第一部分 控制工程仿真实例
引言—控制工程建模与分析方法
eÅýñ^) Ø1î eÅ"dòÕýñLCÙ5¥8†Ÿýñ/Œ b “1î e
Å"d41 p aÉ›!9 a/ a› as a_©Vñ b
ñ ôRP¨ Ô×E„ÿ bWE b Ø‚„) ØZE # ZëVLŸ eÅžd
L? e?V?M e?? M e?V ??"d e?? ? ?"d e?V??"d
eÅžÛ"d eÅ© b eÅýñ¥‹¨S¶* ùö1^ý<3ÁVñ ÂÄý aÈ0 a
8Á aÈ aó "d„p aÏ’©ª Ÿ ”Zž <5 Ø aìg?à aŒY5Å a
3þ eÅ a öÜ6?é5× b
»B†sÊ |/wî eÅ aF-14¤ à eÅ a ÚÇO"d eÅ„Ï’(ÿ eÅ
? ???MATLAB Â… eÅýñÉ›!9Ð_‡ b ÏMATLAB4? e??
ñý Q ?¹ eÅýñ_‡4y „s4µ 復ûyN ºL è -n5
É›º b
eÅýñý Q¿vÿ ? Vs¹†s y „s b
一、建模
¹ eÅ"dÉ›!9Ð_‡n5³1y ë eÅ"d¥ ˜ eÅýñý Q4
¹Õy ZT b
1．状态方程形式
?Z??T ?T A1îU¾Õ™T MLŸ"dÏP¨ L¿ eÅ
!9 b
d
d
x
AxBu
t
yCxDu
A1
T?
x^?Mu?ysY?{ ??{M ,,ABC?D µa’¥»” b
eÅýñý QÏy ëÿZñ¥ 7^ ss/ñ½óBñy L è b
R= 2.0; % Ohms
L= 0.5 %; Henrys
MATLAB 工程仿真与应用 30 例
2
Km = .015; % torque constant
Kb = .015; % emf constant
Kf = 0.2; % Nms
J= 0.02; % kg.m^2
A = [-R/L -Kb/L; Km/J -Kf/J];
B = [1/L; 0];
C = [0 1];
D = [0];
sys_dc = ss(A,B,C,D)
MATLAB 7›{ Æñ½ € V çn4/¥ ¥€'÷ÏsžM‹ñ
?/]i????
a =
x1 x2
x1 -4 -0.03
x2 0.75 -10
b =
u1
x1 2
x2 0
c =
x1 x2
y1 0 1
d =
u1
y1 0
2．传递函数形式
.?f?? ?T A2îU b.¾f” †M"dÏP¨ ^ÜÅ eÅö1
¨¥™T b
2
2
10
s
Hs
ss
A2
T?
s¹ ˜ ô ˜®MÐØ01¿ çl V•n¯Mf”M1€! b
eÅýñý QÏy ëÿZñ¥ 7^ tf MATLAB 7??{ ?
sys_tf = tf(sys_dc) % sys_dc ^B???y ???Z? ?
' V??
Transfer functio
《MATLAB工程仿真与应用30例》.pdf