Linear And Non Linear Analysis - stepper motor

LINEAR AND NON LINEAR ANALYSIS

The linear and nonlinear analysis of the motor performance with respect to the torque produced by the rotor of the motor is explained.

Let

Tm be the motor torque produced by the rotor in Nm

J be the inertia of the rotor and load combination in kgm2    

ω be the angular velocity of the rotor        

D be the damping coefficient or viscous frictional coefficient 

Tf  be the frictional load torque independent of the speed      

Θs be the step angle in radians        

F be the stepping rate in steps/sec or pps  

Frictional load torque Tf = K θ        

According to rotor dynamics  

Tm=―J*dω/dt+Dω+Tf ………….. (2.30)

Also θs=θ=ωt=step angle       

ω=θs/t=f θs …………..(2.31)

where f=1/t ………….(2.32)

By putting ω=f θs

Tm=J *d/dt(f θs )+D(f θs )+Tf …………..(2.33)

θs=360/mNr is fixed for a particular type of motor      

S o θs can be considered as constant

Therefore Tm=J θs* d/dt(f)+D θs(f)+Tf      ………….(2.34)

In equation 2.47 if viscous friction constant is neglected the equation will be a linear equation, the corresponding acceleration will be nonlinear and the equation will be nonlinear which given rise to nonlinear analysis.

Linear acceleration on linear analysis

If the damping coefficient is neglected D=0

The expression for motor torque becomes

Tm=―J*dω/dt+Tf                                                               ………………(2.35)

Tm-Tf= J*dω/dt

(Tm-Tf)/J=          dω/dt

dω=((Tm-Tf)/J)dt                                                             ……………….. (2.36)

Integrating

ω=((Tm-Tf)/J)dt+ω1      ………………..    (2.37)

Where

ω1=Integration constant

Mathematically ω1 is the constant of integration but it indicates the initial angular velocity of the motor before the occurrence of acceleration.

Therefore ω=θs f and ω1= θs f1       

Substituting  ω and ω1 in equation 2.50   

((Tm-Tf)/J)t+ θs f1= θsf ………………..(2.38)

Dividing throughout by θs we get    

((Tm-Tf)/J θs)t+ f1=f    

Therefore stepping rate f=((Tm-Tf)/J θs)t+ f1      ……………(2.39)

And Tf = K θ       

Nonlinear (exponential) acceleration on Nonlinear analysis

Considering the torque produced by the motor

Tm=jθs df/dt +Dθsf+Tf  …….(2.40)

(Tm-Tf)= jθs df/dt +Dθsf        

Dividing throughout by jθs We get  

(df/dt)+(D/J)f-(Tm-Tf /j θs )=0

(or) (df/dt)+(D/J)f=(Tm-Tf /j θs )       …. (2.41)

The above eqn. 2.54 is of the form  

(dy/dx)+py=Q Which have the solution of

ye∫pdx =∫Q e∫pdx+C      …..………….(2.42 )

Here y=f; x=t; p=(D/j) and Q=(Tm-Tf)/jθs =constant   

fe∫D/J dt=∫(Tm-Tf)/jθse∫D/J dt+C      …………………(2.43)

fe∫D/J t=∫(Tm-Tf)/jθse∫D/J t+C ………….…….(2.44)

fe∫D/J t=(Tm-Tf)/jθs(e∫D/J t/(D/J))+C         ………………..(2.45)

where C is the integration constant  

To find C substituting initial condition at t=0; f=f(0)=f1f1e0==(Tm-

Tf)/jθs(1/(D/J))+C …………..…..(2.46)

f1===(Tm-Tf)/Jθs(J/D))+C     

…………..…...(2.47)    

f1=(Tm-Tf)/Dθs+C       

………….…..(2.48)      

C= f1-(Tm-Tf)/Dθs        .

……………...(2.49)      

Substituting eqn. (2.62) in eqn. (2.58)       

f e(D/J)t=(Tm-Tf)/Jθs(J/D)e(D/J)t+( f1-(Tm-Tf)/Dθs)     …………….….(2.50)

f e(D/J)t=(Tm-Tf)/Dθs e(D/J)t+( f1-(Tm-Tf)/Dθs) ………..……( 2.51)

Dividing throughout by e(D/J)t we get      

F=Tm-Tf/Dθs                                                       +(f1-Tm-Tf/Dθs)e-D/j t                    ……………(2.52)

Stepping frequency f= Tm-Tf/Dθs    +(f1-Tm-Tf/Dθs)e-D/j t

The above equation is a nonlinear exponential equation which gives rise to nonlinear acceleration of the rotor of the motor.