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UNSW MMAN3200 Notes.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} {\large MMAN3200 Notes} \begin{align*} \text{\bf Linear}&\text{ \bf Time Invariant (LTI):}\\ &\frac{d^ny(t)}{dt^n}=\sum_{i=1}^{n-1}a_i\cdot\frac{d^iy(t)}{dt^i}+\sum_{i=1}^m b_i\cdot\frac{d^ix(t)}{dt^i}+a_0\cdot y(t)+b_0\cdot x(t)\\ % &\left\{a_i\right\}_{i=0}^{n-1},\quad\left\{b_i\right\}_{i=0}^m\quad\text{are independent of $x$ and $y$ (but technically can be time dependent as an ODE).}\\ &\quad\text{If in addition, they are independent of $t$ (time) as well, the system is LTI.}\\ \\ \text{\bf Super}&\text{\bf position Principle:}\\ &\text{If $y_1(t)$ and $y_2(t)$ are the responses of $x_1(t)$ and $x_2(t)$ respectively, we know that $x(t)=x_1(t)+x_2(t)$ gives a}\\ &\text{response of $y(t)=y_1(t)+y_2(t)$.\quad Generally,\quad$x(t)=\sum_{k=1}^Nc_k\cdot x_k(t)$\quad gives the response\quad$y(t)=\sum_{k=1}^Nc_k\cdot y_k(t)$.}\\ % &\text{Recall the Fourier Series:}\quad f(x)=\sum_{n=-\infty}^\infty c_n~e^{inx}~,\qquad \text{where}\quad c_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)~e^{-inx}~dx.\quad\text{(MATH2019)}\\ &x(t)=\sum_{k=-\infty}^\infty c_k\cdot\phi_k(t),\quad\text{where}\quad\phi_k(t)=e^{j\cdot\frac{2\pi}{T}\cdot k\cdot t}. \quad\text{(It can be viewed that $x=\frac{2\pi}{T}t$)}\\ % &\text{To generalise to integrals of distributions:}\quad x(t)=\int_{\mu=a}^bc(\mu)\cdot x_\mu(t)\cdot d\mu,\\ &\text{which will give the response}\quad y(t)=\int_{\mu=a}^bc(\mu)\cdot y_\mu(t)\cdot d\mu,\quad \text{where $y_\mu(t)$ is the response of input $x_\mu(t)$.}\\ \\ \text{\bf Fourier}&\text{ \bf Transformation:}\\ &x(t)=\int_{-\infty}^{+\infty}x[\omega]\cdot e^{j\omega t}\cdot d\omega,\quad\text{where $x[\omega]$ is the Fourier Transform of $x(t)$.}\quad \text{It can be interpreted that $x[\omega]$}\\ &\text{``amplifies'' frequency $\omega$ in the continuous range from $-\infty$ to $+\infty$, and the result of the integral is the function $x(t)$.}\\ \\ &\text{Let $H(\omega)$ be the ratio of the input and response at frequency $\omega$. The Frequency Response is $H(\omega)\cdot e^{j\omega t}$,}\\ &\text{which is the response to the ``pure'' harmonic function $e^{j\omega t}$ (without being ``amplified'' by $x[\omega]$).}\\ % &\text{The response to $x(t)$ is therefore}\quad y(t)=\int_{-\infty}^{+\infty}x[\omega]\cdot H(\omega)\cdot e^{j\omega t}\cdot d\omega\\ \\ \text{\bf Convolu}&\text{\bf tion:}\\ &\text{Run a function backward, shifted by $t$, then times it with another function and integrate the result over the domain.}\\ &\boxed{(x*y)(t)=\int_{-\infty}^{+\infty}x(\tau)\cdot y(t-\tau)~d\tau.}\qquad \text{As}\quad\int_{-\infty}^{+\infty}x(\tau)\cdot y(t-\tau)~d\tau=\int_{-\infty}^{+\infty}y(\tau)\cdot x(t-\tau)~d\tau,\quad\boxed{x*y=y*x.}\\ &\text{convolution theorem for laplace transform:}\quad\mathcal{L}(x*y)=\mathcal{L}(x)\cdot\mathcal{L}(y).\\ &\text{If}\quad z(t)=x(t)*y(t),\quad\text{then}\quad z[\omega]=x[\omega]\cdot y[\omega].\qquad \text{Also, if}\quad z(t)=x(t)\cdot y(t),\quad\text{then}\quad z[\omega]=x[\omega]*y[\omega].\\ % &\text{Impulse response: Let $\delta(t)$ be $M$ (magnitude) when $t\in[-\epsilon,+\epsilon]$, and zero otherwise, and also}\quad \int_{-\infty}^{+\infty}\delta(t)~dt=1.\\ &\text{Sampling:}\quad\lim_{\epsilon\to 0}\int_{-\infty}^{+\infty}x(t)\cdot\delta(t-a)~dt=x(a).\quad\text{(The limit is often implicit.)}\quad \text{Since $\delta$ is even, $\delta(t-\tau)=\delta(\tau-t)$.}\\ &\text{Putting $t$ as $\tau$ and $a$ as $t$, we will have}\quad x(t)=\int_{-\infty}^{+\infty}x(\tau)\cdot\delta(t-\tau)~d\tau=(x*\delta)(t).\\ &\text{The response to $x(t)$ will be $y(t)=(x*h)(t)$, where $h(t)$ is the response to impulse $\delta(t)$.\quad Also}\quad \mathcal{L}(y)=\mathcal{L}(x)\cdot\mathcal{L}(h).\\ % % %&x(t)=\int_{-\infty}^{+\infty}x[s]~e^{j\omega t}~ds,\quad %x(t)*y(t)=\int_{-\infty}^{+\infty}\left(\int_{-\infty}^{+\infty}x[s]~e^{jst}~ds\right)\cdot\left(\int_{-\infty}^{+\infty}y[s]~e^{js(t-\tau)}~ds\right)~d\tau. %\\ %\\ %&\mathcal{L}(x)=X(s)=\int_0^\infty e^{-st}x(t)dt %=\int_0^\infty e^{-st}\left(\int_{-\infty}^{+\infty}x[\omega]~e^{j\omega t}d\omega\right)dt %=\int_0^\infty\int_{-\infty}^{+\infty}x[\omega]~e^{j\omega t}~e^{-st}~d\omega~dt\\ %&\qquad=\int_{-\infty}^{+\infty}\int_0^\infty x[\omega]~e^{j\omega t}~e^{-st}~dt~d\omega\\ \end{align*} % % % \begin{align*} \text{\bf Impe}&\text{\bf dence:}\quad\text{Input $=$ Impedance $\times$ Output.}\\ &\text{Electricity:}\\ &v=R~i,~~V(s)=[R]\cdot J(s)~~\ldots~~\text{Impedance is }R.\\ &v=\frac{1}{C}\int i~dt,~~V(s)=\left[\frac{1}{Cs}\right]\cdot J(s).\\ &v=L\frac{di}{dt},~~V(s)=[Ls]\cdot J(s).\\ \\ &\text{Heat:}\\ &T_1-T_2=\left[-\frac{d}{KA}\right]\cdot q,\quad q=h_c A(T_1-T_2),~~ \text{where the surface convective coefficient of heat transfer }h_c=-\frac{K}{d}.\\ &q=C_t~\frac{dT}{dt},\quad\text{where the thermal capacitance }C_t=\rho cV.\\ &T=\frac{1}{C_t}\int q~dt,\quad T(s)=\left[\frac{1}{C_t}\right]\cdot Q(s).\\ \\ &\text{Fluid:}\\ &P_1-P_2=[R_f]~q.\\ &\text{The bulk modulus }\beta\equiv\rho_0\cdot\frac{p-p_0}{\rho-\rho_0}=\rho_0\cdot\frac{\dot{p}}{\dot{\rho}}~~.\quad \dot{p}=\beta\cdot\frac{\dot{\rho}}{\rho_0}=\frac{\beta}{\rho_0}\cdot\frac{\dot{m}}{V},\quad \dot{p}=\left[\frac{\beta}{V\rho_0}\right]\dot{m}.\quad \text{($\dot{m}$ is mass flow rate.)}\\ &\text{Gas:}\quad pV=mRT,\quad \dot{p}=\left[\frac{RT}{V}\right]\dot{m}.\\ &\text{Inertia (liquid of density $\rho$ through a tube of length $l$ and cross section $A$ with pressure $p_1$ and $p_2$ at two ends):}\\ &\text{Volumetric flow rate }q=vA,\quad F=ma,\quad (p_1-p_2)A=(\rho Al)\ddot{l}=(\rho Al)\dot{v}=(\rho Al)\frac{\dot{q}}{A}=\rho l\dot{q}.\quad \therefore\dot{p}=\left[\frac{\rho l}{A}\right]\dot{q}.\\ \\ \text{\bf Exam}&\text{\bf ples}\\ &\text{In a mass spring system with an external force: }F=M\ddot{x}+kx.\quad F(s)=M(s^2X(s)-sx_0-\dot{x_0})+kX(s).\\ &\text{If the system is initially at its neutral point at rest, $x_0=0$ and $\dot{x_0}=0$.}\quad F(s)=X(s)(Ms^2+k).\\ &\text{The transfer function (Output/Input): }G(s)=\frac{X(s)}{F(s)}=\frac{1}{Ms^2+k.}\\ &\text{If $F$ is a unit impulse }\delta(t), F(s)=\mathcal{L}\{\delta(t)\}=1.\quad X(s)=\frac{1}{Ms^2+k},\\ &x(t)=\mathcal{L}^{-1}\left\{\frac{1}{Ms^2+k}\right\} =\mathcal{L}^{-1}\left\{\frac{1}{\sqrt{Mk}}\cdot\frac{\sqrt{\frac{k}{M}}}{s^2+\frac{k}{M}}\right\} =\frac{1}{\sqrt{Mk}}\cdot\sin\left(\sqrt{\frac{k}{M}}\cdot t\right).\\ \\ &\text{Let $A=\frac{1}{M}$, $a=\sqrt{-\frac{k}{M}}$ and $b=-\sqrt{-\frac{k}{M}}$.}\quad ab=\frac{k}{M}.\\ &\text{Non-dimensional form: }G(s) =\frac{\frac{1}{k}}{\frac{M}{k}s^2+1} =\frac{\frac{1}{k}}{\left(\sqrt{-\frac{M}{k}}s+1\right)\left(-\sqrt{-\frac{M}{k}}s+1\right)} =\frac{\frac{A}{ab}}{\left(\frac{s}{a}+1\right)\left(\frac{s}{b}+1\right)}.\\ &\text{Normalised form: }G(s) =\frac{\frac{1}{M}}{\left(s+\sqrt{-\frac{k}{M}}\right)\left(s-\sqrt{-\frac{k}{M}}\right)} =\frac{A}{(s+a)(s+b)}.\\ % %&\text{Steady state gain: }\lim_{t\to\infty}\frac{x(t)}{f(t)} %=\lim_{s\to 0}\frac{sX(s)}{\delta(s)} %=\lim_{s\to 0}\frac{sX(s)}{s} %=\lim_{s\to 0}G(s)F(s) %=\lim_{s\to 0}\left(\frac{A}{(s+a)(s+b)}\cdot 1\right) %=\frac{A}{ab} %=\frac{1}{k}.\\ % \end{align*} \end{document}