Sliding Mode Control (SMC)
The sliding film control theory is a control theory based on modern control theory. Its core is the Lyapunov function . The core of the sliding film control is to establish a sliding mode surface, pull the controlled system down the sliding mode surface, and make the system Moving along the sliding surface, the advantage of synovial film control is that it ignores external disturbances and uncertain parameters, and adopts a more violent way to achieve the control purpose, but this kind of violence also brings some problems, that is, the difference between positive and negative signals High-frequency switching, general hardware cannot perform high-frequency switching of signals, so some other methods are needed to avoid this problem, and the high-frequency switching of the model will cause the output signal to oscillate, causing the system The back and forth oscillation between the die surfaces cannot be eliminated, which is also a problem of synovial film control.
advantage
Sliding modal can be designed
insensitive to disturbance
shortcoming
The hardware cannot adapt to high-frequency signal switching
Output signal oscillation caused by high-frequency signal switching
system modeling
We can establish a simple second-order system equation of state
x ˙ 1 = x 2 x ˙ 2 = u \begin{align} \dot x_1 &= x_2 \nonumber \\ \dot x_2 &= u \nonumber \\ \end {align}x˙1x˙2=x2=u
Our control goal is very clear, that is, we hope that x 1 = 0 , x 2 = 0 x_1 = 0,x_2=0x1=0,x2=0
design sliding surface
s = c x 1 + x 2 s=cx_1+x_2 s=cx1+x2
There is a question here, what is the sliding surface, why is it designed like this, why is it not like another, in fact, this involves a question is what is the target of our control, is x 1 = 0 , x 2 = 0 x_1 = 0,x_2=0x1=0,x2=0 , then ifs = 0 s=0s=0呢
{ c x 1 + x 2 = 0 x ˙ 1 = x 2 ⇒ c x 1 + x ˙ 1 = 0 ⇒ { x 1 = x 1 ( 0 ) e − c t x 2 = − c x 1 ( 0 ) e − c t \begin{equation} \begin{cases} cx_1 + x_2 = 0 \\ \dot x_1 = x_2 \\ \end{cases} \Rightarrow cx_1+\dot x_1 = 0 \Rightarrow \begin{cases} x_1 = x_1(0)e^{-ct} \\ x_2 = -cx_1(0)e^{-ct} \\ \end{cases} \nonumber \end{equation} {
cx1+x2=0x˙1=x2⇒cx1+x˙1=0⇒{
x1=x1(0)e−ctx2=−cx1(0)e−ct
It can be seen that the state quantity will eventually tend to 0, and it tends to 0 exponentially. ccThe larger c , the faster the speed. So ifs = cx 1 + c 2 = 0 s=cx_1+c_2=0s=cx1+c2=0 , then the state of the system will tend to zero along the sliding surface, (s = 0 s=0s=0 is called the sliding surface)
design reaching law
It says above that if s = 0 s=0s=The 0 state variable will eventually tend to 0, how to ensures = 0 s=0s=0 , this is the control rateuuu needs to guarantee the content
s ˙ = cx ˙ 1 + x ˙ 2 = cx 2 + u \dot s = c \dot x_1 + \dot x_2 = cx_2+us˙=cx˙1+x˙2=cx2+The law of reaching u
refers tos ˙ \dot ss˙ , the reaching law generally has the following designs
{ s ˙ = − ε sgn ( s ) , ε > 0 s ˙ = − ε sgn ( s ) − ks , ε > 0 , k > 0 s ˙ = − k ∣ s ∣ α sgn ( s ) , 0 < α < 1 \begin {cases} \dot s = - \varepsilon sgn(s), \varepsilon > 0 \\ \dot s = - \varepsilon sgn(s)-ks , \varepsilon > 0 , k>0\\ \dot s = - k|s|^{\alpha}sgn(s), 0 < \alpha < 1 \end{cases}⎩
⎨
⎧s˙=−εsgn(s),e>0s˙=−εsgn(s)−ks,e>0,k>0s˙=−k∣s∣αsgn(s),0<a<1
s g n ( s ) = { 1 , s > 0 − 1 , s < 0 sgn(s) = \begin{cases} 1,s>0 \\ -1,s<0 \\ \end{cases} sgn(s)={ 1,s>0−1,s<0
According to the above reaching law, we can obtain the control quantity uuu (choose the first control rate).
u = − cx 2 − ε sgn ( s ) u = -cx_2-\varepsilon sgn(s)u=−cx2−ε s g n ( s )
We impose a control quantityuuu can ensure that the system is finally stable at the origin.
proof of validity
In the control principle, the Lyapunov function is used to judge the stability of the system. For the system state equation s ˙ = cx 2 + u \dot s = cx_2+us˙=cx2+u , our goal at this time is to pull the system down to the vicinity of the sliding mode surface, and the control target issss , forsss if there exists a continuous functionVVV satisfies the following two formulas, then the system will be at the equilibrium points = 0 s=0s=0 is stable, that is,lim t → ∞ V = 0 {\lim\limits_{t \to \infty}V = 0}t→∞limV=0
lim ∣ s ∣ → ∞ V = ∞ {\lim\limits_{|s| \to \infty}V = \infty} ∣s∣→∞limV=∞
V ˙ < 0 f o r s ≠ 0 \dot V < 0 \ for \ s \ne 0 V˙<0 for s=0
The way we prove it is to let V = 1 2 s 2 V= \frac {1} {2} s ^ 2V=21s2 , obviously we satisfy the first condition, we haveVVV is derived,
V ˙ = ss ˙ = − s ε sgn ( s ) = − ε ∣ s ∣ < 0 \dot V = s \dot s = -s \varepsilon sgn(s) = - \varepsilon|s| < 0V˙=ss˙=−sεsgn(s)=−ε∣s∣<0
also satisfies the second condition, so the final system will be stable near the sliding surface, which means that the two variables will also be stable at the position where the sliding surface expects them to be stable, that is, the zero point.
infinite time problem
The above analysis seems to be impeccable, but it is actually useless, because we finally come to the conclusion that when the time tends to infinity, the state of the system will tend to 0. Is this useful? No, because the infinite time is too Horrible, it doesn’t make sense if the system is not stable after a person dies, so we must require him to be reachable in a finite time, so we modify a second condition V of Lyapunov ˙ ≤
− α V 1 2 \dot V \le - \alpha V ^ {\frac {1} {2}}V˙≤−αV21
For this improved condition, the variable can be separated and then integrated
d V dt ≤ − α V 1 2 V − 1 2 d V ≤ − α dt ∫ 0 t V − 1 2 d V ≤ ∫ 0 t − α dt V 1 2 ( t ) − V 1 2 ( 0 ) ≤ − 1 2 α t V 1 2 ( t ) ≤ − 1 2 α t + V 1 2 ( 0 ) \begin {align} \frac {\text d V} {\text dt} &\le - \alpha V ^ {\frac {1} {2}} \nonumber\\ V ^ {- \frac {1} {2}} \text d V &\le - \alpha \text dt \nonumber\\ \int^{t}_{0} V ^ {- \frac {1} {2}} \text d V &\le \int^{t}_{0} - \alpha \text dt \nonumber\\ V ^ {\frac {1} {2}} (t) - V ^ {\frac {1} {2}} (0) &\le - \frac {1} {2} \alpha t \nonumber\\ V ^ {\frac {1} {2}} (t) &\le - \frac {1} {2} \alpha t + V ^ {\frac {1} {2}} (0) \nonumber \\ \end{align}dtdVV−21dV∫0tV−21dVV21(t)−V21(0)V21(t)≤−αV21≤−αdt≤∫0t−αdt≤−21t _≤−21t _+V21(0)
According to the above equation, it can be seen that VVV will reach stability in a finite time, and the final stable time is
tr ≤ 2 V 1 2 ( 0 ) α t_r \le \frac {2V^ \frac {1} {2} (0)} {\alpha}tr≤a2V _21(0)
Because of the change of Lyapunov condition, the controller uuu Let us consider the integral function
{ V ̇ = ss ̇ = − s ε sgn ( s ) = − ε ∣ s ∣ V = 1 2 s 2 V ̇ ≤ − α V 1 2 ⇒ V ̇ = − ε ∣ s ∣ ≤ − α s 2 ⇒ ε ≥ α 2 \begin{cases} \dot V = s \dot s = -s \varepsilon sgn(s) = - \varepsilon|s|\\ V = \frac {1} {2}; s ^ 2 \\ \dot V \le - \alpha V ^ {\frac{1} {2}} \end{cases} \Rightarrow \dot V = - \varepsilon|s| \le-\alpha\frac{s}{\sqrt{2}}\Rightarrow\varepsilon\ge\frac{\alpha}{\sqrt{2}}⎩
⎨
⎧V˙=ss˙=−sεsgn(s)=−ε∣s∣V=21s2V˙≤−αV21⇒V˙=−ε∣s∣≤− a2s⇒e≥2a
That is, for the ε \varepsilon that was arbitrarily specified beforeε adds a control condition
interference problem
The above discussion is actually based on an assumption that there is no interference, and the control without interference is very easy to do, and it has no practical significance. Here we add the interference term to the state equation. We mentioned before that the synovial membrane method is not sensitive to interference. , here we explain in principle why the synovium method is not sensitive to disturbances.
The state equation
x ˙ 1 = x 2 x ˙ 2 = u + d \begin{align} \dot x_1 &= x_2 \nonumber\\ \dot x_2 &= u + d \nonumber\\ \end{align }x˙1x˙2=x2=u+d
This has no effect on our design of the synovial surface, our synovial surface is as follows
s = cx 1 + x 2 s = cx_1+x_2s=cx1+x2
Our reaching law design also does not change
s ˙ = − ε sgn ( s ) \dot s = - \varepsilon sgn(s)s˙=− ε s g n ( s )
our control quantityuuu does not change
u = − ε sgn ( s ) − cx 2 u = - \varepsilon sgn(s) - cx_2u=−εsgn(s)−cx2
s ˙ = c x ˙ 1 + x ˙ 2 = c x 2 + u + d \begin{align} \dot s &= c \dot x_1 + \dot x_2 \nonumber\\ &=cx_2 + u + d \nonumber\\ \end{align} s˙=cx˙1+x˙2=cx2+u+d
The equivalent of the equivalent quantity is given by
V = 1 2 s 2 V ̇ = ss ̇ = s ( cx 2 + u + d ) = s ( − ε sgn ( s ) + d ) ≤ − ε ∣ s ∣ + sd ≤ − ε ∣ s ∣ + s L ≤ ∣ s ∣ ( ε − L ) \begin{align} V &= \frac {1} {2} s ^ 2 \nonumber\\ \dot V &= s \dot s \nonumber\\ &= s (cx_2 + u + d) \nonumber\\ &= s (- \itempsilon sgn(s) + d) \nonumber\\ & \le -\itemepsilon|s| + sd \nonumber\\ & \le -\varepsilon|s| + sL \nonumber\\ & \le |s|(\varepsilon - L) \nonumber\\ \end{align}VV˙=21s2=ss˙=s(cx2+u+d)=s(−εsgn(s)+d)≤−ε∣s∣+sd≤−ε∣s∣+sL≤∣ s ∣ ( e−L)
where LLL is interferenceddd by
V ̇ ≤ − α V 1 2 − ε ∣ s ∣ + s L ≤ − α s 2 − ε ∣ s ∣ ≤ − α s 2 − s L ε ∣ s ∣ ≥ α s 2 + s L ε ≥ sgn ( s ) ( α 2 + L ) ε ≥ ( α 2 + L ) \begin{align} \dot V &\le - \alpha V ^ {\frac{1} {2}} \nonumber\\ - \varepsilon|s| + sL & \le -\alpha \frac{s}{\sqrt{2}} \nonumber\\ -\valuepsilon|s| & \le -\alpha \frac{s}{\sqrt {2}} - sL \nonumber\\ \valuepsilon|s| & \ge \alpha \frac{s}{\sqrt{2}} + sL \nonumber\\varepsilon &\ge sgn(s)(\frac{\alpha}{\sqrt{2}} + L)\ nonumber\\\varepsilon & \ge(\frac{\alpha}{\sqrt{2}}+L)\nonumber\\\end{align}V˙−ε∣s∣+sL−ε∣s∣ε∣s∣ee≤−αV21≤− a2s≤− a2s−sL≥a2s+sL≥sgn(s)(2a+L)≥(2a+L)
So we directly proved that when our disturbance has an upper bound, our synovial membrane parameter $\varepsilon $ only needs to meet the above conditions to converge to the vicinity of the synovial membrane surface at an exponential convergence speed.
Example of a Synovial Membrane Design Methodology for a Three-Order System
The model of the third-order system is as follows
x ˙ 1 = x 2 x ˙ 2 = x 3 x ˙ 3 = f ( x ) + g ( x ) u \begin {align} \dot x_1 &= x_2 \nonumber\\ \dot x_2 &= x_3 \nonumber\\ \dot x_3 &= f(x) + g(x)u \nonumber\\ \end {align}x˙1x˙2x˙3=x2=x3=f(x)+g(x)u
Suppose, we expect x 1 x_1x1The target is x 1 d x_{1d}x1 d, note that here is different from the previous one, the control target here is no longer 0
e 1 = x 1 − x 1 de 2 = e ˙ 1 = x ˙ 1 − x ˙ 1 d = x 2 − x ˙ 1 de 3 = e ˙ 2 = x ¨ 1 − x ¨ 1 d = x 3 − x ¨ 1 d \begin{align} e_1 &= x_1 - x_{1d} \nonumber\\ e_2 &= \dot e_1 = \dot x_1 - \dot x_{1d} = x_2 - \dot x_{1d} \nonumber\\ e_3 &= \dot e_2 = \ddot x_1 - \ddot x_{1d} = x_3 - \ddot x_{1d} \nonumber\\ \end{ align}e1e2e3=x1−x1 d=e˙1=x˙1−x˙1 d=x2−x˙1 d=e˙2=x¨1−x¨1 d=x3−x¨1 d
Design sliding surface
s = c 1 e 1 + c 2 e 2 + e 3 s = c_1 e_1 + c_2 e_2 + e_3s=c1e1+c2e2+e3
Design Lyapunov function
V = 1 2 s 2 V = \frac{1}{2} s ^ 2V=21s2.
The equation for the equation
V ̇ = ss ̇ = s ( c 1 e ̇ 1 + c 2 e ̇ 2 + e ̇ 3 ) = s ( c 1 e 2 + c 2 e 3 + x 3 − x ̈ 1 d ( 3 ) ) = s ( c 1 and 2 + c 2 and 3 + x 3 − f ( x ) + g ( x ) u − x 1 d ( 3 ) ) = s ( Γ − f ( x ) + g ( x ) u − x 1 d ( 3 ) ) = sg ( x ) ( Γ − f ( x ) − x 1 d ( 3 ) g ( x ) + u ) \begin{align} \dot V &= s \dot s \nonumber\\ &= s (c_1 \dot e_1 + c_2 \dot e_2 + \dot e_3) \nonumber\\ &= s (c_1 e_2 + c_2 e_3 + x_3 - \ddot x^{(3) }_{1d}) \nonumber\\ &= s (c_1 e_2 + c_2 e_3 + x_3 - f(x) + g(x)u - x^{(3)}_{1d}) \nonumber\\ & = s (\Gamma - f(x) + g(x)u - x^{(3)}_{1d}) \nonumber\\ &= sg(x)(\frac {\Gamma - f(x) - x^{(3)}_{1d}} {g(x)} + u) \nonumber\\ end{align}V˙=ss˙=s(c1e˙1+c2e˙2+e˙3)=s(c1e2+c2e3+x3−x¨1 d(3))=s(c1e2+c2e3+x3−f(x)+g(x)u−x1 d(3))=s ( C−f(x)+g(x)u−x1 d(3))=s g ( x ) (g(x)C−f(x)−x1 d(3)+u)
Here we design reaching law
s ˙ = − ε sgn ( s ) = Γ − f ( x ) + g ( x ) u − x 1 d ( 3 ) \dot s = - \varepsilon sgn(s) = \Gamma - f(x) + g(x)u - x^{(3)}_{1d}s˙=−εsgn(s)=C−f(x)+g(x)u−x1 d(3)
Get the control amount uuu
u = − ε sgn ( s ) − Γ + f ( x ) + x 1 d ( 3 ) g ( x ) u = \frac {-\varepsilon sgn(s) - \Gamma + f(x) + x^ {(3)}_{1d}} {g(x)}u=g(x)−εsgn(s)−C+f(x)+x1 d(3)
Bringing in the Lyapunov function,
V ˙ = − s ε sgn ( s ) = − ε ∣ s ∣ ≤ 0 \dot V = -s \varepsilon sgn(s) = -\varepsilon |s| \le 0V˙=−sεsgn(s)=−ε∣s∣≤0
Here we can see that the system will be stable, if you need to control the time to reach stability, limitε \varepsilonε can