Dynamics of a Single Step of a Stepper Motor
Abstract Generated abstract
This paper analyzes the single-step dynamics of a multiphase reactive stepper motor under simplifying assumptions that reduce the electromechanical model to a nonlinear rotor equation with a cubic approximation of motor torque. It derives analytical and semi-analytical expressions for step time and rotor motion under several load types, including constant resisting torque, dry friction, viscous friction, and combined loads, using elliptic integrals, approximate piecewise torque replacement, asymptotic methods, and graphical amplitude constructions. The results provide practical formulas for estimating step execution time, oscillation damping, and load influence, with stated agreement close to graphical calculation methods and, in one constant-load approximation, an error not exceeding about 5 percent over the considered range.
Full Text
DYNAMICS OF A SINGLE STEP OF A STEPPER MOTOR
V. V. Zhulovyan
(Submitted by the scientific seminar of the Department of Electrical Machines and General Electrical Engineering)
The dynamics of a single step of an \(m\)-phase reactive stepper motor (S.M.) in the general case is described by a system of \(m+1\) equations
\[ \begin{gathered} U_1=i_1R_1+\frac{d\psi_1}{dt},\\ \cdots\\ U_m=i_mR_m+\frac{d\psi_m}{dt},\\ I^2\frac{d^2(\Theta)}{dt^2}+\lambda\frac{d\Theta}{dt}+M_\theta(i_1,i_2,\ldots,i_m\Theta)-M_{\mathrm{н}}=0. \end{gathered} \tag{1} \]
Since each of the equations included in (1) is nonlinear, obtaining a solution in general form does not appear possible.
Let us make a number of assumptions:
- The current in the control windings rises instantaneously.
- The inductance of the control windings does not depend on the angular position of the rotor.
These assumptions are quite valid when there are significant ballast resistances in the circuits of the control windings.
- We approximate the motor torque by a cubic parabola [1]
\[ M_\theta=M_m\left[a(2p\Theta)-b(2p\Theta)^3\right], \tag{2} \]
where \(M_m\) is the maximum value of the torque developed by the motor;
\(a\) and \(b\) are the coefficients of the parabola, determined from a calculated or experimentally obtained torque curve;
\(\Theta\) is the geometrical angle of misalignment between the magnetic axes of the rotor and stator;
\(p\) is the number of pairs of rotor teeth.
Under the assumptions made, the dynamics of a single step of the S.M. will be determined only by the equation of motion of the rotor
\[ I\frac{d^2\Theta}{dt^2}+\lambda\frac{d\Theta}{dt}+M_m\left[a(2p\Theta)-b(2p\Theta)^3\right]-M_{\mathrm{н}}=0, \tag{3} \]
where
\[ I \]
is the reduced moment of inertia;
\[ \lambda \]
is the coefficient of viscous friction of the load;
\[ M_{\mathrm{n}}=M_{\mathrm{t}}+R\,\operatorname{sign}\left|\frac{d\Theta}{dt}\right| \]
is the load torque, consisting of a constant resisting torque and dry friction.
Introducing the new variable \(x=2p\Theta\) and the relative time
\[ \tau=\sqrt{\frac{2pM_m}{I}}\cdot t, \]
we obtain
\[ \frac{d^{2}x}{d\tau^{2}}+2\delta\frac{dx}{d\tau}+\alpha x-bx^{3}-\mu_{\mathrm{n}}=0, \tag{4} \]
where
\[ \delta=\frac{\lambda}{2\sqrt{2pIM_m}}; \qquad \mu_{\mathrm{n}}=\beta+\kappa\,\operatorname{sign}\left|\frac{dx}{d\tau}\right|; \]
\[ \beta=\frac{M_{\mathrm{t}}}{M_m}; \qquad \kappa=\frac{R}{M_m}. \]
- Let us consider the solution of equation (4) for the case when the load is determined only by the constant resisting torque \(\mu_{\mathrm{n}}=\beta\). For \(\beta=0\) (ideal no-load operation), after a number of transformations we obtain
\[ \tau_x=F(\Theta,\varphi)\cdot\frac{\Theta_{\mathrm{sh}}}{v_0}, \tag{5} \]
where \(F(\Theta,\varphi)\) is the normal elliptic integral of the first kind, whose value is determined, depending on the values of \(\Theta,\varphi\), from the corresponding tables:
\[ \sin\varphi=\frac{x}{\Theta_{\mathrm{sh}}}; \qquad \sin\Theta=\Theta_{\mathrm{sh}}\sqrt{\frac{b/a}{2-\Theta_{\mathrm{sh}}^{2}b/a}}; \]
\[ v_0=\Theta_{\mathrm{sh}}\sqrt{a\left(1-\frac{b}{2a}\Theta_{\mathrm{sh}}^{2}\right)}; \qquad \Theta_{\mathrm{sh}}=\frac{2\pi}{m}. \]
The step execution time is expressed through the complete elliptic integral of the first kind, since for \(x=0\), \(\varphi=\dfrac{\pi}{2}\),
\[ \tau_{\mathrm{sh}}=\frac{\Theta_{\mathrm{sh}}}{v_0}K\Theta. \tag{6} \]
For \(\beta\ne0\), after a number of transformations we obtain
\[ \tau_{\mathrm{sh}}= \int_{0}^{x} \frac{dx} {\sqrt{2}\sqrt{\dfrac{a}{2}\left(\Theta_{\mathrm{sh}}^{2}-x^{2}\right) -\dfrac{b}{4}\left(\Theta_{\mathrm{sh}}^{4}-x^{4}\right) -\beta\left(\Theta_{\mathrm{sh}}-x\right)}}. \tag{7} \]
Determination of the elliptic integral (6) is a rather cumbersome task, since it is associated with calculating the roots of the fourth-degree polynomial under the integral sign.
To simplify the solution, we replace the resulting torque curve
\[ M_p=ax-bx^{3}-\beta \]
by two curves (Fig. 1):
\[ a_1x_1-b_1x_1^{3}=ax-bx^{3}-\beta \quad\text{for } x_1\ge0; \]
\[ a_2x_1-b_2x_1^{3}=ax-bx^{3}+\beta \quad\text{for } x_1\le0, \tag{8} \]
where
\[ x_1=x-x_0. \]
The coefficients \(a_1,\ a_2,\ b_1,\ b_2\) are determined from the following expressions:
\[ a_1=-\frac{x_{1\mathrm{н}}^{\prime\,2}(1-\beta)} {x_{1\mathrm{н}}(x_{1\mathrm{н}}^{\prime\,2}-x_{1m}^{\prime\,2})}; \qquad b_1=\frac{1-\beta} {x_{1m}(x_{1\mathrm{н}}^{\prime\,2}-x_{1m}^{\prime\,2})}; \]
\[ a_2=\frac{x_{1\mathrm{н}}^{\prime\,2}(1+\beta)} {x_{1m}'(x_{1\mathrm{н}}^{\prime\,2}-x_{1m}^{\prime\,2})}; \qquad b_2=\frac{1+\beta} {x_{1m}'(x_{1\mathrm{н}}^{\prime\,2}-x_{1m}^{\prime\,2})}, \tag{9} \]
Fig. 1.
where
\[ x_{1m}=x_m-x_0;\qquad x_{1m}'=x_m+x_0;\qquad x_{1\mathrm{н}}=x_{\mathrm{н}}-x_0;\qquad x_{1\mathrm{н}}'=x_{\mathrm{н}}+x_0; \]
\[ x_m=\sqrt{\frac{a}{3b}}. \]
The values \(x_0,\ x_{\mathrm{н}},\ x_{\mathrm{н}}'\) are determined from the solution of the equation
\[ ax-bx^3-\beta=0. \]
Taking (7) into account, the equation of motion will be
\[ \frac{d^2x_1}{d\tau^2}+a_1x_1-b_1x_1^3=0 \qquad \text{for } x_1\geqslant 0, \]
\[ \frac{d^2x_1}{d\tau^2}+a_2x_1-b_1x_1^3=0 \qquad \text{for } x_1\leqslant 0. \tag{10} \]
Hence, after a number of transformations, we obtain
\[ \tau_x=\frac{\Theta_{\mathrm{ш}}}{v_0}F(\Theta_1,\varphi_1) \qquad \text{for } x_1\geqslant 0; \]
\[ \tau_x=\frac{x_{\mathrm{н}}}{v_0}F(\Theta_2,\varphi_2) \qquad \text{for } x_1\leqslant 0, \tag{11} \]
where
\[ \sin \varphi_1=\frac{x_1}{\Theta_{\mathrm{sh}}}; \qquad \sin \Theta_{\mathrm{sh}}\cdot \sqrt{\frac{b_1/\alpha_1}{2-\Theta_{\mathrm{sh}}^{2}b_1/\alpha}}; \]
\[ \sin \varphi_2=\frac{x_1}{x_{\mathrm{p}}}; \qquad \sin \Theta_2=x_{\mathrm{p}} \sqrt{\frac{b_2/\alpha_2}{2-x_1^{2}b_2/\alpha_2}}, \]
\(x_{\mathrm{p}}\) is the value of the first half-oscillation, which can be determined either graphically (Fig. 3) or from the solution of the equation
\[ \frac{v_0^2}{2}= \frac{\alpha_2 x_1^2}{2} \left(1-\frac{b_2 x_1^2}{2}\right). \]
The difference between the exact solution (7) and the approximate value (11) does not exceed 5% over the entire range of possible values of \(\beta\), both for three-phase and for four-phase stepper motors.
The step completion time is
\[ \tau_{\mathrm{sh}}=\frac{\Theta_{\mathrm{sh}}}{v_0}\,K\cdot(\Theta_1). \tag{12} \]
Fig. 2. Dependence of the step completion time on the value of \(\beta\).
Figure 2 gives the dependences of the step completion time on the magnitude of the load.
- When the motor operates against a load of the dry-friction type, it is more convenient to write the equation as
\[ \frac{d^2x}{dt^2}+\alpha x-bx^3-k=0 \qquad \text{for } \frac{dx}{d\tau}<0, \]
\[ \frac{d^2x}{dt^2}+\alpha x-bx^3+k=0 \qquad \text{for } \frac{dx}{d\tau}>0. \tag{13} \]
The rate of decay of oscillations as a function of the magnitude of dry friction is most simply determined graphically [2]. The procedure for determining the amplitudes of the half-oscillations is shown in Fig. 3, where
\[ W=\frac{a x^2}{2}\left(1-\frac{b x^2}{2}\right) \]
is the curve of potential energy.
Fig. 3. Determination of the sequence of amplitudes of the half-oscillations.
Since the velocity changes sign at the end of each half-oscillation, the solution of equation (13) can be carried out analogously to the case \(\beta \ne 0\). Then, for the first half-oscillation, one may write
\[ \tau_x=\frac{\Theta_{\mathrm{ш}}}{v_{01}}\,F(\Theta_1,\varphi_1) \quad \text{for } x_1 \ge 0, \]
\[ \tau_x=\frac{x_{1\mathrm{п}}}{v_{01}}\,F(\Theta_2,\varphi_2) \quad \text{for } x_1 \le 0. \tag{14} \]
For the second half-oscillation,
\[ \tau_x=\frac{x'_{1\mathrm{п}}}{v_{02}}\,F(\Theta_3,\varphi_3) \quad \text{for } x''_1 \le 0, \]
\[ \tau_x=\frac{x''_{2\mathrm{п}}}{v_{02}}\,F(\Theta_4,\varphi_4) \quad \text{for } x''_1 \ge 0 \tag{14} \]
and so on.
The values \(x_{1\mathrm{p}}, x'_{1\mathrm{p}} \ldots x_{n\mathrm{p}}\) are determined from the construction of the sequence of amplitudes; for example, \(x_{1\mathrm{p}} = x_1 + x_0\).
The dependence of the step response time on the magnitude of dry friction will have the same form as on the magnitude of \(\beta\).
- When the motor operates with a combined load consisting of a constant resisting torque and dry friction, the equation of motion will be
\[ \begin{aligned} \frac{d^2 x}{d\tau^2} + ax - bx^3 - \beta - \kappa &= 0 \qquad \text{for } \frac{dx}{d\tau} < 0,\\ \frac{d^2 x}{d\tau^2} + ax - bx^3 - \beta + \kappa &= 0 \qquad \text{for } \frac{dx}{d\tau} > 0. \end{aligned} \tag{15} \]
The solution for each half-oscillation will look analogous to (14). The procedure for determining the sequence of half-oscillation amplitudes for this case is shown in Fig. 4.
Fig. 4. Determination of the sequence of half-oscillation amplitudes.
It is useful to note that, with the graphical method presented for determining amplitudes, one can solve the inverse problem, namely: determine the required value of dry friction in order to obtain the desired rate of damping of rotor oscillations. Such a problem arises in those cases when it is necessary to determine the minimum value of dry friction in order to prevent a “loss” of a step at a control frequency close to the natural frequency of oscillation of the rotor.
- When the motor operates into a load of the viscous-friction type, the equation of motion will be
\[ \frac{d^{2}x}{d\tau^{2}}+2\delta\frac{dx}{d\tau}+\alpha x-bx^{3}=0. \tag{16} \]
In the first approximation we shall seek the solution in the form [3]
\[ x=a\cdot \sin\psi . \tag{17} \]
The values of \(a\) and \(\psi\) are determined from the system of equations
\[ \frac{da}{d\tau}=-\delta a-\frac{3}{4}\cdot \frac{\delta\cdot b\cdot a^{3}}{2\omega_{0}^{2}+\delta^{2}}; \]
\[ \frac{d\psi}{d\tau}=\omega_{0}-\frac{3ba^{2}}{8\omega_{0}}+ \frac{3\delta^{2}ba^{2}}{4\omega_{0}(2\omega_{0}^{2}+\delta^{2})}, \tag{18} \]
where
\[ \omega_{0}=\sqrt{\alpha-\delta^{2}}. \]
Integrating and substituting into (17), we obtain
\[ x=\frac{A e^{-\delta\tau}}{\sqrt{1-A^{2}B^{2}e^{-2\delta\tau}}}\cdot \sin\left( \omega_{0}\tau-\gamma\ln \frac{1-A^{2}B^{2}e^{-2\delta\tau}}{1-A^{2}B^{2}} +\varphi_{0} \right), \tag{19} \]
where
\[ A^{2}=\frac{a_{0}^{2}}{1+B^{2}a_{0}^{2}};\qquad B^{2}=\frac{3b}{4(2\omega_{0}^{2}+\delta^{2})};\qquad \gamma=\frac{1}{2\delta B^{2}}\left(\frac{3b}{8\omega_{0}}+ \frac{B^{2}\delta^{2}}{\omega_{0}^{2}}\right). \]
The values of \(a_{0}\) and \(\varphi_{0}\) are determined from the initial conditions for a unit step, with
\[ \tau=0;\quad x=\Theta_{\mathrm{sh}};\quad \frac{dx}{d\tau}=0; \]
\[ x=\Theta_{\mathrm{sh}}=a_{0}\cdot \sin\varphi_{0}. \]
\[ \frac{dx}{d\tau}=0=a_{0}\left[ B^{2}\delta\sin\varphi_{0}+ \left(\frac{3b}{8\omega_{0}}+\frac{B^{2}\delta^{2}}{\omega_{0}^{2}}\right)\cdot \cos\varphi_{0} \right]+ \]
\[ +(\delta\sin\varphi_{0}-\omega_{0}\cdot \cos\varphi_{0}). \tag{20} \]
The solution of this system of equations is most simply carried out graphically.
The solution obtained is more general than that given in [4], since it is suitable for finite values of viscous friction.
For \(\delta=0\) we obtain
\[ x=\Theta_{\mathrm{sh}}\cos\left[\omega_{0}\tau\left(1-\frac{3b\Theta_{\mathrm{sh}}^{2}}{8\omega_{0}^{2}}\right)\right]. \tag{21} \]
The step response time is
\[ \tau_{\mathrm{sh}}= \frac{\pi}{ 2\omega_{0}\left(1-\dfrac{3b\Theta_{\mathrm{sh}}^{2}}{8\omega_{0}^{2}}\right) }. \tag{22} \]
The dependence of the step response time on the magnitude of viscous friction for a three-phase stepper motor is shown in Fig. 5.
The character of the rotor motion for various values of viscous friction for a three-phase stepper motor is shown in Fig. 6. For comparison, the dependence \(x(\tau)\) (dashed curve) is given for \(2\delta=0.6\), obtained
by solving equation (16) by the delta method [5], which has greater accuracy than the Schäfer method or the method of M. L. Bykhovskii, since it requires a smaller number of auxiliary constructions to obtain the solution.
Fig. 5. Dependence of \(x(\tau)\) on the magnitude of viscous friction.
The sequence of amplitudes as a function of the magnitude of viscous friction can be determined graphically [2] by plotting an approximate value of the function
Fig. 6. Dependence of the step-settling time on the magnitude of viscous friction.
\[ P(x)=\frac{\pi x\delta}{\omega_0}\cdot \frac{1-\dfrac{3bx^2}{16(2a-\delta^2)}} {2-\dfrac{3bx^2(2a-3\delta^2)}{32\omega_0^2(2a-\delta^2)}}. \tag{23} \]
Using (23), the sequence of amplitudes for finite values of viscous friction can be determined with sufficient accuracy, since expressions (18) were used in its derivation.
In Fig. 7 the function \(P(x)\) is plotted for the value \(2\delta=0.6\), and the procedure for determining the sequence of amplitudes is shown.
Fig. 7. Determination of the sequence of amplitudes.
- When the motor operates under a load consisting of viscous friction and a constant resisting torque, the equation of motion will be
\[ \frac{d^{2}x}{d\tau^{2}}+2\delta\frac{dx}{d\tau}+\alpha\cdot x-bx^{3}-\beta=0 . \tag{24} \]
In this case, the solution in the first approximation proves insufficient; obtaining the solution in the second and third approximations is associated with considerable difficulties and, most importantly, does not provide satisfactory accuracy.
Let us replace equation (23) by two equations
\[ \frac{d^{2}x_{1}}{d\tau^{2}}+2\delta\frac{dx_{1}}{d\tau}+\alpha_{1}\cdot x_{1}-b_{1}\cdot x_{1}^{3}=0 \quad \text{for } x_{1}\geqslant 0; \]
\[ \frac{d^{2}x_{1}}{d\tau^{2}}+2\delta\frac{dx_{1}}{d\tau}+\alpha_{2}\cdot x_{1}-b_{2}x_{1}^{3}=0 \quad \text{for } x_{1}\leqslant 0, \tag{25} \]
where the values \(\alpha_{1}, \alpha_{2}, b_{1}, b_{2}\) are determined from (9).
The solution of each equation in the first approximation is written as
\[ x_{1}= \frac{A_{1}\cdot e^{-\delta\tau}} {\sqrt{1-A_{1}^{2}B_{1}e^{-2\delta\tau}}} \sin\left( \omega_{1}\cdot\tau-\gamma_{1}\ln \frac{1-A_{1}^{2}B_{1}^{2}e^{-2\delta\tau}} {1-A_{1}^{2}B_{1}^{2}} +\varphi_{01} \right) \tag{26} \]
for \(x_{1}>0\).
The obtained solutions must be matched to one another.
In Fig. 8, calculated and graphically determined dependences \(x_{1}(\tau)\) are presented for the values \(\beta=0.4\) and \(2\delta=0.4\). The figure shows good agreement between the calculated and graphically determined dependence \(x(\tau)\) for a four-phase stepper motor. To improve the accuracy of the solution for the case of a three-phase stepper motor, a correction coefficient equal to \((1+\beta)\) was found. Taking the correction coefficient into account, solution (26) is written as
\[ x_{1}= \frac{A_{1}\cdot e^{-\delta\tau}} {\sqrt{1-A_{1}^{2}B_{1}^{2}e^{-\delta\tau}}} \cdot \sin\left[ \omega_{1}\cdot\tau-(1+\beta)\gamma_{1}\ln \frac{1-A_{1}^{2}B_{1}^{2}\cdot e^{-2\delta\tau}} {1-A_{1}^{2}\cdot B_{1}^{2}} +\varphi_{01} \right] \]
\[ \text{for } x_{1}\geqslant 0; \tag{27} \]
\[ x_1=\frac{A_2 e^{-\delta \tau}}{\sqrt{1-A_2^2 B_2^2 e^{-2\tau}}}\cdot \sin\left[\omega_2 \tau-(1+\beta)\gamma_2 \ln \frac{1-A_2^2 B_2^2 e^{-2\delta\tau}}{1-A_2^2 B_2^2} +\varphi_{02}\right] \]
for \(x_1 \le 0\),
where the values \(A_1\) and \(\varphi_{01}\) are determined from the initial conditions without taking the correction factor into account.
Fig. 8. Dependence \(x(\tau)\) for the case \(\beta=0.4\) and \(2\delta=0.4\).
For \(\delta=0\), the step execution time for a four-phase stepper motor is
\[ \tau_{\mathrm{sh}}= \frac{\pi}{2\omega_1\left(1-\dfrac{3b}{8\omega^2}\cdot \Theta_{\mathrm{sh}}^2\right)}, \tag{28} \]
for a three-phase stepper motor,
\[ \tau_{\mathrm{sh}}= \frac{\pi}{2\omega_1\left[1-\dfrac{(1+\beta)3b}{8\omega^2}\cdot \Theta_{\mathrm{sh}}^2\right]}. \tag{29} \]
If the results obtained from (11) are regarded as exact, then it can be shown that, with the aid of (28) and (29), the step execution time can be determined for the values \(\beta=0.6\) and \(\beta=0.4\), respectively.
Conclusion
The analytical relations obtained make it possible, quite simply, to determine the influence of various types of load on the character of rotor motion with an accuracy only slightly inferior to graphical calculation methods; to determine the time of execution of a step or of some part of a step; and, by means of additional graphical constructions, to determine the damping rate of oscillations when the motor operates with a load of the viscous-friction type.
REFERENCES
-
D. S. Borisov, A. E. Kobrinskii. On the dynamics of a stepper electric motor. Izvestiya Akademii Nauk, Energetika i Avtomatika, No. 6, 1960.
-
G. Kauderer. Nonlinear Mechanics. Foreign Literature Publishing House, Moscow, 1961.
-
E. P. Popov, I. P. Pal’tov. Approximate Methods for the Investigation of Nonlinear Automatic Systems. Fizmatgiz, 1960.
-
N. N. Bogolyubov, Yu. A. Mitropol’skii. Asymptotic Methods in the Theory of Nonlinear Oscillations. Fizmatgiz, 1958.
-
V. Kaningem. Introduction to the Theory of Nonlinear Systems. Gosenergoizdat, 1962.