# Resonance

Formally, the small-amplitude resonance can be computed as follows: First, recall Newton's law for angular acceleration:

T = µ AWhere:

T -- torque applied to rotorWe assume that, for small amplitudes, the torque on the rotor can be approximated as a linear function of the displacement from the equilibrium position. Therefore, Hooke's law applies:

µ -- moment of inertia of rotor and load

A -- angular acceleration, in radians per second per second

T = -where:kθ

We can equate the two formulas for the torque to get:k-- the "spring constant" of the system, in torque units per radian

θ -- angular position of rotor, in radians

µ A = -Note that acceleration is the second derivitive of position with respect to time:kθ

A = dso we can rewrite this the above in differential equation form:^{2}θ/dt^{2}

dTo solve this, recall that, for:^{2}θ/dt^{2}= -(k/µ) θ

f(The derivitives are:t) =a sin bt

df(Note that, throughout this discussion, we assumed that the rotor is resonating. Therefore, it has an equation of motion something like:t)/dt=ab cos bt

d^{2}f(t)/dt^{2}= -ab^{2}sin bt= -b^{2}f(t)

θ =This is an admissable solution to the above differential equation if we agree that:a sin(2πft)

a= angular amplitude of resonance

f= resonant frequency

Solving for the resonant frequencyb= 2πf

b^{2}=k/µ

*f*as a function of

*k*and µ, we get:

It is crucial to note that it is the moment of inertia of the rotor plus any coupled load that matters. The moment of the rotor, in isolation, is irrelevant! Some motor data sheets include information on resonance, but if any load is coupled to the rotor, the resonant frequency will change!f= (k/µ )^{0.5}/ 2π

In practice, this oscillation can cause significant problems when the stepping rate is anywhere near a resonant frequency of the system; the result frequently appears as random and uncontrollable motion.

**Resonance and the Ideal Motor**

Up to this point, we have dealt only with the small-angle
spring constant *k* for the system. This can be
measured experimentally, but if the motor's torque versus
position curve is sinusoidal, it is also a simple function
of the motor's holding torque. Recall that:

T = -The small angle spring constanthsin( ((π / 2) / S) θ )

*k*is the negative derivitive of T at the origin.

Substituting this into the formula for frequency, we get:k= -dT / dθ = - (-h((π / 2) / S) cos( 0 ) ) = (π / 2)(h/ S)

Given that the holding torque and resonant frequency of the system are easily measured, the easiest way to determine the moment of inertia of the moving parts in a system driven by a stepping motor is indirectly from the above relationship!f= ( (π / 2)(h/ S) / µ )^{0.5}/ 2π = (h/ ( 8π µ S ) )^{0.5}

µ =For practical purposes, it is usually not the torque or the moment of inertia that matters, but rather, the maximum sustainable acceleration that matters! Conveniently, this is a simple function of the resonant frequency! Starting with the Newton's law for angular acceleration:h/ ( 8πf^{ 2}S )

A = T / µWe can substitute the above formula for the moment of inertia as a function of resonant frequency, and then substitute the maximum sustainable running torque as a function of the holding torque to get:

A = (Measuring acceleration in steps per second squared instead of in radians per second squared, this simplifies to:h/ ( 2^{0.5}) ) / (h/ ( 8πf^{ 2}S ) ) = 8π Sf^{ 2}/ (2^{0.5})

AThus, for an ideal motor with a sinusoidal torque versus rotor position function, the maximum acceleration in steps per second squared is a trivial function of the resonant frequency of the motor and rigidly coupled load!_{steps}= A / S = 8πf^{ 2}/ (2^{0.5})

For a two-winding permanent-magnet or variable-reluctance motor, with an ideal sinusoidal torque-versus-position characteristic, the two-winding holding torque is a simple function of the single-winding holding torque:

Where:h_{2}= 2^{0.5}h_{1}

Substituting this into the formula for resonant frequency, we can find the ratios of the resonant frequencies in these two operating modes:h_{1}-- single-winding holding torque

h_{2}-- two-winding holding torque

This relationship only holds if the torque provided by the motor does not vary appreciably as the stepping rate varies between these two frequencies.f_{1}= (h_{1}/ ... )^{0.5}

f_{2}= (h_{2}/ ... )^{0.5}= ( 2^{0.5}h_{1}/ ... )^{0.5}= 2^{0.25}(h_{1}/ ... )^{0.5}= 2^{0.25}f_{1}= 1.189...f_{1}

In general, as will be discussed later, the available torque will tend to remain relatively constant up until some cutoff stepping rate, and then it will fall. Therefore, this relationship only holds if the resonant frequencies are below this cutoff stepping rate. At stepping rates above the cutoff rate, the two frequencies will be closer to each other!