版權(quán)說明:本文檔由用戶提供并上傳,收益歸屬內(nèi)容提供方,若內(nèi)容存在侵權(quán),請進(jìn)行舉報(bào)或認(rèn)領(lǐng)
文檔簡介
1、1,Chapter 5 Root Locus Method,5.1 The Root Locus Concept 5.2 The Root Locus Procedure 5.3 Examples for Drawing Root Locus 5.4 Parameter Design by the Root Locus Method 5.5 Relationship between Performance and the distributing of close-loop zeros and poles 5.6 Compensation by Using Root Locus Method
2、5.7 Summary 5.8 Three-term (PID) Controllers,2,Relative stability and transient performance are directly related to the location of the roots. It is necessary to adjust one or more system parameters to obtain suitable root locations. The above two facts make it worthwhile to determine how the roots
3、migrate(移動) about the s-plane as the parameters are varied.,5.1 The Root Locus Concept,3,The root locus method was introduced by Evans in 1948. It is a graphical method for sketching the locus of roots in the s-plane as a parameter is varied. It provides an approximate sketch that can be used to obt
4、ain qualitative(定性的) information concerning the stability and performance of the system.,4,Example 5.1: A second-order system,The figure is shown the root locus as K changes from 0 to infinity.,5,Definition of Root Locus The root locus is the path of the roots of the characteristic equation traced o
5、ut in the s-plane as a system parameter is changed.,6,5.2 The Root Locus Procedure,Magnitude and Angle Requirement Procedure for drawing root locus,7,Magnitude and Angle Requirement,Root locus (negative feedback) magnitude requirement angle requirement,8,9,10,Note: When plotting the root locus, only
6、 the angle requirement is the sufficient and necessary condition (because s is varied with K ). The magnitude requirement is used to determine K for a given root s1. The angle requirement is used to verify a test point s1 as a root location.,11,Procedure for Drawing Root Locus,Step 1:Prepare the roo
7、t locus sketch.,(a) Write the characteristic equation so that the parameter of interest, K, appears as multiplier.,(b) Factor P(s) in terms of n poles and m zeros.,12,(c) Locate the open-loop poles and zeros of P(s) in the s-plane with selected symbols. The locus of the roots of 1+KP(s)=0 begins at
8、the open-loop poles and ends at the open-loop zeros as K increases from 0 to infinity. If nm, there are n-m branches of the root locus approaching the n-m zeros at infinity.,13,Proving:,At the starting point of the root locus: K=0,14,At the end point of the root locus: K,and the characteristic equat
9、ion can be written as,when,15,(d) The number of separate loci is n.,(e) The root loci are symmetrical with the respect to the horizontal real axis because complex roots must appear as pairs of complex conjugate roots.,16,Step 2:Locate the Root Locus on the Real Axis,The root locus on the real axis a
10、lways lies in a section of the real axis to the left of an odd number of poles and zeros.,17,It can be ascertained by the angle requirement.,18,Step 3: The Asymptotes of the Root Locus The loci proceed to the zeros at infinity along asymptotes centered at (asymptote centroid) and with angles .,19,s時(shí)
11、,上式可化為: 為漸近線公式。有n-m個(gè)解,表示n-m條直線,經(jīng)過公共點(diǎn)( , j0)。 可得直線夾角滿足: 將 代入特征方程,可推導(dǎo)得:,20,Example 5.1: A second-order system,21,Example 5.2,22,Step 4: Determine where the root locus crosses the imaginary axis. The point where the locus crosses the imaginary axis, we have s=jw into the characteristic equation and sol
12、ving for w.,23,Example 5.3 A unity feedback system,24,Note: The actual point at which the root locus crosses the imaginary axis can also be readily evaluated by utilizing the Routh-Hurwitz criterion.,25,Example 5.3 A unity feedback system,26,Step 5: Determine the breakaway point of the root locus Th
13、e root locus left the real axis at a breakaway point. The locus leaves the real axis where there are a multiplicity of roots, typically two.,27,28,Breakaway point,29,Assume the breakaway point s=d: Method 1:,30,Method 2:,31,Example 5.4 A feedback system,32,Example 5.5 Feedback systems,33,The angle b
14、etween the direction of emergence (or entry) of q coincident poles (or zeros) on the real axis(根軌跡離開或進(jìn)入實(shí)軸上q重極點(diǎn)(或零點(diǎn))方向之間的夾角),34,Example 5.6 Feedback systems,35,補(bǔ)充說明: 一般來說,實(shí)軸上兩相鄰開環(huán)極點(diǎn)之間若有根軌跡,則必有分離點(diǎn);實(shí)軸上兩相鄰開環(huán)零點(diǎn)(包括無窮遠(yuǎn)處零點(diǎn))之間若有根軌跡,則必有會合點(diǎn);而實(shí)軸上開環(huán)極點(diǎn)和開環(huán)零點(diǎn)之間,或者分離點(diǎn)和會合點(diǎn)同時(shí)存在,或者都沒有。,36,Step 6:Determine the angle of
15、departure of the locus from complex poles and the angle of arrival at complex zeros, using the phase criterion. The angle of departure from complex poles is given by (2k+1)(angles of the vectors from all other open-loop poles to the pole in question) + (angles of the vectors from the open-loop zeros
16、 to the complex pole in question).,37,Proving:,38,The angle of arrival at a complex zero may be found from the same rule and then the sign changed to produce the final result.,39,Proving:,40,Example 5.7 Feedback systems,41,42,Step 7: Complete the root locus sktech Rule 1:The sum of the closed-loop p
17、oles If there are at least two more open-loop poles than open-loop zeros, the sum of the closed-loop poles is constant, independent of K, and equal to the sum of the open-loop poles. (如果開環(huán)極點(diǎn)比開環(huán)零點(diǎn)至少多2個(gè),閉環(huán)極點(diǎn)的和為一不依賴于K的常數(shù),且等于開環(huán)極點(diǎn)的和。),43,閉環(huán)極點(diǎn)的和與積 設(shè)閉環(huán)系統(tǒng)特征方程: 則有以下特征根與多項(xiàng)式系數(shù)的關(guān)系: 對于穩(wěn)定系統(tǒng),有:,44,例: 系統(tǒng)根軌跡與虛軸交于: 由
18、于閉環(huán)特征方程: 則: 因此:,45,可利用相角條件計(jì)算閉環(huán)極點(diǎn)位置 例:求閉環(huán)特種根的實(shí)部為-4時(shí)的值。,46,Rule 2:The gain at a selected root location sx The gain Kx at a selected root sx, x=1,2,n, on the locus is obtained by joining the point to all open-loop poles and zeros and measuring the length of each line . The gain is given by,47,Example 5
19、.8,48,At the breakaway point s=-2.6, Gain K is,49,Summary of the Root Locus plotting: Table 7.2 at page 424,50,5.3 Examples for drawing root locus,Example 5.9: Plot the root locus of the following system,51,Step 1: find the poles and zeros of GH and plot them.,52,Step 2: draw locus on the real axis,
20、53,Step 3: calculate asymptote angles and center of asymptotes:,Since n-m=1, there is only one asymptote at 180, corresponding to the real axis.,54,Step 4: Not necessary. Step 5: Determine the breakaway point on the real axis,55,56,Example 5.10:,Drawing root locus,57,5. The points crossing the imagi
21、nary axis,58,59,60,Example 5.11:,Drawing root locus,61,5. The points crossing the imaginary axis,62,63,64,Generalized root locus 1. Zero-degree root locus System contains inner positive-feedback-loop Or for K=0 - 2. Parameter root locus Equivalent unity feedback transform,65,1. Zero-degree root locu
22、s The close-loop characteristic equation:,magnitude requirement angle requirement,66,Rules for drawing zero-degree root locus,1. Number of root locus branches is the number of characteristic roots. 2. The symmetry of root locus. 3. The start point and end point. 4. The locus on the real axis has eve
23、n number of real poles and zeros on its right. 5. The Asymptotes of the Root Locus .,67,6. The breakaway point of the root locus 7. The angle of departure of the locus from complex poles and the angle of arrival at complex zeros. 8. The root locus crossing the imaginary axis.,68,Example 5.12: A posi
24、tive-feedback system:,Drawing the zero-degree root locus,69,70,Example 5.13: Feedback system,Drawing the parameter root locus.,2. Parameter root locus,71,72,73,5.4 Parameter Design by the Root Locus Method,The root locus method can be readily extended to the investigation of two or more parameters.,
25、Example 5.14: The closed-loop characteristic equation is with two parameters:,74,The effect of varying b is determined from the root locus equation:,First of all, let b=0, and therefore we evaluate the effect of varying a by the equation:,Rewritten as:,75,The sketch of the root locus is shown in the
26、 figure.,Let a=a1, the root locus equation of the closed-loop system becomes:,76,The root locus as b varies is shown in the figure.,77,As a be selected to different value, the root locus of b is changed as shown in the figure.,78,5.5 Relationship between Performance and the Distributing of Close-loop Zeros and Poles,2. If there is a closed-loop zero close to a pole, then the residue at this
溫馨提示
- 1. 本站所有資源如無特殊說明,都需要本地電腦安裝OFFICE2007和PDF閱讀器。圖紙軟件為CAD,CAXA,PROE,UG,SolidWorks等.壓縮文件請下載最新的WinRAR軟件解壓。
- 2. 本站的文檔不包含任何第三方提供的附件圖紙等,如果需要附件,請聯(lián)系上傳者。文件的所有權(quán)益歸上傳用戶所有。
- 3. 本站RAR壓縮包中若帶圖紙,網(wǎng)頁內(nèi)容里面會有圖紙預(yù)覽,若沒有圖紙預(yù)覽就沒有圖紙。
- 4. 未經(jīng)權(quán)益所有人同意不得將文件中的內(nèi)容挪作商業(yè)或盈利用途。
- 5. 人人文庫網(wǎng)僅提供信息存儲空間,僅對用戶上傳內(nèi)容的表現(xiàn)方式做保護(hù)處理,對用戶上傳分享的文檔內(nèi)容本身不做任何修改或編輯,并不能對任何下載內(nèi)容負(fù)責(zé)。
- 6. 下載文件中如有侵權(quán)或不適當(dāng)內(nèi)容,請與我們聯(lián)系,我們立即糾正。
- 7. 本站不保證下載資源的準(zhǔn)確性、安全性和完整性, 同時(shí)也不承擔(dān)用戶因使用這些下載資源對自己和他人造成任何形式的傷害或損失。
最新文檔
- 2025年榆林市住房公積金管理中心招聘備考題庫(19人)及答案詳解(考點(diǎn)梳理)
- 2025江蘇經(jīng)貿(mào)職業(yè)技術(shù)學(xué)院招聘26人備考題庫(第二批)及答案詳解(易錯(cuò)題)
- 2025年嘉興海寧市教育系統(tǒng)事業(yè)單位赴天津職業(yè)技術(shù)師范大學(xué)招聘教師4人備考題庫有答案詳解
- 2025中國農(nóng)業(yè)大學(xué)鄧愛華教授誠聘博士后備考題庫及答案詳解(易錯(cuò)題)
- 2026四川自貢醫(yī)元健康管理有限責(zé)任公司招聘工作人員11人備考題庫及答案詳解(易錯(cuò)題)
- 2026江西吉安市峽江縣選調(diào)峽江籍在外事業(yè)單位工作人員返鄉(xiāng)工作備考題庫及完整答案詳解1套
- 2026云南紅河州檢驗(yàn)檢測院招聘編外人員1人備考題庫及答案詳解(易錯(cuò)題)
- 2026江蘇泰州市興化市人才儲備中心招募見習(xí)人員備考題庫(第1號)含答案詳解
- 2025國航股份廣東分公司休息室就業(yè)見習(xí)崗招聘2人備考題庫及答案詳解1套
- 2026年1月云南麗江市消防救援支隊(duì)招錄政府專職消防員4人備考題庫及答案詳解1套
- 高速防滑防凍安全知識培訓(xùn)課件
- 電氣線路安全知識培訓(xùn)課件
- 瑞馬唑侖病例分享
- T-CSER-015-2023 場地環(huán)境信息地球物理探測技術(shù)指南
- 2025至2030中國背板連接器行業(yè)發(fā)展趨勢分析與未來投資戰(zhàn)略咨詢研究報(bào)告
- T/CCMA 0173-2023流動式起重機(jī)用高性能平衡閥
- GB/T 18910.103-2025液晶顯示器件第10-3部分:環(huán)境、耐久性和機(jī)械試驗(yàn)方法玻璃強(qiáng)度和可靠性
- 勵志類的美文欣賞范文(4篇)
- 廣東省廣州市白云區(qū)2024-2025學(xué)年六年級(上)期末語文試卷(有答案)
- GB/T 45166-2024無損檢測紅外熱成像檢測總則
- 山東省菏澤市東明縣2024-2025學(xué)年七年級上學(xué)期考試生物試題
評論
0/150
提交評論