% !TeX program = xelatex % !TeX encoding = utf8 % !TeX root = komfour_zf.tex %% TODO: publish to CTAN \documentclass[twocolumn, margin=small]{tex/hsrzf} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Packages %% TODO: publish to CTAN \usepackage{tex/hsrstud} %% Language configuration \usepackage{polyglossia} \setdefaultlanguage[variant=uk]{english} %% Math \usepackage{amsmath} \usepackage{amsthm} %% Layout \usepackage{enumitem} %% License configuration \usepackage[ type={CC}, modifier={by-nc-sa}, version={4.0}, lang={english}, ]{doclicense} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Metadata \course{Electrical Engineering} \module{KomFour} \semester{Spring Semester 2020} \authoremail{npross@hsr.ch} \author{Naoki Pross -- \texttt{\theauthoremail}} \title{Cheat sheets for \texttt{\themodule}} \date{\thesemester} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Macros and settings %% Equal by definition \newcommand\defeq{\overset{\mathrm{def.}}{=}} %% number sets \newcommand\Nset{\mathbb{N}} \newcommand\Zset{\mathbb{Z}} \newcommand\Qset{\mathbb{Q}} \newcommand\Rset{\mathbb{R}} \newcommand\Cset{\mathbb{C}} %% Missing operators \DeclareMathOperator\sgn{sgn} %% Complex operators \DeclareMathOperator\cjs{cjs} \newcommand\cjsl[1]{\cos #1 + j\sin #1} \newcommand\ej[1]{e^{j#1}} \newcommand\conj[1]{\overline{#1}} \newcommand\len[1]{\lvert#1\rvert} \renewcommand\Re{\operatorname{Re}} \renewcommand\Im{\operatorname{Im}} %% Theorems \newtheoremstyle{komfourzf} % name of the style to be used {\topsep} {\topsep} {} {0pt} {\bfseries} {.} { } { } \theoremstyle{komfourzf} \newtheorem{theorem}{Theorem} \newtheorem{definition}{Definition} \newtheorem{lemma}{Lemma} \setlist[description]{% align=right, labelwidth=2cm, leftmargin=!, % format={\normalfont\itshape}} \setlist[itemize]{% align=right, labelwidth=5mm, leftmargin=!} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Document \begin{document} \section{Complex Numbers} \begin{definition}[Complex Unit and Zero] \[ j \defeq +\sqrt{-1} \iff j^2 = -1 \] \[ 1 = (1,0) \quad 0 = (0,0) \quad j = (0,1) \] \end{definition} \begin{definition}[Negation and Sum] Let \(z, w \in \Cset\) \[ -z = (-z_1, -z_2) \quad z \oplus w = (z_1 + w_1, z_2 + w_2) \] \end{definition} \begin{lemma} The complex numbers form an additive group. Let \(z, w, v \in \Cset\), we have \begin{description}[leftmargin=3cm] \item[Identity] \(z + 0 = z\) \item[Commutativity] \(z + w = w + z\) \item[Associativity] \(z + (w + v) = (z + w) + v\) \item[Inverse property] \(z + (-z) = (-z) + z = 0\) \end{description} \end{lemma} \begin{definition}[Multiplication] Let \(z, w \in \Cset\) \[ (a,b) \odot (c,d) = (ac - bd, ad + bc) \] \end{definition} \begin{lemma} The complex numbers form a commutative ring. Let \(z,w,v \in\Cset\) \begin{description}[leftmargin=3cm] \item[Identity] \(1\cdot z = z\) \item[Commutativity] \(z \cdot w = w \cdot z\) \item[Associativity] \(z (w v) = (z w) v\) \item[Distributivity] \(z (w + v) = zw + zv\) \end{description} \end{lemma} \begin{definition}[Real and imaginary part and conjugation] Let \(z = a + jb\). The \emph{real} part of \(z\) is \(\Re(z) = a\), similarly the \emph{imaginary} part is \(\Im(z) = b\). We can thus define the \emph{complex conjugate} \(\conj{z}\) of \(z\) to be \[ z = \Re(z) + j\Im(z) \quad \conj{z} = \Re(z) - j\Im(z) \] \end{definition} \begin{definition}[Absolute value] If \(z = a + jb\) we define the \emph{absolute value} \(\len{z} = \sqrt{a^2 + b^2}\) \end{definition} \begin{lemma}[Properties of absolute value] Let \(z,w\in\Cset\). We have \(z\conj{z} = \len{z}^2\) and as a consequence \(\len{zw} = \len{z}\cdot\len{w}\) and \(\len{\conj{z}} = \len{z}\). In addition we have the inequalities \begin{align*} -\len{z} \leq &\Re(z) \leq \len{z} & \len{z} &\leq \len{\Re(z)} + \len{\Im(z)} \\ -\len{z} \leq &\Im(z) \leq \len{z} & \len{z + w} &\leq \len{z} + \len{w} \end{align*} The last one is the \emph{triangle inequality}. Notice that \(\len{z} \in\Rset^+_0\). \end{lemma} \begin{definition}[Reciprocal and quotients] If \(z\) is a non-zero complex number we define the \emph{reciprocal} \(z^{-1}\) of \(z\) to be \(z^{-1} = \len{z}^{-2}\conj{z}\). If \(z = 0\) the reciprocal \(0^{-1}\) is left undefined. It is now possible to define \(z/w = zw^{-1}\) with \(z,w \in\Cset\) and \(w \neq 0\). \end{definition} \begin{lemma}[Properties of conjugation] Let \(z,w \in\Cset\). \(\conj{z} = z\) iff \(z \in \Rset\) and \(\conj{z} = \conj{w}\) iff \(z = w\). Furthermore: \begin{align*} \conj{\conj{z}} &= z & \conj{z \pm w} &= \conj{z} \pm \conj{w} & \Re(z) &= (z + \conj{z})/2 \\ \conj{z\cdot w} &= \conj{z}\cdot\conj{w} & \conj{z/w} &= \conj{z}/\conj{w} & \Im(z) &= (z - \conj{z})/2j \end{align*} \end{lemma} \begin{definition}[Argument and polar notation] An alternative representation of a complex number \(z = a + jb\) is its \emph{polar form} \(z = r \angle \phi\), where \(r = \len{z}\) and \(\phi = \arg{z}\). \begin{align*} a &= r\cos\phi & b &= r \sin\phi & r &= \sqrt{z\conj{z}} \end{align*} For \(a = 0\) we define \(\phi = \lim_{a\to 0} \arctan(b/a) = \pm\pi/2\) and otherwise \begin{align*} \phi = \arg(z) &= \begin{cases} \arctan(b/a) & a > 0 \\ \arctan(b/a) + \pi & a < 0 \end{cases} \\ &= \begin{cases} \arccos(a/r) & b \geq 0 \\ -\arccos(b/r) & b < 0 \\ \end{cases} \end{align*} Another variant of this notation is \[ z = r\cjs\phi = r(\cos\phi + j\sin\phi) \] \end{definition} \begin{lemma}[Arithmetic in polar notation] Let \(z,w\in\Cset\) then the product \(zw\) has \[ \len{zw} = \len{z}\cdot\len{w} \quad \arg(zw) = \arg z + \arg w \] Similarly the quotient \(z/w\) follows \[ \len{z/w} = \len{z}/\len{w} \quad \arg(z/w) = \arg z - \arg w \] Lastly from the product we see that for \(k \in \Nset\) \[ \len{z^k} = \len{z}^k \quad \arg{z^k} = k \arg{z} \] \end{lemma} \begin{theorem}[De Moivre's formula] Let \(n \in\Nset\) \[ \left(\cos\phi + j\sin\phi\right)^n = \cos(n\phi) + j\sin(n\phi) \] As a consequence with the binomial formula \((a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^n\), recalling that \(\binom{n}{k} = n!/(k!(n-k)!)\) (Pascal's triangle), we have \begin{align*} \sin(nx)&=\sum _{k=0}^{n}{\binom {n}{k}}(\cos x)^{k}\,(\sin x)^{n-k}\,\sin {\frac {(n-k)\pi }{2}}\\ \cos(nx)&=\sum _{k=0}^{n}{\binom {n}{k}}(\cos x)^{k}\,(\sin x)^{n-k}\,\cos {\frac {(n-k)\pi }{2}} \end{align*} \end{theorem} \section{Complex valued functions} \begin{definition}[Function in \(\Cset\)] Let \(f: \mathbb{D} \to \mathbb{W}\) with both \(\mathbb{D}, \mathbb{W} \subseteq \Cset\) that maps \(z = (a + jb) \mapsto w = (u + jv)\), then \(u = \Re f(z)\) and \(v = \Im f(z)\). If \(f\) is a bijection with inverse \(f^{-1}\), then \(a = \Re f^{-1}(w), b = \Im f^{-1}(w)\). \end{definition} \begin{definition}[Differentiation in \(\Cset\)] Let \(f\) be a function of \(z\) and \(h \in \Cset\). We have the limit \[ \lim_{\len{h} \to 0} \frac{f(z_0 + h) - f(z_0)}{h} = f'(z_0) \] to define the \emph{derivative} of \(f\) at the point \(z_0\). \end{definition} \begin{lemma}[Local dilation and rotation] Let \(f\) be a differentiable function in \(\Cset\). If \(f'(z) \neq 0\) everywhere, then \(f\) is a conformal map (i.e. preserves angles) with local dilation of \(\len{f'(z)}\) and rotation of \(\arg f'(z)\) \end{lemma} \begin{definition}[Linear function] \end{definition} \begin{definition}[Monomial and \(n\)-th root] Let \(w = z^n\) be a monomial of degree \(n\in\Nset\). Using the polar notation we see that \((r\angle \phi)^n = r^n \angle (n\phi)\). Because \(r\angle\phi = r\angle(\phi+2\pi)\) there cannot be a bijection between \(w\) and \(z\), if we want to define an inverse function \(z = \sqrt[n]{w}\) we get many values with the form \[ z_k = \sqrt[n]{r}\angle(\phi + k2\pi)/n \qquad 0 \leq k < n \] This fact implies that in general for \(z,u \in\Cset\) \(\sqrt[n]{zu} \neq \sqrt[n]{z}\sqrt[n]{u}\), as the relationship holds only for \emph{some} values of \(\sqrt[n]{z} \text{ and } \sqrt[n]{u}\). \end{definition} \begin{theorem}[Roots of a polynomial] Every complex polynomial of degree \(n\) has always \(n\) roots in \(\Cset\). \end{theorem} \begin{theorem} Every complex polynomial of degree \(n\) with coefficients can be \emph{uniquely} rewritten in term of its roots. \[ P(z) = \sum_{k=0}^n a_k z^k = a_n \prod_{k=0}^{n} (z - z_k) \] \end{theorem} \begin{theorem}[Polynomal with real coefficients] The roots of a polynomial with real coefficients of degree \(n\), always come in conjugate complex pairs of \(r\) and \(\conj{r}\). That is because \[ (z - r)(z - \conj{r}) = z^2 - 2\Re(r)z + \len{z}^2 \] \end{theorem} \begin{lemma} From the previous theorem follows that a polynomial with real coefficients of \emph{odd} degree, has \emph{always} at least one real solution because \(r \in\Rset \iff r = \conj{r}\). \end{lemma} \begin{theorem} All roots of a polynomial \(p(z) = \sum_{k=0}^n a_k z^k\) are inside of the open disk centered at the origin of radius \(\sum_{k=0}^n \len{a_k / a_n}\). \end{theorem} \begin{theorem}[Cardano's cubic formula] % TODO \end{theorem} \begin{definition}[Exponential] If \(z\) is a complex number we define the exponential function \(e^z\) by its convergent power series \[ e^z = \sum_{n=0}^\infty \frac{z^n}{n!} \] \end{definition} \begin{theorem}[Euler's formula] By setting the argument of the exponential function to \(jt\) for some \(t \in\Rset\) we can reorder the power series to be a sum of the power series of \(j\sin\) and \(\cos\), and thus define \[ e^{jt} = \cos t + j\sin t = \cjs t = 1\angle t \] \end{theorem} \begin{lemma}[Rules for exponents] Let \(a,b \in\Cset\) and \(k\in\Zset\), we can show that \[ e^a e^b = e^{a+b} \quad e^a / e^b = e^{a-b} \quad \left(e^a \right)^k = e^{ak} \] \end{lemma} \begin{definition}[Trigonometric functions] When \(z\) is a complex number we define \begin{align*} \cos z &= \frac{e^{jz} + e^{-jz}}{2} & \sin z &= \frac{e^{jz} - e^{-jz}}{2j} \end{align*} like the (real) hyperbolic trigonometric functions \begin{align*} \cosh z &= \left( e^z + e^{-z} \right)/2 & \sinh z &= \left( e^z - e^{-z} \right)/2 \end{align*} Notice that the sinus function is point symmetric to \(\pi/2\), because \(\sin(\pi/2 - z) = \sin(\pi/2 + z)\). \end{definition} \begin{lemma}[Some trigonometric identities] Let \(x,a,b \in\Rset\) and \(\alpha,\beta \in\Cset\) \begin{align*} \sinh(jx) &= j\sin(x) \qquad \cosh(jx) = \cos(x) \\ \sin(a + jb) &= \sin(a)\cosh(b) + j\cos(a)\sinh(b) \\ \cos(a + jb) &= \cos(a)\cosh(b) + j\sin(a)\sinh(b) \\ 2\sin(\alpha)\sin(\beta) &= \cos(\alpha - \beta) - \cos(\alpha + \beta) \\ 2\sin(\alpha)\cos(\beta) &= \sin(\alpha - \beta) + \sin(\alpha + \beta) \end{align*} \end{lemma} \begin{lemma}[Superposition of sinuses] Let \(s(t) = A\sin(\omega t + \varphi)\) be a sinusoidal wave. We can rewrite \(s\) in complex form with \[ s(t) = \Im\left(Ae^{j(\omega t + \varphi)}\right) = \Im Ae^{j\varphi}\cdot e^{j\omega t} \] If we now wish to sum \(N\) sinusoids with the same frequency \(\omega\), we can set % TODO: Satz 18 \end{lemma} \begin{definition}[Logarithm] Because \(w = e^z\) defined from \(\Cset \to \Cset\) is not a bijection (\(e^{z + 2\pi j} = e^z\)), unless we restrict the imaginary part of the domain to \((\pi, \pi]\), we get only an equivalence relationship because \[ \ln\left[\len{w} e^{j(\phi + k2\pi)}\right] = \ln\len{w} + j(\phi + k2\pi) \] where \(k \in\Zset\). Similarly for \(z,w\in\Cset\) \begin{align*} \ln(w) &\equiv z &\pmod{2\pi j} \\ \ln(w^k) &\equiv k\ln(w) &\pmod{2\pi j} \\ \ln(zw) &\equiv \ln(z) + \ln(w) &\pmod{2\pi j} \\ \ln(z/w) &\equiv \ln(z) - \ln(w) &\pmod{2\pi j} \end{align*} \end{definition} \begin{lemma}[General exponentiation] So far we have only exponentiation for an exponent \(k\in\Zset\), by adding \(m \in\Nset\) we can define the quotient \(k/m \in\Qset\) that together with \(z\in\Cset\) gives \begin{align*} z^{k/m} &= e^{\ln(z) k/m} \\ &= \exp\big((\ln\len{z} + j(\arg z + 2\pi n)) k/m\big) \\ &= \exp\big(\ln\len{z}\cdot k/m)\exp((\arg z + 2\pi n)jk/m\big) \\ &= \len{z}^{k/m}\exp\big((\arg z + 2\pi n)jk/m\big)= \sqrt[m]{z^k} \end{align*} like in the reals, except that we have \(m\) values because of the \(m\)-th root. If we let \(w \in\Cset\) the expression \(z^w\) cannot be equal to an unique value because \begin{align*} z^w = e^{w \ln z} &= \exp\big( w (\ln\len{z} + j \arg{z} + 2\pi nj)\big) \\ &= e^{w(\ln\len{z} + j\arg z)} e^{w2\pi nj} \end{align*} instead it is said to be \emph{multivalued}. This means that there are no general exponentiation rules. \end{lemma} \section{Fourier Series} \begin{definition}[Real trigonometric polynomial] Let \(\omega = 2\pi/T \in\Rset\) and \(A_n, B_n\) be sequences in \(\Rset\). We define a \emph{real trigonometric polynomial} of degree \(N\) to be \[ \tau_N(t) = \frac{A_0}{2} + \sum_{n=1}^N A_n \cos(n\omega t) + B_n \sin(n\omega t) \] \end{definition} \begin{lemma}[Orthogonality of the basis functions] Let \(m,n \in\Nset_0\) \begin{align*} \int\limits_0^T \cos(m\omega t)\cos(n\omega t) &= \begin{cases} T & m = n = 0 \\ T/2 & m = n > 0 \\ 0 & m \neq n \end{cases} \\ \int\limits_0^T \sin(m\omega t)\sin(n\omega t) &= \begin{cases} T/2 & m = n \wedge n \neq 0 \\ 0 & m \neq n \\ 0 & m = 0 \vee n = 0 \end{cases} \\ \int\limits_0^T \cos(m\omega t)\sin(n\omega t) &= 0 \end{align*} \end{lemma} \begin{definition} We denote with \(\Omega\) the space of real valued, \(T\)-periodic, piecewise continuous functions, that have only a finite number of discontinuities, in which both the right and left limit exist, within the interval \([0,T)\). \end{definition} \begin{theorem}[Fourier coefficients] For any \(f\in\Omega\) we can now define the \emph{Fourier coefficients} \begin{align*} a_n &= \frac{2}{T}\int\limits_0^T f(t)\cos(n\omega t)\di{t} & a_0 &= \frac{2}{T}\int\limits_0^T f(t)\di{t} \\ b_n &= \frac{2}{T}\int\limits_0^T f(t)\sin(n\omega t)\di{t} & b_0 &= 0 \end{align*} Worth noting are the special cases when \(n=0\). \end{theorem} \begin{definition}[Fourier Polynomial] We can now use the Fourier coefficients as sequences for a trigonometric polynomial to obtain a \emph{Fourier Polynomial} \[ S_N(t) = \frac{a_0}{2} + \sum_{n=1}^N a_n\cos(n\omega t) + b_n\sin(n\omega t) \] \end{definition} \begin{lemma} A trigonometric polynomial has the smallest distance (by the \(L^2\) metric) from a function \(f\in\Omega\), iff \(A_n = a_n\) and \(B_n = b_n\), in other words iff it is a Fourier Polynomial. \end{lemma} \begin{definition}[Fourier Series] We can finally define the \emph{Fourier Series} to be the infinite Fourier Polynomial, by letting \(N\to\infty\) \[ S(t) = \frac{a_0}{2} + \sum_{n=1}^N a_n\cos(n\omega t) + b_n\sin(n\omega t) \] \end{definition} \begin{theorem}[Fourier coefficients of even and odd functions] Recall that a function is said to be \emph{even} if \(f(-x) = f(x)\) or \emph{odd} if \(f(-x) = -f(x)\). We can show that if a function is \begin{itemize} \item odd, then \(b_n = 0\) for all \(n\), and \[ a_n = \frac{4}{T}\int\limits_0^{T/2} f(t)\cos(n\omega t)\di{t} \] \item even, then \(a_n = 0\) for all \(n\), and \[ b_n = \frac{4}{T}\int\limits_0^{T/2} f(t)\sin(n\omega t)\di{t} \] \end{itemize} \end{theorem} \begin{lemma}[Linearity of Fourier coefficients] Recall that linearity means \(L(\mu x + \lambda y) = \(\mu L(x) + \lambda L(y)\). We then let \(f,g \in\Omega\) be functions with Fourier series and \(h = \mu f + \lambda g\) where \(\mu,\lambda\in\Rset\) are constants. By denoting with \(a_n^{(f)}\) the Fourier coefficient \(a_n\) of the function \(f\), and similarly with \(b_n^{(f)}\), it is easily shown that \begin{align*} a_n^{(h)} &= \mu a_n^{(f)} + \lambda a_n^{(g)} & b_n^{(h)} &= \mu b_n^{(f)} + \lambda b_n^{(g)} \end{align*} \end{lemma} \begin{lemma}[Fourier coefficients after time dilation] Let \(f\in\Omega\) be a function with a Fourier Series and \(g(t) = f(rt)\) with \(0 \neq r \in\Rset\). It follows that \(a_n^{(g)} = a_n^{(f)}\) and \(b_n^{(g)} = \sgn(r) \cdot b_n^{(f)}\). \end{lemma} \begin{lemma}[Fourier coefficients after time translation] Let \(f\in\Omega\) be a function with a Fourier Series and \(g(t) = f(t + \tau)\) with \(\tau\in\Rset\). It follows that \begin{align*} a_n^{(g)} &= \cos(n\omega \tau)\cdot a_n^{(f)} + \sin(n\omega \tau)\cdot b_n^{(f)} & n &\geq 0\\ b_n^{(g)} &= -\sin(n\omega \tau)\cdot a_n^{(f)} + \cos(n\omega \tau)\cdot b_n^{(f)} & n &> 0 \end{align*} \end{lemma} \begin{theorem}[Fourier theorem] For any \(f\in\Omega\) the Fourier series of \(f\) converges in \(L^2\) metric to \(f\). \[ \lim_{N\to\infty} \left\lVert \frac{a_0}{2} + \sum_{n=0}^N a_n \cos(n\omega t) + b_n \sin(n\omega t) - f(t) \right\rVert = 0 \] \end{theorem} \begin{theorem}[Plancherel Parselval theorem] Let \(f\in\Omega\) with a Fourier Series with coefficients \(a_n\) and \(b_n\). \[ \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n^2 + b_n^2\right) \leqq \frac{2}{T} \int\limits_0^T \lvert f(t)\rvert^2 \di{t} = \left\lVert f \right\rVert^2 \] \end{theorem} \begin{theorem} Both sequences \(a_n, b_n\) for the Fourier coefficients of a function \(f\in\Omega\) converge to zero. \begin{align*} \lim_{n\to\infty} a_n &= \lim_{n\to\infty} \frac{2}{T} \int\limits_0^T f(t) \cos(n\omega t) \di{t} = 0 \\ \lim_{n\to\infty} b_n &= \lim_{n\to\infty} \frac{2}{T} \int\limits_0^T f(t) \sin(n\omega t) \di{t} = 0 \end{align*} \end{theorem} \begin{theorem}[Rate of convergence of Fourier coefficients] If \(f\) is a \(T\)-periodic, \((m-2)\) times differentiable, continuous function. And if its \((m-1)\)-th derivative is pieceweise monotonous and \(\in \Omega\), then there exists a constant \(c \in\Rset\) such that \[ \len{a_n} \leq \frac{c}{n^m} \qquad \len{b_n} \leq \frac{c}{n^m} \qquad m,n\in\Nset \] \end{theorem} \begin{theorem}[Integration and differentiation of the Fourier series] It is possible to show from the previous theorem (and others before) that when \(m\geq 2\) the Fouriers converges \emph{uniformly}. This means that it is possible to integrate or differentiate the series term by term. \[ f'(t) = \sum_{n=1}^\infty b_n n\omega\cos(n\omega t) - a_n n\omega\sin(n\omega t) \] and \begin{align*} \int\limits_0^t f(\tau) \di{\tau} &= \left(\sum_{n=1}^\infty \frac{b_n}{n\omega} \right) + \frac{a_0}{2} t \\ &+ \left(\sum_{n=1}^\infty \frac{a_n}{n\omega}\sin(n\omega t) - \frac{b_n}{n\omega}\cos(n\omega t) \right) \end{align*} \end{theorem} \section{Fourier Transform} \section{License} \doclicenseThis \end{document} % vim: set et ts=2 sw=2 spelllang=en spell linebreak :