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System Dynamics · Module 3 of 10

Electrical and Electromechanical Systems

An RLC circuit is a mass-spring-damper in disguise: the same second-order equation, the same natural frequency and damping. The DC motor joins the electrical and mechanical worlds into one coupled system.

01

Readiness check

This module models circuits dynamically. Tick only what you can do closed-notes.

  • Recall the element laws for R, L, and C.
  • Write Kirchhoff's voltage law around a loop.
  • Read ωn and ζ from a second-order equation.
  • Recall the time constant τ = L/R or RC.
  • State that a motor converts electrical to mechanical power.
0 or 1 weak itemsContinue with this module.
2 weak itemsRevisit capacitors and inductors in Electrical Circuits, Module 4.
3 or more weak itemsReview the second-order form in Module 2.
02

The core idea

A series RLC circuit obeys the same second-order equation as a mass-spring-damper, with the same natural frequency and damping ratio. A DC motor couples a circuit to a rotor, and its behaviour splits into a fast electrical and a slow mechanical time constant.

L q̈ + R q̇ + q/C = v(t)ωn = 1/√(LC), ζ = (R/2)√(C/L)τe = L/R, τm = R J/(Kt Kb)

Applying Kirchhoff's voltage law to a series RLC loop gives L q̈ + R q̇ + q/C = v(t), where q is charge. This is the exact electrical twin of m ẍ + c ẋ + k x = F: inductance plays the role of mass, resistance the damper, and the inverse of capacitance the spring. So the natural frequency is ωn = 1/√(LC) and the damping ratio ζ = (R/2)√(C/L). The DC motor is the bridge between domains: applied voltage drives a current through the armature inductance and resistance, the current makes torque (τ = Kt i), and the spinning rotor makes a back-EMF (vb = Kb ω). The motor has two characteristic times, a fast electrical one τe = L/R and a slow mechanical one τm, and in SI units Kt equals Kb.

The skill works when: you map L, R, C to mass, damper, spring and read the dynamics from the analogy.
The skill breaks down when: the capacitor is treated as inertia, or the two motor time constants are confused.
The concept. The series RLC loop is the electrical mass-spring-damper. Inductance stores energy like a mass, the capacitor like a spring, and the resistor dissipates like a damper, giving the same second-order dynamics.
03

The skills, taught in order

Five skills carry the second-order form into circuits and then into the coupled DC motor.

3.1 The RLC circuit as a dynamic system

Kirchhoff's voltage law around a series RLC loop gives L q̈ + R q̇ + q/C = v(t). The same equation in current form has a derivative on the source. Either way it is second order, with two energy stores: the inductor and the capacitor.

3.2 Electrical natural frequency and damping

Matching the standard form, the natural frequency is ωn = 1/√(LC) and the damping ratio is ζ = (R/2)√(C/L). A small resistance gives a lightly damped, ringing circuit; a large one overdamps it, exactly as in mechanics.

MechanicalElectrical (series)Role
mass minductance Linertia / energy store
damper cresistance Rdissipation
stiffness k1/capacitance, 1/Crestoring / energy store
force Fvoltage veffort input

The force-voltage analogy. Each mechanical element has an exact electrical counterpart, so one analysis serves both.

3.3 The DC motor model

The armature circuit is L di/dt + R i = v − Kbω; the torque is τ = Kt i, and the rotor obeys J ω̇ + b ω = Kt i. Back-EMF couples the mechanical speed back into the electrical loop, making the motor a genuinely coupled system.

3.4 Electrical and mechanical time constants

The motor has two characteristic times: a fast electrical one τe = L/R, set by the armature, and a slow mechanical one τm = R J/(Kt Kb), set by the rotor inertia. Their wide separation lets the electrical transient be treated as instantaneous in many control models.

3.5 Energy conversion and Kt = Kb

In consistent SI units the torque constant Kt (N·m/A) equals the back-EMF constant Kb (V·s/rad), because the electrical power vb i delivered to the rotor equals the mechanical power τ ω. This identity is a statement of energy conservation in the coupling.

Engineering connection: every servo and actuator in Robotics and Mechatronics rests on this motor model and its two time constants.

04

Worked example 1: an RLC circuit as a second-order system

A series RLC circuit has L = 0.5 H, C = 2 µF, and R = 250 Ω. Find the natural frequency and the damping ratio.

Figure 1. The charge or current rings and decays just like an underdamped mechanical system. The resistance sets the damping; the inductance and capacitance set the frequency.
  1. ProblemFind ωn and ζ for the RLC circuit in Figure 1.
  2. Given / findL = 0.5 H, C = 2 µF, R = 250 Ω. Find ωn and ζ.
  3. AssumptionsIdeal linear elements; series loop, so the standard second-order form holds.
  4. Modelωn = 1/√(LC); ζ = (R/2)√(C/L) by analogy with the mechanical system.
  5. Equationsωn = 1/√(LC)ζ = (R/2)√(C/L)
  6. SolveLC = 0.5 × 2×10−6 = 1×10−6, so ωn = 1/√(10−6) = 1000 rad/s. ζ = (250/2)√(2×10−6/0.5) = 125 × √(4×10−6) = 125 × 0.002 = 0.25.
  7. Checkζ = 0.25 is below 1, so the circuit is underdamped and rings, as Figure 1 shows. The same √(LC) that set resonance in AC circuits sets the natural frequency here.
  8. ConclusionThe circuit is a second-order system with ωn = 1000 rad/s and ζ = 0.25, identical in form to a lightly damped mechanical oscillator.
Result. ωn = 1000 rad/s, ζ = 0.25.
05

Worked example 2: DC motor time constants

A DC motor has armature inductance L = 0.01 H, resistance R = 2 Ω, rotor inertia J = 0.002 kg·m², and constants Kt = Kb = 0.1 (SI). Find the electrical and mechanical time constants.

Figure 2. The motor has a fast electrical transient in the armature and a slow mechanical transient in the rotor. The 80-to-1 separation lets the electrical dynamics often be ignored in control models.
  1. ProblemFind τe and τm for the DC motor in Figure 2.
  2. Given / findL = 0.01 H, R = 2 Ω, J = 0.002 kg·m², Kt = Kb = 0.1. Find τe and τm.
  3. AssumptionsNegligible mechanical friction relative to the back-EMF term; linear motor.
  4. Modelτe = L/R from the armature; τm = R J/(Kt Kb) from the rotor and coupling.
  5. Equationsτe = L/Rτm = R J/(Kt Kb)
  6. Solveτe = 0.01/2 = 0.005 s = 5 ms. τm = (2 × 0.002)/(0.1 × 0.1) = 0.004/0.01 = 0.4 s.
  7. Checkτm/τe = 0.4/0.005 = 80, a wide separation, so the current settles long before the speed does. Units: τm = (Ω·kg·m²)/(N·m/A · V·s/rad) reduces to seconds.
  8. ConclusionThe electrical transient (5 ms) is 80 times faster than the mechanical one (0.4 s), so a speed-control model can often treat the current as reaching its value instantly.
Result. τe = 5 ms, τm = 0.4 s.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Capacitor as inertiaωn built with C in the wrong place"Which element stores like a spring?"1/C is the stiffness; L is the inertia.
Confusing the two time constantsElectrical and mechanical times swapped"Is this L/R or R J/(KtKb)?"τe is armature, τm is rotor.
Kt not equal to KbEnergy balance does not close"Am I in consistent SI units?"In SI, Kt = Kb numerically.
Ignoring back-EMFMotor modeled as a plain RL circuit"Does speed feed back into the loop?"Include the Kbω term; it couples the domains.
07

Practice ladder

Level 1 · Direct skill

A series RLC has L = 0.1 H and C = 10 µF. Find the natural frequency.

Show answer

ωn = 1/√(LC) = 1/√(10−6) = 1000 rad/s.

Level 2 · Mixed concept

That same circuit has R = 40 Ω. Find the damping ratio.

Show answer

ζ = (R/2)√(C/L) = 20 × √(10−5/0.1) = 20 × √(10−4) = 20 × 0.01 = 0.2.

Level 3 · Independent problem

A motor has L = 0.005 H, R = 1 Ω, J = 0.001 kg·m², Kt = Kb = 0.05. Find both time constants and their ratio.

Show answer

τe = 0.005/1 = 5 ms. τm = (1 × 0.001)/(0.05 × 0.05) = 0.001/0.0025 = 0.4 s. Ratio τm/τe = 80.

Transfer task | Real engineering

Explain why a speed controller for this motor can usually ignore the armature inductance, and when that simplification would fail.

What good work looks like

Because τe is far smaller than τm, the current reaches its value almost instantly on the mechanical timescale, so L can be dropped. It fails when the control bandwidth approaches 1/τe, where the electrical dynamics matter again.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I mapped L, R, C to the right mechanical elements."
"Give me three RLC circuits; I will find ωn and ζ for each."
"Solve this motor for me." Writing the coupled equations is the skill.
"Which time constant is which?" Deriving each from its physics is the point.

Portfolio task

Take a real RLC circuit or a small DC motor, compute its natural frequency and damping, or its two time constants, and compare to a measured or datasheet value.

Must include: the element values, a computed ωn and ζ or τe and τm, and a comparison.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the series RLC equation.

L q̈ + R q̇ + q/C = v(t), the electrical mass-spring-damper.

2. Give ωn and ζ for the RLC.

ωn = 1/√(LC) and ζ = (R/2)√(C/L).

3. What couples the DC motor's two domains?

Torque τ = Kt i one way and back-EMF vb = Kb ω the other.

4. Give the two motor time constants.

τe = L/R (electrical) and τm = R J/(Kt Kb) (mechanical).

5. Why does Kt = Kb in SI?

Energy conservation: electrical power into the rotor equals the mechanical power out.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive ωn and ζ for an RLC from KVL.
+3 daysFind the time constants of three motors.
+7 daysCarry the modeling into fluid and thermal systems, Module 4.
+30 daysReuse the motor model when sizing an actuator.
10

Textbook mapping

This module follows Karnopp, Margolis, and Rosenberg, System Dynamics: Modeling, Simulation, and Control of Mechatronic Systems, 5th edition. Use these references to read further.

Topic in this moduleWhere to read more
Electrical elements and RLC circuitsKarnopp, Margolis & Rosenberg, Chapter 2
The DC motor and electromechanical couplingKarnopp, Margolis & Rosenberg, Chapter 5
Electrical and mechanical time constantsKarnopp, Margolis & Rosenberg, Chapter 5

Chapter numbers refer to the 5th edition. The circuit and motor relations are standard, so any recent edition will align closely.