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
$$ e^{At}=\mathcal{L}^{-1}[(sI-A)^{-1}] $$Matrix Exponential
Diagonalizable
If $A$ is diagonalizable: $A=QDQ^{-1}$, then $A^k=QD^kQ^{-1}$. Using Taylor series, we can get
$$ e^{At}=Q\begin{bmatrix} e^{\lambda_1 t} & & \\ & \ddots & \\ & & e^{\lambda_n t} \end{bmatrix}Q^{-1} $$where $\lambda_1, \cdots, \lambda_n$ are the eigenvalues of $A$.
Jordan Normal Form
If $A$ is not diagonalizable, it can be transformed to the Jordan normal form.
If $\lambda_i$ has multiplicity $k$, then there are $n-\operatorname{rank}(A-\lambda_iI)$ Jordan blocks associated with $\lambda_i$. For example, if $\lambda_i$ has multiplicity 3, the matrix can have either 1 or 2 Jordan blocks:
$$ \begin{pmatrix} \lambda_i & 1 & 0 \\ 0 & \lambda_i &1 \\ 0 & 0 & \lambda_i \end{pmatrix} \qquad \begin{pmatrix} \lambda_i & 0 & 0 \\ 0 & \lambda_i &1 \\ 0 & 0 & \lambda_i \end{pmatrix} $$How to find $Q$ with the example of 1 Jordan block:
$$ \begin{cases} (A-\lambda_iI)q_1=0 \\ (A-\lambda_iI)q_2=q_1 \\ (A-\lambda_iI)q_3=q_2 \end{cases} $$Let $A_i$ be a size $n$ Jordan block associated with $\lambda_{i}$, then we have $(A_i-\lambda_{i} I)^n=0$.
Using this property, the Taylor series becomes a finite sum:
$$ e^{At}=e^{\lambda t + (A -\lambda I)t} = e^{\lambda t} \left[ I+t(A-\lambda I)+\frac{t^2}{2}(A-\lambda I)^2 +\cdots + \frac{t^{n-1}}{(n-1)!}(A-\lambda I)^{n-1} \right] $$Time-Variant 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
$$ \Phi(t,t_0)=\hat{X}(t)\hat{X}^{-1}(t_0) $$which 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 is
$$ \begin{align*} X&=\Phi(t,t_0)X(t_0)+\int_{t_0}^t \Phi(t,\tau)B(\tau)U(\tau)d\tau \\ Y&=C(t)X(t)+D(t)U(t) \\ &=C(t)\Phi(t,t_0)X(t_0)+\int_{t_0}^t [C(t)\Phi(t,\tau)B(\tau)+D(t)\delta(t-\tau)]U(\tau)d\tau \end{align*} $$If $X(t_0)=0$, then
$$ Y=\int_{t_0}^t G(t,\tau)U(\tau)d\tau $$which is the impulse response for an LTV system, compared to an 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 a zero-order hold, we have another discretization. From
$$ X(t)=e^{A(t-t_0)}X(t_0)+\int_{t_0}^t e^{A(t-\tau)}Bu(\tau)d\tau $$setting $t=(k+1)T$ and $t_0=kT$ gives
$$ \begin{split} X_{k+1}&=e^{AT}X_{k}+\int_0^T e^{A\tau}d\tau Bu_{k} \\ &=A_dX_{k}+B_du_{k} \end{split} $$From this, the solution for $X_{k}$ is
$$ X_{k}=A_d^kX_{0}+\sum_{m=0}^{k-1}A_d^{(k-1)-m}B_du_{m} $$Stability
Transfer function does not change under transformation.
- equivalent system $\implies$ same TF
- same TF $\centernot\implies$ equivalent system
BIBO stable and steady state response
For a constant input $u(t)=a$:
$$ \lim_{t\to\infty} y(t)=aG(s)|_{s=0} $$For a sinusoidal input $u(t)=A\cos(\omega t)$:
$$ \lim_{t\to\infty} y(t)=AM\cos(\omega t+\varphi) $$where
$$ \begin{dcases} M =\lvert G(s) \rvert_{s=j\omega} \\ \varphi =\angle G(s)|_{s=j\omega} \end{dcases} $$
| LTI | LTV | |
|---|---|---|
| BIBO stable | $\int_0^\infty\lvert g(t) \rvert dt<\infty$ $g(t)\to 0$ as $t\to\infty$ is a necessary but not sufficient condition. All poles of the transfer function have negative real parts. BIBO stable $\centernot\implies$ asymptotically stable. | $\int_{t_0}^t \lVert C(t)\Phi(t,\tau)B(\tau) \rVert d\tau<\infty$ and $\lVert D(t) \rVert<\infty$. |
| marginally stable | Eigenvalues of $A$ have non-positive real parts; those with zero real part must be simple roots. | $\lVert \Phi(t,t_0) \rVert<\infty$ for all $t_0,t$ such that $t>t_0$. |
| asymptotically stable | Eigenvalues of $A$ have negative real parts. Asymptotically stable $\implies$ BIBO stable. | $\lVert \Phi(t,t_0) \rVert \to 0 \text{ as } t\to\infty$. Invariant under Lyapunov transformation. Asymptotically stable $\centernot\implies$ BIBO stable. |
Discrete time systems behave like continuous time systems, except:
- $g_k$ is absolutely summable $\iff$ $g(t)$ is absolutely integrable.
- All poles have magnitude less than 1 $\iff$ all poles of the continuous time system have negative real parts.
- Eigenvalues of $A_d$ have magnitude $\leq 1$ $\iff$ Eigenvalues of $A$ have non-positive real parts.
Controllability and Observability
Controllability: For any initial state $x_{0}$ and any target state $x_f$, there exists an input $u(t)$ over a finite time interval $[0, T]$ that transfers $x_{0}$ to $x_f$.
$$ \mathcal{C}=[B,AB,A^2B,\dots,A^{n-1}B] $$Observability: The ability to determine the initial state $x_{0}$ from measurements of the output $y(t)$ and input $u(t)$ over a finite time interval.
$$ \mathcal{O}=\begin{bmatrix} C \\ CA \\ \vdots \\ CA^{n-1} \end{bmatrix} $$The controllability of $(A,B)$ is equivalent to the observability of $(A^T,B^T)$, and vice versa.
Continuous Time Controllability
$$ W_c(t)=\int_0^t e^{A\tau}BB'e^{A'\tau}d\tau $$must be nonsingular for the system to be controllable. The input can be found by
$$ u=-B'e^{A'(t_f-t)}W_c^{-1}(t_f) [e^{At_f}X_0-X_f] $$- If $W_c(t)$ is nonsingular for any time $t>0$, it is nonsingular for all time $t>0$.
- The time to achieve state transfer can be arbitrarily small, but may require a large control input.
The controllability Gramian is
$$ W_c=\lim_{t\to\infty} W_c(t) $$When $A$ is a Hurwitz matrix (asymptotically stable), the Lyapunov equation can be used to solve for $W_c$:
$$ AW_c+W_cA'=-BB' $$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 $\operatorname{rank}(\mathcal{C})=n_1$, choose
$$ P^{-1}=\begin{bmatrix} q_1 & q_2 & \cdots & q_{n_1} & \cdots & q_n \end{bmatrix} $$where the first $n_1$ columns are independent columns from $\mathcal{C}$, and the remaining $n-n_1$ columns are chosen such that $P$ is 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 $\operatorname{rank}(\mathcal{O})=n_2$, choose
$$ P=\begin{bmatrix} p_1 \\ p_2 \\ \vdots \\ p_{n_2} \\ \vdots \\ p_n \end{bmatrix} $$to transform the system to
$$ \bar{A}=\begin{bmatrix} \bar{A}_o & 0 \\ \bar{A}_{21} & \bar{A}_{\bar{o}} \end{bmatrix} \qquad \bar{B}=\begin{bmatrix} \bar{B}_o \\ \bar{B}_{\bar{o}} \end{bmatrix} \qquad \bar{C}=\begin{bmatrix} \bar{C}_o & 0 \end{bmatrix} $$Combining these two, we can get the Kalman decomposition:
$$ \frac{d}{dt}\bar{X}=\begin{bmatrix} \bar{A}_{co} & 0 & \bar{A}_{13} & 0 \\ \bar{A}_{21} & \bar{A}_{c\bar{o}} & \bar{A}_{23} & \bar{A}_{24} \\ 0 & 0 & \bar{A}_{\bar{c}o} & 0 \\ 0 & 0 & \bar{A}_{43} & \bar{A}_{\bar{c}\bar{o}} \end{bmatrix} \begin{bmatrix} \bar{x}_{co} \\ \bar{x}_{c\bar{o}} \\ \bar{x}_{\bar{c}o} \\ \bar{x}_{\bar{c}\bar{o}} \end{bmatrix} + \begin{bmatrix} \bar{B}_{co} \\ \bar{B}_{c\bar{o}} \\ 0 \\ 0 \end{bmatrix}u \\ \ \\ \bar{Y}=\begin{bmatrix} \bar{C}_{co} & 0 & \bar{C}_{\bar{c}o} & 0 \end{bmatrix} \bar{X}+Du $$The transfer function is determined solely by the controllable and observable subsystem ($\bar{X}_{co}$).
Jordan Form Criteria
Hautus lemma: if $[A-\lambda I, B]$ has full row rank for every eigenvalue $\lambda$ of $A$, then the system is controllable.
After the system is transformed into Jordan forms:
- Controllability: the rows of $\bar{B}$ corresponding to the last row of each Jordan block associated with eigenvalue $\lambda_i$ must be linearly independent.
- Observability: the columns of $\bar{C}$ corresponding to the first column of each Jordan block associated with eigenvalue $\lambda_i$ must be linearly independent.
If the system is SISO:
- Controllability: each eigenvalue $\lambda_i$ can only have one Jordan block, and every entry of $\bar{B}$ corresponding to the last row of each Jordan block is non-zero.
- Observability: each eigenvalue $\lambda_i$ can only have one Jordan block, and every entry of $\bar{C}$ corresponding to the first column of each Jordan block is non-zero.
Sampling
If the continuous time system is controllable, then the discrete time system obtained by zero-order hold is controllable if
$$ \abs{\operatorname{Im}(\lambda_i-\lambda_j)}\ne \frac{2\pi m}{T}, \quad \forall m=1,2,\dots, $$when $\operatorname{Re}(\lambda_i-\lambda_j)=0$. Where $T$ is the sampling time.
If the continuous time system is uncontrollable, the discrete time system is also uncontrollable.
LTV Systems
Jordan form theorem not applicable to LTV system.
Controllability: A system is controllable on $[t_0, t_1]$ if the controllability Gramian is nonsingular
$$ W_c(t_0,t_1)=\int_{t_0}^{t_1}\Phi(t_1,\tau)B(\tau)B'(\tau)\Phi'(t_1,\tau)d\tau $$Alternatively, $[A(t), B(t)]$ is controllable on $[t_0, t_f]$ if there exists $t_1 \in [t_0, t_f]$ such that the rank of
$$ [M_0(t_1),\dots,M_{n-1}(t_1)] $$is $n$, where
$$ \begin{dcases} M_0(t)=B(t) \\ M_{m+1}(t)=-A(t)M_m(t)+\frac{d}{dt}M_m(t) \end{dcases} $$Observability: A system is observable on $[t_0, t_1]$ if the observability Gramian is nonsingular.
$$ W_o(t_0,t_1)=\int_{t_0}^{t_1}\Phi'(\tau,t_0)C'(\tau)C(\tau)\Phi(\tau,t_0)d\tau $$Alternatively, $[C(t), A(t)]$ is observable on $[t_0, t_f]$ if there exists $t_1 \in [t_0, t_f]$ such that the rank of
$$ \begin{bmatrix} N_0(t_1)\\ \vdots \\ N_{n-1}(t_1) \end{bmatrix} $$is $n$, where
$$ \begin{dcases} N_0(t)=C(t) \\ N_{m+1}(t)=N_m(t)A(t)+\frac{d}{dt}N_m(t) \end{dcases} $$Controllability of $[A(t),B(t)]$ is dual to observability of $[-A'(t), C'(t)]$.
Controllable Form
Let the characteristic polynomial of $A$ (for an $n=4$ system) be $\det(sI-A)=s^4+\alpha_1 s^3+\alpha_2 s^2+\alpha_3 s+\alpha_4$. The system can be transformed using $\bar{X}=PX$ where
$$ P^{-1}= \begin{bmatrix} B & AB & A^2B & A^3B \end{bmatrix} \begin{bmatrix} 1 & \alpha_1 & \alpha_2 & \alpha_3 \\ 0 & 1 & \alpha_1 & \alpha_2 \\ 0 & 0 & 1 & \alpha_1 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$into the controllable form
$$ \bar{A}=\begin{bmatrix} -\alpha_1 & -\alpha_2 & -\alpha_3 & -\alpha_4 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix} \qquad \bar{B}=\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \qquad \bar{C}=\begin{bmatrix} \beta_1 & \beta_2 & \beta_3 & \beta_4 \end{bmatrix} $$The transfer function is
$$ G(s)=\frac{\beta_1s^3+\beta_2s^2+\beta_3s+\beta_4}{s^4+\alpha_1s^3+\alpha_2s^2+\alpha_3s+\alpha_4} $$Pole Placement
- Find the original and desired characteristic polynomial $\Delta s, \bar{\Delta}s$ to decide the coefficients $\alpha, \bar{\alpha}$
- Find $P$ that can transform the system into controllable form.
- Use control law $K=\bar{K}P$ where $\bar{K}=\begin{bmatrix} \bar{\alpha}_1-\alpha_1 & \bar{\alpha}_2-\alpha_2 & \bar{\alpha}_3-\alpha_3 & \bar{\alpha}_4-\alpha_4\end{bmatrix}$
so we have
$$ \begin{gather*} u(t)=-\bar{K}\bar{X}(t)+r(t)=-KX(t)+r(t) \\ \dot{\bar{X}}=[\bar{A}-\bar{B}\bar{K}]\bar{X}(t)+\bar{B}r(t) \end{gather*} $$