% !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*} \sin(x + \pi/2) &= \cos(x) \qquad \cos(x - \pi/2) = \sin(x) \\ \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 \(\omega\), the resulting sinusoid \(A\sin(\omega t + \varphi)\) has \[ A = \left\lvert \sum_{n=1}^N A_n e^{j\varphi_n} \right\lvert \quad \varphi = \arg\sum_{n=1}^N A_n e^{j\varphi_n} \] \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 \(N\to\infty\) \[ S(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n\cos(n\omega t) + b_n\sin(n\omega t) \] \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} \begin{theorem}[Dirichlet pointwise convergence] Let \(f\in\Omega\), then it is known that its Fourier series converges to \[ \lim_{\epsilon\to 0}\frac{f(t-\epsilon) + f(t+\epsilon)}{2} \] for every \(t\), where the left and right derivative \(f'(t-\epsilon)\), \(f'(t+\epsilon)\), with \(\epsilon\to 0\), exist. In the special case where \(f\) is continuous at \(t\), and the derivatives exist, there the Fourier series converges exactly to \(f(t)\), i.e. the value of the function at \(t\). \end{theorem} \begin{definition}[Complex representation of the Fourier coefficients] By letting \(n\in\Zset\) and \[ c_n = \conj{c_{-n}} = \frac{a_n - jb_n}{2} = \frac{1}{T}\int\limits_0^T f(t) e^{-jn\omega t} \di{t} \] using a notational trick for negative indices. We can compactly write a Fourier series or polynomial as \[ S(t) = \sum_{n=-\infty}^\infty c_n e^{jn\omega t} \] \end{definition} \begin{theorem}[Complex Fourier coefficients of even and odd functions] By the definition of \(c_n\) and the previous similar theorem for the real coefficients, it is clear that when a function \(f\in\Omega\) is \emph{even}, then \(\Im(c_k) = 0\), whereas when \(f\) is \emph{odd} \(\Re(c_k) = 0\) (\(k\in\Zset\)). \end{theorem} \begin{theorem}[Complex Fourier coeffients after time translation] Similarly to the previous theorem, we can now compactly write that if \(f\in\Omega\) has a Fourier series with coefficients \(c_k^{(f)}\), and \(g(t) = f(t + \tau)\), then \[ c_k^{(g)} = e^{jk\omega \tau} c_k^{(f)} \qquad k \in \Zset \] \end{theorem} %% TODO: fourier transform \section{License} \doclicenseThis \end{document} % vim: set et ts=2 sw=2 spelllang=en spell linebreak :