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\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}

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\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} \(\conj{z}\).
  \[
    z = \Re(z) + j\Im(z)
    \quad
    \conj{z} = \Re(z) - j\Im(z)
  \]
\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 \(\len{z} \in\Rset^+_0\), \(z\conj{z} = \len{z}^2\) 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}.
\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}[More 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}[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 \(\phi = \arg{z} = r(\cos\phi + j\sin\phi)\).
  \begin{align*}
    a &= r\cos\phi &
    b &= r \sin\phi &
    r &= \sqrt{z\conj{z}}
  \end{align*}
  For \(a = 0\) we define \(\phi = \lim_{a\to 0} \arctan(b/a)\) 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 of \emph{odd} degree, has 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} Let \(a,b \in\Cset\) and \(k\in\Zset\)
  \[
    e^a e^b = e^{a+b} \quad
    e^a / e^b = e^{a-b} \quad
    \left(e^a \right)^k = e^{ak}
  \]
\end{lemma}

\begin{definition}[Trigonometric functions]
\end{definition}

\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 instead of 1 or 2. 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}.
\end{lemma}

\section{Fourier Series}

\section{Fourier Transform}

\section{License}
\doclicenseThis

\end{document}
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