DC Machine Modeling: Transfer Functions & State-Space Models

Recap from Lecture 2B

Coupled electromechanical equations:

Dynamic Model

\[\begin{aligned} V &= R_a I_a + L_a\frac{dI_a}{dt} + K_e\omega_m\\ J\frac{d\omega_m}{dt} &= K_t I_a - B\omega_m - T_L \end{aligned}\]

Today's focus:

  • Transform time-domain equations to frequency domain
  • Derive transfer functions for system analysis
  • Develop complete block diagram representation
  • State-space modeling approach
  • Stability analysis and system characteristics

Laplace Transform Approach

Why Use Laplace Transforms?

Advantages of frequency domain analysis:

Time Domain:

  • Differential equations
  • Complex to solve
  • Hard to visualize system behavior
  • Numerical methods often needed

Frequency Domain:

  • Algebraic equations
  • Easier manipulation
  • Clear system properties
  • Transfer functions reveal characteristics

Laplace Transform

\[\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty f(t)e^{-st}dt\]

where \(s = \sigma + j\omega\) is the complex frequency variable

Key property: \(\mathcal{L}\left\{\dfrac{df}{dt}\right\} = sF(s) - f(0^-)\)

Laplace Transform of Motor Equations

Armature equation:

\[V = R_a I_a + L_a\frac{dI_a}{dt} + K_e\omega_m\]

Taking Laplace transform (assuming zero initial conditions):

\[V(s) = R_a I_a(s) + L_a sI_a(s) + K_e\Omega_m(s)\]
\[\boxed{V(s) = (R_a + sL_a)I_a(s) + K_e\Omega_m(s)}\]

Mechanical equation:

\[J\frac{d\omega_m}{dt} = K_t I_a - B\omega_m - T_L\]

Taking Laplace transform:

\[Js\Omega_m(s) = K_t I_a(s) - B\Omega_m(s) - T_L(s)\]
\[\boxed{(Js + B)\Omega_m(s) = K_t I_a(s) - T_L(s)}\]

Solving for Armature Current

From armature equation:

\[V(s) = (R_a + sL_a)I_a(s) + K_e\Omega_m(s)\]

Solving for \(I_a(s)\):

\[(R_a + sL_a)I_a(s) = V(s) - K_e\Omega_m(s)\]

Current Expression

\[\boxed{I_a(s) = \frac{V(s) - K_e\Omega_m(s)}{R_a + sL_a}}\]

This can be written as:

\[I_a(s) = \frac{1}{R_a + sL_a}V(s) - \frac{K_e}{R_a + sL_a}\Omega_m(s)\]

Electrical impedance: \(Z_a(s) = R_a + sL_a\)

Solving for Speed

From mechanical equation:

\[(Js + B)\Omega_m(s) = K_t I_a(s) - T_L(s)\]

Solving for \(\Omega_m(s)\):

\[\Omega_m(s) = \frac{K_t I_a(s) - T_L(s)}{Js + B}\]

Speed Expression

\[\boxed{\Omega_m(s) = \frac{K_t}{Js + B}I_a(s) - \frac{1}{Js + B}T_L(s)}\]

Interpretation:

  • First term: Speed response to electromagnetic torque
  • Second term: Speed response to load disturbance

Mechanical impedance: \(Z_m(s) = Js + B\)

Block Diagram Development

Block Diagram: Electrical Subsystem

From \(I_a(s) = \dfrac{V(s) - K_e\Omega_m(s)}{R_a + sL_a}\)

Electrical Subsystem Block Diagram
Electrical Subsystem Block Diagram

Interpretation:

  • Input voltage \(V(s)\) drives the armature circuit
  • Back-EMF \(K_e\Omega_m(s)\) opposes the applied voltage
  • Net voltage across impedance determines current

Block Diagram: Mechanical Subsystem

From \(\Omega_m(s) = \dfrac{K_t I_a(s) - T_L(s)}{Js + B}\)

Mechanical Subsystem Block Diagram
Mechanical Subsystem Block Diagram

Interpretation:

  • Armature current produces electromagnetic torque \(T_e = K_t I_a\)
  • Net torque (electromagnetic minus load) accelerates the rotor
  • Mechanical dynamics relate net torque to speed

Complete Block Diagram

Combining electrical and mechanical subsystems:

Complete Block Diagram
Complete Block Diagram

Key observation: The back-EMF provides negative feedback, which is crucial for stability and speed regulation.

Transfer Function Derivation

Transfer Function: Voltage to Speed

Goal: Find \(G_v(s) = \dfrac{\Omega_m(s)}{V(s)}\) with \(T_L(s) = 0\)

From the two equations:

\[\begin{aligned} V(s) &= (R_a + sL_a)I_a(s) + K_e\Omega_m(s) \\ (Js + B)\Omega_m(s) &= K_t I_a(s) \end{aligned}\]

From equation (2):

\[I_a(s) = \frac{(Js + B)\Omega_m(s)}{K_t}\]

Substituting into equation (1):

\[V(s) = (R_a + sL_a)\frac{(Js + B)\Omega_m(s)}{K_t} + K_e\Omega_m(s)\]
\[V(s) = \left[\frac{(R_a + sL_a)(Js + B)}{K_t} + K_e\right]\Omega_m(s)\]

Transfer Function: Simplification

Expanding the numerator:

\[\frac{(R_a + sL_a)(Js + B)}{K_t} + K_e = \frac{R_aJs + R_aB + JL_as^2 + BL_as}{K_t} + K_e\]
\[= \frac{JL_as^2 + (BL_a + JR_a)s + BR_a}{K_t} + K_e\]
\[= \frac{JL_as^2 + (BL_a + JR_a)s + BR_a + K_eK_t}{K_t}\]

Voltage-to-Speed Transfer Function

\[\boxed{G_v(s) = \frac{\Omega_m(s)}{V(s)} = \frac{K_t}{JL_as^2 + (BL_a + JR_a)s + (BR_a + K_eK_t)}}\]

Transfer Function: Load Torque to Speed

Goal: Find \(G_L(s) = \dfrac{\Omega_m(s)}{T_L(s)}\) with \(V(s) = 0\)

From the equations with \(V = 0\):

\[\begin{aligned} 0 &= (R_a + sL_a)I_a(s) + K_e\Omega_m(s)\\ (Js + B)\Omega_m(s) &= K_t I_a(s) - T_L(s) \end{aligned}\]

From first equation:

\[I_a(s) = -\frac{K_e\Omega_m(s)}{R_a + sL_a}\]

Substituting into second equation:

\[(Js + B)\Omega_m(s) = -\frac{K_tK_e\Omega_m(s)}{R_a + sL_a} - T_L(s)\]

Load-to-Speed Transfer Function

\[\boxed{G_L(s) = \frac{\Omega_m(s)}{T_L(s)} = \frac{-(R_a + sL_a)}{JL_as^2 + (BL_a + JR_a)s + (BR_a + K_eK_t)}}\]

Second-Order System Form

Standard second-order form:

\[G(s) = \frac{K\omega_n^2}{s^2 + 2\zeta\omega_ns + \omega_n^2}\]

Comparing with our transfer function:

\[G_v(s) = \frac{K_t}{JL_as^2 + (BL_a + JR_a)s + (BR_a + K_eK_t)}\]

System Parameters

Natural frequency:

\[\boxed{\omega_n = \sqrt{\frac{BR_a + K_eK_t}{JL_a}}}\]

Damping ratio:

\[\boxed{\zeta = \frac{BL_a + JR_a}{2\sqrt{JL_a(BR_a + K_eK_t)}}}\]

DC gain:

\[\boxed{K_{dc} = \frac{K_t}{BR_a + K_eK_t}}\]

Physical Interpretation of Parameters

Natural frequency \(\omega_n\):

  • Characterizes how fast the system responds
  • Higher \(\omega_n\) means faster response
  • Depends on electrical and mechanical parameters

Damping ratio \(\zeta\):

  • \(\zeta < 1\): Underdamped (oscillatory response)
  • \(\zeta = 1\): Critically damped
  • \(\zeta > 1\): Overdamped (slow, non-oscillatory)
  • Most DC motors are overdamped

DC gain \(K_{dc}\):

  • Steady-state speed per unit voltage
  • Units: (rad/s)/V
  • Important for speed control design

Time Constants

For many DC motors, we can identify two time constants:

Electrical Time Constant

\[\boxed{\tau_e = \frac{L_a}{R_a}}\]

Determines how fast the current responds to voltage changes

Mechanical Time Constant

\[\boxed{\tau_m = \frac{J}{B}}\]

Determines how fast the speed responds to torque changes

Typical relationship: \(\tau_m \gg \tau_e\)

The mechanical response is much slower than the electrical response

State-Space Modeling

Why State-Space Models?

Advantages over transfer functions:

  • Can handle multiple inputs and outputs (MIMO systems)
  • Works for time-varying and nonlinear systems
  • Direct access to all internal states
  • Foundation for modern control design
  • Better for computer implementation

State-Space Form

\[\begin{aligned} \dot{\mathbf{x}} &= \mathbf{A}\mathbf{x} + \mathbf{B}\mathbf{u}\\ \mathbf{y} &= \mathbf{C}\mathbf{x} + \mathbf{D}\mathbf{u} \end{aligned}\]

where:

  • \(\mathbf{x}\): State vector
  • \(\mathbf{u}\): Input vector
  • \(\mathbf{y}\): Output vector
  • \(\mathbf{A, B, C, D}\): System matrices

State Selection

Choose states that completely describe the system:

State Variables

\[\mathbf{x} = \begin{bmatrix} I_a \\ \omega_m \end{bmatrix}\]
  • \(I_a\): Armature current (A)
  • \(\omega_m\): Motor speed (rad/s)

Input Variables

\[\mathbf{u} = \begin{bmatrix} V \\ T_L \end{bmatrix}\]
  • \(V\): Applied voltage (V)
  • \(T_L\): Load torque (N·m)

Output: Typically \(y = \omega_m\) (speed measurement)

Deriving State Equations

Starting from the dynamic equations:

\[\begin{aligned} V &= R_a I_a + L_a\frac{dI_a}{dt} + K_e\omega_m\\ J\frac{d\omega_m}{dt} &= K_t I_a - B\omega_m - T_L \end{aligned}\]

Solving for derivatives:

\[\frac{dI_a}{dt} = -\frac{R_a}{L_a}I_a - \frac{K_e}{L_a}\omega_m + \frac{1}{L_a}V\]
\[\frac{d\omega_m}{dt} = \frac{K_t}{J}I_a - \frac{B}{J}\omega_m - \frac{1}{J}T_L\]

State-Space Representation

In matrix form:

State Equation

\[\begin{bmatrix} \dfrac{dI_a}{dt} \\ \dfrac{d\omega_m}{dt} \end{bmatrix} = \begin{bmatrix} -\dfrac{R_a}{L_a} & -\dfrac{K_e}{L_a} \\ \dfrac{K_t}{J} & -\dfrac{B}{J} \end{bmatrix} \begin{bmatrix} I_a \\ \omega_m \end{bmatrix} + \begin{bmatrix} \dfrac{1}{L_a} & 0 \\ 0 & -\dfrac{1}{J} \end{bmatrix} \begin{bmatrix} V \\ T_L \end{bmatrix}\]

Output Equation

If output is speed:

\[y = \begin{bmatrix} 0 & 1 \end{bmatrix} \begin{bmatrix} I_a \\ \omega_m \end{bmatrix} + \begin{bmatrix} 0 & 0 \end{bmatrix} \begin{bmatrix} V \\ T_L \end{bmatrix}\]

System Matrices

Identifying the matrices:

State Matrix A

\[\mathbf{A} = \begin{bmatrix} -\dfrac{R_a}{L_a} & -\dfrac{K_e}{L_a} \\ \dfrac{K_t}{J} & -\dfrac{B}{J} \end{bmatrix}\]

Describes system dynamics

Input Matrix B

\[\mathbf{B} = \begin{bmatrix} \dfrac{1}{L_a} & 0 \\ 0 & -\dfrac{1}{J} \end{bmatrix}\]

Describes how inputs affect states

Output Matrix C

\[\mathbf{C} = \begin{bmatrix} 0 & 1 \end{bmatrix}\]

Selects speed as output

Feedthrough Matrix D

\[\mathbf{D} = \begin{bmatrix} 0 & 0 \end{bmatrix}\]

No direct feedthrough from input to output

Stability Analysis

Eigenvalues and Stability

System stability determined by eigenvalues of \(\mathbf{A}\):

\[\det(s\mathbf{I} - \mathbf{A}) = 0\]
\[\det\begin{bmatrix} s + \dfrac{R_a}{L_a} & \dfrac{K_e}{L_a} \\ -\dfrac{K_t}{J} & s + \dfrac{B}{J} \end{bmatrix} = 0\]

Expanding the determinant:

\[s^2 + s\left(\frac{R_a}{L_a} + \frac{B}{J}\right) + \frac{BR_a + K_eK_t}{JL_a} = 0\]

Stability Condition

System is stable if all eigenvalues have negative real parts

For DC motor: Always stable since all parameters are positive

Pole Locations

The characteristic equation is:

\[JL_as^2 + (BL_a + JR_a)s + (BR_a + K_eK_t) = 0\]

This is identical to the denominator of our transfer function!

Poles determine system response:

  • Real, negative poles → Exponential decay
  • Complex conjugate poles → Damped oscillation
  • Location determines speed of response

Step Response Characteristics

Key performance metrics:

  • Rise time \(t_r\): Time to reach 90% of final value
  • Settling time \(t_s\): Time to stay within 2% of final value
  • Peak time \(t_p\): Time to first peak (if oscillatory)
  • Overshoot \(M_p\): Maximum overshoot percentage
Step Response
Step Response Characteristics

Most DC motors exhibit overdamped response: smooth approach to steady-state without oscillation.

Steady-State Analysis

At steady-state, all derivatives are zero:

From the state equations with \(\dfrac{dI_a}{dt} = 0\) and \(\dfrac{d\omega_m}{dt} = 0\):

\[\begin{aligned} 0 &= -\frac{R_a}{L_a}I_{a,ss} - \frac{K_e}{L_a}\omega_{m,ss} + \frac{1}{L_a}V_{ss}\\ 0 &= \frac{K_t}{J}I_{a,ss} - \frac{B}{J}\omega_{m,ss} - \frac{1}{J}T_{L,ss} \end{aligned}\]

Solving for steady-state speed:

\[R_a I_{a,ss} = V_{ss} - K_e\omega_{m,ss}\]
\[K_t I_{a,ss} = B\omega_{m,ss} + T_{L,ss}\]

Steady-State Speed

\[\boxed{\omega_{m,ss} = \frac{K_t V_{ss} - R_a T_{L,ss}}{BR_a + K_e K_t}}\]

Speed Regulation

Speed regulation quantifies how speed changes with load:

Speed Regulation Definition

\[\text{Speed Regulation} = \frac{\omega_{m,nl} - \omega_{m,fl}}{\omega_{m,fl}} \times 100\%\]

where nl = no-load, fl = full-load

From steady-state equation:

\[\Delta\omega_m = -\frac{R_a}{BR_a + K_e K_t}\Delta T_L\]

Speed drop per unit load torque:

\[\frac{d\omega_m}{dT_L} = -\frac{R_a}{BR_a + K_e K_t}\]

Design Insight

For better speed regulation (less speed drop with load):

  • Increase \(K_e\) and \(K_t\) (stronger magnetic field)
  • Decrease \(R_a\) (lower armature resistance)

Example Problem

Example: DC Motor Parameter Analysis

Given a separately-excited DC motor with parameters:

  • Armature resistance: \(R_a = 2~\Omega\)
  • Armature inductance: \(L_a = 0.01\) H
  • Back-EMF constant: \(K_e = K_t = 0.5\) V/(rad/s) = 0.5 N\(\cdot\)m/A
  • Moment of inertia: \(J = 0.02\) kg\(\cdot\)m\(^2\)
  • Viscous friction: \(B = 0.001\) N\(\cdot\)m\(\cdot\)s/rad

Tasks:

  1. Calculate the time constants \(\tau_e\) and \(\tau_m\)
  2. Determine the natural frequency \(\omega_n\) and damping ratio \(\zeta\)
  3. Find the steady-state speed for \(V = 100\) V and \(T_L = 5\) N\(\cdot\)m
  4. Calculate the DC gain \(K_{dc}\)

Example: Solution

1. Time constants:

\[\begin{aligned} \tau_e &= \frac{L_a}{R_a} = \frac{0.01}{2} = \textcolor{accentcolor}{\textbf{0.005~s = 5~ms}}\\ \tau_m &= \frac{J}{B} = \frac{0.02}{0.001} = \textcolor{accentcolor}{\textbf{20~s}} \end{aligned}\]

Note: \(\tau_m \gg \tau_e\) (mechanical response much slower)

2. Natural frequency and damping ratio:

\[\begin{aligned} \omega_n &= \sqrt{\frac{BR_a + K_e K_t}{JL_a}} = \sqrt{\frac{0.001 \times 2 + 0.5^2}{0.02 \times 0.01}}\\ &= \sqrt{\frac{0.002 + 0.25}{0.0002}} = \sqrt{1260} = \textcolor{accentcolor}{\textbf{35.5~rad/s}} \end{aligned}\]
\[\begin{aligned} \zeta &= \frac{BL_a + JR_a}{2\sqrt{JL_a(BR_a + K_e K_t)}} = \frac{0.001 \times 0.01 + 0.02 \times 2}{2\sqrt{0.02 \times 0.01 \times 0.252}}\\ &= \frac{0.00001 + 0.04}{2\sqrt{0.0000504}} = \frac{0.04001}{0.01418} = \textcolor{accentcolor}{\textbf{2.82}} \end{aligned}\]

System is heavily overdamped (no oscillations expected).

3. Steady-state speed:

\[\begin{aligned} \omega_{m,ss} &= \frac{K_t V_{ss} - R_a T_{L,ss}}{BR_a + K_e K_t}\\ &= \frac{0.5 \times 100 - 2 \times 5}{0.001 \times 2 + 0.5^2}\\ &= \frac{50 - 10}{0.002 + 0.25} = \frac{40}{0.252}\\ &= \textcolor{accentcolor}{\textbf{158.73~rad/s}} = \textcolor{accentcolor}{\textbf{1515.8~rpm}} \end{aligned}\]

4. DC gain:

\[\begin{aligned} K_{dc} &= \frac{K_t}{BR_a + K_e K_t} = \frac{0.5}{0.001 \times 2 + 0.5^2}\\ &= \frac{0.5}{0.252} = \textcolor{accentcolor}{\textbf{1.984~(rad/s)/V}} \end{aligned}\]

This means the motor speed increases by approximately 2 rad/s for every 1 V increase in applied voltage at steady-state.

Summary: Key Takeaways

Transfer function models:

  • Voltage-to-speed: \(G_v(s) = \dfrac{K_t}{JL_a s^2 + (BL_a + JR_a)s + (BR_a + K_e K_t)}\)
  • Load-to-speed: \(G_L(s) = \dfrac{-(R_a + sL_a)}{JL_a s^2 + (BL_a + JR_a)s + (BR_a + K_e K_t)}\)
  • Second-order system with parameters \(\omega_n\) and \(\zeta\)

State-space model:

  • States: \([I_a, \omega_m]^T\) (current and speed)
  • Inputs: \([V, T_L]^T\) (voltage and load torque)
  • Complete description of system dynamics
  • Suitable for modern control design

System characteristics:

  • Inherently stable (all poles in left half-plane)
  • Usually overdamped (\(\tau_m \gg \tau_e\))
  • Natural speed regulation due to back-EMF feedback