Modal Feedback Control
1. The concept of modality
A mode is a certain sub-motion in the system (relative to the overall motion), and the overall dynamic characteristics of the system are composed of sub-motions (it can be compared to the concept of vector base in space).
Obviously, for a certain system, its motion is determined by the poles of the characteristic equation, and different poles (such as si = λ i s_i = \lambda _isi=li) in the time domain corresponds to C ie λ it C_i e^{\lambda _i t}CieliThe motion component of t . Therefore, in the time domain,the mode can be simply understood as the poleλ i \lambda_iliThe corresponding motion mode e λ ite^{\lambda _i t}elit。
It is worth noting that, in general, if the characteristic equation of the system is
Δ ( s ) = ∏ i = 1 n ( s − λ i ) \Delta (s) = \prod _{i=1} ^n \left( s - \lambda_i \right)D ( s )=i=1∏n(s−li) then its motion in the time domain can be expressed as
x ( t ) = ∑ i = 1 n C ie λ it = ∑ i = 1 nxi ( t ) x(t) = \sum _{i=1} ^n C_i e^{\lambda _i t} = \sum _{i=1} ^n x_i (t)x(t)=i=1∑nCielit=i=1∑nxi( t ) Among themC i C_iCiis a coefficient, but it can also be understood as different sub-motion modes in the whole motion x ( t ) x(t)The weightaccounted for in x ( t ) . Therefore,when discussing the modality itself, we are talking about e λ ite^{\lambda _i t}elit , butC ie λ it C_i e^{\lambda _i t}Cielit。
2. The idea and control quantity of modal feedback controluuu design
The idea of modal control can be expressed in one sentence: by designing the gain matrix KK in the feedback loopK such that the system has the desired mode distribution (ie, has the desired pole distribution). Similar to the idea of compensation-compensation is to add zero-pole points to the system to realize the operation of mathematical transfer function zero-pole point cancellation, so that the system has the desired zero-pole points.
Generally, the dynamic characteristics of the system can be expressed by the following equation
{ x ˙ = A x + B uy = C x (1) \begin{cases} \dot x = Ax + Bu \\ y = Cx \tag{1} \ end{cases}{
x˙=Ax+B uy=Cx( 1 ) This form is also called the Cauchy form.
任u = − K y (2) u=-Ky \tag{2}u=−Ky(2)则 x ˙ = A x + B u = A x + B ( − K y ) = A x − B K y = A X − B K ( C x ) = ( A − B K C ) x = A ~ x ⟶ A ~ = A − B K C (3) \begin{aligned} \dot x &= Ax + Bu = Ax + B \left( -Ky \right) = Ax - BKy = AX - BK \left( Cx \right) \\ &= \left( A - BKC \right) x = \tilde A x \longrightarrow \tilde A = A-BKC \tag{3} \end{aligned} x˙=Ax+B u=Ax+B(−Ky)=Ax−B Ky=AX−BK(Cx)=(A−B K C )x=A~x⟶A~=A−B K C( 3 ) By means of formula (2),uuu is eliminated, and only the state quantity xxis obtainedFormula (3) of x . It can be seen that formula (2) is to output the systemyyy goes through a gainKKAfter K , as inputuuu direct negative feedback to the input, which is the meaning of "feedback" in modal feedback control.
When the output is the state quantity, that is, y = xy=xy=x ,C = IC = IC=I,
u = − K x , A ~ = A − B K x ˙ = A ~ x = ( A − B K ) x (4) u = - K x, \\ \tilde A = A - BK \\ \dot x = \tilde A x = \left( A - BK \right) x \tag{4} u=−Kx,A~=A−BKx˙=A~x=(A−BK)x( 4 )
For the original equation (1), you can write its characteristic equation
Δ ( s ) = det ( s I − A ) \Delta (s) = \det \left( sI - A \right)D ( s )=the( s I−A ) Then also, for adding feedback controlu = − K xu=-Kxu=− K x system (4), its characteristic equation is
Δ ( s ) = det ( s I − A ~ ) = det ( s I − A + BK ) \Delta (s) = \det \left( sI - \tilde A \right) = \det \left( sI - A + BK \right)D ( s )=the( s I−A~)=the( s I−A+B K ) If there is a desired poleλ i ∗ \lambda_i^*li∗, then the desired characteristic equation of the system is
Δ ∗ ( s ) = ∏ i = 1 n ( s − λ i ∗ ) \Delta^* (s) = \prod _{i=1} ^n \left( s - \lambda_i^* \right)D∗(s)=i=1∏n(s−li∗) Then just makeΔ ( s ) = Δ ∗ ( s ) \Delta (s) = \Delta^* (s)D ( s )=D∗(s)，即
det ( s I − A + B K ) = ∏ i = 1 n ( s − λ i ∗ ) (5) \det \left( sI - A + BK \right) = \prod _{i=1} ^n \left( s - \lambda_i^* \right) \tag{5} the( s I−A+BK)=i=1∏n(s−li∗)( 5 ) Reverse solveKKK , the negative feedback path can be obtained. And this useu = − K xu=-Kxu=− The form of control by K x is called modal feedback control.
3. Example of modal feedback control
设有系统
{ x ˙ 1 = − 1 T x 1 + 1 T u x ˙ 2 = x 1 \begin{cases} \dot x_1 = - \frac{1}{T} x_1 + \frac{1}{T} u \\ \dot x_2 = x_1 \end{cases} {
x˙1=−T1x1+T1ux˙2=x1Design the modal controller for it so that it has desired poles λ 1 ∗ , λ 2 ∗ \lambda_1^*, \lambda_2^*l1∗,l2∗。
From the Cauchy form of the system,
A = [ − 1 T 0 1 0 ] , B = [ 1 T 0 ] A = \left[ \begin{matrix} -\frac{1}{T} & 0 \ \ 1 & 0 \end{matrix} \right], B = \left[ \begin{matrix} \frac{1}{T} \\ 0 \end{matrix} \right]A=[−T1100],B=[T10] Design the modal feedback controller asu = − K xu=-Kxu=−Kx，由于 x ∈ R 2 × 2 , u ∈ R 1 × 1 x \in \mathbb{R}^{2 \times 2}, u \in \mathbb{R}^{1 \times 1} x∈R2×2,u∈R1×1，故 K ∈ R 1 × 2 K \in \mathbb{R}^{1 \times 2} K∈R1×2，即
K = [ K 1 K 2 ] K = \left[ \begin{matrix} K_1 & K_2 \end{matrix} \right] K=[K1K2] then
det ( s I − A + B K ) = det ( [ s 0 0 s ] − [ − 1 T 0 1 0 ] + [ 1 T 0 ] ⋅ [ K 1 K 2 ] ) = det ( [ s + 1 T 0 − 1 s ] + 1 T [ K 1 K 2 0 0 ] ) = ∣ s + K 1 + 1 T K 2 T − 1 s ∣ = s 2 + K 1 + 1 T s + K 2 T \begin{aligned} \det \left( sI - A + BK \right) &= \det \left( \left[ \begin{matrix} s & 0 \\ 0 & s \end{matrix} \right] - \left[ \begin{matrix} -\frac{1}{T} & 0 \\ 1 & 0 \end{matrix} \right] + \left[ \begin{matrix} \frac{1}{T} \\ 0 \end{matrix} \right] \cdot \left[ \begin{matrix} K_1 & K_2 \end{matrix} \right] \right) \\ &= \det \left( \left[ \begin{matrix} s + \frac{1}{T} & 0 \\ -1 & s \end{matrix} \right] + \frac{1}{T} \left[ \begin{matrix} K_1 & K_2 \\ 0 & 0 \end{matrix} \right] \right) \\ &= \Bigg\lvert \begin{matrix} s + \frac{K_1 + 1}{T} & \frac{K_2}{T} \\ -1 & s \end{matrix} \Bigg\rvert \\ &= s^2 + \frac{K_1 + 1}{T} s + \frac{K_2}{T} \end{aligned} the( s I−A+BK)=the([s00s]−[−T1100]+[T10]⋅[K1K2])=the([s+T1−10s]+T1[K10K20])=
s+TK1+1−1TK2s
=s2+TK1+1s+TK2则
s 2 + K 1 + 1 T s + K 2 T = ( s − λ 1 ∗ ) ( s − λ 2 ∗ ) = s 2 − ( λ 1 ∗ + λ 2 ∗ ) s + λ 1 ∗ λ 2 ∗ s^2 + \frac{K_1 + 1}{T} s + \frac{K_2}{T} = \left( s - \lambda_1^* \right) \left( s - \lambda_2^* \right) = s^2 - \left( \lambda_1^* + \lambda_2^*\right) s + \lambda_1^* \lambda_2^* s2+TK1+1s+TK2=(s−l1∗)(s−l2∗)=s2−( l1∗+l2∗)s+l1∗l2∗The infinitive is
{ K 1 + 1 T = − ( λ 1 ∗ + λ 2 ∗ ) K 2 T = λ 1 ∗ λ 2 ∗ \begin{cases} \frac{K_1 + 1}{T} = - \left ( \lambda_1^* + \lambda_2^*\right) \\ \frac{K_2}{T} = \lambda_1^* \lambda_2^*\end{cases}{
TK1+1=−( l1∗+l2∗)TK2=l1∗l2∗Then
K 1 = − T ( λ 1 ∗ + λ 2 ∗ ) − 1 , K 2 = T λ 1 ∗ λ 2 ∗ K_1 = -T \left( \lambda_1^* + \lambda_2^*\right) - , \quad K_2 = T \lambda_1^* \lambda_2^*K1=−T( l1∗+l2∗)−1,K2=T λ1∗l2∗Then, the gain matrix K on the negative feedback path K = [ K 1 K 2 ] K = \left[ \begin{matrix} K_1 & K_2 \end{matrix} \right]K=[K1K2] can make the system have desired polesλ 1 ∗ , λ 2 ∗ \lambda_1^*, \lambda_2^*l1∗,l2∗。