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1、ch8 stability in the frequency domainmain content mapping contours in the s-plane the nyquist criterion relative stability and nyquist criterion closed-loop frequency response system bandwidth examples and simulation summaryintroductionfeedback control systemhow to determine the stability ? roots ev

2、aluation routh-hurwitz criterion (absolute/relative stability) root locus nyquist criterion ( absolute/relative stability) 8.1 mapping contours in the s-plane contour map conformal mapping s-plane f(s)-planea contour mapping that retains the angles on the s-plane on the f(s)-plane figure 1 mapping f

3、(s)=2s+1figure 2 mapping for f(s)=s/(s+2) figure 3 mapping for f(s)=s/(s+1/2) principle of the argument if a contour in the s-plane encircles z zeros and p poles of f(s) and does not pass through any poles or zeros of f(s) and the traversal is in the clockwise direction along the contour, the corres

4、ponding contour in f(s)-plane encircles the origin of f(s)-plane n=z-p times in the clockwise direction. sfthe rules of contour clockwise traversal of a contour is assumed to be positive. encirclement is defined by the rule : “ clockwise and eyes right ” or “ counterclockwise and eyes left ” figure

5、4 evaluation of the net angle of f(s)-plane2*pi*n=2*pi*z-2*pi*p figure 5 evaluation of the net angle of f(s)-planen=z-p=3-1=2 figure 6 evaluation of the net angle of f(s)-planen=z-p=0-1=-1 8.2 the nyquist criterion figure 7 single-loop feedback control systemthe closed-loop characteristic equation:0

6、)()()()(1)(11mkkniisssskshsgsffigure 8 nyquist contour in s-planenpzthe number of zeros of f(s) within the contour is : snyquist stability criterion a feedback system is stable if and only if the contour in the l(s)-plane does not encircle the (-1,0) point when the number of poles of l(s) in the rig

7、ht-hand s-plane is zero (p=0) a feedback system is stable if and only if ,for the contour , the number of counterclockwise encirclements of the (-1,0) point is equal to the number of poles of l(s) in the right-hand s-plane.application of nyquist criterion example 1: system with 2 real poles example

8、2: system with a pole at origin example 3: system with 3 poles example 4: system with 2 poles at origin example 5: system with a pole in rhp example 6: system with a zero in rhp refer to p506-515figure 9 nyquist contour and mapping for gh(s)example 1example 2figure 10 nyquist contour and mapping for

9、 gh(s)figure 11 nyquist contour and mapping for gh(s)example 3figure 12 nyquist contour for gh(s) when (a) k=1, (b) k=2, (c) k=3example 4figure 13 nyquist contour for gh(s) example 6figure 10 nyquist diagram and mapping for gh(j w)/k稳定性分析的补充举例稳定性分析的补充举例 开环传递函数不含积分环节开环传递函数不含积分环节例例1 1：系统开环传递函数为系统开环传递函

10、数为试判定闭环系统的稳定性。试判定闭环系统的稳定性。1)()(tskshsg0p0n0npz闭环系统稳定闭环系统稳定例例2 2：系统开环传递函数为系统开环传递函数为试判定闭环系统的稳定性。试判定闭环系统的稳定性。) 1)(1()()(21ststkshsg0p0n0npz闭环系统稳定闭环系统稳定例例3 3：系统开环传递函数为系统开环传递函数为试判定闭环系统的稳定性。试判定闭环系统的稳定性。)5)(2)(1(200)()(sssshsg0p2n2npz闭环系统不稳定闭环系统不稳定-1.59-1v开环传递函数含积分环节开环传递函数含积分环节需对开环幅相曲线作修正：需对开环幅相曲线作修正：从从=0=0+ +处，逆时针补画处，逆时针补画9090半径为无穷大的圆弧。半径为无穷大的圆弧。例例4 4：系统开环传递函数为系统开环传递函数为试判定闭环系统的稳定性。试判定闭环系统的稳定性。)4)(2(200)()(sssshsg2npz闭环系统不稳定闭环系统不稳定例例5 5：系统开环传递函数为系统开环传递函数为 试判定闭环系统的稳定性。试判定闭环系统的稳定性。) 1()()(2tsskshsg0p2