Cheat Sheet

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\[\begin{split}\begin{aligned} (a^2-b^2)&=(a-b)(a+b)\\ (a^3-b^3)&=(a-b)(a^2+ab+b^2)\\ (a^4-b^4)&=(a-b)(a+b)(a^2+b^2)\\\end{aligned}\end{split}\]

\[\begin{split}\begin{aligned} e^{a}e^{b}&=e^{a+b}\\ (e^{x})^n&=e^{2x}\\\end{aligned}\end{split}\]

\[\begin{split}\begin{aligned} (A+B)^T&=A^T+B^T\\ (AB)^T&=B^TA^T\\ AB^T&=BA^T\\ (A^T)^{-1}&=(A^{-1})^T\\\end{aligned}\end{split}\]

\[\begin{split}\begin{aligned} det(A^T)&=det(A)\\\end{aligned}\end{split}\]

Also, if A is square, then its eigenvalues are equal to the eigenvalues of its transpose. Additionally, if the matrix is also differentiable and nonsingular

\[ \begin{align}\begin{aligned} \begin{aligned} \frac{d}{dt}(P^{-1}(t)=-P^{-1}(t)\dot{P}(t)P^{-1}(t)\end{aligned}\\which can be found with;\end{aligned}\end{align} \]

\frac{d}{dt}(P(t)P^{-1}(t))=\frac{d}{dt}(I)=0

\[\begin{aligned} A^T&=A\end{aligned}\]

\[\begin{aligned} A^T=-A\end{aligned}\]

\[\begin{split}A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\end{split}\]

\[\begin{split}A^{-1}= \frac{1}{\lvert A\rvert} \begin{bmatrix*}[r] a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{bmatrix*} \,.\end{split}\]

\[\begin{split}\left|B\right|=det(B)=\left|\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right|\end{split}\]

\[\begin{split}=a\left|\begin{array}{cc}e&f\\h&i\end{array}\right|-b\left|\begin{array}{cc}d&f\\g&i\end{array}\right|+c\left|\begin{array}{cc}d&e\\g&h\end{array}\right|\end{split}\]

\[\begin{split}\begin{aligned} sin(2a)&=2cos(a)sin(a)\\ cos(2a)&=1-2sin^2(a)=2cos^2(a)-1\\\end{aligned}\end{split}\]

States that the derivative of the integral is (for constant limits of integration):

\[ \begin{align}\begin{aligned} \begin{aligned} \frac{d}{dt}(\int_{a}^{b} f(x,t) \,dt)&=\int_{a}^{b} \frac{\partial}{\partial x}f(x,t)\,dt\end{aligned}\\States that the derivative of the integral is (for limits of\end{aligned}\end{align} \]

integration that are not constant)

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} &\frac{d}{dt}(\int_{f(a)}^{f(b)} f(x,t) \,dt)=\\ &=f(x,b(x))\frac{d}{dx}b(x)-f(x,a(x))\frac{d}{dx}a(x)+\int_{f(a)}^{f(b)} \frac{\partial}{\partial x}f(x,t)\,dt\end{aligned}\end{split}\\Note: :math:`t` is a variable in the integration limit, so the above\end{aligned}\end{align} \]

formula must be used to derive \dot{P}(t).

\[\begin{split}\begin{aligned} \int x^n\,dx &= \frac{1}{n+1}x^{n+1}\\ \int \frac{1}{x}\,dx &= \ln |x|\\ \int u\dot{v}\,dx &= uv - \int v du\\ \int e^x\,dx &= e^x \\ \int a^x\,dx &= \frac{1}{\ln a} a^x\\ \int \ln x\,dx &= x \ln x - x\\ \int \sin x\,dx &= -\cos x\\ \int \cos x\,dx &= \sin x\\ \int \tan x\,dx &= \ln |\sec x|\\ \int \sec x\,dx &= \ln |\sec x + \tan x|\\ \int \sec^2 x\,dx &= \tan x\\ \int \sec(x) \tan(x)\,dx &= \sec x\\\end{aligned}\end{split}\]

\[\begin{aligned} dxdy=rdrd\theta\end{aligned}\]

don’t forget to change the limits of integration!

\[\begin{split}\begin{aligned} \frac{d}{dt}(tan^-1(x))&=\frac{1}{1+x^2}\\\end{aligned}\end{split}\]

\[\begin{split}\begin{aligned} f(t)&={\ensuremath{\mathcal{L}{\left[f(t)\right]}}}=F(s)\\ 1&=\dfrac{1}{s}\\ \delta(t)&=1\\ \delta(t-t_0)&=e^{-st_0}\\ f'(t)&=sF(s) - f(0) \\ f^{n}(t)&=s^nF(s) - s^{(n-1)} f(0) - \cdots - f^{(n-1)}(0)\\ t^n (n=0,1,2,\dots)&=\dfrac{n!}{s^{n+1}}\\ \sin kt&=\dfrac{k}{s^2+k^2}\\ \cos kt&=\dfrac{s}{s^2+k^2} \\ e^{at}&=\dfrac{1}{s-a}\\ t^ne^{at}&=\dfrac{n!}{(s-a)^{n+1}}\\ e^{at}\sin kt&=\dfrac{k}{(s-a)^2+k^2}\\ e^{at}\cos kt&=\dfrac{s-a}{(s-a)^2+k^2}\\ t\sin kt&=\dfrac{2ks}{(s^2+k^2)^2}\\ t\cos kt&=\dfrac{s^2-k^2}{(s^2+k^2)^2} \\\end{aligned}\end{split}\]

First translation theorom:

\[\begin{split}\begin{aligned} {\ensuremath{\mathcal{L}{\left[e^{at}f(t)\right]}}}&=F(s-a)\\\end{aligned}\end{split}\]

\[\begin{split}\begin{aligned} \int_{-\infty}^{\infty} f(x) \delta(x-a) \,dx&=f(a)\\ \int_{-\infty+\epsilon}^{\infty+\epsilon} f(x) \delta(x-a) \,dx&=f(a), \epsilon>0\\ \delta(x-a)&=0\\\end{aligned}\end{split}\]

Given some nonlinear system of the form:

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \dot{x}(t)&=f(x(t),u(t),t)\\ y(t)&=g(x(t),t)\end{aligned}\end{split}\\A linearization can be performed about nominal trajectories\end{aligned}\end{align} \]

x^0(t), u^0(t), and u^0(t) (shorthand is 0) by defining the jacobian of w.r.t. the states evaluated at the nominal trajectories (0) as:

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} A(t)= \frac{\partial f}{\partial x}\Bigr|_0= \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \dots & \frac{\partial f_1}{\partial x_n} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \dots & \frac{\partial f_2}{\partial x_n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial f_n}{\partial x_1} & \frac{\partial f_n}{\partial x_2} & \dots & \frac{\partial f_n}{\partial x_n} \\ \end{bmatrix}\Bigr|_0\end{aligned}\end{split}\\As well as the jacobian of w.r.t. the controls evaluated at the nominal\end{aligned}\end{align} \]

trajectories (0) as:

\[\begin{split}\begin{aligned} B(t)= \frac{\partial f}{\partial u}\Bigr|_0= \begin{bmatrix} \frac{\partial f_1}{\partial u_1} & \frac{\partial f_1}{\partial u_2} & \dots & \frac{\partial f_1}{\partial u_m} \\ \frac{\partial f_2}{\partial u_1} & \frac{\partial f_2}{\partial u_2} & \dots & \frac{\partial f_2}{\partial u_m} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial f_n}{\partial u_1} & \frac{\partial f_n}{\partial u_2} & \dots & \frac{\partial f_n}{\partial u_m} \\ \end{bmatrix}\Bigr|_0\end{aligned}\end{split}\]

Finally, the output equation may need to be linearized about 0 as well

\[\begin{split}\begin{aligned} C(t)= \frac{\partial g}{\partial x}\Bigr|_0= \begin{bmatrix} \frac{\partial g_1}{\partial x_1} & \frac{\partial g_1}{\partial x_2} & \dots & \frac{\partial g_1}{\partial x_n} \\ \frac{\partial g_2}{\partial x_1} & \frac{\partial g_2}{\partial x_2} & \dots & \frac{\partial g_2}{\partial x_n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial f_n}{\partial x_1} & \frac{\partial g_n}{\partial x_2} & \dots & \frac{\partial g_n}{\partial x_n} \\ \end{bmatrix}\Bigr|_0\end{aligned}\end{split}\]

Then the perterbation variables are defined as:

\[\begin{split}\begin{aligned} \delta x(t)=x(t)-x^0(t)\\ \delta u(t)=u(t)-u^0(t)\\ \delta y(t)=y(t)-y^0(t)\end{aligned}\end{split}\]

The final linearized system is:

\[\begin{split}\begin{aligned} \delta \dot{x}(t)&=A(t)\delta x(t)+B(t)\delta u(t)\\ \delta y(t)&=C(t)\delta x(t)\end{aligned}\end{split}\]

\[\begin{split}\begin{aligned} \dot{x}(t)&=A(t)x(t)\\ x(t)&=\Phi(t,t_0)x_0\\ \Phi(t,t_0)&=x(t)x(t_0)^-1\\\end{aligned}\end{split}\]

The system is

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \dot{x}(t)&=A(t)x(t)+B(t)u(t) \nonumber\\ y(t)&=C(t)x(t)\nonumber\\ x(t_0)&=x_0 \label{eq:sys1}\end{aligned}\end{split}\\where the solution is,\end{aligned}\end{align} \]

\[\begin{split}\begin{aligned} x(t)&=\Phi(t,t_0)x_0+\int_{t_0}^t \Phi(t,\tau)B(\tau)u(\tau)\,d\tau \label{eq:sol1}\\ y(t)&=C(t)\Phi(t,t_0)x_0+\int_{t_0}^t \underbrace{C(t)\Phi(t,\tau)B(\tau)}_{(G(t,\tau))} u(\tau)\,d\tau \nonumber\end{aligned}\end{split}\]

The above two equations are called the variation of constants formulas. They contain two terms, the first term is the free response which is due to x_0 and the second term is the forced resonse due to the input u(t). Additionally, the impulse response is defined as

\[\begin{aligned} G(t,\tau)=C(t)\Phi(t,t_0)B(\tau), \tau \leq t \label{eq:impulse_response}\end{aligned}\]

The basic equation is:

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \Phi(t,\tau)&=e^{A(t-\tau)}\\\end{aligned}\end{split}\\Which can be calculated using:\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} {\ensuremath{\mathcal{L}^{-1}{\left[(sI-A)^{-1}\right]}}}&=e^{At}\\\end{aligned}\end{split}\\After this, :math:`\tau` must be added in, which can be done with by\end{aligned}\end{align} \]

taking the inverse of \Phi(\tau,0) to get \Phi(0,\tau) and then multiplying by \Phi(t,0) as:

\[\begin{split}\begin{aligned} \Phi(t,0)\Phi(0,\tau)&=\Phi(t,\tau)\\\end{aligned}\end{split}\]

The STM is the unique solution to

\[ \begin{align}\begin{aligned} \begin{aligned} \frac{\partial}{\partial t}(\Phi(t,t_0))=A(t)\Phi(t,t_0)\end{aligned}\\with inital conditions :math:`\Phi(t_0,t_0) =I`.\end{aligned}\end{align} \]

To solve:

• multiply the above matrices out
• take the Laplace Transform of each element in the matrix
• solve the algebreic equation for each \Phi_{i,i}(s)
• take the inverse Laplace transfrom to find \Phi_{i,i}(t)

\[\begin{split}\begin{aligned} \Phi(t_0,t_0) &=I\\ \Phi(t,t_0)^-1 &=\Phi(t_0,t)\\ \Phi(t,t_0)&=\Phi(t,t_1)\Phi(t_1,t_0)\\\end{aligned}\end{split}\]

The system Eqn. [eq:sys1] is stable if, given x(t_0)=x_0, x(t) (Eqn. [eq:sol1]) is bounded is bounded \forall t \geq t_0. In this case,

\[lim_{t\to\infty} x(t)\]

may not go to zero. Stability can also be determined by looking at each (i,j) component of the STM as:

\[\begin{aligned} |\Phi_{ij}(t,t_0)| \leq k < \infty, \forall t_0 \leq t\end{aligned}\]

The system Eqn. [eq:sys1] is asymtotically stable if, given x(t_0)=x_0 x(t), x(t) (Eqn. [eq:sol1]) decays to zero, that is:

\[lim_{t\to\infty} x(t) = 0\]

The system, Eqn. [eq:sys1], is BIBO stable if when x_0=0, the forced output response y(y) to every bounded input u(t) is bounded. This can be determined as:

\[ \begin{align}\begin{aligned} \begin{aligned} \int_{-\infty}^{t} |G_{ij}(t,\tau)| d\tau \leq k < infty\end{aligned}\\where, :math:`G(t,\tau)` was defined in Eqn. [eq:impulse\_response] and\end{aligned}\end{align} \]

the above equation requires that G(t,\tau) is “absolutely integrable.”

Can we estimate a unique x(t_0)=x_0, given u(t) and y(t) over the time interval [t_0,t_1]? If we have x_0 we can solve

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} x(t)&=\Phi(t,t_0)x_0+\int_{t_0}^t \Phi(t,\tau)B(\tau)u(\tau)\,d\tau\\\end{aligned}\end{split}\\The state is unobservable for the unforced system (:math:`u(t)=0`), if:\end{aligned}\end{align} \]

\[\begin{aligned} y(t)=C(t)\Phi(t,t_0)x_0=0\end{aligned}\]

The observability matrix for a LTI system is:

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \mathcal{O}= \begin{bmatrix} C\\ CA\\ \vdots \\ CA^{n-1} \end{bmatrix}\end{aligned}\end{split}\\if the :math:`rank(\mathcal{O})=n` then the system is observable. The\end{aligned}\end{align} \]

unobservable states are in the null space of the observability matrix, i.e. \mathcal{O}x_0=0

Observability Gramian

\[ \begin{align}\begin{aligned} \begin{aligned} M(t_0,t_1)=\int_{t_0}^{t_1} \Phi^T(t,t_0)C^T(t)C(t)\Phi^T(t_0,t)dt\end{aligned}\\a state :math:`x_0=x(t_0)` unobservable at time :math:`t_0` iff\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned} \begin{aligned} M(t_0,t_1)x_0=0,\forall t_1>t_0\end{aligned}\\so, the unobservable states are in the null-space of the Observability\end{aligned}\end{align} \]

Gramian. If the only solution that lives in the null-space is the zero vector x(t_0)=0, then the system is completely observable. Also, note that ther is no need to carry out the complete integral to see that the Observability Gramian will have unobservable states in its null space, i.e. integration does not change the form of the matrix.

A system is controllable if we can find a u(t) that drives the state x(t) from x_0 in finite time t_f.

The observability matrix for a LTI system is:

\[ \begin{align}\begin{aligned} \begin{aligned} \mathcal{C}= \begin{bmatrix} B \; BA\; \dots\; BA^{n-1} \end{bmatrix}\end{aligned}\\if the :math:`rank(\mathcal{C})=n` then the system is controllable.\end{aligned}\end{align} \]

Recall that the rank of a matrix is the number of linearly independent columns.

Controllability Gramian

\[ \begin{align}\begin{aligned} \begin{aligned} W(t_0,t_1)=\int_{t_0}^{t_1} \Phi^T(t,t_0)B(t)B^T(t)\Phi^T(t_0,t)dt\end{aligned}\\this matrix is always symetric and positive definite. The system is\end{aligned}\end{align} \]

completly controllable if there exists t_1>t_0: W(t_0,t_1)>0. If the system is controllable at t_0, then one control that drives the state to the origin is:

\[\begin{aligned} u_0(t)=-B^T(t)\Phi^T(t_0,t)W^{-1}(t_0,t_1)x_0\end{aligned}\]

Given a LTV system as

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \dot{x}(t)&=A(t)x(t)+B(t)u(t)\\ y(t)&=C(t)x(t)\end{aligned}\end{split}\\its dual system is\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \dot{x}(t)&=-A^T(t)x(t)+C^T(t)u(t)\\ y(t)&=B^T(t)x(t)\end{aligned}\end{split}\\The controllable (or uncontrollable) states of one system are the\end{aligned}\end{align} \]

observable (or unobservable) states of the other system.

If x is a random vector with a PDF f(x) and g:{\mathbb{R}}_n \rightarrow {\mathbb{R}}_m is a function of x

\[\begin{split}\begin{aligned} E[g(x)]&=\int_{{\mathbb{R}}_n} g(x)f(x)\,dx \in {\mathbb{R}}_m\\\end{aligned}\end{split}\]

\[\begin{aligned} \bar{x}&=E[x]=\int_{{\mathbb{R}}_n} xf(x)\,dx \in {\mathbb{R}}_n\end{aligned}\]

\[\begin{split}\begin{aligned} P_{xx}&=E[(x-\bar{x})(x-\bar{x})^T]\\ &=\int_{{\mathbb{R}}_n} (x-\bar{x})(x-\bar{x})^T f(x)\,dx \in {\mathbb{R}}_{nxn}\\\end{aligned}\end{split}\]

Note: the covariance matrix is symmetric as well as positive semidefinite, so

\[ \begin{align}\begin{aligned} \begin{aligned} \forall v \in {\mathbb{R}}_n, v^TP_xv\geq0\end{aligned}\\Thus,\end{aligned}\end{align} \]

\[\begin{split}\begin{aligned} v^TP_xv&=\int_{R_n} v^T (x-\bar{x})(x-\bar{x})^Tvf(x)\,dx\\ &=\int_{R_n} (x-\bar{x})^2f(x)\,dx \geq 0\end{aligned}\end{split}\]

If x1 and x2 are subvectors of x

\[\begin{split}\begin{aligned} P_{x_1x_2}&=E[(x_1-\bar{x_1})(x_2-\bar{x_2})^T]\\ &=\int_{{\mathbb{R}}_n} (x_1-\bar{x_1})(x_2-\bar{x_2})^T f(x)\,dx \in {\mathbb{R}}_{n_1xn_2}\\\end{aligned}\end{split}\]

\[\begin{aligned} P^+=((P^-)^{-1}+C^TR^{-1}C)^{-1}\end{aligned}\]

The relative likelyhood that a random variable x will take on vaules on a given interval.

\[\begin{aligned} Area=P(a \leq x \leq b) = \int_a^b f_x(x) dx\end{aligned}\]

If you integrate over the PDF over the entire range then it must equal 1 and the PDF must always be greater than 0.

\[\begin{aligned} \int_{{\mathbb{R}}} f_x dx=1 \forall x, f_x(x) \geq 0\end{aligned}\]

For a PDF uniformly distributed over [a,b], f(x)=constant=c. c can then be determined by \int_a^b c dx=1 which results in f(x)=\frac{1}{b-a}.

Given the joint PDF f_x(x)=f_x(x_1,x_2), the marginal PDF of x_1 is:

\[\begin{aligned} f_{x_1}(x_1)=\int_{-\infty}^{\infty} f_x(x_1,x_2) dx_2\end{aligned}\]

\phi_x(s) is useful to compute the PDF for x and the Gaussian distribution. The expected value can be used to calculate the characteristic function as:

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \phi_x(s)&=E[e^{jx^Ts}]=\int_{{\mathbb{R}}_{n}} e^{jx^Ts}f(x)\,dx\\\end{aligned}\end{split}\\where :math:`j^2=-1` and :math:`s` is a complex vector of order\end{aligned}\end{align} \]

n

The statistal properties of x are equivalently specified by PDF f(x) or by the characteristic function \phi_x(s).

Similar to a Fourier Transform \phi_x(s) can be put back into the time domain with:

\[\begin{split}\begin{aligned} f(x)&=\frac{1}{(2 \pi)^n}\int_{{\mathbb{R}}_{n}} e^{jx^Ts} \phi_x(s),ds\\\end{aligned}\end{split}\]

If the following equations are true, then the PDF’s are independent.

\[\begin{split}\begin{aligned} f(x,y)&=f_x(x)f_y(y)\\ \phi_{xy}(s,r)&=\phi_x(s)\phi_y(r)\\\end{aligned}\end{split}\]

If x and y are independent, the conditional density function and conditional mean satisfy:

\[\begin{split}\begin{aligned} f(x|y)&=\frac{f(x,y)}{f_y(y)}=\frac{f_x(x)f_y(y)}{f_y(y)}=f_x(x)\\ E[x|y]&=\int_{{\mathbb{R}}_{n}} xf_x(x)\,dx=E[x]=\bar{x}\end{aligned}\end{split}\]

If x and y are independent, then they are uncorrelated. But, if they are uncorrelated, they may not be independent!

If x and y are uncorrelated, then they satisfy:

\[\begin{aligned} E[xy^T]=E[x]E[y^T]\end{aligned}\]

If x and y are uncorrelated, the cross-covariances must be zero:

\[\begin{split}\begin{aligned} P_{xy}&=E[(x-\bar{x})(y-\bar{y})^t]\\ &=E[xy^T-x\bar{y}^T-\bar{x}y+\bar{x}\bar{y}^T]\\ &=E[xy^T]-E[x]\bar{y}^T-\bar{x}E[y^T]+\bar{x}\bar{y}^T\\ &=E[xy^T]-\bar{x}\bar{y}^T-\bar{x}\bar{y}^T+\bar{x}\bar{y}^T\\ &=E[xy^T]-\bar{x}\bar{y}^T=0\\\end{aligned}\end{split}\]

Why model the probablity densit function as a Gaussian (or normal) distribution:

• provides a good statistical model for many natural phenomina
• computationally tractable because the statistical properties are described completely by first (mean, \bar{x}) and second (variance, P) moments
• normality is preserved through linear transforms (both static and dynamic)

A random vector x \in {\mathbb{R}}_n is Gaussian distributed or normal if the characteristic function has the form:

\[ \begin{align}\begin{aligned} \begin{aligned} \phi_x(s)=e^{j\bar{x}^Ts-\frac{1}{2}s^TPs}\end{aligned}\\where :math:`s \in \, C_n`, :math:`\bar{x}=E[x]`, and\end{aligned}\end{align} \]

P=[(x-\bar{x})(x-\bar{x})^T].

• For such a random vector, we use the notation x=N(\bar{x},P).

• Two vectors x and y are jointly Gaussian distributed if (x^T,y^T) is Gaussian.

• When a random vector x is Gaussian and it’s covariance matrix P_x is nonsingular, it’s PDF can be evaluated with:

  \[\begin{aligned} f(x)=\frac{e^{-\frac{1}{2}(x-\bar{x})^TP_x^{-1}(x-\bar{x})}}{\sqrt{(2\pi)^ndet(P_x)}}\end{aligned}\]

If P_x is singular, then the above equation will not work, but the characteristic function can define the Gaussian distribution indirectly.

In a random process, we are looking at a family of random vectors (x(t),t\in I) indexed by time.

\[\begin{aligned} \bar{x}(t)&=E[x(t)], t \in I\end{aligned}\]

\[\begin{aligned} P(t,\tau)=E[(x(t)-\bar{x}(t))(x(t)-\bar{x}(t))^T],t \in I\end{aligned}\]

For two random processes x(t) and y(t)

\[\begin{aligned} P_{xy}(t,\tau)=E[(x(t)-\bar{x}(t))(y(t)-\bar{y}(t))^T],t \in I\end{aligned}\]

Which satisfies:

\[\begin{aligned} P(t)=E[x(t)x^T(t)]-\bar{x}(t)\bar{x}^T(t)\end{aligned}\]

Which satisfies:

\[\begin{aligned} P_{xy}(t)=E[x(t)y^T(t)]-\bar{x}(t)\bar{y}^T(t)\end{aligned}\]

todo..

todo..

This section combines the ideas of Gaussian distribution and random process

• a random process is x(t) is Gaussian if all of the vectors x_1(t),…x_n(t) are jointly Gaussian

• a Gaussian random process is white if the vectors x(t_1), …x(t_m) are independent, otherwise it is colored

• for a Gaussian and white process, the covariance kernal satisfies

  \[\begin{split}\begin{aligned} P(t,\tau)&=0,t \neq \tau\\ P(t,\tau)&=Q(t)\delta(t-\tau)\end{aligned}\end{split}\]

• a random process is Markov if

  \[\begin{aligned} &f(x(t_m)|x(t_{m-1}),...,x(t_1))=f(x(t_m)|x(t_{m-1}))\end{aligned}\]

• a random process is Gauss-Markov if it is both Gauss and Markov

The standard model is:

\[\begin{split}\begin{aligned} \dot{x}(t)&=A(t)x(t)+B(t)u(t)+w(t), t \geq t_0\\ y(t)&=C(t)x(t)+v(t)\label{eq:sys2}\end{aligned}\end{split}\]

1. the intial condition x(t_0) is Gaussian

   \[\begin{aligned} x(t_0)&=N(\bar{x}(t_0),P(t_0))\end{aligned}\]

2. the disturbance w(t) is a zero-mean, Gaussian, white process that is independent of x(t_0)

   \[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} E[w(t)]&=0\\ E[w(t)w^T(\tau)]&=\underbrace{R_w(t)\delta(t-\tau)}_{covariance\;kernal}\\ E[w(t)(x(t_0-\bar{x}(t_0))^T]&=0\end{aligned}\end{split}\\**EX:** in the case of a :math:`2X2` system, if there is some\end{aligned}\end{align} \]

   covariance \sigma_w given for the second state variable then

   \[\begin{split}\begin{aligned} R_w(t)= \begin{bmatrix} 0 &0\\ 0 &\sigma_w \end{bmatrix}\end{aligned}\end{split}\]

3. the measurment noinse v(t) is a zero-mean, Gaussian, white process that is independent of x(t_0)

   \[\begin{split}\begin{aligned} E[v(t)]&=0\\ E[v(t)v^T(\tau)]&=R_v(t)\delta(t-\tau)\\ E[v(t)(x(t_0-\bar{x}(t_0))^T]&=0\end{aligned}\end{split}\]

4. the processes v(t) and w(t) are uncorrelated

   \[\begin{aligned} E[w(t)v^T(t)]=0\end{aligned}\]

With the standard model and assumptions, the process x(t) is Markov.

Recall the variation of consants formula:

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} x(t)&=\Phi(t,t_0)x(t_0)+\int_{t_0}^{t} \Phi(t,\tau) B(\tau)u(\tau)d\tau +...\\ &+ \int_{t_0}^{t}\Phi(t,\tau)w(\tau)d\tau \end{aligned}\end{split}\\Then,\end{aligned}\end{align} \]

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} x(t_m)&=\Phi(t_m,t_{m-1})x(t_{m-1})+...\\ &+ \int_{t_{m-1}}^{t_m} \Phi(t_m,\tau)(B(\tau)u(\tau)+ w(\tau))d\tau \end{aligned}\end{split}\\Notice that the results does not depend on :math:`x(\tau)`,\end{aligned}\end{align} \]

\tau < t_{m-1}.

Also refered to as the expected value.

\[\begin{aligned} \bar{x}(t)=\Phi(t,t_0)\bar{x}(t_0)+\int_{t_0}^{t} \Phi(t,\tau) B(\tau)u(\tau)d\tau \end{aligned}\]

\[\begin{aligned} \bar{y}(t)=E[C(t)x(t)+v(t)]=C(t)\bar{x}(t)\end{aligned}\]

Using x(t)-\bar{x}(t) we can define:

\[ \begin{align}\begin{aligned} \begin{aligned} P(t)=\Phi(t,t_0)P(t_0)\Phi^T(t,t_0)+\int_{t_0}^{t} \Phi(t,\tau) R_w(\tau)\Phi^T(t,\tau)d\tau \end{aligned}\\The above equation also satisfies the Lyapunov Equation.\end{aligned}\end{align} \]

For the output y(t):

\[\begin{aligned} P_y(t)=C(t)P(t)C^T(t)+R_v(t)\end{aligned}\]

with the standard model and assumptions, the covariance matrix satisfies:

\[ \begin{align}\begin{aligned} \begin{aligned} \dot{P}(t)&=A(t)P(t)+P(t)A^T(t)+R_w(t)\end{aligned}\\Which can be derived using the **Leibniz Integral Rule**.\end{aligned}\end{align} \]

To find the steady state covariance matrix:

• set \dot{P}(t)=0
• then solve for P_{ss}
  • this will require a computer!

For the state x(t):

\[\begin{aligned} P_x(t,\tau)=\Phi(t,t_0)P(t_0)\Phi^T(t,t_0)+\int_{t_0}^{t} \Phi(t,\sigma) R_w(\sigma)\Phi^T(t,\sigma)d\sigma \end{aligned}\]

For the output y(t):

\[\begin{split}\begin{aligned} P_y(t,\tau)=C(t)\Phi(t,t_0)P(t_0)\Phi^T(t,t_0)C^T(\tau)+ \\ \int_{t_0}^{t} C(t)\Phi(t,\sigma) R_w(\sigma)\Phi^T(t,\sigma)C^T(\tau)d\sigma +R_v(t)\delta(t-\tau) \end{aligned}\end{split}\]

After evaluating \frac{f(x,y)}{f_y(y)}, the result is PDF of a Guassian vector defined as:

\[ \begin{align}\begin{aligned} \begin{aligned} f(x|y)=\frac{e^{-\frac{1}{2}(x-E^T[x|y])^TP^{-1}_{x|y}(x-E[x|y])}}{\sqrt{(2\pi)^ndet(P_{x|y}})}\end{aligned}\\with a mean and covariance defined as follows:\end{aligned}\end{align} \]

\[\begin{split}\begin{aligned} E[x|y]&=\bar{x}+P_{xy}P^{-1}_y(y-\bar{y})\\\end{aligned}\end{split}\]

\[\begin{aligned} P_{x|y}&=P_x+P_{xy}P^{-1}_yP_{yx}\end{aligned}\]

Basic estimation procedure is:

1. determine \hat{x}^-=\bar{x} which is an estimate of the state
   • based off of state equations
   • affected by state uncertainty, w(t)
2. collect measurments from output equation y(t)
3. update estimate of state (\hat{x}^-) based off of new info from y(t)
   • based off of output equation
   • affected by sensor uncertainty, v(t)

Measurments are incorporated simultaneously.

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \hat{x}^+&=\hat{x}^-+k(z-C\hat{x}^-)\\ P^+&=P^-C^T(CP^-C^T+R)^{-1}\end{aligned}\end{split}\\where, :math:`k=P^-C^T(CP^-C^T+R)^{-1}` or in an algebraically\end{aligned}\end{align} \]

equivalent form,

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \hat{x}^+&=(P^+(P^-)^{-1})\hat{x}^-+(P^+C^TR^{-1})z\\ P^+&=((P^-)^{-1}+C^TR^{-1}C)^{-1}\end{aligned}\end{split}\\where :math:`R` is the covariance of the sensor measurment from\end{aligned}\end{align} \]

y(t)

also note: z=y

Measurments are incorporated recursively.