第8讲现代控制系统分析设计

1.控制系统模型 ............................................................................................................................................. 3

例8_1 ...................................................................................................................................................... 3 例8_2 ...................................................................................................................................................... 4 例8_3 ...................................................................................................................................................... 6 例8_4 ...................................................................................................................................................... 8 例8_5 ...................................................................................................................................................... 9 例8_6 .....................................................................................................................................................10 例8_7 .....................................................................................................................................................11 2．控制系统的时域分析 .............................................................................................................................13

例8_8 .....................................................................................................................................................13 例8_9 .....................................................................................................................................................14 例8_10 ...................................................................................................................................................15 例8_11 ...................................................................................................................................................16 例8_12 ...................................................................................................................................................17 例8_13 ...................................................................................................................................................19 例8_14 ...................................................................................................................................................20 例8_15 ...................................................................................................................................................22 3．控制系统的根轨迹 .................................................................................................................................23

例8_16 ...................................................................................................................................................23 例8_17 ...................................................................................................................................................24 例8_18 ...................................................................................................................................................25 例8_19 ...................................................................................................................................................26 4．控制系统的频域分析 .....................................................................................................................28 例8_20 ...................................................................................................................................................28 例8_21 ...................................................................................................................................................30 例8_22 ...................................................................................................................................................31 例8_23 ...................................................................................................................................................32 例8_24 ...................................................................................................................................................33 例8_25 ...................................................................................................................................................34 例8_26 ...................................................................................................................................................35 例8_27 ...................................................................................................................................................37 例8_28 ...................................................................................................................................................38 例8_29 ...................................................................................................................................................40 例8_30 ...................................................................................................................................................42 例8_31 ...................................................................................................................................................43 例8_32 ...................................................................................................................................................45 例8_33 ...................................................................................................................................................47 例8_34 ...................................................................................................................................................49 例8_35 ...................................................................................................................................................50 例8_36 ...................................................................................................................................................51 例8_37 ...................................................................................................................................................52 例8_38 ...................................................................................................................................................53 例8_39 ...................................................................................................................................................54 例8_40 ...................................................................................................................................................56 例8_41 ...................................................................................................................................................57 例8_42 ...................................................................................................................................................58 6．最优控制器设计 .....................................................................................................................................59

例8_43 ...................................................................................................................................................59 例8_44 ...................................................................................................................................................60

例8_45 ...................................................................................................................................................61 例8_46 ...................................................................................................................................................62 例8_47 ...................................................................................................................................................64 例8_48 ...................................................................................................................................................66 例8_49 ...................................................................................................................................................68 例8_50 ...................................................................................................................................................70 例8_51 ...................................................................................................................................................71 例8_52 ...................................................................................................................................................73

1.控制系统模型 例8_1

设系统的零极点增益模型为：

H(s)?6(s?3)(s?1)(s?2)(s?5)

求系统的传递函数及状态空间模型。 解： k=6; z=[-3];

p=[-1,-2,-5];

[num,den]=zp2tf(z,p,k); [a,b,c,d]=zp2ss(z,p,k); printsys(num,den); printsys(a,b,c,d);

num/den =

6 s + 18

----------------------- s^3 + 8 s^2 + 17 s + 10 a =

x1 x2 x3 x1 -1.00000 0 0 x2 2.00000 -7.00000 -3.16228 x3 0 3.16228 0 b =

u1 x1 1.00000 x2 1.00000 x3 0 c =

x1 x2 x3 y1 0 0 1.89737 d =

u1 y1 0

例8_2

给定离散系统的状态空间方程，

?00???2.8?1.4?1???1.4??0?000??x(k)???u(k)?x(k?1)?? ???1.8?0.3?1.4?0.6??1??????000.60???0???y(k)??0001?x(k)?求传递函数模型和零极点模型，并判断其稳定性。 解：

a=[-2.8 -1.4 0 0; 1.4 0 0 0; -1.8 -0.3 -1.4 -0.6; 0 0 0.6 0;]; b=[1;0; 1; 0]; c=[0 0 0 1]; d=[0];

[num,den]=ss2tf(a,b,c,d); [z,p,k]=ss2zp(a,b,c,d) pzmap(p,z);

title('Pole-Zero Map');

z =

-0.5000 + 1.1358i -0.5000 - 1.1358i p =

-1.4000 + 0.0000i -1.4000 - 0.0000i -1.0606 -0.3394 k =

0. 6000

回目录

Pole-Zero Map1.510.5Imag Axis0-0.5-1-1.5-1.4-1.2-1-0.8-0.6Real Axis-0.4-0.20 回目录

8_3

已知两个系统

???x??01??0?1????1?2??x1???1?u1

?y1??13?x?1?u1???x??01???x?0?2?2?????1?31?u2

?y?1?4?x??2?2求按串联、并联、单位负反馈、单位正反馈连接时的系统状态方程。 解：

a1=[0,1;-1,-2]; b1=[0;1]; c1=[1,3]; d1=[1];

a2=[0,1;-1,-3]; b2=[0;1]; c2=[1,4]; d2=[0];

sys1=ss(a1,b1,c1,d1); sys2=ss(a2,b2,c2,d2);

title='Series state space matrix' sys=series(sys2,sys1)

title='parallel state space matrix'

[a,b,c,d]=parallel(a1,b1,c1,d1,a2,b2,c2,d2) title='Negative Feedback state space matrix' [a,b,c,d]=feedback(a1,b1,c1,d1,a2,b2,c2,d2) title='positive Feedback state space matrix' [a,b,c,d]=feedback(a1,b1,c1,d1,a2,b2,c2,d2,+1)

title =

Series state space matrix a =

x1 x2 x3 x4 x1 0 1 0 0 x2 -1 -2 1 4 x3 0 0 0 1 x4 0 0 -1 -3 b =

u1 x1 0 x2 0 x3 0 x4 1 c =

x1 x2 x3 x4

例 y1 1 3 1 4 d =

u1 y1 0

Continuous-time model. title =

parallel state space matrix a =

0 1 0 0 -1 -2 0 0 0 0 0 1 0 0 -1 -3 b = 0 1 0 1 c =

1 3 1 4 d = 1 title =

Negative Feedback state space matrix a =

0 1 0 0 -1 -2 -1 -4 0 0 0 1 1 3 -2 -7 b = 0 1 0 1 c =

1 3 -1 -4 d = 1 title =

positive Feedback state space matrix a =

0 1 0 0 -1 -2 1 4 0 0 0 1 1 3 0 1 b = 0 1 0 1 c =

1 3 1 4 d = 1 回目录

例8_4

时不变系统

???31??11??x?x???1?3??11?u????? ??11??y???x?1?1???判别系统的能控性和能观性。

解：

a=[-3,1;1,-3]; b=[1,1;1,1]; c=[1,1;1,-1]; cam=ctrb(a,b); rcam=rank(cam) oam=obsv(a,c); roam=rank(oam) rcam = 1 roam =

2

回目录

例8_5

线性系统

H(s)?当?分别取－1，0，＋1 时，判别系统的可控性和可观性，并求出相应的状态方程。 解：

for alph=[-1:1]; alph

num=[1,alph];

den=[1 10 27 18];

[a b c d]=tf2ss(num,den); cam=ctrb(a,b); rcam=rank(cam) oam=obsv(a,c); coam=rank(oam) end;

alph = -1 rcam = 3 coam = 3 alph = 0 rcam = 3 coam = 3 alph = 1 rcam = 3 coam = 2

回目录

s??

s3?10s2?27s?18

例8_6

利用Lyapunov方程确定线性时不变系统

?1??x1?2x2?x ???x2?x1?4x2的稳定性。 解：a??1?2????

1?4??令q=I，求解Lyapunov方程

ap?paT??q

然后确定p的正定性来证实系统的稳定性。

a=[-1,-2;1,-4]; q=[1,0;0,1]; p=lyap(a,q); detp=det(p)

detp =

0.0567

回目录

例8_7

对连续模型

H(s)?6(s?3)

(s?1)(s?2)(s?5)采用MATLAB提供的各种函数进行离散化处理，得到稍有差异的状态方程。 解：

disp('Contunue system'); k=6; z=[-3];

p=[-1,-2,-5];

[a,b,c,d]=zp2ss(z,p,k); %discrete system t=0.1;

disp('Discrete system --using c2d'); [a1,b1]=c2d(a,b,t)

disp('Discrete system--using c2dm with zoh'); [a2,b2,c2,d2]=c2dm(a,b,c,d,t,'zoh')

disp('Discrete system--using c2dm with foh'); [a3,b3,c3,d3]=c2dm(a,b,c,d,t,'foh')

disp('Discrete system--using c2dm with Tustin'); [a4,b4,c4,d4]=c2dm(a,b,c,d,t,'tustin')

Contunue system

Discrete system --using c2d a1 =

0.9048 0 0 0.1338 0.4651 -0.2237 0.0243 0.2237 0.9602 b1 =

0.0952 0.0784 0.0135

Discrete system--using c2dm with zoh a2 =

0.9048 0 0 0.1338 0.4651 -0.2237 0.0243 0.2237 0.9602 b2 =

0.0952 0.0784 0.0135 c2 =

0 0 1.8974 d2 = 0

Discrete system--using c2dm with foh a3 =

0.9048 0 0 0.1338 0.4651 -0.2237 0.0243 0.2237 0.9602 b3 =

0.0906

0.0611 0.0240 c3 =

0 0 1.8974 d3 =

0.0089

Discrete system--using c2dm with Tustin a4 =

0.9048 0 0 0.1385 0.4545 -0.2300 0.0219 0.2300 0.9636 b4 =

0.9524 0.7965 0.1259 c4 =

0.0021 0.0218 0.1863 d4 =

0.0119 回目录

2．控制系统的时域分析 例8_8

典型二阶系统

2?n H(s)?22s?2??ns??n其中?n为自然频率（无阻尼振荡频率），?为相对阻尼系数。试绘制出?n?6，?分别为

0.1,0.2,…,1.2,2.0时的单位阶跃响应。 解： wn=6;

kosi=[0.1:0.1:1.0,2.0]; figure(1); hold on;

for kos=kosi num=wn.^2;

den=[1,2*kos*wn,wn.^2]; step(num,den) end;

%title('Step response'); hold off;

Step ResponseFrom: U(1)1.81.61.41.2AmplitudeTo: Y(1)10.80.60.40.200246810Time (sec.) 回目录

例8_9

对典型二阶系统

2?n H(s)?22s?2??ns??n绘制出当??0.7，?n取2，4，6，8，10，12时的单位阶跃响应。

w=[2:2:12]; kos=0.7; figure(1); hold on; for wn=w

num=wn.^2;

den=[1,2*kos*wn,wn.^2]; step(num,den) end;

%title('Step response'); hold off;

Step ResponseFrom: U(1)1.41.21AmplitudeTo: Y(1)0.80.60.40.2000.511.522.533.54Time (sec.) 回目录

例8_10

求三阶系统

5(s2?5s?6)H(s)?3 2s?6s?10s?8的单位阶跃响应。 解：

num=[5 25 30]; den=[1 6 10 8]; figure

step(num,den);

title('Step response'); hold off; Step responseFrom: U(1)4.543.53AmplitudeTo: Y(1)2.521.510.500123456Time (sec.) 回目录

例8_11

求典型二阶系统

2?n H(s)?22s?2??ns??n当??0.7，?n?6时的单位冲激响应。

解： wn=6; kos=0.7; figure(1); num=wn.^2;

den=[1,2*kos*wn,wn.^2]; figure

impulse(num,den)

title('Impulse response');

Impulse responseFrom: U(1)32.52AmplitudeTo: Y(1)1.510.50-0.500.511.5Time (sec.) 回目录

例8_12

有高阶系统

????1.6?0.900??1????0.9000??0??x?????0.40.5?0.5?2.45??x????u????002.450??1? ??y??1111?x??0?求单位阶跃响应，单位冲激响应激零输入响应（设初始状态x0??1解：

a=[2.25,-5,-1.25,-0.5;2.25,-4.25,-1.25,-0.25; 0.25,-0.5,-1.25,-1;1.25,-1.75,-0.25,-0.75]; b=[4,6;2,4;2,2;0,2]; c=[0,0,0,1;0,2,0,2]; d=zeros(2,2); figure;

subplot(2,2,1); step(a,b,c,d);

axis([0,6,-0.5,2.5]); %title('Step Response'); subplot(2,2,2); impulse(a,b,c,d);

%title('Impulse response'); subplot(2,2,3);

%axis([0,6,-0.5,2.5]); x0=[1;1;1;-1];

initial(a,b,c,d,x0);

%title('Intial response'); subplot(2,2,4); pzmap(a,b,c,d);

%title('Pole-Zero Map') ;

1?1?T）1

Step ResponseFrom: U(1)2To: Y(1)Impulse ResponseFrom: U(1)2To: Y(1)From: U(2)From: U(2)Amplitude010500481204812Amplitude10-2200To: Y(2)To: Y(2)-200481204812Time (sec.)Initial Condition Results 1To: Y(1)Time (sec.)Pole zero map10.5Imag AxisAmplitude0-120-20510150-0.5-1-1.5To: Y(2)Time (sec.)-1-0.5Real Axis0

回目录

例8_13

多输入多输出系统

??5?1.25?2.25??2.25?4.25?1.25?x?????0.25?0.5?1.25???1.25?1.75?0.25???000y???020???0.5??46??24??0.25??x???u?1??22???? ?0.75??02?1?x2??求单位阶跃响应和单位冲激响应。

解：

a=[2.25,-5,-1.25,-0.5;2.25,-4.25,-1.25,-0.25; 0.25,-0.5,-1.25,-1;1.25,-1.75,-0.25,-0.75]; b=[4,6;2,4;2,2;0,2]; c=[0,0,0,1;0,2,0,2]; d=zeros(2,2); figure(1);

step(a,b,c,d); figure(2);

impulse(a,b,c,d); Impulse ResponseFrom: U(1)2From: U(2)To: Y(1)10Amplitude-11510To: Y(2)50-50481204812Time (sec.)回目录

例8_14

连续系统

?00???1.6?0.9?1???0.9??0?000??x???u????x?0.5?0.5?2.45??0.4?1? ?????002.450???0???y??1111?x?以t=0.5s采样周期，采用双线性变换算法转换为离散系统，然后求出离散系统的单位阶跃响应、单位冲激响应及零输入响应（设初始状态x0T??111?1?）。

解：

a1=[-1.6,-0.9,0,0;0.9,0,0,0;

0.4,0.5,-5.0,-2.45;0,0,2.45,0]; b1=[1;0;1;0]; c1=[1,1,1,1]; d1=[0]; t=0.5;

[a,b,c,d]=c2dm(a1,b1,c1,d1,t,'tustin'); figure(1);

subplot(2,2,1); dstep(a,b,c,d); subplot(2,2,2); dimpulse(a,b,c,d); subplot(2,2,3); x0=[1;1;1;-1];

dinitial(a,b,c,d,x0); axis([0 6 0.5 2.5]); subplot(2,2,4);

[z,p,k]=ss2zp(a,b,c,d,1); zplane(z,p);

Step ResponseFrom: U(1)20.6Impulse ResponseFrom: U(1)AmplitudeAmplitude1.50.4To: Y(1)10.50To: Y(1)0.20-0.205101520250369121518Time (sec.)Initial Condition Results 1Time (sec.)1Imaginary Part0102030Amplitude0.50-0.5-1-1-0.500.5Real Part1To: Y(1)0.50-0.5Time (sec.) 回目录

例8_15

离散二阶系统

H(z)?0.632 2z?1.368z?0.568求当输入幅值为±1的方波信号时系统的输出响应。 解：

num=0.632;

den=[1 ,-1.368,0.568];

u1=[ones(1,50),-1*ones(1,50)]; u=[u1,u1,u1]; figure(1);

dlsim(num,den,u);

%title('Discrete System Simulation');

Step ResponseFrom: U(1)20.6Impulse ResponseFrom: U(1)AmplitudeAmplitude1.50.4To: Y(1)10.50To: Y(1)0.20-0.205101520250369121518Time (sec.)Initial Condition Results 1Time (sec.)Linear Simulation Results 100AmplitudeTo: Y(1)To: Y(1)0.5Amplitude500-50-1000-0.501020300100200Time (sec.)Time (sec.)回目录 3．控制系统的根轨迹 例8_16

设开环系统

H(s)?k(3s?1)

s(2s?1)绘制出通过单位负反馈构成的闭环系统的根轨迹。 解：

num=[3 1]; den=[2 1 0];

rlocus(num,den);

title('Root locus');

Root locus10.80.60.40.2Imag Axis0-0.2-0.4-0.6-0.8-1-1-0.8-0.6-0.4-0.200.2Real Axis0.40.60.81 回目录

例8_17

设开环系统

H(s)?k(s?5)

s(s?2)(s?3)绘制出闭环系统的根轨迹，并确定交点处的增益k。 解：

num=[1 5];

den=[1 5 6 0]; rlocus(num,den);

%title('Root locus'); [k,p]=rlocfind(num,den) gtext('k=0.5'); 108642Imag Axis0-2-4-6-8-10-6-5-4-3-2Real Axis-1012回目录

例8_18

已知开环系统传递函数

H(s)?k

s4?16s3?36s2?80s绘制出闭环系统的根轨迹。 解： num=[1];

den=[1 16 36 80 0]; rlocus(num,den);

%title('Root locus'); 15105Imag Axis0-5-10-15-15-10-5Real Axis05 回目录

例8_19

已知开环系统传递函数

H(s)?k(s?2) 22(s?4s?3)解：

num=[1 2]; den1=[1 4 3];

den=conv(den1,den1); figure(1);

rlocus(num,den);

%title('Root locus'); [k,p]=rlocfind(num,den) %Checking the stability k=55;

figure(2); num1=k*[1 2]; den=[1 4 3];

den1=conv(den,den);

[num,den]=cloop(num1,den1,-1); impulse(num,den);

%title('Impulse Response (k=55)'); %Checking the stability figure(3); k=k+1; k=56;

num1=k*[1 2]; den=[1 4 3];

den1=conv(den,den);

[num,den]=cloop(num1,den1,-1); impulse(num,den);

%title('Impulse Response (k=56)');

Error in ==> D:\\MATLABR11\\toolbox\\control\\rlocfind.m

On line 58 ==> [k,poles] = rlocfind(tf(a,b),varargin{:});

642Imag Axis0-2-4-6-4-3-2-1Real Axis012回目录 4．控制系统的频域分析 例8_20

典型二阶系统

2?n H(s)?22s?2??ns??n绘制出?取不同值时的Bode图。

wn=6;

kosi=[0.1:0.1:1.0]; w=logspace(-1,1,100); figure(1); num=[wn.^2]; for kos=kosi

den=[1 2*kos*wn wn.^2];

[mag,pha,w1]=bode(num,den,w); subplot(2,1,1); hold on;

semilogx(w1,mag); subplot(2,1,2); hold on;

semilogx(w1,pha); end;

subplot(2,1,1); grid on;

xlabel('Frequency (rad/sec)'); ylabel('Gain dB'); subplot(2,1,2); grid on;

xlabel('Frequency (rad/sec)'); ylabel('Phase deg'); hold off;

6Gain dB4200123456Frequency (rad/sec)789100-50-100-150-200Phase deg0123456Frequency (rad/sec)78910 回目录

例8_21

有系统

H(s)?绘制出系统的Bode图。 k=100; z=[-4];

p=[0 -0.5 -50 -50];

[num,den]=zp2tf(z,p,k); %subplot(2,1,1); grid on;

%subplot(2,1,2); %grid on;

bode(num,den);

title('Bode plot'); 6100(s?4)s(s?0.5)(s?50)2

Gain dB4200123456Frequency (rad/sec)Bode plot78910Magnitude (dB)Phase (deg)2000-200-90-180-270-101210101010Frequency (Hz)回目录

例8_22

系统传递函数模型为

H(s)?s?1?0.5se 3(s?2)求出有理传递函数的频率响应，然后在同一张图上绘出以四阶pade近似表示的系统频率响应。 num=[1 1];

den=conv([1 2],conv([1,2],[1,2])); w=logspace(-1,2); t=0.5;

[m1,p1]=bode(num,den,w); p1=p1-t*w'*180/pi; [n2,d2]=pade(t,4); numt=conv(n2,num); dent=conv(den,d2);

[m2,p2]=bode(numt,dent,w); subplot(2,1,1);

semilogx(w,20*log10(m1),w,20*log10(m2),'r--'); subplot(2,1,2);

semilogx(w,p1,w,p2,'r--'); subplot(2,1,1); grid on;

title('Bode plot');

xlabel('Frequency (rad/sec)'); ylabel('Gain dB'); subplot(2,1,2); grid on;

xlabel('Frequency (rad/sec)');

ylabel('Phase deg'); Bode plot0-20Gain dB-40-60-80-100-110010010Frequency (rad/sec)1102Phase deg-1000-2000-3000-4000-11010010Frequency (rad/sec)1102 回目录

例8_23

离散二阶系统

H(z)?0.632

z2?1.368z?0.568求系统的离散Bode图 num=0.632;

den=[1 ,-1.368,0.568]; grid on;

dbode(num,den,0.1);

title('Discrete Bode Plot'); Discrete Bode PlotFrom: U(1)2010Phase (deg); Magnitude (dB)0-10-201000To: Y(1)-100-200-300-400110Frequency (rad/sec) 回目录

例8_24

开环系统

H(s)?50

(s?5)(s?2)绘制出系统的Nyquist曲线，并判别系统的稳定性，最后求出闭环系统的单位冲激响应。 k=50; z=[];

p=[-1,-5,2];

[num,den]=zp2tf(z,p,k); figure(1);

nyquist(num,den);

title('Nyquist Plot'); figure(2);

[num1,den1]=cloop(num,den); impulse(num1,den1);

title('Impulse Response');

Impulse ResponseFrom: U(1)1050-5AmplitudeTo: Y(1)-10-15-20-25-30-3500.511.522.5Time (sec.) 回目录

例8_25

开环系统

H(s)?50

(s?1)(s?5)(s?2)绘制出系统的Nyquist曲线，并判别系统的稳定性，最后求出闭环系统的单位冲激响应。 k=50; z=[];

p=[-1,-5,2];

[num,den]=zp2tf(z,p,k); figure(1);

nyquist(num,den);

title('Nyquist Plot'); figure(2);

[num1,den1]=cloop(num,den); impulse(num1,den1);

title('Impulse Response'); Impulse ResponseFrom: U(1)1050-5AmplitudeTo: Y(1)-10-15-20-25-30-3500.511.522.5Time (sec.) 回目录

例8_26

线性系统状态空间模型

?100??0?0???0??0?011??x???u????x?001??0?0? ??????62.5?213.8?204.2?54???1???y??1562187500x??绘制出系统的Bode图和Nyquist曲线，判断系统的稳定性，并绘制闭环系统的单位冲激响应进行验证。

a=[0,1,0,0;0,0,1,0;0,0,0,1;-62.5,-213.8,-204.2,-54]; b=[0,0,0,1]';

c=[1562,1875,0,0]; d=0;

w=logspace(-1,1); figure(1);

bode(a,b,c,d,1,w); title('Bode Plot'); figure(2);

nyquist(a,b,c,d,1,w); title('Nyquist plot'); [z,p,k]=ss2zp(a,b,c,d); p

figure(3);

[a1,b1,c1,d1]=cloop(a,b,c,d); impulse(a1,b1,c1,d1);

title('Impulse Response');

p =

-50.0011 -2.4969 -1.0027 -0.4992

Impulse ResponseFrom: U(1)5432AmplitudeTo: Y(1)10-1-2-300.511.522.533.544.55Time (sec.) 回目录

例8_27

离散系统

H(z)?0.632 2z?1.368z?0.568绘制出系统的Nyquist曲线，判别闭环系统的稳定性，并绘制出闭环系统的单位冲激响应。 num=0.632;

den=[1 ,-1.368,0.568]; [z,p,k]=tf2zp(num,den); p

dnyquist(num,den,0.1);

title('Discrete Nyquist Plot'); figure(2)

[num1,den1]=cloop(num,den); dimpulse(num1,den1);

title('Discrete Impulse Response');

p =

0.6840 + 0.3165i 0.6840 - 0.3165i Discrete Impulse ResponseFrom: U(1)8642AmplitudeTo: Y(1)0-2-4-6-8051015202530Time (sec.)

回目录

例8_28

一多环系统，

G(s)?r(s)16.7s (0.85s?1)(0.25s?1)(0.0625s?1)y(s)10GainG(s)其结构如图，试用Nyquist曲线判断系统的稳定性。

k1=16.7/0.0125; z1=[0];

p1=[-1.25 -4 -16];

[num1,den1]=zp2tf(z1,p1,k1); [num,den]=cloop(num1,den1); [z,p,k]=tf2zp(num,den); p

figure(1);

nyquist(num,den);

title('Nyquist Plot'); figure(2)

[num2,den2]=cloop(num,den); impulse(num2,den2);

title('Impulse Response');

p =

-10.5969 +36.2148i -10.5969 -36.2148i -0.0562

Impulse ResponseFrom: U(1)201510AmplitudeTo: Y(1)50-5-10-1500.10.20.30.40.50.6Time (sec.) 回目录

例8_29

非线性系统

??ax?bu?x??u?f(?) ???cx?其中

??2.1?1.87?0.2030.894??1??1.0?0?000??b???，c??00?1?2? a??100??0?0?????0010???0?k1??f(?)?k2?

当?k1,k2?的（1/2,2/3）或[1,2]时，判断非线性系统的绝对稳定性。

a=[-2.1,-1.87,-0.203,0.894;1,0,0,0;0,1,0,0;0,0,1,0]; b=[1,0,0,0]'; c=[0,0,-1,-2]; d=0;

[num,den]=ss2tf(a,b,c,d); num=-1*num;

[z,p,k]=tf2ss(num,den); p

figure(1);

nyquist(num,den);

title('Nyquist Plot');

p = 1 0 0 0

Nyquist PlotFrom: U(1)0.40.30.2Imaginary Axis0.1To: Y(1)0-0.1-0.2-0.3-0.4-2.5-2-1.5-1-0.50Real Axis 回目录

4． 极点配置和观测器设计

例8_30

系统开环传递函数为

G(s)?16.7s

(0.8s?1)(0.25s?1)(0.0625s?1)绘制出Nickols图。 k1=16.7/0.0125; z1=[0];

p1=[-1.25 -4 -16];

[num,den]=zp2tf(z1,p1,k1); figure(1);

nichols(num,den);

axis([-360,0,-40,40]); ngrid('New');

title('Nickols Plot'); Nickols PlotFrom: U(1)40200Open-Loop Gain (dB)-20To: Y(1)-40-60-80-100-200-150-100-50050100Open-Loop Phase (deg) 回目录

例8_31

被控对象

H(s)?10

(s?1)(s?2)(s?3)设计反馈控制器u=-kx，使闭环系统的极点为?1??2?j23，?2??2?j23，

?3?10。

%Pole Placement--using transformation matrix %

disp('Pole placement--using transformation matrix'); a=[0,1,0;0,0,1;-6,-11,-6]; b=[0;0;10]; %Step1

cam=ctrb(a,b);

disp('The rank of controllability matrix'); rc=rank(cam); %Step2

beta=poly(a); %Step3

a1=beta(2); a2=beta(3); a3=beta(4);

w=[a2,a1,1;a1,1,0;1,0,0]; t=cam*w; %Step4

j=[-2+2*sqrt(3)*i, 0, 0; 0,-2-2*sqrt(3)*i,0; 0,0,-10]; alph=poly(j); aa1=alph(2); aa2=alph(3); aa3=alph(4); %Step 5

k=[aa3-a3,aa2-a2,aa1-a1]*(inv(t)) %

%Pole Placement ---using Ackermann's formula %

disp('Pole Placement ---using Ackermann''s formula'); a=[0,1,0;0,0,1;-6,-11,-6]; b=[0;0;10]; %Step1

cam=ctrb(a,b);

disp('The rank of controllability matrix'); rc=rank(cam); %Step2

j=[-2+2*sqrt(3)*i, 0, 0; 0,-2-2*sqrt(3)*i,0; 0,0,-10]; alph=poly(j); %Step3

phi=polyvalm(alph,a); %step 4

k=[0,0,1]*(inv(cam))*phi %

%Pole Placement --using place function in Matlab

%

disp('Pole Placement --using place function in Matlab'); p=[-2+2*sqrt(3)*i,-2-2*sqrt(3)*i,-10]; k=place(a,b,p)

Pole placement--using transformation matrix The rank of controllability matrix k =

15.4000 4.5000 0.8000

Pole Placement ---using Ackermann's formula The rank of controllability matrix k =

15.4000 4.5000 0.8000

Pole Placement --using place function in Matlab k =

15.4000 4.5000 0.8000 回目录

例8_32

含积分环节的类型1伺服系统，设对象为

H(s)?1

s(s?1)(s?2)设计控制器u=-kx+k1r，使闭环系统具有极点?2?j23，－10。

%Pole Placement--using transformation matrix %

disp('Pole placement--using transformation matrix');

disp('Pole Placement --using place function in Matlab'); a=[0,1,0;0,0,1;0,-2,-3]; b=[0;0;1]; c=[1 0 0]; d=[0];

disp('The rank of controllability matrix'); rc=rank(ctrb(a,b))

p=[-2+2*sqrt(3)*i,-2-2*sqrt(3)*i,-10]; k=place(a,b,p) a1=a-b*k; b1=b*k(1); c1=c; d1=d;

figure(1);

step(a1,b1,c1,d1);

title('Step Response of Designed Servoi System'); figure(2);

[y,x,t]=step(a1,b1,c1,d1); plot(t,x);

title('Step Response curves for x1,x1,x3'); grid on;

Pole placement--using transformation matrix

Pole Placement --using place function in Matlab The rank of controllability matrix rc = 3

place: ndigits= 15 k =

160.0000 54.0000 11.0000

Step Response curves for x1,x1,x31050-500.511.522.53 回目录

例8_33

不含积分环节的类型1伺服系统设计。对车载倒立摆系统实现控制，使小车位置作为输出的闭环系统具有极点：?1?j3，－5，－5，－5，－5。 根据牛顿第二定律，可得线性化模型：

???(M?m)g??u?Ml? ???Mx?u?mg??

Tsign of inverted pendulum control system %

% Using pole placement Ackermann %

%Modelling

a=[0,1,0,0;20.601,0,0,0;0,0,0,1;-0.4905,0,0,0]; b=[0;-1;0;0.5]; c=[0,0,1,0]; d=[0];

a1=[a,zeros(4,1);-c,0]; b1=[b;0];

%Check the controllability

disp('The rank of controllability matrix'); rc=rank(ctrb(a1,b1))

p=[-1+sqrt(3)*i,-1-sqrt(3)*i,-5,-5,-5]; k=acker(a1,b1,p) %Step Response k1=k(1:4); ki=-k(5);

ac=[a-b*k1,b*ki;-c,0]; bc=[0,0,0,0,1]'; cc=[c,0]; dc=[0]; figure(1);

v=[0,5,-0.4,1.4]; step(ac,bc,cc,dc); axis(v);

title('Step Response of inverted pendulum System'); xlabel('Sec');

ylabel('Output y=x3'); figure(2);

[y,x,t]=step(ac,bc,cc,dc); plot(t,x);

legend('x1','x2','x3','x4','x5');

title('Step Response curves for x1,x1,x3,x4,x5'); xlabel('Sec');

ylabel('x1,x2,x3,x4,x5'); grid on;

The rank of controllability matrix rc = 5 k =

-157.6336 -35.3733 -56.0652 -36.7466 50.9684

Step Response curves for x1,x1,x3,x4,x51.41.210.8x1,x2,x3,x4,x5x1x2x3x4x50.60.40.20-0.2-0.40123Sec456 回目录

例8_34

开环系统

??ax?bu?x ?y?cx?其中

?0??0?，c??100? ????1??设计全阶状态观测器，使观测器得闭环极点为?2?j23，－5。

Tsign of full-order state observer %

a=[0,1,0;0,0,1;-6,-11,-6]; b=[0;0;1]; c=[1,0,0]; d=[0];

%Check the controllability

disp('The rank of observability matrix'); rc=rank(obsv(a,c))

% Constrt the dual system (a1,b1,c1) a1=a'; b1=c'; c1=b';

%Silve K using poles placement -Ackermann p=[-2+2*sqrt(3)*i,-2-2*sqrt(3)*i,-5]; k=acker(a1,b1,p);

%Obtain the state of obser gain ke

ke=k'

The rank of observability matrix rc = 3 ke =

3.0000 7.0000 -1.0000 回目录

10??0a??001?，b??????6?11?6??

例8_35

被控对象

1??0?0?x???x??1?u

20.60????y??10?x设计调节器使闭环极点为?1.8?j2.4，而起状态不可测，因此设计状态观测器使其闭环极点

为－8，－8。

Tsign of regulator and state observer %

a=[0,1;20.6,0]; b=[0;1]; c=[1,0]; d=[0];

%Check the controllability and observability disp('The rank of controllability matrix'); rc=rank(ctrb(a,b))

disp('The rank of observability matrix'); ro=rank(obsv(a,c)) Tsign the regulator

p=[-1.8+2.4*i,-1.8-2.4*i]; k=acker(a,b,p)

Tsign the state observer a1=a';b1=c';c1=b'; p1=[-8,-8];

k1=acker(a1,b1,p1);

ke=k1'

The rank of controllability matrix rc = 2

The rank of observability matrix ro = 2 k =

29.6000 3.6000 ke =

16.0000

84.6000 回目录

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