A = −5, and v_p = −5 cos 10,000t V, Q times the source amplitude and 90° behind it.
4. Fit the constants. With v = −5 cos 10,000t + e^−1000t(K₁ cos 9950t + K₂ sin 9950t), v(0) = 0 gives K₁ = 5, and dv/dt(0) = −1000K₁ + 9950K₂ = 0 gives K₂ = 0.503.
5. State the result. For t ≥ 0, with t in seconds, v(t) = −5 cos 10,000t + e^−1000t(5 cos 9950t + 0.503 sin 9950t) V.

Because 9950 rad/s is so close to 10,000 rad/s, the decaying terms at first nearly cancel the steady one, so the amplitude starts to build at about 5,000 V/s, as in the undamped case, reaches about 3.2 V after 1 ms, and stays within 1 percent of 5 V after about 4.6 ms. Without the resistor the roots become ±j10,000, the trial solution needs the factor t, and v(t) = −5000t cos 10,000t + 0.5 sin 10,000t V grows by 5 V every millisecond.

A Two-Node Thermal Network

A transistor dissipating P = 20 W sits on a heat sink. Model the package as a node with thermal capacitance C₁ = 2 J/K, about that of 5 g of copper, joined to the heat sink by R₁ = 1.0 K/W, and the heat sink as a node with C₂ = 90 J/K, about that of 100 g of aluminum, joined to the air by R₂ = 1.5 K/W. The values are illustrative, and the model is a two-node Cauer ladder. Find the temperature rises above ambient, θ₁ and θ₂, after the power switches on at t = 0.

1. Write the node equations. Heat flow balances at each node as current does in an RC circuit: C₁ dθ₁/dt = P − (θ₁ − θ₂)/R₁ and C₂ dθ₂/dt = (θ₁ − θ₂)/R₁ − θ₂/R₂.
2. Put them in matrix form. With t in seconds, dθ/dt = Aθ + [P/C₁; 0], where A = [−0.5, 0.5; 0.01111, −0.01852].
3. Solve the characteristic equation. det(λI − A) = λ² + 0.51852λ + 0.0037037 = 0 gives λ₁ = −0.007244 s⁻¹ and λ₂ = −0.5113 s⁻¹, time constants of 138.0 s and 1.956 s. The constant term equals 1/(R₁C₁R₂C₂), a quick check.
4. Find the particular solution. In the steady state all 20 W flows through both resistances, so θ₂ = PR₂ = 30 K and θ₁ = P(R₁ + R₂) = 50 K.
5. Fit the constants. Each rise is its final value plus multiples of e^λ₁t and e^λ₂t in the proportions set by the eigenvectors of A, and requiring both rises to start at zero gives the result below, in kelvins.

θ₁(t) = 50 − 30.88e^−t/138.0 − 19.12e^−t/1.956

θ₂(t) = 30 − 30.43e^−t/138.0 + 0.43e^−t/1.956

Both rises start at zero, θ₁ with the slope P/C₁ = 10 K/s and θ₂ with zero slope, since no heat has yet reached the heat sink. The fast mode, which barely touches θ₂, is the transistor heating relative to its heat sink: within 10 s, θ₁ − θ₂ reaches 19.5 K of an eventual 20 K. The slow mode is the assembly warming as a whole, with θ₁ at 30.0 K after 60 s, 46.5 K after 300 s, and 49.6 K after 600 s.

The stiffness ratio, 0.5113/0.007244, is about 71. Forward Euler is stable only for h < 2 × 1.956 s = 3.91 s; at h = 3.5 s its early temperatures swing from step to step with errors up to 18 K, and at h = 4 s it diverges. Over the first 600 s, backward Euler at h = 30 s takes 20 steps, never oscillates, and ends within 0.21 K. The trapezoidal rule at h = 10 s ends within 0.001 K, but its first steps put θ₁ − θ₂ at 28.1, 15.9, 21.3, and 19.0 K against true values of 19.5 to 19.7 K: trap ringing of the 2 s mode. A third node for the silicon die, with an assumed 0.02 J/K and 0.3 K/W, would add a 6 ms mode and push the ratio above 20,000.

DC Motor Speed After a Voltage Step

Take the permanent-magnet DC motor whose model Control System Modeling and Block Diagrams derives: armature resistance R = 1 Ω and inductance L = 1 mH, torque and back-EMF constants K_t = K_e = 0.05 in SI units, inertia J = 1.25 × 10⁻⁴ kg·m², and negligible friction and load torque. At rest, it is connected to 24 V at t = 0. Find the speed ω(t) and the current i(t).

1. Combine the equations. The armature gives L di/dt = v − Ri − K_eω, and the rotor gives J dω/dt = K_ti. Substituting i = (J/K_t) dω/dt gives d²ω/dt² + (R/L) dω/dt + [K_tK_e/(LJ)]ω = K_tv/(LJ), or d²ω/dt² + 1000 dω/dt + 20,000ω = 9.6 × 10⁶.
2. Find the roots. λ² + 1000λ + 20,000 = 0 gives λ = −500 ± √230,000, or λ₁ = −20.42 s⁻¹ and λ₂ = −979.58 s⁻¹, distinct real roots with time constants of 49.0 ms and 1.02 ms.
3. Find the particular solution. The constant ω_p = 9.6 × 10⁶/20,000 = 480 rad/s is the speed at which the back EMF equals the applied voltage.
4. Fit the constants. The motor starts with ω(0) = 0 and no current, so dω/dt(0) = K_ti(0)/J = 0. Then K₁ + K₂ = −480 and λ₁K₁ + λ₂K₂ = 0, so K₁ = −490.22 and K₂ = 10.22.
5. State the results. For t ≥ 0, ω(t) = 480 − 490.22e^−20.42t + 10.22e^−979.58t rad/s, and i(t) = (J/K_t) dω/dt = 25.02(e^−20.42t − e^−979.58t) A.

The current rises at v/L = 24,000 A/s, peaks at 22.56 A after 4.04 ms, just short of the 24 A stall current, and decays as the back EMF builds. The speed reaches 63.2 percent of 480 rad/s after 50.0 ms and stays within 2 percent of it after 193 ms. With roots a factor of 48 apart, the model is mildly stiff: forward Euler needs steps under 2.04 ms, although the speed changes over tens of milliseconds.