ME 547 Review
State-Space Model
$$ \begin{gather*} \dot{X}=AX+Bu \\ Y=CX+Du \end{gather*} $$Transfer function
$$ G(s)=\frac{Y(s)}{U(s)}=C(sI-A)^{-1}B+D $$Impulse response
$$ g(t)=\mathcal{L}^{-1}[G(s)] $$Output
$$ \begin{gather*} y(t)=Ce^{At}X(0)+\int_0^t g(t-\tau)u(\tau)d\tau \\ y(t)=Ce^{A(t-t_0)}X(0)+\int_{t_0}^t g(t-\tau)u(\tau)d\tau \end{gather*} $$where
$$ e^{At}=\mathcal{L}^{-1}[(sI-A)^{-1}] $$Finding $e^{At}$
Diagonalizable
$$ A=QDQ^{-1} $$then
$$ A^k=QD^kQ^{-1} $$ $$ e^{At}=Q\begin{bmatrix} e^{a_1t} & & \\ & \ddots & \\ & & e^{a_nt} \end{bmatrix}Q^{-1} $$Jordan Forms
$\lambda_i$ has multiplicity $k$, then there are $n-\operatorname{rank}(A-\lambda_iI)$ Jordan block associated with $\lambda_i$. For example, $\lambda_i$ has multiplicity $3$ and $1$ Jordan block, then the matrix is
$$ \begin{pmatrix} \lambda_i & 1 & 0 \\ 0 & \lambda_i &1 \\ 0 & 0 & \lambda_i \end{pmatrix} $$When finding $Q$
$$ \require{mathtools} \require{centernot} \DeclarePairedDelimiters\norm{\lVert}{\rVert} \DeclarePairedDelimiters\abs{\lvert}{\rvert} \DeclareMathOperator{\eig}{eig} \begin{dcases} (A-\lambda_iI)q_1=0 \\ (A-\lambda_iI)q_2=q_1 \\ (A-\lambda_iI)q_3=q_2 \end{dcases} $$$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
$$ \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix} $$have $\lambda=2,n=2$
then $(A_i-\lambda I)^n=0$.
With this equation we can get
$$ \begin{gather*} e^{At}=e^{\lambda t}I+te^{\lambda t}(A-\lambda I)+\frac{t^2 e^{\lambda t}}{2}(A-\lambda I)^2+\frac{t^3e^{\lambda t}}{3!}(A-\lambda I)^3+\cdots \\ A^k=\lambda^kI+k\lambda^{k-1}(A-\lambda I)+\frac{k(k-1)}{2}(A-\lambda I)^2\lambda^{k-2}+\cdots \end{gather*} $$Time-Varying System
Given $n$ independent initial states, find the fundamental matrix of $\dot{X}=AX$
$$ \hat{X}=[X_1(t),X_2(t),\dots,X_n(t)] $$then form the state transition matrix
$$ \Phi(t,t_0)=\hat{X}(t)\hat{X}^{-1}(t_0) $$it satisfies
$$ \begin{dcases} \frac{\partial}{\partial t}\Phi(t,t_0)=A(t)\Phi(t,t_0) \\ \Phi(t_0,t_0)=I \end{dcases} $$The solution
$$ \begin{align*} X&=\Phi(t,t_0)X(t_0)+\int_{t_0}^t \Phi(t,\tau)B(\tau)U(\tau)d\tau \\ Y&=C(t)X(t)+D(t)U(t) \\ &=C(t)\Phi(t,t_0)X(t_0)+\int_{t_0}^t [C(t)\Phi(t,\tau)B(\tau)+D(t)\delta(t-\tau)]U(\tau)d\tau \end{align*} $$If $X(t_0)=0$, then
$$ Y=\int_{t_0}^t G(t,\tau)U(\tau)d\tau $$which is the impulse response to LTV system, compared to LTI system
$$ y=\int_{t_0}^t [Ce^{A(t-\tau)}B+D\delta(t-\tau)]u(\tau)d\tau $$Discrete-Time System
$$ \begin{align*} X[k+1]&=(I+AT)X[k]+BTu[k] \\ y[k]&=CX[k]+Du[k] \end{align*} $$With zero-hold we have another discretization. From
$$ X(t)=e^{A(t-t_0)}X(t_0)+\int_{t_0}^t e^{A(t-\tau)}Bu(\tau)d\tau $$make $t=(k+1)T, t_0=kT$
$$ \begin{split} X[k+1]&=e^{AT}X[k]+\int_0^T e^{A\tau}d\tau Bu[k] \\ &=A_dX[k]+B_du[k] \end{split} $$From this we have
$$ X[k]=A_d^kX[0]+\sum_{m=0}^{k-1}A_d^{(k-1)-m}B_du[m] $$Stability
Transfer function does not change under transformation.
equivalent system $\implies$ same TF
same TF $\centernot\implies$ equivalent system
BIBO state response
$u(t)=a$.
$$ \lim_{t\to\infty} y(t)=aG(s)|_{s=0} $$$u(t)=A\cos(\omega t)$
$$ \lim_{t\to\infty} y(t)=AM\cos(\omega t+\varphi) $$where
$$ \begin{dcases} M =\abs{G(s)}_{s=j\omega} \\ \varphi =\angle G(s)|_{s=j\omega} \end{dcases} $$
| LTI | LTV | |
|---|---|---|
| BIBO | $\int_0^\infty\abs{g(t)}<\infty$ RF: all poles on LHP RF: $g(t)\to 0$ as $t\to\infty$ $\centernot\implies$ asymptotically stable unchanged | $\int_0^\infty\abs{g(t,\tau)}<\infty$ $\begin{cases}\norm{D(t)}<\infty \\ \int_{t_0}^t\norm{C(t)\Phi(t,\tau)B(\tau)}d\tau<\infty \end{cases}$ unchanged |
| marginally | $\eig(A)$ non-positive. $0$ real part are simple root unchanged | $\norm{\Phi(t,t_0)}<\infty$ for any $t_0,t$ that $t>t_0$ unchanged |
| asymptotically | $\eig(A)$ negative $\implies$ BIBO unchanged | $\begin{cases}\norm{\Phi(t,t_0) |
Discrete system just like Continuous system except
- $g[k]$ absolute summable $\iff$ $g(t)$ absolute integrable
- all poles magnitude$<1$ $\iff$ all poles on LHP
- $\abs{\eig(A)}\le 1$ $\iff$ $\eig(A)$ 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 $x(n)$.
$$ \mathcal{C}=[B,AB,A^2B,\dots,A^{n-1}B] $$Observability: Estimate $x(0)$ when $u$ and $y$ are known.
$$ \mathcal{O}=\begin{bmatrix} C \\ CA \\ \vdots \\ CA^{n-1} \end{bmatrix} $$controllability of $(A,B)$ is the same as the observability of $[A^T,B^T]$ and vice versa.
Continuous Time
$$ W_c(t)=\int_0^t e^{A\tau}BB'e^{A'\tau}d\tau $$has to be non-singular.
Input is
$$ u=-B'e^{A'(t_f-t)}W_c^{-1}(t_f) [e^{At_f}X_0-X_f] $$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
$$ W_c=\lim_{t\to\infty} W_c(t) $$When $A$ is Hurwitz, Lyapunov equation can be used to solve for $W_c$
$$ AW_c+W_cA'=-BB' $$If the system is stable, $W_c$ is invertible $\implies$ $W_c(t)$ is invertible.
$[A-\lambda I, B]$ has full row rank for every eigenvalue $\lambda$ of $A$.
controllability is invariant under equivalence transformation.
Kalman Decomposition
Let $\bar{X}=PX$ then
$$ \begin{gather*} \bar{A}=PAP^{-1}\qquad \bar{B}=PB\qquad \bar{C}=CP^{-1}\qquad \bar{D}=D \\ \bar{\mathcal{C}}=P\mathcal{C}\qquad \bar{\mathcal{O}}=\mathcal{O}P^{-1} \end{gather*} $$If $\rho(\mathcal{C})=n_1$, choose
$$ P^{-1}=\begin{bmatrix} q_1 & q_2 & \cdots & q_{n_1} & \cdots & q_n \end{bmatrix} $$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
$$ \bar{A}=\begin{bmatrix} \bar{A}_c & \bar{A}_{12} \\ 0 & \bar{A}_{\bar{c}} \end{bmatrix} \qquad \bar{B}=\begin{bmatrix} \bar{B}_c\\ 0 \end{bmatrix} \qquad \bar{C}=\begin{bmatrix} \bar{C}_c & \bar{C}_{\bar{c}} \end{bmatrix} $$If $\rho(\mathcal{O})=n_2$, choose
$$ P=\begin{bmatrix} p_1 \\ p_2 \\ \vdots \\ p_{n_2} \\ \vdots \\ p_n \end{bmatrix} $$change the system to
$$ \bar{A}=\begin{bmatrix} \bar{A}_o & 0 \\ \bar{A}_{21} & \bar{A}_{\bar{o}} \end{bmatrix} \qquad \bar{B}=\begin{bmatrix} \bar{B}_o \\ \bar{B}_{\bar{o}} \end{bmatrix} \qquad \bar{C}=\begin{bmatrix} \bar{C}_o & 0 \end{bmatrix} $$Combine these two we can get
$$ \frac{d}{dt}\bar{X}=\begin{bmatrix} \bar{A}_{co} & 0 & \bar{A}_{13} & 0 \\ \bar{A}_{21} & \bar{A}_{c\bar{o}} & \bar{A}_{23} & \bar{A}_{24} \\ 0 & 0 & \bar{A}_{\bar{c}o} & 0 \\ 0 & 0 & \bar{A}_{43} & \bar{A}_{\bar{c}\bar{o}} \end{bmatrix} \begin{bmatrix} \bar{x}_{co} \\ \bar{x}_{c\bar{o}} \\ \bar{x}_{\bar{c}o} \\ \bar{x}_{\bar{c}\bar{o}} \end{bmatrix} + \begin{bmatrix} \bar{B}_{co} \\ \bar{B}_{c\bar{o}} \\ 0 \\ 0 \end{bmatrix}u \\ \ \\ \bar{Y}=\begin{bmatrix} \bar{C}_{co} & 0 & \bar{C}_{\bar{c}o} & 0 \end{bmatrix} \bar{X}+Du $$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
$$ \abs{\operatorname{Im}(\lambda_i-\lambda_j)}\ne \frac{2\pi m}{T} $$when $\operatorname{Re}(\lambda_i-\lambda_j)=0$. 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:
$$ W_c(t_0,t_1)=\int_{t_0}^{t_1}\Phi(t_1,\tau)B(\tau)B'(\tau)\Phi'(t_1,\tau)d\tau $$ $$ \begin{dcases} M_0(t)=B(t) \\ M_{m+1}(t)=-A(t)M_m(t)+\frac{d}{dt}M_m(t) \end{dcases} $$there exists $t_1>t_0$ that
$$ \operatorname{rank}[M_0(t_1),\dots,M_{n-1}(t_1)]=n $$Observability:
$$ W_c(t_0,t_1)=\int_{t_0}^{t_1}\Phi'(\tau,t_0)C'(\tau)C(\tau)\Phi(\tau,t_0)d\tau $$Jordan form theorem not applicable to LTV system.
$$ \begin{dcases} N_0(t)=C(t) \\ N_{m+1}(t)=N_m(t)A(t)+\frac{d}{dt}N_m(t) \end{dcases} $$there exists $t_1>t_0$ that
$$ rank\begin{bmatrix} N_0(t_1)\\ \vdots \\ N_{n-1}(t_1) \end{bmatrix} =n $$$[A(t),B(t)]$ controllable $\iff$ $[-A'(t),B'(t)]$ observable.
Controllable Form
Let
$$ P^{-1}=\begin{bmatrix} B & AB & A^2B & A^3B \end{bmatrix} \begin{bmatrix} 1 & \alpha_1 & \alpha_2 & \alpha_3 \\ 0 & 1 & \alpha_1 & \alpha_2 \\ 0 & 0 & 1 & \alpha_1 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$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
$$ \bar{A}=\begin{bmatrix} -\alpha_1 & -\alpha_1 & -\alpha_3 & -\alpha_4 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix} \qquad \bar{B}=\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \qquad \bar{C}=\begin{bmatrix} \beta_1 & \beta_2 & \beta_3 & \beta_4 \end{bmatrix} $$This is the controllable form, it has transfer function
$$ G(s)=\frac{\beta_1s^3+\beta_2s^2+\beta_3s+\beta_4}{s^4+\alpha_1s^3+\alpha_2s^2+\alpha_3s+\alpha_4} $$Pole Placement
$$ \Delta s\to\alpha\\ \bar{\Delta} s\to\bar{\alpha} $$Let
$$ P^{-1}=C\begin{bmatrix} 1 & \alpha_1 & \alpha_2 & \alpha_3 \\ 0 & 1 & \alpha_1 & \alpha_2 \\ 0 & 0 & 1 & \alpha_1 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$ $$ \begin{split} K&=\begin{bmatrix} \bar{\alpha}_1-\alpha_1 & \bar{\alpha}_2-\alpha_2 & \bar{\alpha}_3-\alpha_3 & \bar{\alpha}_4-\alpha_4 \end{bmatrix} P \\ &=\bar{K}P \end{split} $$Then we have
$$ \begin{gather*} u(t)=-\bar{K}\bar{X}(t)+r(t)=-KX(t)+r(t) \\ \dot{\bar{X}}=[\bar{A}-\bar{B}\bar{K}]\bar{X}(t)+\bar{B}r(t) \end{gather*} $$