1、Are Servo Shaft linear motors difficult to integrate into a machine?Not difficult, just a little different. The Servo Shaft linear motor is simpler to install, as it replaces the ball screw, nut,end bearings, motor mount, couplings, and rotary motor. Alignment of the Linear Shaft Motor is not critical (even for high performance packages) and consists of mainly ensuring there is some clearance between the forcer and shaft over the entire travel. Inoservo will assist with selection of suitable components.2、Are versions of the SSLM available for use in waterproof, vacuum or clean rooms？Yes, the Servo Shaft linear motor can be built for a variety of operating environments. To determine if and which Servo Shaft linear motor is suitable for a specific application, an applications engineer must review the specifications.3、Can a Servo Shaft linear motor be mounted vertically?Yes, a Servo Shaft linear motor provides the same performance when mounted vertically or horizontally. However, it is recommended that a vertically mounted Servo Shaft linear motor be counterbalanced.4、Can more than one forcer be used with a single shaft?Yes, more than one forcer can be used in conjunction with a single shaft as long as the forcers do not physically interfere with each other. Two forcers may also be tied together and driven with one drive to double the output force.5、Can the shaft of the Servo Shaft linear motor transmit a rotary force?Yes, it is possible. To determine which Servo Shaft linear motor is most suitable for your specific application, an applications engineer must review the specifications.6、Do magnets ever lose their magnetism over time?The Servo Shaft linear motors use a rare earth magnet, which will maintain their strength for 99 years. However, when operating at high temperatures (>150°C), these rare earth magnets can lose strength. Lower temperatures have no effect the magnets as long as frost does not form in the air gap.7、Do standard rotary motor electronics work with linear motors?The Servo Shaft linear motor is designed to operate with most off-the-shelf motor controls and drives.Basically, the Li Servo Shaft linear motor uses the same electric circuit as other linear motors and rotary servo motors.8、Does only one forcer need to have the halls or do both need halls?In an application where two coils are connected to the same drive, the same coil of each drive must be above the same magnet in order to run. (See drawing below) This is why when the second forcer is flipped the U and V leads must also be flipped. As such only one of the two coils needs to have halls.9、How accurate are Servo Shaft linear motors?By eliminating the conversion of rotary to linear motion, a major source of positioning error is removed. This results in high performance and accuracy. While the Servo Shaft linear motor itself does not have inherent resolution, position accuracy is ultimately determined by the linear encoder feedback accuracy and the core stuffiness of the Servo Shaft linear motor. Testing has shown that with encoder resolutions less then 10nm, the Servo Shaft linear motor。
专利名称：Linear drive motor multiple carrier controlsystem发明人：Prucher, Bryan P.申请号：EP85110711申请日：19850826公开号：EP0188657A3公开日：19870616专利内容由知识产权出版社提供专利附图：摘要：Disclosed is a multiple carrier transportation system. The system employs a plurality of linear induction motors positioned along a track. A plurality of carrier units are used for moving work pieces along the track. Each carrier is mounted with a reaction plate or secondary. The reaction plate passes over the linear induction motors and reacts with the electromagnetic field generated therein. The electromagnetic energy induced from the linear induction motors Into the secondary produces a mechanical thrust which propels the carrier along the track. A control system utilizIng a series of freedback loops and a computer is used to monitor and coordinate the movement of the carriers in the system. Attached to each carrier is a position indicator. The Indicator relays position information to position sensors. The sensors relay position information to a controlmodule or computer which determines the velocity and acceleration of the carrier. These values are then compared to a set of velocity and acceleration curves stored in memory. Deviations between the actual and desired speeds are corrected for by varying the power input to the linear induction motors. The control modules or computers are interconnected and receive Information and instruction from a central data computer. The position, velocity, and acceleration of each carrier in the system is thus monitored and controlled.申请人：CHRYSLER CORPORATION更多信息请下载全文后查看。
CHAPTER 44.1(a)For convergence of the steepest-descent algorithm:where is the largest eigenvalue of the correlation matrix R . We are givenThe two eigenvalues of R areλ1 = 0.5λ2 = 1.5Hence . The step-size parameter µ must therefore satisfy the conditionWe may thus choose µ = 1.0.(b)From Eq. 4.9 of the text,With µ = 1 andwe therefore have0µ2λmax-----------<<λmax R 10.50.51=λmax 1.5=0µ21.5------<< 1.334=w n 1+()w n ()µp Rw n ()–[]+=p 0.50.25=w n 1+()w n ()0.50.2510.50.51w n ()–+=That is,Equivalently, we may write(c)To investigate the effect of varying the step-size parameter µon the trajectory,we findit convenient to work with v (n )rather than with w (n ).The k th natural mode of the steepest descent algorithm is described bySpecifically,For the initial condition, we haveFrom the solution to Problem 2.2:100110.50.51– w n ()0.50.25–=00.5–0.5–0w n ()0.50.25–=w 1n +1()w 2n +1()00.5–0.5–1w 1n ()w 2n ()0.50.25–=w 1n +1()0.5–w 2n ()0.5–=w 2n +1()0.5–w 1n ()0.25–=νk n +1()1µλk –()νk n (),=k 12,=ν1n +1()10.5µ–()ν1n ()=ν2n +1()1 1.5µ–()ν2n ()=v 0()ν10()ν20()=Q H w o=Hence,That is,For n > 0, we have,k = 1,2Hence,Solution - This represents an oscillatory trajectory.w o 0.5=Q 12------111–1=v 0()12------11–110.5=12------0.50.5=ν10()ν20()0.52------==νk n ()1µλk –()n νk 0()=ν1n ()10.5µ–()n ν10()=ν2n ()1 1.5µ–()n ν20()=µ1=ν1n ()0.5()n ν10()=ν2n ()0.5–()n ν20()=This second solution represents a damped trajectory.The transition from a damped to an oscillatory trajector occurs atSpecifically, for 0 <µ < 0.667, the trajectory is damped. On the other hand, for0.667 <µ < 1.334 the trajectory is oscillatory.4.2We are given(a)The correlation matrix R =r (0).Hence,λmax =r (0)Correspondingly,(b)Time constant of the filter is (c)µ0.1=ν1n ()0.59()nν10()=ν2n ()0.85()n ν20()=µ11.5------0.667==J w ()J min r 0()w w o –()2+=µmax 2λmax -----------2r 0()----------==τ11µλ1---------≈1µr 0()-------------=Slope = 2r (0)(w - w o )J (w )0w o w wJ min4.3(a)There is a single mode with eigenvalue λ1 =r (0), and q 1 = 1, Hence,where v 1(n ) =q 1(w o -w (n )) = (w o -w (n ))(b)4.4The estimation error e (n ) equalswhere d (n )is the desired response,w (n )is the tap-weight vector,and u (n )is the tap-input vector. Hence, the gradient of the instantaneous squared error equals4.5Consider the approximation to the inverse of the correlation matrix:where µ is a positive constant bounded in value aswhere is the largest eigenvalue of R .Note that according to this approximation,we have R -1(1) =µI . Correspondingly, we may approximate the optimum Wiener solution as∂J ∂w------2r 0()w w o –()=J n ()J min λ1v 1n ()2+=∂J n ()∂λ1-------------v 1n ()2w o w n ()–()2==e n ()d n ()w Hn ()u n ()–=∇ˆJ n ()∂∂w -------e n ()2[]∂∂w -------e n ()e *n ()[]==e n ()∂e *n ()∂w ----------------e *n ()∂e n ()∂w-------------+=2e *n ()u n ()–2u n ()d *n ()2u n ()u Hn ()w n ()+==R 1–n +1()µI µR –()k k =0n ∑≈0µ2λmax-----------<<λmaxIn the second term, put k =i +1 or i =k -1:(1)where, in the second line, we have used the fact thatHence, rearranging Eq. (1):which is the standard formula for the steepest descent algorithm.4.6For stability of the steepest-descent algorithm, we therefore requireTo satisfy this requirement,the step-size parameter µshould be positive,since .Hence,the steepest-descent algorithm becomes unstable when the step-size parameter is negative.w n +1()R 1–n +1()p=µI µR –()k pk =0n∑≈µp µI µR –()k pk =1n∑+=w n +1()µp µI µR –()I µR –()i pk =1n∑+=µp I µR –()w n ()+=µI µR –()ip i =0n∑w n ()=w n +1()w n ()µp Rw n ()–[]+=J w n +1()()J w n ()()12--µg n ()2–=J w n +1()()J w n ()()<µg n ()20>4.7LetBy definition,Hence,But, the estimation error isTherefore,At the minimum point of the error-performance surface,the correction is zero.Hence, at this point, we havewhich is a restatement of the principle of orthogonality.4.8We are givenFor n = 1, we have δw n +1()w n +1()w n ()–=µp Rw n ()–[]=p E u n ()d *n ()[]=R E u n ()u Hn ()[]=δw n +1()µE u n ()d *n ()[]E u n ()u H n ()[]w n ()–{}=µE u n ()d *n ()u H n ()w n (){}[]=e n ()d n ()w H n ()u n ()–=δw n +1()µE u n ()e *n ()[]=δw n +1()E u n ()e *n ()[]0=J approx n ()J 0()J ∞()–[]en τ⁄–J ∞()+=Solving for e 1/τ, we getTaking natural logarithms:or4.9(a)The cross-correlation between the “desired response”u (n ) and the tap input u (n -1) isThe mean-square value of the tap input u (n -1) isHence,For the recursive computation of the parameter a , we note that a (n ) = -w (n ); hence(b)The error-performance surface is defined byJ approx 1()J 1()=J 0()J ∞()–[]e1τ⁄–J ∞()+=e 1τ⁄J 0()J ∞()–J 1()J ∞()–-----------------------------=1τ--J 0()J ∞()–J 1()J ∞()–-----------------------------ln =τJ 1()J ∞()–J 0()J ∞()–-----------------------------ln =p E u n ()u n -1()[]r 1()==E u 2n -1()[]r 0()=w n +1()w n ()µr 1()r 0()w n ()–()+=a n +1()a n ()µr 1()r 0()a n ()+[]–=J n ()r 0()2r 1()w n ()–r 0()w 2n ()+=r 0()2r 1()a n ()r 0()a 2n ()++=The corresponding plot of the error performance surface is therefore(c)The condition on the step-size parameter is4.10The second-order AR process u (n ) is described by the difference equation(1)Hence,and the AR parameters equalAccordingly, we write the Yule-Walker equations as(2)(3)Eqs. (1), (2) and (3) yield r (0)aJ (n )-r (1)/r (0)r (0) - -------r 2(1)r (0)00µ2r 0()----------<<u n ()0.5u n -1()–u n -2()v n ()++=w 10.5w 2,–1==a 10.5a 2,1–==r 0()r 1()r 1()r 0()0.5–1r 1()r 2()=σv 2a k r k ()k =02∑=1a 0r 0()a 1r 1()a 2r 2()++=r 0()0.5r 1()r 2()–+=r (0) = 0r (1) = 1r (2) = -1/2Hence,The eigenvalues of R are -1, +1. For convergence of the steepest descent algorithm:where λmax is the largest eigenvalue of the correlation matrix R . Hence, with λmax = 1.0 <µ < 24.11(1)Hence,Accordingly, we may write as (2)r (0) = 0r (1) = -2/3Therefore,R 0110=0µ2λmax-----------<<u n -2()0.5u n -1()u n ()v n ()–+=w 11w 20.5==Rw b rB *=r 0()r 1()r 1()r 0()10.5r 2()r 1()=σv 2a k r k ()k =02∑=1r 0()r 1()–0.5r 2()–=R 023⁄–23⁄–0=The eigenvalues of R are λ = -2/3, 2/3For convergence of the steepest descent algorithm:4.12The third-order AR process u (n ) is described by the difference equationHencew 1 = -0.5,w 2 = -0.5,w 3 = 0.5and the AR parameters equala 1 = 0.5,a 2 = 0.5,a 3 = -0.5Accordingly, we write the Yule-Walker equations as(3)Equations (1), (2) and (3) yieldr (0) = 1r (1) = -1/2r (2) = -1/2r (3) = +10µ2λmax-----------<<0µ223⁄---------<<0µ3<<u n ()0.5u n -1()–0.5u n -2()–0.5u n -3()v n ()++=r 0()r 1()r 2()r 1()r 0()r 1()r 2()r 1()r 0()0.5–0.5–0.5r 1()r 2()r 3()=σv 2a k r k ()k =03∑=1r 0()a 1r 1()a 2r 2()a 33()+++=(a)The correlation matrix is(b)The eigenvalues of R are 0, 1.5, 1.5(c)For convergence of the steepest descent algorithm, we requireHence, with λmax = 1.5, we have4.13(a)The first-order MA process is described by the difference equationTaking the z -transform of both sides:U (z ) = (1 - 0.2z -1)V (z )Accordingly,for which we inferw 1 = -0.2,w 2 = -0.04R 112⁄–12⁄–12⁄–112⁄–12⁄–12⁄–1=0µ2λmax-----------<<0µ21.5------<<u n ()v n ()0.2v n -1()–=U z ()V z ()-----------110.2z 1––()1–----------------------------------=110.2()z 1–0.04z2–++------------------------------------------------------≈u n ()0.2u n -1()0.04u n -2()++v n ()=u n ()0.2u n -1()–0.04u n -2()–v n ()+=Correespondingly, the AR parameters equala 1 = 0.2,a 2 = 0.04Yule-Walker equations are(1)(2)(3)Equations (1), (2) and (3) yieldr (0) = 1.0401r (1) = -0.2000r (2) = -0.0016The correlation matrix isThe eigenvalues of R are = 1.2401, 0.8401,λmax = 1.2401For convergence of the steepest descent algorithm:That is,(b)r 0()r 1()r 1()r 0()0.2–0.04–r 1()r 2()=σv 2a k r k ()k =03∑=1r 0()a 1r 1()a 2r 2()++=R r 0()r 1()R 1()R 0() 1.04010.2000–0.2000– 1.0401==0µ21.2401---------------<<0µ 1.6127<<U z()V z ()-----------110.2z 1––()1–----------------------------------=Accordingly,w 1 = -0.2,w 2 = -0.04,w 3 = -0.008The AR parameters area 1 = 0.2,a 2 = 0.04,a 3 = 0.008The Yule-Walker equations are(4)(5)(6)Equations (4), (5) and (6) yieldr (0) = 1.04r (1) = -0.2r (2) = 0.0r (3) = -0.003The correlation matrix and its eigenvalues areλ = 0.7572, 1.04 1.3228,λmax = 1.3228For convergence of the steepest descent algorithm:110.2z 1–0.04z 2–0.008z 3–+++-----------------------------------------------------------------------------≈u n ()0.2u n -1()0.04u n -2()0.008u n -3()+++v n ()=r 0()r 1()r 2()r 1()r 0()r 1()r 2()r 1()r 0()0.2–0.04–0.008r 1()r 2()r 3()=σv 2a k r k ()k =03∑=1r 0()a 1r 1()a 2r 2()a 3r 3()+++=R 1.040.2–00.2– 1.040.2–00.2– 1.04=(c)When the system is approximated by a second-order AR model,When it is approximated by a third-order AR model,These results show that the upperbound on the step-size parameters becomes tighter as the approximating model order is increased.4.14(a)Take z -transforms of both sides:U (z ) = -0.5z -1U (z ) +V (z ) - 0.2z -1V (z )Approximation to third-order yieldsAccordingly,0µ21.3228---------------<<0µ 1.5119<<0µ 1.6127<<0µ 1.5119<<u n ()0.5u n -1()–v n ()0.2v n -1()–+=U z ()V z ()-----------10.2z 1––10.5z1–+------------------------=110.5z 1–+()10.2z 1––()1–----------------------------------------------------------------=U z ()V z ()-----------110.5z 1–+()10.2z 1–0.04z 2–0.08z 3–+++()-------------------------------------------------------------------------------------------------------------=110.7z 1–0.14z 2–0.028z3–+++-----------------------------------------------------------------------------≈u n ()0.7u n -1()0.14u n -2()0.028u n -3()+++v n ()=u n ()0.7u n -1()0.14u n -2()–0.028u n -3()–v n ()+–=w 1 = -0.7,w 2 = -0.14,w 3 = -0.028The AR parameters area 1 = 0.7,a 2 = 0.14,a 3 = 0.028The Yule-Walker equations are(1)(2)(3)Equations (1), (2), and (3) yieldr (0) = 1.6554r (1) = -1.0292r (2) = 0.5175r (3) = -0.2646The correlation matrix (b)The eigenvalues of R are 0.4358, 1.1379, 3.3925,λmax = 3.3925(c)For convergence of the steepest descent algorithm we require r 0()r 1()r 2()r 1()r 0()r 1()r 2()r 1()r 0()0.7–0.14–0.028r 1()r 2()r 3()=σv 2a k r k ()k =03∑=r 0()a 1r 1()a 2r 2()a 3r 3()+++1=R r 0()r 1()r 2()r 1()r 0()r 1()r 2()r 1()r 0() 1.6554 1.0292–0.51751.0292– 1.6554 1.0292–0.5175 1.0292– 1.6554==0µ2λmax-----------λmax ,<< 3.3925=0µ23.3925---------------<<0µ0.589<<4.15(a)Substract from both sidesStarting with the initial value and iterating:From Eq. (2.52) of the text:(1)The transient behavior of Newton’s algorithm is thus characterized by a single exponential whose corresponding time constant is(2)where .(b)Taking the logarithm of both sides of Eq. (2):When ,. Hencew n +1()1µ–()w n ()µR 1–p+=1µ–()w n ()µw o (R 1–p ,+w o )==w o w n +1()w o –1µ–()w n ()µw o w o–+=1µ–()w n ()w o –()=w o w n ()w o –1µ–()n w 0()w o –()=J n ()J min w w o –()H R w w o –()+=J min 1µ–()n w 0()w o –[]H R 1µ–()n w 0()w o –[]+=J min 1µ–()2n w 0()w o –[]H R w 0()w o –[]+=J min 1µ–()2n J 0()J –min ()+=1µ–()2k e k τ⁄–=µ1«2k 1µ–()ln k τ⁄–=µ1«1µ–()µ–≈lnfrom which we readily deduce the time constant,.2k µk τ⁄–≈–τ12µ------=µ1«。
