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

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