Generalized modal control example
In the article Generalized Modal Control , the author introduced the ideas and algorithms of generalized modal control. This article will give a simple example of generalized modal control to deepen understanding.
1. Original system
Take the transfer function of the original open-loop system as
G ( s ) = 2 15 s 2 + s G ( {\rm s} ) = \frac{2}{15 {\rm s}^2 + {\rm s} }G(s)=15s2+s2It can be seen that it has a zero pole and is critically stable. When the external input is a step signal, the system will directly become an unstable system. Therefore, we might as well look directly at its closed-loop unit negative feedback system W ( s ) W ( {\rm s} )W(s)：
W ( s ) = G ( s ) 1 + G ( s ) = 2 15 s 2 + s + 2 W ( {\rm s} ) = \frac{G ({\rm s}) }{1 + G( {\rm s}) } = \frac{2}{15 {\rm s}^2 + {\rm s} + 2} W(s)=1+G(s)G(s)=15s2+s+22Its step response is:
It can be seen that the original closed-loop system tends to be stable, but its shortcomings are obvious: too much oscillation, very large overshoot, too long adjustment time, and obvious resonance of the entire system , which obviously cannot be applied in practice.
2. Regulator design
The original system is a 2-order system, then the lowest order of the regulator should be ν = n − 1 = 1 \nu = n-1 = 1n=n−1=Level 1 . Let's just design a first-order regulator:
R ( s ) = Q ( s ) P ( s ) = q 1 s + q 0 s + p 0 R({\rm s}) = \frac{Q({\rm s })} {P({\rm s})} = \frac{q_1 {\rm s} + q_0}{
{\rm s} + p_0}R(s)=P(s)Q(s)=s+p0q1s+q0After passing the regulator into the feedforward path, the closed-loop transfer function of the system obtained is
W c ( s ) = R ( s ) G ( s ) 1 + R ( s ) G ( s ) = q 1 s + q 0 s + p 0 2 15 s 2 + s 1 + q 1 s + q 0 s + p 0 2 15 s 2 + s = 2 ( q 1 s + q 0 ) s ( 15 s + 1 ) ( s + p 0 ) + 2 ( q 1 s + q 0 ) = 2 ( q 1 s + q 0 ) 15 s 3 + ( 15 p 0 + 1 ) s 2 + ( p 0 + 2 q 1 ) s + 2 q 0 \begin {aligned} W_c({\rm s}) &= \frac{R({\rm s}) G ({\rm s}) }{1 + R({\rm s}) G({\rm s })} \\ &= \frac{ \frac{q_1 {\rm s} + q_0}{ {
\rm s} + p_0} \frac{2}{15 {\rm s}^2 + {\rm s } } }{1 + \frac{q_1 {\rm s} + q_0}{
{\rm s} + p_0} \frac{2}{15 {\rm s}^2 + {\rm s} } } \\ &= \frac{2 \left( q_1 {\rm s} + q_0 \right)}{ {\rm s} ( 15{\rm s} + 1) \left( {\rm s} + p_0 \right) + 2 \left( q_1 {\rm s} + q_0 \right) } \\ &= \frac{2 \left( q_1 {\rm s} + q_0 \right)}{ 15{\rm s}^3 + \left( 15p_0 + 1 \right){\rm s}^2 + \left( p_0 + 2q_1 \right){\rm s} + 2q_0 } \end{aligned} Wc(s)=1+R(s)G(s)R(s)G(s)=1+s+p0q1s+q015s2+s2s+p0q1s+q015s2+s2=s(15s+1)(s+p0)+2(q1s+q0)2(q1s+q0)=15s3+(15p0+1)s2+(p0+2q1)s+2q02(q1s+q0)Assume that the desired system has 3 poles: − 0.1 , − 0.2 , − 0.2 -0.1, -0.2, -0.2−0.1,−0.2,− 0.2 , thenthe characteristic equation ofthe expected system
(ie, the denominator of the transfer function) is: ( s + 0.1 ) ( s + 0.2 ) 2 ( {\rm s} + 0.1 ) ( {\rm s} + 0.2 )^2(s+0.1)(s+0.2)2 Therefore, in order for the adjusted system to achieve the desired system indicators, theadjusted system should have the desired pole distribution, that is: the denominators of the transfer functions should be the same:
15 s 3 + ( 15 p 0 + 1 ) s 2 + ( p 0 + 2 q 1 ) s + 2 q 0 = ( s + 0.1 ) ( s + 0.2 ) 2 15{\rm s}^3 + \left( 15p_0 + 1 \right){\rm s}^ 2 + \left( p_0 + 2q_1 \right){\rm s} + 2q_0 = ( {\rm s} + 0.1 ) ( {\rm s} + 0.2 )^215s3+(15p0+1)s2+(p0+2q1)s+2q0=(s+0.1)(s+0.2)2 By comparing the coefficients, we can directly obtain
p 0 = 13 30 , q 1 = 23 60 , q 0 = 3 100 , p_0 = \frac{13}{30}, \quad q_1 = \frac{23}{60}, \quad q_0 = \frac{3}{100},p0=3013,q1=6023,q0=1003, R ( s ) = q 1 s + q 0 s + p 0 = 11.5 s + 0.9 30 s + 13 R({\rm s}) = \frac{q_1 {\rm s} + q_0}{
{\rm s} + p_0} = \frac{11.5 {\rm s} + 0.9}{30{\rm s} + 13} R(s)=s+p0q1s+q0=30s+1311.5s+0.9
3. System after adding regulator
The closed-loop transfer function after adding the regulator is
W c ( s ) = 2 ( q 1 s + q 0 ) 15 s 3 + ( 15 p 0 + 1 ) s 2 + ( p 0 + 2 q 1 ) s + 2 q 0 = 23 s + 1.8 450 s 3 + 225 s 2 + 36 s + 1.8 W_c({\rm s}) = \frac{2 \left( q_1 {\rm s} + q_0 \right)}{ 15{\ rm s}^3 + \left( 15p_0 + 1 \right){\rm s}^2 + \left( p_0 + 2q_1 \right){\rm s} + 2q_0 } = \frac{ 23{\rm s} +1.8 }{ 450{\rm s}^3 +225{\rm s}^2 + 36{\rm s} + 1.8 }Wc(s)=15s3+(15p0+1)s2+(p0+2q1)s+2q02(q1s+q0)=450s3+225 s2+36s+1.823 s+1.8The comparison chart between the step response of this system and the original system is as follows:
The red line in the above figure is the adjusted closed-loop system, and the blue line is the original closed-loop system.
It can be seen that after adding the designed first-order regulator, the oscillation of the system disappears, the adjustment time is greatly reduced, the overshoot is almost zero, the response speed is accelerated, and the control effect is very good, which can be applied in practice.