FOUNDATION · CHAPTER 3 OF 8

From a physical system to a transfer function

This chapter turns hardware into equations you can transform: masses, springs, dampers, gears and a DC motor, then a curved law replaced by its tangent at an operating point. A transfer function exists only once the physics is written as a linear model.

Updated 13 Oct 2026

What you will be able to do

You will write the force or torque balance of a mechanical system and turn it into a transfer function. You will move a load's inertia across a gear train and say why the factor is squared. You will build the DC motor model from two physical laws and see why its two constants are equal. You will linearise a curved law about an operating point and say what that costs you.

Before you start

Chapter 2, The Laplace transform as a working tool
The second law of motion; torque as force times radius; moment of inertia; arc length $r θ$
The force on a current-carrying conductor, $F = B l i$ ; the law of induction (emf equals rate of change of flux)
Power as force times speed, torque times angular speed, voltage times current
Derivatives of $sin θ$ and $h$ ; the tangent line to a curve; complex roots of a quadratic

Springs, masses, dampers and gears

Every model here starts with a balance. The forces on a mass add up to its mass times acceleration. The torques on a rotor add up to its moment of inertia times angular acceleration. An ideal spring pushes back in proportion to its stretch, a viscous damper in proportion to the speed of one end relative to the other. Write the balance in time, then transform it with zero initial conditions, so each $d / d t$ becomes a factor $s$ . The output over the input is $G (s)$ .

MASS, SPRING AND DAMPER

m \overset{x}{¨} = f - b \overset{x}{˙} - k x \Rightarrow G (s) = \frac{X ( s )}{F ( s )} = \frac{1}{m s ^{2} + b s + k}

Constant mass $m$ (kg), ideal spring $k$ (N/m), viscous damper $b$ (N·s/m), force $f$ (N) applied at $t = 0$ , zero initial conditions. The transform is Chapter 2's, from $0^{-}$ ; for a signal zero before $t = 0$ the bilateral form gives the same. $x$ (m) is measured from equilibrium, so a constant weight only shifts the equilibrium.

Rotation has the same shape: $J \ddot{θ} = τ - B \dot{θ} - k_{θ} θ$ , with moment of inertia $J$ (kg·m²), rotational damping $B$ (N·m·s/rad), torsional stiffness $k_{θ}$ (N·m/rad), torque $τ$ (N·m) and angle $θ$ (rad). A pair of meshing gears joins two shafts. At the contact both gears move through the same arc, $r_{1} θ_{1} = r_{2} θ_{2}$ . Meshing gears share a tooth size, so tooth count is proportional to radius, and the angles are in the inverse ratio of the tooth counts. An ideal train wastes no power.

GEAR RATIO

n = \frac{N _{1}}{N _{2}} = \frac{θ _{2}}{θ _{1}} = \frac{ω _{2}}{ω _{1}}, τ_{1} ω_{1} = τ_{2} ω_{2} \Rightarrow τ_{2} = \frac{τ _{1}}{n}

Ideal, rigid gears: no slip, backlash or loss. Shaft 1 (motor side) has $N_{1}$ teeth, shaft 2 (load side) $N_{2}$ . $θ$ in rad, $ω$ in rad/s; $τ_{1}$ (applied at shaft 1) and $τ_{2}$ (delivered at shaft 2) in N·m.

REFLECTED INERTIA

J_{e q} = J_{m} + n^{2} J_{L}, B_{e q} = B_{m} + n^{2} B_{L}

Ideal gears. $J_{m}$ , $B_{m}$ : motor shaft (rotor and its gear); $J_{L}$ , $B_{L}$ : load shaft (load and its gear); kg·m² and N·m·s/rad. The load needs $τ_{2} = J_{L} \ddot{θ}_{2}$ . The motor shaft supplies $τ_{1} = n τ_{2}$ , and $\ddot{θ}_{2} = n \ddot{θ}_{1}$ , so $τ_{1} = n^{2} J_{L} \ddot{θ}_{1}$ . The same steps on $B_{L} \dot{θ}_{2}$ give $n^{2} B_{L}$ .

WATCH OUT FOR

Using $n$ when you need $n^{2}$ , or $1/ n$ when you need $n$ . Test: a reducing train ( $n < 1$ , load slower) must make the load look lighter to the motor, so the factor has to be below one.

WORKED EXAMPLE

A 2 kg mass slides on a frictionless horizontal surface, held by a spring of 50 N/m and a viscous damper of 8 N·s/m. A force $f (t)$ is applied from $t = 0$ . Find $G (s) = X (s) / F (s)$ , its poles, and the final displacement for a 10 N step.

Solution

Balance the forces. The spring and damper oppose the motion, so $2 \overset{x}{¨} = f - 8 \overset{x}{˙} - 50 x$ , that is $2 \overset{x}{¨} + 8 \overset{x}{˙} + 50 x = f$ .
Transform with zero initial conditions, because $G (s)$ is defined that way: $(2 s^{2} + 8 s + 50) X = F$ , so $G (s) = \frac{1}{2 s ^{2} + 8 s + 50}$ .
The poles are the roots of $2 s^{2} + 8 s + 50 = 0$ , or $s^{2} + 4 s + 25 = 0$ after dividing by 2: $s = - 2 \pm j 21 \approx - 2 \pm j 4.58$ . Both lie in the LHP, the open left half-plane (real part below zero). The RHP is the open right half-plane; the $j ω$ axis is in neither.
With both poles in the LHP, the final value theorem holds for a step. $G (0) = 1/50$ m/N, so a 10 N step ends at $0.2$ m. Check: the spring then pushes back $50 \times 0.2 = 10$ N, balancing the force.

Answer: $G (s) = 1/ (2 s^{2} + 8 s + 50)$ , poles $- 2 \pm j 4.58$ , final displacement 0.2 m.

The DC motor

A motor needs one law each way between its electrical and mechanical sides. Take $N$ conductors in series, each of length $l$ (m) at radius $r$ (m) in a uniform air-gap flux density $B_{g}$ (T; $B$ is damping). Carrying current $i$ , each feels a force $B_{g} l i$ at radius $r$ : a torque $B_{g} l r i$ . Turning at $ω$ , each moves at $r ω$ and sweeps flux at the rate $B_{g} l r ω$ , which the law of induction makes its emf. A commutator, a rotary switch that reverses each conductor's current as it passes from one pole to the next, keeps all $N$ torques and emfs adding. The field is constant; a voltage $v$ (V) drives an armature of resistance $R$ (Ω) and inductance $L$ (H).

TORQUE AND BACK EMF

τ_{m} = (N B_{g} l r) i = K_{t} i, e_{b} = (N B_{g} l r) ω = K_{b} ω

Uniform field; all $N$ conductors in series and kept in step by the commutator. $τ_{m}$ in N·m, $i$ in A, $e_{b}$ in V, $ω$ in rad/s. Both constants equal $N B_{g} l r$ in SI units.

EQUAL CONSTANTS FROM POWER BALANCE

e_{b} i = τ_{m} ω \Rightarrow K_{b} ω i = K_{t} i ω \Rightarrow K_{b} = K_{t} = K

Lossless conversion: the electrical power absorbed by the back emf equals the mechanical power delivered, for every $i$ and $ω$ ; the losses sit in $R$ and $B$ . $K$ is in N·m/A, the same as V·s/rad. In rpm or kgf·cm the two numbers differ.

MOTOR TRANSFER FUNCTION

\frac{Ω ( s )}{V ( s )} = \frac{K}{( L s + R ) ( J s + B ) + K ^{2}}

Constant field, no saturation, no load torque, zero initial conditions. From $L d i / d t + R i = v - K ω$ and $J \overset{ω}{˙} + B ω = K i$ : put $I = (V - K Ω) / (L s + R)$ into $(J s + B) Ω = K I$ . $J$ (kg·m²) and $B$ (N·m·s/rad) include the reflected load. If $L \approx 0$ : $K / (R J s + R B + K^{2})$ .

WATCH OUT FOR

Dropping the back emf. Test: with $B = 0$ , dropping it leaves $K / ((L s + R) J s)$ , a pole at 0, so a step makes the speed grow forever. With it, the speed settles where $K ω = v$ and $i = 0$ .

WORKED EXAMPLE

A motor has $R = 2 Ω$ , $K = 0.5$ N·m/A, $L \approx 0$ , negligible friction and rotor inertia 0.5 kg·m². It drives an 8 kg·m² load through gears with 20 teeth on the motor shaft and 80 on the load shaft. Find $Ω (s) / V (s)$ and the final motor speed for a 12 V step.

Solution

Find the ratio: $n = 20/80 = 0.25$ . By the reflected-inertia result, the motor supplies $n$ times the load's torque and the load accelerates at $n$ times the motor's rate, so the two factors multiply: the load looks like $n^{2} J_{L} = 0.0625 \times 8 = 0.5$ kg·m². So $J_{e q} = 0.5 + 0.5 = 1.0$ kg·m².
With $B = 0$ and $L \approx 0$ , the motor transfer function reduces to $K / (R J_{e q} s + K^{2}) = 0.5/ (2 s + 0.25)$ .
Divide top and bottom by 0.25 to read the gain and time constant: $Ω/ V = 2/ (8 s + 1)$ , a gain of 2 (rad/s)/V and a time constant of 8 s.
The pole at $- 0.125$ is in the LHP, so the final value theorem holds: $ω = 2 \times 12 = 24$ rad/s. Check: $K ω = 0.5 \times 24 = 12$ V equals $v$ , so the current is zero, as it must be with no friction. The load turns at $nω = 6$ rad/s.

Answer: $Ω/ V = 2/ (8 s + 1)$ ; final motor speed 24 rad/s (load 6 rad/s).

Linearisation about an operating point

Many physical laws are curved. Gravity on a pendulum goes as $sin θ$ , and flow out of a tank goes as $h$ . A transfer function needs a linear equation. So you choose a steady operating point, write each variable as its steady value plus a small deviation, and replace each curve by its tangent there. $G (s)$ then relates deviations only: add the operating values back to get the real totals.

FIRST-ORDER LINEARISATION

y = f (x) \approx f (x_{0}) + f^{'} (x_{0}) (x - x_{0}), δ y \approx f^{'} (x_{0}) δ x

Needs $f$ differentiable at $x_{0}$ and $δ x = x - x_{0}$ small. If $f^{''}$ exists, the leading dropped term is $\frac{1}{2} f^{''} (x_{0}) δ x^{2}$ . The slope $f^{'} (x_{0})$ depends on $x_{0}$ , so a new operating point in general gives a new $G (s)$ .

PENDULUM ABOUT AN OPERATING ANGLE

m l^{2} δ \ddot{θ} = δ τ - m g l cos θ_{0} δ θ \Rightarrow \frac{δ Θ ( s )}{δ τ ( s )} = \frac{1}{m l ^{2} s ^{2} + m g l cos θ _{0}}

Point mass $m$ (kg) on a light rod of length $l$ (m), angle $θ$ (rad) from hanging down, pivot torque $τ$ (N·m) acting to increase $θ$ , $g$ in m/s², no friction: $m l^{2} \ddot{θ} = τ - m g l sin θ$ . Steady state: $τ_{0} = m g l sin θ_{0}$ . Subtract it, and use $sin θ \approx sin θ_{0} + cos θ_{0} δ θ$ .

Operating angle $θ_{0}$	Stiffness $m g l cos θ_{0}$	Poles
0 (hanging down)	$+ m g l$	$\pm j g / l$
$π /2$ (horizontal)	0	double pole at 0
$π$ (upright)	$- m g l$	$\pm g / l$ (one in the RHP)

Tank of area $A$ (m²), level $h$ (m), inflow $q_{in}$ (m³/s), outflow $c h$ through an orifice, $c$ in m $^{2.5}$ /s: $A \dot{h} = q_{in} - c h$ .
At the operating point, $q_{0} = c h_{0}$ . The slope of $c h$ there is $c / (2 h_{0})$ .
Subtract the steady balance: $A δ \dot{h} = δ q_{in} - \frac{c}{2 h _{0}} δ h$ . Put $R_{h} = 2 h_{0} / c$ (s/m²).
Transform: $\frac{δ H}{δ Q _{in}} = \frac{R _{h}}{A R _{h} s + 1}$ . The time constant $A R_{h} = 2 A h_{0} / c$ grows with the operating level.

What is lost is easy to state. The model is the tangent, so it is right only near the operating point; the table below shows the error of using $θ$ for $sin θ$ about $θ_{0} = 0$ . Superposition holds for the deviations, not for the totals. Behaviour far away is missing: the linear pendulum about $θ_{0} = 0$ never swings over the top. And the tank's slope $c / (2 h_{0})$ grows without limit as $h_{0} \to 0$ , so its model fails as the tank empties.

$θ$ (rad)	$sin θ$	Relative error of using $θ$
0.1	0.0998	0.17%
0.5	0.4794	4.3%
1.0	0.8415	18.8%

Check yourself

CHECK YOURSELF 1

A motor shaft gear has 25 teeth and meshes with a load gear of 100 teeth. The motor-side inertia is 0.5 kg·m² and the load inertia is 16 kg·m². Gears are ideal. What is the equivalent inertia at the motor shaft, in kg·m²?

Answer and solution

Answer: $1.5$ (accepted range $\pm 0.01$ )

The ratio is $n = N_{1} / N_{2} = 25/100 = 0.25$ .
The reflected load inertia is $n^{2} J_{L} = 0.0625 \times 16 = 1$ kg·m².
$J_{e q} = 0.5 + 1 = 1.5$ kg·m².

CHECK YOURSELF 2

Why are the torque constant $K_{t}$ and the back emf constant $K_{b}$ equal in SI units?

(A)
Because the armature resistance is the same in both laws
(B)
Because lossless conversion requires $e_{b} i = τ_{m} ω$ for every current and speed
(C)
Because the inductance cancels in the transfer function
(D)
Because the field flux is always one tesla

Answer and solution

Answer: (B)

Electrical power absorbed by the back emf is $K_{b} ω i$ . Mechanical power delivered is $K_{t} iω$ .
With no loss in the conversion these are equal for all $i$ and $ω$ , so $K_{b} = K_{t}$ .
Resistance, inductance and flux density do not enter this argument.

CHECK YOURSELF 3One or more options are correct

Which statements about first-order linearisation about an operating point are true?

(A)
For a twice-differentiable $f$ , the dropped term is second order in the deviation, so the error shrinks faster than the deviation does
(B)
The resulting transfer function describes deviations from the operating point, not the totals
(C)
The same $G (s)$ applies at every operating point of any nonlinear system
(D)
A function with a corner at the operating point, such as $∣ x ∣$ at $0$ , has no tangent there, so the method fails

Answer and solution

Answer: (A), (B), (D)

A is true: the leading dropped term is $\frac{1}{2} f^{''} (x_{0}) δ x^{2}$ .
B is true: $δ y = f^{'} (x_{0}) δ x$ relates deviations.
C is false: the slope $f^{'} (x_{0})$ changes with $x_{0}$ ; the pendulum's stiffness $m g l cos θ_{0}$ is one case.
D is true: no finite slope exists at a corner.

CHECK YOURSELF 4

A tank of area $A = 3$ m² drains through an orifice with outflow $q = c h$ , $c = 0.2$ m $^{2.5}$ /s ( $h$ in m, $q$ in m³/s). Linearise about a steady level $h_{0} = 9$ m. What is the time constant of $δ H / δ Q_{in}$ , in seconds?

Answer and solution

Answer: $90$ (accepted range $\pm 0.5$ )

$R_{h} = 2 h_{0} / c = 2 \times 3/0.2 = 30$ s/m².
The time constant is $A R_{h} = 3 \times 30 = 90$ s.

From a physical system to a transfer function

What you will be able to do

Before you start

Springs, masses, dampers and gears

The DC motor

Linearisation about an operating point

Check yourself

Used in these GATE topics