1.25 ------------------------- s^2 + 52.36s + 1250.00，这是一个双轴步兵车云台的传递函数，你能给我它的PID控制器matlab代码吗

当你使用PID控制器时，需要先将传递函数转换成标准形式，为此，我们可以使用MATLAB中的tf函数。然后，我们可以使用MATLAB中的pidtune函数进行控制器参数的自动调整。以下是示例代码：

% 双轴步兵车云台传递函数
sys = tf([1.25], [1, 52.36, 1250]);

% PID控制器
C = pidtune(sys, 'pid');

% 反馈系统
T = feedback(C * sys, 1);

% 仿真
t = 0:0.01:10;
step(T, t);

在以上代码中，我们使用pidtune函数来自动调整PID控制器的参数。最后，我们使用feedback函数构建反馈系统，并使用step函数进行仿真。你可以根据需要进行调整。

相关问题

用lingo写代码maxz=1.15x4A+1.40 x2c +1.25 x3B +1.06 x5D (xiA,XiB,Xic,XiD ≥ 0( i = 1, 2, 3, 4, 5) X3B ≤ 40000 X2c ≤ 30000 s,ti -1.15x3A-1.06 x4D + X5D=0 -1.15x2a-1.06 xD +X4A + X4D=0 -1.15x1A-1.06 x2D+X3A+X3B+X3D=0 -1.06x1D+X2A+X2c+X2D=0(X1A+X1D=100000

on startMovie
global maxz,x1A,x2c,x3B,x4A,x4D,x5D,x1D,x2A,x2D,x3A,x3D

-- set initial values
maxz = 0
x1A = 100000
x2c = 0
x3B = 0
x4A = 0
x4D = 0
x5D = 0
x1D = 100000
x2A = 0
x2D = 0
x3A = 0
x3D = 0

-- create simplex table
set simplexTable to [[1.15,0,0,1.06,-1,0,0,0,0,0,0],
[0,1.40,0,0,0,1,-1,0,0,0,0],
[0,0,1.25,0,0,0,0,0,-1,0,0],
[-1.15,0,0,-1.06,0,0,0,0,0,1,0],
[-1.15,0,0,0,0,0,0,-1,-1,0,1],
[0,0,0,0,0,0,1,0,0,0,0]]

-- set constraints
set constraints to [x1A,x2c,x3B,x4A,x4D,x5D,x1D,x2A,x2D,x3A,x3D]

-- while there are negative coefficients in the objective function row
repeat while oneOrMore(count(if(x<0,x,0)) in simplexTable[1]) do

-- find entering variable and pivot column
set minColValue to 9999999
repeat with i = 1 to count(simplexTable[1])
if simplexTable[1][i] < 0 then
set colValue to -1 * simplexTable[1][i] / simplexTable[i+1][oneOrMore(count(if(x>0,1,0)) in simplexTable[i+1])]
if colValue < minColValue then
set minColValue to colValue
set pivotCol to i
end if
end if
end repeat

-- find exiting variable and pivot row
set minRowValue to 9999999
repeat with i = 2 to count(simplexTable)
if simplexTable[i][pivotCol] > 0 then
set rowValue to simplexTable[i][1] / simplexTable[i][pivotCol]
if rowValue < minRowValue then
set minRowValue to rowValue
set pivotRow to i
end if
end if
end repeat

-- perform pivot operation
set pivotValue to simplexTable[pivotRow][pivotCol]
repeat with i = 1 to count(simplexTable[pivotRow])
set simplexTable[pivotRow][i] to simplexTable[pivotRow][i] / pivotValue
end repeat
repeat with i = 1 to count(simplexTable)
if i != pivotRow then
set rowMultiplier to simplexTable[i][pivotCol]
repeat with j = 1 to count(simplexTable[i])
set simplexTable[i][j] to simplexTable[i][j] - rowMultiplier * simplexTable[pivotRow][j]
end repeat
end if
end repeat

-- update solution
set constraints[pivotRow-1] to pivotCol
end repeat

-- set final values
set maxz to simplexTable[1][1] * -1
set x1A to constraints[0]
set x2c to constraints[1]
set x3B to constraints[2]
set x4A to constraints[3]
set x4D to constraints[4]
set x5D to constraints[5]
set x1D to constraints[6]
set x2A to constraints[7]
set x2D to constraints[8]
set x3A to constraints[9]
set x3D to constraints[10]

end on

用matlab写代码maxz=1.15x4A+1.40 x2c +1.25 x3B +1.06 x5D (xiA,XiB,Xic,XiD ≥ 0( i = 1, 2, 3, 4, 5) X3B ≤ 40000 X2c ≤ 30000 s,ti -1.15x3A-1.06 x4D + X5D=0 -1.15x2a-1.06 xD +X4A + X4D=0 -1.15x1A-1.06 x2D+X3A+X3B+X3D=0 -1.06x1D+X2A+X2c+X2D=0(X1A+X1D=100000

x = optimvar('x',5,'LowerBound',0);
maxz = 1.15*x(1) + 1.40*x(2) + 1.25*x(3) + 1.06*x(5);
s = optimproblem('Objective',maxz);
s.constraints.c1 = x(3) <= 40000;
s.constraints.c2 = x(2) <= 30000;
s.constraints.c3 = -1.15*x(3) - 1.06*x(4) + x(5) == 0;
s.constraints.c4 = -1.15*x(2) - 1.06*x(4) + x(1) == 0;
s.constraints.c5 = -1.15*x(1) - 1.06*x(2) + x(3) + x(4) == 0;
s.constraints.c6 = -1.06*x(5) + x(1) + x(2) == 100000;
[xsol,fval,exitflag,output] = solve(s);
disp(xsol.x); % display the solution of x
disp(fval); % display the optimal objective value