Linear System

ME 547 Review


State-Space Model

\[ \displaylines{\dot{X}=AX+Bu \\ Y=CX+Du } \]

Transfer function \[ \displaylines{G(s)=\dfrac{Y(s)}{U(s)}=C(sI-A)^{-1}B+D } \] Impulse response \[ \displaylines{g(t)=\mathcal{L}^{-1}[G(s)] } \]

Output \[ \displaylines{y(t)=Ce^{At}X(0)+\int_0^t g(t-\tau)u(\tau)d\tau \\ y(t)=Ce^{A(t-t_0)}X(0)+\int_{t_0}^t g(t-\tau)u(\tau)d\tau } \] where \[ \displaylines{e^{At}=\mathcal{L}^{-1}[(sI-A)^{-1}] } \]

Finding \(e^{At}\)

Diagonalizable

\[ \displaylines{A=QDQ^{-1} } \]

then \[ \displaylines{A^k=QD^kQ^{-1} } \]

\[ \displaylines{e^{At}=Q\begin{bmatrix} e^{a_1t} & & \\ & \ddots & \\ & & e^{a_nt} \end{bmatrix}Q^{-1} } \]

Jordan Forms

\(\lambda_i\) has multiplicity \(k\), then there are \(n-rank(A-\lambda_iI)\) Jordan block associated with \(\lambda_i\). For example, \(\lambda_i\) has multiplicity \(3\) and \(1\) Jordan block, then the matrix is \[ \displaylines{\begin{pmatrix} \lambda_i & 1 & 0 \\ 0 & \lambda_i &1 \\ 0 & 0 & \lambda_i \end{pmatrix} } \] When finding \(Q\) \[ \displaylines{\begin{cases} (A-\lambda_iI)q_1=0 \\ (A-\lambda_iI)q_2=q_1 \\ (A-\lambda_iI)q_3=q_2 \end{cases} } \]

\(e^{At}\) and \(A^k\)

\(A_i\) be a Jordan block associated with \(\lambda\). Let \(n\) be the size of largest Jordan block. For example, the matrix \[ \displaylines{\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix} } \] have \(\lambda=2,n=2\)

then \((A_i-\lambda I)^n=0\).

With this equation we can get \[ \displaylines{e^{At}=e^{\lambda t}I+te^{\lambda t}(A-\lambda I)+\dfrac{t^2 e^{\lambda t}}{2}(A-\lambda I)^2+\dfrac{t^3e^{\lambda t}}{3!}(A-\lambda I)^3+\cdots \\ A^k=\lambda^kI+k\lambda^{k-1}(A-\lambda I)+\dfrac{k(k-1)}{2}(A-\lambda I)^2\lambda^{k-2}+\cdots } \]

Time-Varying System

Given \(n\) independent initial states, find the fundamental matrix of \(\dot{X}=AX\) \[ \displaylines{\hat{X}=[X_1(t),X_2(t),\dots,X_n(t)] } \] then form the state transition matrix \[ \displaylines{\Phi(t,t_0)=\hat{X}(t)\hat{X}^{-1}(t_0) } \] it satisfies \[ \displaylines{\begin{cases} \dfrac{\partial}{\partial t}\Phi(t,t_0)=A(t)\Phi(t,t_0) \\ \Phi(t_0,t_0)=I \end{cases} } \] The solution \[ \displaylines{\begin{aligned} X&=\Phi(t,t_0)X(t_0)+\int_{t_0}^t \Phi(t,\tau)B(\tau)U(\tau)d\tau \\ Y&=C(t)X(t)+D(t)U(t) \\ &=C(t)\Phi(t,t_0)X(t_0)+\int_{t_0}^t [C(t)\Phi(t,\tau)B(\tau)+D(t)\delta(t-\tau)]U(\tau)d\tau \end{aligned} } \] If \(X(t_0)=0\), then \[ \displaylines{Y=\int_{t_0}^t G(t,\tau)U(\tau)d\tau } \] which is the impulse response to LTV system, compared to LTI system \[ \displaylines{y=\int_{t_0}^t [Ce^{A(t-\tau)}B+D\delta(t-\tau)]u(\tau)d\tau } \]

Discrete-Time System

\[ \displaylines{\begin{aligned} X[k+1]&=(I+AT)X[k]+BTu[k] \\ y[k]&=CX[k]+Du[k] \end{aligned} } \]

With zero-hold we have another discretization. From \[ \displaylines{X(t)=e^{A(t-t_0)}X(t_0)+\int_{t_0}^t e^{A(t-\tau)}Bu(\tau)d\tau } \] make \(t=(k+1)T, t_0=kT\) \[ \displaylines{\begin{aligned} X[k+1]&=e^{AT}X[k]+\int_0^T e^{A\tau}d\tau Bu[k] \\ &=A_dX[k]+B_du[k] \end{aligned} } \] From this we have \[ \displaylines{X[k]=A_d^kX[0]+\sum_{m=0}^{k-1}A_d^{(k-1)-m}B_du[m] } \]

Stability

Transfer function does not change under transformation.

equivalent system \(\Rightarrow\) same TF

same TF \(\nRightarrow\) equivalent system

BIBO state response

  1. \(u(t)=a\). \[ \displaylines{\lim_{t\rightarrow\infty} y(t)=aG(s)|_{s=0} } \]

  2. \(u(t)=A\cos(\omega t)\) \[ \displaylines{\lim_{t\rightarrow\infty} y(t)=AM\cos(\omega t+\varphi) } \] where \[ \displaylines{\begin{cases} M=|G(s)|_{s=j\omega} \\ \varphi\, \ =\angle G(s)|_{s=j\omega} \end{cases} } \]

LTI LTV
BIBO \(\int_0^\infty|g(t)|<\infty\)
RF: all poles on LHP
RF: \(g(t)\rightarrow 0\) as \(t\rightarrow\infty\)
\(\nRightarrow\) asymptotically stable
unchanged
\(\int_0^\infty|g(t,\tau)|<\infty\)
\(\begin{cases}||D(t)||<\infty \\ \int_{t_0}^t||C(t)\Phi(t,\tau)B(\tau)||d\tau<\infty \end{cases}\)
unchanged
marginally \(eig(A)\) non-positive. \(0\) real part are simple root
unchanged
\(||\Phi(t,t_0||<\infty\) for any \(t_0,t\) that \(t>t_0\)
unchanged
asymptotically \(eig(A)\) negative
\(\Rightarrow\) BIBO
unchanged
\(\begin{cases}||\Phi(t,t_0)||<\infty \\ ||\Phi(t,t_0)||\rightarrow 0 \text{ as } t\rightarrow\infty \end{cases}\)
\(\nRightarrow\) BIBO
unchanged under Lyapunov transformation: \(P(t),P^{-1}(t)\) bounded

Discrete system just like Continuous system except

  1. \(g[k]\) absolute summable \(\longleftrightarrow\) \(g(t)\) absolute integrable
  2. all poles magnitude\(<1\) \(\longleftrightarrow\) all poles on LHP
  3. \(|eig(A)|\le 1\) \(\longleftrightarrow\) \(eig(A)\) non-positive

Controllability & Observability

Controllability: For any \(x(0)\) and \(x(n)\), there exist an input of finite time length that transfer \(x(0)\) to \(x(n)\). \[ \displaylines{\mathcal{C}=[B,AB,A^2B,\dots,A^{n-1}B] } \] Observability: Estimate \(x(0)\) when \(u\) and \(y\) are known. \[ \displaylines{\mathcal{O}=\begin{bmatrix} C \\ CA \\ \vdots \\ CA^{n-1} \end{bmatrix} } \] controllability of \((A,B)\) is the same as the observability of \([A^T,B^T]\) and vice versa.

Continuous Time

\[ \displaylines{W_c(t)=\int_0^t e^{A\tau}BB'e^{A'\tau}d\tau } \]

has to be non-singular.

Input is \[ \displaylines{u=-B'e^{A'(t_f-t)}W_c^{-1}(t_f)[e^{At_f}X_0-X_f] } \] If \(W_c(t)\) is nonsingular for any time \(t>0\), it is nonsingular for any time \(t>0\).

Time can be arbitrarily small, but input could get large.

Controllability grammiar is \[ \displaylines{W_c=\lim_{t\to\infty} W_c(t) } \] When \(A\) is Hurwitz, Lyapunov equation can be used to solve for \(W_c\) \[ \displaylines{AW_c+W_cA'=-BB' } \] If the system is stable, \(W_c\) is invertible \(\Longrightarrow\) \(W_c(t)\) is invertible.


\(\begin{bmatrix} A-\lambda I & B \end{bmatrix}\) has full row rank for every eigenvalue \(\lambda\) of \(A\).


controllability is invariant under equivalence transformation.

Kalman Decomposition

Let \(\bar{X}=PX\) then \[ \displaylines{\bar{A}=PAP^{-1}\qquad \bar{B}=PB\qquad \bar{C}=CP^{-1}\qquad \bar{D}=D \\ \bar{\mathcal{C}}=P\mathcal{C}\qquad \bar{\mathcal{O}}=\mathcal{O}P^{-1} } \]
If \(\rho(\mathcal{C})=n_1\), choose \[ \displaylines{P^{-1}=\begin{bmatrix} q_1 & q_2 & \cdots & q_{n_1} & \cdots & q_n \end{bmatrix} } \] where the first \(n_1\) column are independent columns from \(\mathcal{C}\), the rest \(n-n_1\) columns are chosen for \(P\) to be invertible. Then we have \[ \displaylines{\bar{A}=\begin{bmatrix} \bar{A}_c & \bar{A}_{12} \\ 0 & \bar{A}_{\bar{c}} \end{bmatrix} \qquad \bar{B}=\begin{bmatrix} \bar{B}_c\\ 0 \end{bmatrix} \qquad \bar{C}=\begin{bmatrix} \bar{C}_c & \bar{C}_{\bar{c}} \end{bmatrix} } \]

If \(\rho(\mathcal{O})=n_2\), choose \[ \displaylines{P=\begin{bmatrix} p_1 \\ p_2 \\ \vdots \\ p_{n_2} \\ \vdots \\ p_n \end{bmatrix} } \] change the system to \[ \displaylines{\bar{A}=\begin{bmatrix} \bar{A}_o & 0 \\ \bar{A}_{21} & \bar{A}_{\bar{o}} \end{bmatrix} \qquad \bar{B}=\begin{bmatrix} \bar{B}_o \\ \bar{B}_{\bar{o}} \end{bmatrix} \qquad \bar{C}=\begin{bmatrix} \bar{C}_o & 0 \end{bmatrix} } \]
Combine these two we can get \[ \displaylines{\dfrac{d}{dt}\bar{X}=\begin{bmatrix} \bar{A}_{co} & 0 & \bar{A}_{13} & 0 \\ \bar{A}_{21} & \bar{A}_{c\bar{o}} & \bar{A}_{23} & \bar{A}_{24} \\ 0 & 0 & \bar{A}_{\bar{c}o} & 0 \\ 0 & 0 & \bar{A}_{43} & \bar{A}_{\bar{c}\bar{o}} \end{bmatrix} \begin{bmatrix} \bar{x}_{co} \\ \bar{x}_{c\bar{o}} \\ \bar{x}_{\bar{c}o} \\ \bar{x}_{\bar{c}\bar{o}} \end{bmatrix} + \begin{bmatrix} \bar{B}_{co} \\ \bar{B}_{c\bar{o}} \\ 0 \\ 0 \end{bmatrix}u \\ \ \\ \bar{Y}=\begin{bmatrix} \bar{C}_{co} & 0 & \bar{C}_{\bar{c}o} & 0 \end{bmatrix} \bar{X}+Du } \] Transfer function is only affected by \(\bar{X}_{co}\).

Jordan Block

Controllability: For each \(\lambda_i\), the last row of \(B\) with each Jordan block should be independent.

Observability: For each \(\lambda_i\), the first column of \(C\) with each Jordan block should be independent.

If SISO, there is only one Jordan block for each \(\lambda_i\), so the last row of \(B\) should be nonzero, the first column of \(C\) should be nonzero.

Sampling

If the continuous system is controllable, then the discrete system is controllable if \[ \displaylines{|\text{Im}(\lambda_i-\lambda_j)|\ne \dfrac{2\pi m}{T} } \] when \(\text{Re}(\lambda_i-\lambda_j)=0\). Where \(T\) is the sampling time.

If the continuous system is not uncontrollable, then discrete system is not uncontrollable.

LTV

Jordan form theorem not applicable to LTV system.

Controllability: \[ \displaylines{W_c(t_0,t_1)=\int_{t_0}^{t_1}\Phi(t_1,\tau)B(\tau)B'(\tau)\Phi'(t_1,\tau)d\tau } \]

\[ \displaylines{\begin{cases} M_0(t)=B(t) \\ M_{m+1}(t)=-A(t)M_m(t)+\dfrac{d}{dt}M_m(t) \end{cases} } \]

there exists \(t_1>t_0\) that \[ \displaylines{rank[M_0(t_1),\dots,M_{n-1}(t_1)]=n } \]
Observability: \[ \displaylines{W_c(t_0,t_1)=\int_{t_0}^{t_1}\Phi'(\tau,t_0)C'(\tau)C(\tau)\Phi(\tau,t_0)d\tau } \] Jordan form theorem not applicable to LTV system. \[ \displaylines{\begin{cases} N_0(t)=C(t) \\ N_{m+1}(t)=N_m(t)A(t)+\dfrac{d}{dt}N_m(t) \end{cases} } \] there exists \(t_1>t_0\) that \[ \displaylines{rank\begin{bmatrix} N_0(t_1)\\ \vdots \\ N_{n-1}(t_1) \end{bmatrix} =n } \] \([A(t),B(t)]\) controllable \(\Longleftrightarrow\) \([-A'(t),B'(t)]\) observable.

Controllable Form

Let \[ \displaylines{P^{-1}=\begin{bmatrix} B & AB & A^2B & A^3B \end{bmatrix} \begin{bmatrix} 1 & \alpha_1 & \alpha_2 & \alpha_3 \\ 0 & 1 & \alpha_1 & \alpha_2 \\ 0 & 0 & 1 & \alpha_1 \\ 0 & 0 & 0 & 1 \end{bmatrix} } \] where \(\alpha\) is the coefficients of \(\det(sI-A)=s^4+\alpha_1s^3+\alpha_2s^2+\alpha_3s+\alpha_4\)

With \(\bar{X}=PX\), then \[ \displaylines{\bar{A}=\begin{bmatrix} -\alpha_1 & -\alpha_1 & -\alpha_3 & -\alpha_4 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix} \qquad \bar{B}=\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \qquad \bar{C}=\begin{bmatrix} \beta_1 & \beta_2 & \beta_3 & \beta_4 \end{bmatrix} } \] This is the controllable form, it has transfer function \[ \displaylines{G(s)=\dfrac{\beta_1s^3+\beta_2s^2+\beta_3s+\beta_4}{s^4+\alpha_1s^3+\alpha_2s^2+\alpha_3s+\alpha_4} } \]

Pole Placement

\[ \displaylines{\Delta s\rightarrow\alpha\\ \bar{\Delta} s\rightarrow\bar{\alpha} } \]

Let \[ \displaylines{P^{-1}=C\begin{bmatrix} 1 & \alpha_1 & \alpha_2 & \alpha_3 \\ 0 & 1 & \alpha_1 & \alpha_2 \\ 0 & 0 & 1 & \alpha_1 \\ 0 & 0 & 0 & 1 \end{bmatrix} } \]

\[ \displaylines{\begin{aligned} K&=\begin{bmatrix} \bar{\alpha}_1-\alpha_1 & \bar{\alpha}_2-\alpha_2 & \bar{\alpha}_3-\alpha_3 & \bar{\alpha}_4-\alpha_4 \end{bmatrix} P \\ &=\bar{K}P \end{aligned} } \]

Then we have \[ \displaylines{u(t)=-\bar{K}\bar{X}(t)+r(t)=-KX(t)+r(t) \\ \dot{\bar{X}}=[\bar{A}-\bar{B}\bar{K}]\bar{X}(t)+\bar{B}r(t) } \]