TW201202875A - Command learning type multi axis synchronous control system used for rigid tapping machine center and method thereof - Google Patents
Command learning type multi axis synchronous control system used for rigid tapping machine center and method thereof Download PDFInfo
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201202875 六、發明說明： 【發明所屬之技術領域】 本發明係關於一種攻牙中心機’尤指一種用於剛性攻牙 中心機之命令學習式多軸同步控制系統。 【先前技術】 攻牙中心機在現代電子產業扮演著重要的角色，有許 f電子產品需要進行大量的鑽孔與高.品質的螺紋，都是依 靠攻牙中心機來完成；以手機為例，巴掌大的面積在產 ，製程上卻需要加工10~30個孔洞，其他還有手錶，pda. . 等。電子產品體積越小，進行加工處理就越困冑，因此攻 牙中心機必須具更穩定且更精準的控制，方符需求。 -般來說’攻牙中心機若能以同步控 動與進給轴之移動（如進刀、攻牙、反轉退刀、移 =等：作)’才能完成一個品質良好的加工螺孔。目前攻 一二=成同步控制旋轉轴與進給轴有二種方式，其 工件==中心機機構進行改良，意即對攻牙刀加工 β… 或是振動等問題加以分析後，改声相 關問題的機構設計，藉以降低同步 文良相 更攻牙中心機的控制器設計，斜 方式疋變 不同的控制器來控制加工過程：：對各種加工工作準備 果的控制n或是控制器參數。、出可以得到好的加工結 以剛性攻牙中心機來 步要求非常嚴苛，只要稍不對於旋轉軸與進給軸的同 裂；致^的螺距就極易使 m 因如此，旋轉轴與進給〜步控制非常重要具: 3 .201202875 同步控制架構大致上有兩種，第一種為主從架構，亦即由 進給軸追隨旋轉軸作動，進給轴進給的命令是由旋轉軸 際位置提供。這種方式的問題是動態延遲，因為通常進^ 轴要追上旋轉軸需要一段時間延遲，追上時旋轉軸又已： ，到了個位置了。因此，這種架構通t利用前饋控制來二 償動態延遲。至於第二種同步控制架構為交又耦合架構， 即同時考量進給軸與旋轉軸動態，直接控制所加工出來的 同步誤差。此種控制架構通常是採用Ρω架構，並以經驗 _或嘴試錯誤方式選擇控制參數，但主要問題在於無法以系 統化方式加以設計之。 請參閱第三圖所示，係為一種剛性攻牙機（5〇)及盆所採 取之主從式同步控制架構，其中該主從式同步控制架構係 包含有一旋轉軸位置檢知器（511)、一進給軸位置檢知写 (521)及-學f控制器其中該學習控制器⑽)控制進 給軸（52)追隨旋轉軸（51)，實現主從同步架構，再由旋轉軸 «給軸位置檢知器（5川(521)得知其位置，由電腦（61)計201202875 VI. Description of the Invention: [Technical Field] The present invention relates to a tapping center machine, particularly a command learning multi-axis synchronous control system for a rigid tapping center machine. [Prior Art] The tapping center machine plays an important role in the modern electronics industry. There are a lot of drilling holes and high-quality threads that need to be done by tapping the center machine. Take the mobile phone as an example. The area of the palm is large, but the process requires 10 to 30 holes, and other watches, pda. The smaller the size of the electronic product, the more difficult it is to process. Therefore, the center of the tapping machine must have more stable and more precise control. - Generally speaking, if the tapping center machine can synchronize the movement of the feed axis and the feed axis (such as feeding, tapping, reversing, shifting, etc.), a good quality machining screw hole can be completed. . At present, there are two ways to synchronously control the rotary axis and the feed axis. The workpiece== central machine mechanism is improved, which means that after the analysis of the tapping knife processing β... or vibration, the sound is related. The mechanism design of the problem is to reduce the controller design of the synchronous center and the tapping center machine, and to change the different controllers in the oblique mode to control the machining process: control of various processing work or controller parameters. It is very demanding to get a good machining knot with a rigid tapping center machine, as long as it is not the same as the same axis of the rotating shaft and the feed shaft; the pitch of the ^ is very easy to make m, so the rotating shaft and The feed-to-step control is very important: 3.201202875 There are two types of synchronous control architectures. The first one is the master-slave architecture, that is, the feed axis follows the rotary axis, and the feed axis feed command is rotated. Interaxial position is provided. The problem with this approach is dynamic delay, because it usually takes a while for the axis to catch up with the axis of rotation. When it catches up, the axis of rotation has been: and it has reached a position. Therefore, this architecture uses feedforward control to compensate for dynamic delays. As for the second synchronous control architecture, the cross-coupling architecture, that is, simultaneously considers the feed axis and the rotary axis dynamics, and directly controls the processed synchronization error. This kind of control architecture usually adopts Ρω architecture and selects control parameters in the form of experience _ or mouth test error, but the main problem is that it cannot be designed in a systematic way. Referring to the third figure, it is a master-slave synchronous control architecture for a rigid tapping machine (5〇) and a basin, wherein the master-slave synchronous control architecture includes a rotating shaft position detector (511). ), a feed axis position detection write (521) and - learn f controller, wherein the learning controller (10)) controls the feed axis (52) to follow the rotation axis (51), realizes the master-slave synchronization architecture, and then the rotation axis «Axis position detector (5 Sichuan (521) knows its position, calculated by computer (61)
I SJ •算出進給轴位置誤差與旋轉軸位置誤差，再將兩者位置誤差 相減後獲得同步誤差，將此一同步誤差輸出至學習控制器 (60)，並經由學習控制來補償進給軸（52)位置誤差期使同 步誤差為零。這是一種控制學習的控制架構，係將學習控制 器架設在内部的控制迴路，計算出進給軸與旋轉軸之間的同 步誤差，在經由該學習控制器進行位置補償。這種控制學習 架構的缺點在於對旋轉軸控制與學習控制無關，對於同步誤 差接近零結果尚差強人意，因此本發明提出命令學習控制， 以更新輪入命令的方式來影響進給軸與旋轉軸的控制，以期 4 201202875 使同步誤差接近零結果有更好的表現。 【發明内容】 為了解決上述在剛性攻牙中心 遇的問題，本發明主要目的係提供一 之命令學習式多軸同步控制系統，能I SJ • Calculate the position error of the feed axis and the position error of the rotary axis, and then subtract the position error of the two to obtain the synchronization error, output this synchronization error to the learning controller (60), and compensate the feed via the learning control. The axis (52) position error period causes the synchronization error to be zero. This is a control structure for controlling learning. The learning controller is placed inside the control loop to calculate the synchronization error between the feed axis and the rotary axis, and the position compensation is performed via the learning controller. The disadvantage of this control learning architecture is that the rotation axis control is independent of the learning control, and the result of the synchronization error approaching zero is still unsatisfactory. Therefore, the present invention proposes a command learning control to update the wheeling command to affect the feed axis and the rotation axis. Control, with a view to 4 201202875, the synchronization error is close to zero and the result is better. SUMMARY OF THE INVENTION In order to solve the above problems encountered in a rigid tapping center, the main object of the present invention is to provide a command learning multi-axis synchronous control system capable of
式多軸同步控制系統包含有： 心機的同步控制架構所遭 ——種用於剛性攻牙中心機 一旋轉軸學習控制單元，係内建一旋轉軸學習程序， 用以產生下次旋轉軸控制命令； 一進給軸學習控制單元，係内建一進給軸學習程序， 用以產生下次進給軸控制命令； 一同步誤計算模組，係透過二位置檢知器分別連接至 剛性攻牙中心機的旋轉軸及進給轴，以獲得目前二轴位 置，以計算出旋轉軸位置誤差及進給軸位置誤差，再以旋 轉軸位置誤差及進給軸位置誤差計算出同步誤差，並輸出 至該旋轉軸及進給軸學習控制單元； 一旋轉轴反饋單元，係連接至該旋轉轴學習控制單 元，以取得本次旋轉軸控制命令後輸出； 一旋轉軸動態單元，係連接至該旋轉軸反饋單元，以 取得本次進給軸控制命令後輪出至旋轉軸； 一進給軸反饋單元，係連接至該進給軸學習控制單 元，以取得本次進給轴控制命令後輸出；及 201202875 一進給軸動態單元，係連接至該進給軸反饋單元，r 取得本次進給軸控制命令後輸出至進給軸。 u 上述本發明係主要藉由對旋轉轴及進給軸同步輸出修 正後的控制命令，而且各控制命令係依照前次控制命令^ 生的同步誤差加以修正之，故可在短時間内修正二軸的同 步誤差為零，獲得最佳的加工品質。 5 【實施方式】 請參閱第一圖所示，係為本發明學習式多軸同步控制系 統的系統架構圖，其包含有： μ 一旋轉軸學習控制單元（1 〇 )，係内建一旋轉軸學習程 序，用以產生本次旋轉軸控制命令； 進給軸學習控制單元（2〇)，係内建一進給軸學習程 序，用以產生本次進給軸控制命令； 一同步誤計算模組（3〇)，係透過二位置檢知器（S1)(S2) 刀別連接至剛性攻牙中心機的旋轉軸及進給軸，以獲得目 刖一轴位置’以計算出旋轉軸位置誤差及進給軸位置誤 差，再以旋轉軸位置誤差及進給軸位置誤差計算出同步誤 差，並輸出至該旋轉軸及進給軸學習控制單元（1〇)(2〇); 一旋轉轴反饋單元（1彳），係連接至該旋轉轴學習控制單 兀（1 〇)，以取得本次旋轉軸控制命令後輸出； 一旋轉軸動態單元（12)，係連接至該旋轉軸反饋單元 0 1)’以取得本次進給軸控制命令後輸出至旋轉軸； m 一進給軸反饋單元（21 )，係連接至該進給軸學習控制單 元（20)，以取得本次進給軸控制命令後輸出；及 6 201202875 一進給軸動態單元（22) ’係連接至該進給軸反饋單元 (21 )，以取得本次進給軸控制命令後輸出至進給軸。 以下謹進一步說明本發明同步控制旋轉軸及進給軸的 控制命令產生流程及學習程序。 首先由旋轉軸及進給軸學習單元（1〇)(2〇)分別自外部 取得旋轉轴及進給轴控制命令’並將此二控制命令分別輸 出至旋轉軸反饋及動態單元（11)(12)，以及進給軸反饋及動 態單兀（21)(22)，再由同步誤差計算模組（3〇)透過旋轉軸及 _ 進給轴的位置檢知器（S1 )(S2)獲得目前二軸的位置誤差，再 依據旋轉軸及進給軸的位置誤差計算出本次同步誤差，並 將本次同步誤差輸出至旋轉軸及進給軸學習單元 (10)(20)，分別執行旋轉軸學習單元程序及進給軸學習單元 程序。意即，該旋轉軸及進給軸學習單元（彳〇)(2〇)依據本次 同步誤差及本次控制命令計算出修正量，於下一控制命令 輸入時，以對應的修正量修正該控制命令，如此重覆執行 直到獲得符合要求的精度。 ® 由上述說明可知’本發明命令學習式多軸同步控制系統 如第二圖所示，係包含有： 取得外部旋轉軸及進給軸控制命令及修正值；其中第 一次執行時旋轉軸及進給軸修正值為零（4〇); 依據旋轉軸及進給轴修正值修正旋轉軸及進給軸控制 命令（41)，· 將此二修正後控制命令分別輸出至旋轉軸及進給軸 (42);The multi-axis synchronous control system includes: a synchronous control architecture of the heart machine - a rotary axis learning control unit for the rigid tapping center machine, and a built-in rotary axis learning program for generating the next rotary axis control Command; a feed axis learning control unit, built-in a feed axis learning program for generating the next feed axis control command; a synchronous miscalculation module, connected to the rigid attack by a two-position detector The rotation axis and the feed axis of the tooth center machine are used to obtain the current two-axis position to calculate the rotation axis position error and the feed axis position error, and then calculate the synchronization error by the rotation axis position error and the feed axis position error, and Output to the rotating shaft and feed axis learning control unit; a rotating shaft feedback unit is connected to the rotating shaft learning control unit to obtain the current rotating shaft control command output; a rotating shaft dynamic unit is connected to the Rotating the axis feedback unit to obtain the current feed axis control command and then rotating to the rotary axis; a feed axis feedback unit connected to the feed axis learning control list To achieve this feed control command output shaft; and a feed shaft 201202875 dynamic element, is connected to the feed line axis feedback means, R & lt obtain this feed shaft outputted to the feed shaft control command. u The above invention mainly outputs the corrected control command by synchronously outputting the rotation axis and the feed axis, and each control command is corrected according to the synchronization error of the previous control command, so that the correction can be corrected in a short time. The axis has zero synchronization error for optimum processing quality. 5 [Embodiment] Please refer to the first figure, which is a system architecture diagram of the learning multi-axis synchronous control system of the present invention, which comprises: μ-rotating axis learning control unit (1 〇), which is built-in and rotated. The axis learning program is used to generate the current rotation axis control command; the feed axis learning control unit (2〇) is a built-in feed axis learning program for generating the current feed axis control command; The module (3〇) is connected to the rotary shaft and the feed axis of the rigid tapping center machine through the two-position detector (S1) (S2) to obtain the target axis position to calculate the rotation axis. Position error and feed axis position error, and then calculate the synchronization error by the rotary axis position error and the feed axis position error, and output to the rotary axis and feed axis learning control unit (1〇) (2〇); The axis feedback unit (1彳) is connected to the rotating axis learning control unit (1 〇) to obtain the output of the rotating axis control command; a rotating shaft dynamic unit (12) is connected to the rotating shaft feedback Unit 0 1) 'to obtain this feed axis control After the command is output to the rotary axis; m a feed axis feedback unit (21) is connected to the feed axis learning control unit (20) to obtain the output of the feed axis control command; and 6 201202875 a feed The axis dynamic unit (22) is connected to the feed axis feedback unit (21) to obtain the feed axis control command and output to the feed axis. In the following, the flow of the control command generation and the learning program for synchronously controlling the rotary shaft and the feed shaft of the present invention will be further explained. First, the rotation axis and the feed axis learning unit (1〇) (2〇) respectively obtain the rotation axis and the feed axis control command 'from the outside and output the two control commands to the rotary axis feedback and dynamic unit (11) respectively ( 12), and the feed axis feedback and dynamic single (21) (22), and then obtained by the synchronous error calculation module (3〇) through the rotary axis and the position detector (S1) of the _ feed axis (S2) At present, the position error of the two axes is calculated according to the position error of the rotation axis and the feed axis, and the synchronization error is output to the rotation axis and the feed axis learning unit (10) (20), respectively. Rotary axis learning unit program and feed axis learning unit program. That is, the rotation axis and the feed axis learning unit (彳〇) (2〇) calculate the correction amount according to the current synchronization error and the current control command, and correct the correction amount with the corresponding correction amount when the next control command is input. The control commands are executed repeatedly until the required accuracy is obtained. ® As can be seen from the above description, the command-learning multi-axis synchronous control system of the present invention, as shown in the second figure, includes: obtaining external rotation axis and feed axis control commands and correction values; wherein the first execution of the rotation axis and The feed axis correction value is zero (4〇); the rotary axis and feed axis control commands are corrected according to the rotation axis and feed axis correction value (41), and the two corrected control commands are output to the rotary axis and feed respectively. Axis (42);
i SI 檢知旋轉軸及進給軸的目前位置，以計算旋轉軸及進 7 201202875 給軸位置誤差（43); 依據旋轉轴及進給軸位置誤差計算其間的同步誤差 (44)； 依據同步誤差計算出下次外部旋轉軸及進給軸控制命 令的修正值（45); 止 回到第一步驟（40)，^到加工結果符纟要求的精度為 本發明係主要藉由對旋轉轴及進給軸同步輪出修正後 的控制命令，而且各㈣命令係錢前次控 同：誤差加以修正之，故可在短時間内修正 = 差為零，獲得最佳的加工品質。 门步誤 參 以：謹進一步說明旋轉軸學習程序及進 的洋細過程，本發明係採用根據線性學習控程序 性攻牙學習控制法則，其中該修正後控制命去’提出剛 習控制式為表示之，即·其中…生學 同步誤差，L為學習增益矩陣。 轴與軸的 前 剛性攻牙學f㈣其控制法則為咖)=〇鳥似 一 “〇，ι，··.，ρ〜Γ 、 -人的同步誤差均可作為可參考的物理量 〜⑼故，m成，陣的形式為 學習效 〇⑴ Mp~Oj l2px] 2px] ‘02 In -P~1*1 Vl.2i SI detects the current position of the rotary axis and the feed axis to calculate the rotary axis and advance 7 201202875 axis position error (43); calculate the synchronization error between the rotary axis and the feed axis position error (44); The error calculates the correction value of the next external rotation axis and feed axis control command (45); back to the first step (40), the precision required by the processing result symbol is the invention mainly by the rotation axis And the feed axis synchronously rotates the corrected control command, and each (4) command is controlled by the same time: the error is corrected, so it can be corrected in a short time = the difference is zero, and the best processing quality is obtained. The doorstep error is involved: I would like to further explain the rotary axis learning program and the in-depth process. The present invention adopts the rule-based learning control rule according to the linear learning control, wherein the modified control command is to propose the just-controlled control method. Expressed, that is, where... the learning synchronization error, L is the learning gain matrix. The front rigid tapping of the shaft and the shaft f (4) The law of control is coffee) = 〇 似 like "〇, ι,··., ρ~Γ, - the synchronization error of human can be used as a reference physical quantity ~ (9), therefore m into, the form of matrix is learning effect (1) Mp~Oj l2px] 2px] '02 In -P~1*1 Vl.2
Vl.p. ^,(1) £m(2) £j-i(p) 以子習增益矩陣L為夫叙 A 4, PX，^ 轉數，參考^ Z抽與 修正下一次得控制命令。 W问步误差來 201202875 由於學習增益矩陣L的選擇將會影響反覆式學習控制的 效能，故以下進一步說明學習增益矩陣[為Vl.p. ^,(1) £m(2) £j-i(p) Use the sub-learning gain matrix L as the argument A 4, PX, ^ number of revolutions, reference ^ Z pumping and correct the next control command. W question error is coming 201202875 Since the selection of the learning gain matrix L will affect the performance of the repeated learning control, the learning gain matrix is further explained below.
L I说 In }p-h\ ^p-1,2 其中~為學習增益。 為了設計上列的學習增益，首先要推導誤差動態，就本 發明的剛性攻牙中心機來說’其同步誤差與輸出的關係為 V = n^y ;其中「/ Γ 0 γ 0 • · · 〇 •争* · 0 ο γ 6是Ζ軸與β軸的同步誤差；尤是0軸導螺桿的節距（Pitch);及 心是Z軸導螺桿的節距（Pitch); 由上述可知，r是一常數矩陣，而此可得知系統的輸7 •ϋ CB 0 · 〇 '^/(0)' 與輸出之間關係為： Sjyii) = CAB CB .·. 0 …CB S/(l) _sjy(p) CAp-xB CAp-2B β/{ρ-\) l + yTCBl0i YTCBlm … yTCBl0p ΣϊΤ〇Α'-'ΒΙη l + jycm …jyCA、-Blip i=0 . «=〇 . Ρ-ϊ ρ一i · ρ , ： ΣνΤ€Αρ-'-!ΒΙη - Ι + Ϋγ^Α^ΒΙ^ ί=0 «=〇 其中/+m 學習控制的收斂條件為此矩陣之所有特徵值大小皆小於1 亦即μ,.(/+17^)^1 Vi•，然而另外需要L決定使得其滿足收斂 件。 201202875 根據理論的分析，一般的線性學習控制方法是能保證在 y趨於無窮時，誤差將收斂為零，但是在收斂到誤差為〇之 前的誤差變化是沒有規律可言，有可能透過學習控制之後誤 差反而越來越大，因此可進一步讓誤差收歛具有單調收斂的 特性，所謂的單調收斂是讓誤差值具有這一次的誤差比上一 次的誤差來的小的特性，就本發明的剛性攻牙中心機來說， 每一次的同步誤差皆比上一次的同步誤差來的小，以下進一 步再對收敛條件加以探討之。L I says In }p-h\ ^p-1,2 where ~ is the learning gain. In order to design the learning gains listed above, the error dynamics must first be derived. For the rigid tapping center of the present invention, the relationship between the synchronization error and the output is V = n^y; where "/ Γ 0 γ 0 • · · 〇•争* · 0 ο γ 6 is the synchronization error between the x-axis and the β-axis; especially the pitch of the 0-axis lead screw; and the center is the pitch of the Z-axis lead screw; r is a constant matrix, and it can be known that the system's input 7 • ϋ CB 0 · 〇 '^ / (0) ' and the output relationship is: Sjyii) = CAB CB ..... 0 ... CB S / (l ) _sjy(p) CAp-xB CAp-2B β/{ρ-\) l + yTCBl0i YTCBlm ... yTCBl0p ΣϊΤ〇Α'-'ΒΙη l + jycm ...jyCA, -Blip i=0 . «=〇. Ρ-ϊ一一i · ρ , : ΣνΤ€Αρ-'-!ΒΙη - Ι + Ϋγ^Α^ΒΙ^ ί=0 «=〇 where /+m The convergence condition of the learning control is that all eigenvalues of this matrix are less than 1 That is, μ,.(/+17^)^1 Vi•, however, L is required to make it satisfy the convergence. 201202875 According to the theoretical analysis, the general linear learning control method can guarantee the error when y tends to infinity. Will converge to zero, but in There is no law about the error change before the error is ,. It is possible that the error is more and more large after learning control, so the error convergence can be further converge to monotonic convergence. The so-called monotonic convergence is to make the error value have this. The error of one time is smaller than that of the previous error. For the rigid tapping center of the present invention, each synchronization error is smaller than the previous synchronization error, and the convergence condition is further discussed below. It.
為了使同步誤差能單調收斂，提 設計來加以達成，其中L矩陣變為 誤差動態則為心=(/+1^)^ ;其中 l + yTCBl0i 0In order to make the synchronization error monotonically converge, the design is achieved, where the L matrix becomes the error dynamics as the heart = (/ +1 ^) ^ ; where l + yTCBl0i 0
In V-1,2 P-hp 2pxp 而In V-1,2 P-hp 2pxp and
0 0 Σ^Αι-ιΒΙη l + rTCBll2 … »=0 Ιί prCA^Bln P±fCA^Bln ... l + fCBlp_hp 條件為μ/ΐμ v/，即為_2<咖;1<<()，而誤差㈣㈣ ij€j~~j-}Q Q-^ 0 = /+ΓΡΙ ;由 Rayleigh-Ritz 不等式得到“ 在A, U吖0么…中，“ 0)<ι，可得 α《 即誤差經過母一次運算後將會 變小，具有單調收斂的特性。所以^必·0<1 單調收斂的特性，如果……± U誤差具備 如果e疋對角線矩陣，而且，誤差 就曰具有早調收斂的特性’即收陣的特 對角線㈣Μ差單料料條件。 、 因此再以遞迴關係來決定L，首先由ν，先決0 0 Σ^Αι-ιΒΙη l + rTCBll2 ... »=0 Ιί prCA^Bln P±fCA^Bln ... l + fCBlp_hp The condition is μ/ΐμ v/, which is _2<Caf;1<<() And the error (4) (4) ij € j~~j-}Q Q-^ 0 = /+ΓΡΙ ; obtained by Rayleigh-Ritz inequality "In A, U吖0..., 0" <ι, available α That is, the error will become smaller after the primary operation, and it has the characteristic of monotonic convergence. Therefore, ^ must · 0; 1 monotonic convergence characteristics, if ... ± U error with e 疋 diagonal matrix, and the error 曰 has the characteristics of early convergence convergence 'that is the special diagonal of the array (four) Μ difference Material conditions. Therefore, the reciprocal relationship is used to determine L, first by ν, the prerequisite
I + TPL 學習控制收斂 m 10 201202875 定對角線到的值，然後再決定2矩陣第一行的L， >1，2,31-1。首先將2矩陣第一行的前兩項展開，分別 為仏、=+ rrCBL 及 Q3、= /CA2Bl0t + yTCAB!u + /CB!2i。 為了讓2矩陣變成對角線矩陣，必須仏1 =G’么=0，··％ =0， 由於先前已經決定了 L，由ft, =/C4S/。, +/C：5/"式便可以決定 L，有了心跟^由⑽/„+/0^式就可以決定μ， 以此類推，就可以決定第一行的（·,，/ = 1,2,3../-1，且以此方法 就可以逐行*定各行的卜以下三角學f控制矩陣求得的學 習增益矩陣L，來進行學習控制即可讓系統的^具備 調收斂的特性。 【圖式簡單說明】I + TPL Learning Control Convergence m 10 201202875 Set the diagonal to the value, and then determine the L, >1, 2, 31-1 of the first row of the 2 matrix. First, expand the first two terms of the first line of the 2 matrix, namely 仏, =+ rrCBL and Q3, = /CA2Bl0t + yTCAB!u + /CB!2i. In order to make the 2 matrix a diagonal matrix, 仏1 = G' =0, ··% =0, since L has been previously determined, by ft, =/C4S/. , + / C: 5 / " can decide L, with the heart and ^ by (10) / „ + / 0 ^ can determine μ, and so on, you can decide the first line (·,, / = 1,2,3../-1, and in this way, the learning gain matrix L obtained by the following trigonometric f control matrix can be determined row by row* to perform learning control to make the system Adjust the characteristics of convergence. [Simplified description]
第一圖 塊圖。 第二圖 程圖。 係本發明命令學習式多轴 係本發明命令學習式多料 同步控制系統的方 同步控制方法的流 第The first picture is a block diagram. The second picture. The present invention is a command-learning multi-axis system. The flow of the synchronous control method of the command-learning multi-material synchronous control system of the present invention is
係既有主從式單轴學習式同步控制系統方塊 【主要元件符號說明】 (10)旋轉軸學習控制單元 (12)旋轉軸動態單元 (21)進給軸反饋單元 (30)同步誤計算模組 (11)旋轉#反饋單元 ()進给軸學習控制單元 (22)it動態單元 (50)加X機 201202875 (51) 旋轉軸 （511)旋轉軸位置檢知器 (52) 進給軸位置檢知器 （521)進給軸控制器 (60)學習控制器 （61)電腦There are two master-slave single-axis learning synchronous control system blocks [Main component symbol description] (10) Rotary axis learning control unit (12) Rotary axis dynamic unit (21) Feed axis feedback unit (30) Synchronous miscalculation mode Group (11) Rotation # Feedback unit () Feed axis learning control unit (22) It dynamic unit (50) plus X machine 201202875 (51) Rotary axis (511) Rotary axis position detector (52) Feed axis position Detector (521) feed axis controller (60) learning controller (61) computer
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| TWI492008B (en) * | 2013-11-01 | 2015-07-11 | Ind Tech Res Inst | Working machine controlling system and method thereof |
| CN107918355A (en) * | 2017-12-15 | 2018-04-17 | 苏州新代数控设备有限公司 | A kind of numerically-controlled machine tool tapping control system and its control method |
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| TWI492008B (en) * | 2013-11-01 | 2015-07-11 | Ind Tech Res Inst | Working machine controlling system and method thereof |
| CN107918355A (en) * | 2017-12-15 | 2018-04-17 | 苏州新代数控设备有限公司 | A kind of numerically-controlled machine tool tapping control system and its control method |
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