Electronics Guide

The Modeling Procedure

1. Choose the signals: the input the controller manipulates, the output it controls, and the disturbances that act on the system.
2. Write the element laws for each resistor, inductor, capacitor, spring, damper, and inertia.
3. Write the interconnection laws: Kirchhoff's laws for circuits, and Newton's second law for each mass or inertia.
4. Linearize each nonlinear law about the operating point.
5. Transform with zero initial conditions, replacing each derivative d/dt by s.
6. Check the units, the DC gain, the high-frequency behavior, and the order.
7. Validate the model against a measured step or frequency response.

Transfer Functions from Differential Equations

A lumped, linear, time-invariant system with input u(t) and output y(t) obeys

a_n dⁿy/dtⁿ + … + a₁ dy/dt + a₀y = b_m d^mu/dt^m + … + b₁ du/dt + b₀u

With every initial condition zero, each derivative becomes a power of s, and the ratio of the output transform to the input transform is the transfer function:

G(s) = Y(s)/U(s) = (b_ms^m + … + b₁s + b₀) / (a_nsⁿ + … + a₁s + a₀)

The response to energy stored at the start is a separate term, which Laplace Transform treats. An inductance L in series with a resistance R, such as the armature of a stalled motor, gives a first example. Kirchhoff's voltage law reads L di/dt + Ri = v, so

I(s)/V(s) = 1/(Ls + R) = (1/R)/(τ_es + 1), with τ_e = L/R

The checks pass. The units are amperes per volt. The DC gain is 1/R, as Ohm's law requires. At high frequency the gain approaches 1/(Ls), because the inductance dominates. The order is one, matching the single energy store; a lumped model's order cannot exceed its number of independent energy stores. A pure time delay t_d, such as a digital controller's computation time, contributes the factor e^−st_d, which is not a ratio of polynomials, so analysis keeps it exactly or replaces it with a rational approximation.

Operational-Amplifier Stages

Analog controllers are often built from inverting operational-amplifier stages, whose ideal transfer function is −Z_f(s)/Z_in(s) for input impedance Z_in and feedback impedance Z_f. A resistor R₁ at the input, with a resistor R₂ in series with a capacitor C in the feedback path, gives

V_out(s)/V_in(s) = −[R₂ + 1/(sC)]/R₁ = −R₂/R₁ − 1/(sR₁C)

This is a proportional-integral controller with proportional gain R₂/R₁ and integral gain 1/(R₁C), inverted in sign. The sign belongs in the model. In a block diagram it becomes a minus sign in the block or at a summing junction, and losing it turns negative feedback into positive feedback.

Loading

Two blocks in cascade multiply only if the second draws too little current to change the first block's output. Passive networks usually break this rule. Take two RC low-pass sections, each with R = 10 kΩ and C = 10 nF, so that each alone has the transfer function 1/(τs + 1) with τ = RC = 100 µs. Connected directly, the second section loads the first, and nodal analysis gives

V_out(s)/V_in(s) = 1/(τ²s² + 3τs + 1)

The product of the separate transfer functions would be 1/(τ²s² + 2τs + 1). The extra τs term is the loading. It splits the double pole at −10,000 s⁻¹ into poles near −3,820 and −26,180 s⁻¹ and lowers the −3 dB frequency from 1,024 Hz to 596 Hz. A unity-gain buffer between the sections restores the product. An operational-amplifier stage has an output impedance near zero wherever its loop gain is high, which is why chains of such stages can be drawn as chains of blocks.

A Mass, Spring, and Damper

A mass M on a spring K with viscous damping B, driven by a force f, obeys M d²x/dt² + B dx/dt + Kx = f, so

X(s)/F(s) = 1/(Ms² + Bs + K)

Matching the denominator to M(s² + 2ζω_ns + ω_n²) gives ω_n = √(K/M) and ζ = B/[2√(KM)]. A 2 kg mass on an 800 N/m spring with B = 16 N·s/m has ω_n = 20 rad/s, ζ = 0.2, and a DC gain, the static compliance 1/K, of 1.25 mm/N.

Gear Trains and Reflected Inertia

An ideal gear train with ratio n, the motor speed divided by the load speed, divides speed by n and multiplies torque by n. Referred to the motor shaft, a load inertia J_L appears as J_L/n², because the load's kinetic energy is ½J_L(ω_m/n)², where ω_m is the motor speed. Load damping and stiffness also divide by n², and load torque divides by n. Through a 10:1 reduction, a 2.0 × 10⁻³ kg·m² load therefore appears at the motor as 2.0 × 10⁻⁵ kg·m², twice the inertia of a 1.0 × 10⁻⁵ kg·m² rotor.

Compliance and Resonance

No shaft or belt is perfectly stiff. When a motor inertia J_m drives a load inertia J_L through a coupling of torsional stiffness K_s, Newton's law for each inertia, with damping ignored, gives the transfer function from motor torque to motor speed:

Ω_m(s)/T_m(s) = (J_Ls² + K_s) / {s[J_mJ_Ls² + K_s(J_m + J_L)]}

The numerator places an antiresonance at ω_ar = √(K_s/J_L), where the load acts as a vibration absorber: in this undamped model, a steady sinusoidal torque at that frequency produces no motor motion. The denominator places a resonance at ω_r = √[K_s(J_m + J_L)/(J_mJ_L)]. With J_m = 1.0 × 10⁻⁴ kg·m², J_L = 3.0 × 10⁻⁴ kg·m², and K_s = 300 N·m/rad, they fall at 1,000 rad/s (159 Hz) and 2,000 rad/s (318 Hz). Well below the antiresonance the inertias move together as J_m + J_L, but a speed loop whose bandwidth approaches it needs the compliant model. Servo Drive Systems describes how drives suppress such resonances.

Why the Torque and Back-EMF Constants Are Equal

In SI units the back-EMF constant K_e and the torque constant K_t are the same number. The electrical power absorbed by the back EMF is e_bi = K_eωi, and the mechanical power the motor develops is T_mω = K_tiω. The coupling itself neither stores nor dissipates energy, because the model assigns losses to R and b and stored energy to L and J. The two powers are therefore equal, and K_t = K_e. The units agree as well: 1 N·m/A = 1 J/A = 1 V·s, and the radian is dimensionless.

Data sheets often hide the equality behind other units. A speed constant in rpm/V is the reciprocal of K_e once rpm is converted to rad/s, and a constant in V/krpm is K_e in V·s/rad multiplied by 1,000 × 2π/60, about 104.7. A torque constant of 0.05 N·m/A therefore corresponds to 5.24 V/krpm, or to a speed constant of 191 rpm/V. The motor maker maxon gives the product of its speed and torque constants as 1 in SI units and as 30,000/π, about 9,549, in rpm/V times mN·m/A.

Block Diagram of the Motor

Down the table, voltage drives current, current makes torque, and torque accelerates the load. The back-EMF path, which returns speed to the voltage junction with a minus sign, is a negative feedback loop built into the physics: as the motor speeds up, the back EMF rises and reduces the current that accelerates it.

Transfer Functions

The forward path from voltage to speed is K_t/[(Ls + R)(Js + b)], and the feedback path is K_e. The feedback rule G/(1 + GH), derived under block-diagram algebra below, gives

Ω(s)/V(s) = K_t / [(Ls + R)(Js + b) + K_tK_e]

The load torque, subtracted at the torque junction, reaches speed through the forward path −1/(Js + b) and the same loop, so

Ω(s)/T_L(s) = −(Ls + R) / [(Ls + R)(Js + b) + K_tK_e]

The two share a denominator, as all transfer functions of a single loop do, and dividing either by s gives the transfer function to position. At DC with b = 0 they combine into the speed-torque line ω = v/K_e − RT_L/(K_tK_e). The motor in the speed-loop example of Time-Domain Response of Control Systems, with R = 1 Ω and K_t = K_e = 0.05 in SI units, therefore runs at 480 rad/s unloaded from 24 V and loses 20 rad/s for each 0.05 N·m of load.

Electrical and Mechanical Time Constants

With friction neglected, b = 0, the speed transfer function becomes

Ω(s)/V(s) = (1/K_e) / (τ_eτ_ms² + τ_ms + 1), with τ_e = L/R and τ_m = RJ/(K_tK_e)

The mechanical time constant τ_m is really electromechanical: it contains R because the back EMF, acting through the armature resistance, damps the rotor like a viscous damper of coefficient K_tK_e/R, and any resistance in the driving source adds to R. The τ_m in maxon's motor data includes only the rotor inertia, as maxon notes, so any load inertia lengthens it.

The denominator splits into two real first-order factors, (τ₁s + 1)(τ₂s + 1), only if τ_m ≥ 4τ_e. Matching coefficients gives τ₁ + τ₂ = τ_m and τ₁τ₂ = τ_eτ_m, so

τ_1,2 = (τ_m/2)[1 ± √(1 − 4τ_e/τ_m)]

Treating the motor as separate lags τ_m and τ_e is therefore an approximation whose error depends only on their ratio.

The constants may be treated separately for rough work once τ_m is about ten times τ_e. Viscous friction changes the denominator to LJs² + (Lb + RJ)s + Rb + K_tK_e, and it is negligible when b is much smaller than K_tK_e/R.

The motor of the speed-loop example has τ_m = 50 ms, which implies J = 1.25 × 10⁻⁴ kg·m². Give it an armature inductance of 1 mH, so that τ_e = 1 ms. The exact poles lie at −20.4 and −979.6 s⁻¹, time constants of 49.0 ms and 1.02 ms, so that example's first-order speed model is accurate to about 2 percent. Remove the flywheel, leaving a rotor inertia of 1.0 × 10⁻⁵ kg·m², and τ_m falls to 4 ms, exactly four times τ_e. The poles merge into a double pole at −500 s⁻¹, and no first-order model describes the motor.

A Motor Driving a Fan

Let the example motor, with J = 1.25 × 10⁻⁴ kg·m² and with inductance and friction neglected, drive a fan whose torque is T_L = k_fω², with k_f = 2.22 × 10⁻⁶ N·m·s²/rad².

1. Operating point. At ω₀ = 300 rad/s the fan needs T_L0 = 0.2 N·m. The motor supplies it with i₀ = 0.2/0.05 = 4 A, which requires v₀ = Ri₀ + K_eω₀ = 4 + 15 = 19 V.
2. Linearize the load. Expanding, k_f(ω₀ + Δω)² = k_fω₀² + 2k_fω₀Δω + k_fΔω². Dropping the last term leaves ΔT_L = b_fΔω, with b_f = 2k_fω₀ = 1.33 × 10⁻³ N·m·s/rad. To small deviations, the fan is a viscous damper.
3. Write and transform the deviation equations. From Δi = (Δv − K_eΔω)/R and J dΔω/dt = K_tΔi − b_fΔω, ΔΩ(s)/ΔV(s) = (K_t/R)/(Js + K_tK_e/R + b_f), a first-order lag with a gain of 13.0 (rad/s)/V and a time constant of 32.6 ms.
4. Check validity. The neglected term k_fΔω² is the fraction Δω/(2ω₀) of the linear term, so a 30 rad/s deviation, 10 percent of the operating speed, is modeled to within about 5 percent.

At ω₀ = 150 rad/s the same fan gives a gain of 15.8 (rad/s)/V and a time constant of 39.5 ms; without the fan the values are 20 (rad/s)/V and 50 ms. A controller tuned at one speed therefore sees a different plant at another, which is why some controllers schedule their gains against the operating point.

An Unstable Operating Point

A rod of length l balanced upright on a pivot, with its mass concentrated at the free end, obeys d²θ/dt² = (g/l) sin θ, with θ measured from vertical. About the upright operating point θ₀ = 0, sin θ ≈ Δθ, and the linear model has poles at ±√(g/l), or ±4.43 s⁻¹ for l = 0.5 m. The right-half-plane pole is the instability a balancing controller must overcome. The small-angle approximation stays within 1 percent up to about 14 degrees.

Linearization fails where a nonlinearity has no useful tangent at the operating point: saturation at a supply rail, dead zones, gear backlash, static friction, and quantization. Simulation handles those, while the linear model still serves the small-signal design.

Reduction Rules

The feedback rule follows from two equations. The junction gives E = X − HY and the block gives Y = GE, so Y = GX − GHY and Y/X = G/(1 + GH). Adjacent summing junctions can be merged or reordered, because addition is commutative. A pickoff point and a summing junction cannot simply trade places: the moved branch would copy a different signal, and it needs a junction of its own to restore the original.

A Reduction Procedure

1. Combine blocks in series and in parallel.
2. Close each loop that contains no pickoff point or summing junction of another loop, innermost first.
3. Where loops interlock, move a pickoff point or summing junction until one loop stands alone, then close it.
4. Repeat until one block remains.
5. Clear the compound fractions, then check the result: its DC gain, its order, and a limiting case such as a feedback gain set to zero.

A 1/G block, created by moving a pickoff point past a block or a summing junction ahead of one, is an algebraic device rather than hardware, and it may be improper. Reduction can also cancel a factor that remains physically present, such as a lightly damped mode matched by a zero, so a reduced transfer function does not show whether every internal signal stays bounded.

From Block Diagram to Graph

1. Create a node for the input, for each summing junction's output, and for each block's output.
2. Replace each block with a branch whose transmittance is the block's transfer function.
3. Join each input of a summing junction to the junction's node with a branch of transmittance +1 or −1.
4. Treat each pickoff point as a node with several outgoing branches.
5. If the input node has incoming branches, add a new source node joined to it by a branch of transmittance 1.

Simple graphs reduce by rules that mirror block algebra: branches in series multiply, and branches in parallel add. A self-loop, a branch of gain L that leaves and reenters one node, disappears if every transmittance entering that node is divided by 1 − L. Each node states one linear equation, x_j = ∑_i t_ijx_i, with t_ij the transmittance from node i to node j, so reducing a graph solves simultaneous linear equations.

Two Interlocking Loops

The first diagram has three blocks in its forward path and two overlapping feedback paths, both subtracted. Its two summing junctions form the errors E₁ and E₂, and its signals obey the equations in the table.

The H₁ loop contains G₁ and G₂, and the H₂ loop contains G₂ and G₃. They share G₂, but neither lies inside the other, so neither can be closed first as drawn.

Direct reduction.

1. Separate the loops. Move the H₁ pickoff point from A past G₃ to Y. To return the same signal, the branch needs the factor 1/G₃, so its transmittance becomes H₁/G₃.
2. Combine in series. Nothing now taps A, so G₂ and G₃ form one block, G₂G₃.
3. Close the inner loop. G₂G₃ with feedback H₂ becomes G₂G₃/(1 + G₂G₃H₂).
4. Combine in series again. With G₁, the forward path becomes G₁G₂G₃/(1 + G₂G₃H₂).
5. Close the outer loop with feedback H₁/G₃. Its loop gain is G₁G₂H₁/(1 + G₂G₃H₂), and multiplying numerator and denominator by 1 + G₂G₃H₂ gives

Y/R = G₁G₂G₃ / (1 + G₂G₃H₂ + G₁G₂H₁)

Mason's formula. The graph has nodes R, E₁, X, E₂, A, and Y, and branches R→E₁ (1), E₁→X (G₁), X→E₂ (1), E₂→A (G₂), A→Y (G₃), A→E₁ (−H₁), and Y→E₂ (−H₂).

1. Forward paths. There is one, R→E₁→X→E₂→A→Y, with P₁ = G₁G₂G₃.
2. Loops. L₁ runs E₁→X→E₂→A→E₁, with gain −G₁G₂H₁. L₂ runs E₂→A→Y→E₂, with gain −G₂G₃H₂.
3. Non-touching loops. None: both loops pass through E₂ and A.
4. Determinants. Δ = 1 − (L₁ + L₂) = 1 + G₁G₂H₁ + G₂G₃H₂. The forward path touches both loops, so Δ₁ = 1.
5. Result. Y/R = P₁Δ₁/Δ = G₁G₂G₃/(1 + G₁G₂H₁ + G₂G₃H₂), the same as direct reduction.

A numerical check. With G₁ = 10, G₂ = 2, G₃ = 5, H₁ = 0.1, and H₂ = 0.2, both results give Y/R = 100/(1 + 2 + 2) = 20. For R = 1 the signals E₁ = 0.6, X = 6, E₂ = 2, A = 4, and Y = 20 satisfy every equation in the table.

Non-Touching Loops and a Feedforward Path

The second diagram cascades two stages. In the first, a junction drives G₁, whose output X returns through H₁. In the second, a junction adds X to a feedforward signal from R through G₃ and drives G₂, whose output Y returns through H₂. Both feedback paths are subtracted.

1. Forward paths. P₁ = G₁G₂ passes through both stages, and P₂ = G₃G₂ passes through the feedforward branch.
2. Loops. L₁ = −G₁H₁ and L₂ = −G₂H₂. They share no node, so they form one non-touching pair.
3. Determinant. Δ = 1 − (L₁ + L₂) + L₁L₂ = 1 + G₁H₁ + G₂H₂ + G₁H₁G₂H₂ = (1 + G₁H₁)(1 + G₂H₂).
4. Cofactors. P₁ touches both loops, so Δ₁ = 1. P₂ bypasses the first stage and does not touch L₁, so Δ₂ = 1 − L₁ = 1 + G₁H₁.
5. Result. Assembling the terms gives

Y/R = [G₁G₂ + G₂G₃(1 + G₁H₁)] / [(1 + G₁H₁)(1 + G₂H₂)]

Direct reduction agrees. The first stage closes to G₁/(1 + G₁H₁); adding G₃ in parallel gives [G₁ + G₃(1 + G₁H₁)]/(1 + G₁H₁); and the closed second stage multiplies that by G₂/(1 + G₂H₂). With G₁ = 10, H₁ = 0.4, G₃ = 0.5, G₂ = 8, and H₂ = 0.125, both methods give (80 + 20)/10 = 10. This example exercises the two features the first lacked: a product term for non-touching loops, and a cofactor that is not 1.

Sensitivity Functions

Two functions of the loop gain summarize what feedback does at each frequency. In a unity-feedback loop, the sensitivity S(s) = 1/[1 + L(s)] maps the reference to the error, and a disturbance added at the plant output to the output. The complementary sensitivity T(s) = L(s)/[1 + L(s)] maps the reference to the output and, with reversed sign, sensor noise to the output. The two always sum to 1. Time-Domain Response of Control Systems develops both, with disturbance rejection, peak sensitivity, and the waterbed effect.

Sensitivity also measures how drift in a component reaches the output. For the closed-loop gain M = G/(1 + GH), the fractional change in M per fractional change in each block is

(dM/M)/(dG/G) = 1/(1 + GH) and (dM/M)/(dH/H) = −GH/(1 + GH)

With G = 100 and H = 0.1, a 1 percent drift in G moves the closed-loop gain by only 0.09 percent, but a 1 percent error in H moves it by 0.91 percent in the opposite direction. Feedback suppresses errors in the forward path and passes errors in the sensor almost unchanged, so a loop can be no more accurate than its measurement.

Designing from the Inside Out

Once a loop is closed, the loop around it sees only its closed-loop transfer function, so the loops are designed from the inside out.

1. Current loop. Take the power stage and current sensor as unity gains. A PI controller C(s) = K_P + K_I/s with K_I/K_P = R/L puts its zero on the armature pole. The loop gain reduces to K_P/(Ls), and the closed loop becomes I/I_ref = 1/(1 + s/ω_ci), with ω_ci = K_P/L. In “A Faster Current Loop Pays Off in Servo Motor Control,” a July 2017 Texas Instruments white paper, Brian Fortman writes that many conventional PI current-loop controllers limit the bandwidth to about 10 percent of the PWM carrier frequency, or 1 kHz on a typical 10 kHz carrier. For the example motor, with L = 1 mH, that bandwidth (ω_ci = 6,283 rad/s) needs K_P = 6.28 V/A. The design treats the back EMF as a slow disturbance or cancels it by feedforward, and the pole cancellation is only as good as the estimate of L/R.
2. Speed loop. With friction neglected, a proportional gain K_ω sets the crossover near K_ωK_t/J. Placing it at one-tenth of the current-loop bandwidth, 628 rad/s, requires K_ω = 1.57 A/(rad/s) and leaves a phase margin of about 84 degrees. Integral action, needed to hold speed against a steady load torque, spends part of that margin.
3. Position loop. A proportional gain sets the crossover several times below the speed-loop bandwidth.

How Much Separation

Inside-out design works only if each inner loop is fast compared with the loop around it. If the inner closed loop behaves as 1/(1 + s/ω_inner), then at the outer loop's crossover frequency ω_outer it adds a phase lag of arctan(ω_outer/ω_inner) and a small loss of gain.

The lag comes out of the outer loop's phase margin. Real inner loops lag more than this first-order picture, because of sampling delay, sensor filtering, and imperfect pole cancellation. The Texas Instruments paper states that in most systems the current-loop bandwidth exceeds that of the speed and position loops “by nearly 10 times or even higher,” and Servo Drive Systems gives the separations used in industrial drives.

Benefits and Costs

• A clamp on the current reference protects the motor and power stage, whatever the outer loops demand.
• The current loop corrects supply-voltage changes and back-EMF variation before they disturb the speed, and it makes the motor act as a torque source.
• Each loop needs its own measurement, and no outer loop can be faster than the loops inside it.
• A clamp that saturates makes the cascade nonlinear, so integrators in the outer controllers need anti-windup protection.

Tools

Graphical tools such as MathWorks Simulink build models from blocks, summing junctions, and signal lines. Scripting libraries apply block algebra to transfer-function objects. MATLAB's Control System Toolbox provides series, parallel, feedback, and connect, and its feedback assumes negative feedback unless given a sign of +1. The python-control library provides series, parallel, feedback, whose sign argument defaults to −1, and interconnect for larger diagrams. A SPICE simulator can run a controller built from behavioral sources against a transistor-level power stage; Circuit Simulation (SPICE) describes those programs.

Algebraic Loops

A loop built only from blocks with direct feedthrough, whose outputs depend on their inputs at the same instant, contains no state to break the circle, so every time step requires solving a simultaneous equation. MathWorks documents that Simulink solves such algebraic loops with a nonlinear solver at each time step and that models containing them can run more slowly. Two remedies are common: reduce the loop by hand, so that a static G and H become the single gain G/(1 + GH), or restore a dynamic element the model left out, such as a sensor's small lag.

Stiff Models

Widely separated time constants make a model stiff. The explicit forward Euler method is numerically unstable for a mode with time constant τ unless the step is shorter than 2τ, so the fastest mode sets the step long after it has died away. The example motor is only mildly stiff: its 1 ms electrical mode limits the step to about 2 ms, while the speed takes about 200 ms to settle. Solvers designed for stiff equations, such as MATLAB's ode15s, handle wider separations more efficiently, and when the fast mode does not matter, the separate time-constant approximation removes it. Ordinary Differential Equations derives that step limit and compares the implicit methods that remove it.

Checking the Model Against Hardware

1. Simulate the linearized and nonlinear models at the same operating point with a small step, and confirm that they agree.
2. Measure the real system's small-step response at that point, and compare it with the simulation after adding the operating-point values back to the deviation variables.
3. Adjust uncertain parameters, such as inertia and friction, until the two match, and record the values and assumptions.
4. Repeat at other operating points and at large amplitude, with the limits active.