In Continuous Time, we use the \emph{Dirac Delta} function. (Technically not a function, and the height is actually infinite)

%\begin{figure}[htp]
%\centering
%\includegraphics{images/dirac/dirac.ps}
%\caption{ The Dirac Delta }
%\label{ fig:dirac }
%\end{figure}

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\rput(3.428,-0.3){ $t$ }
	
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In Discrete Time, we use the \emph{Kronecker Delta} function.

%\begin{figure}[htp]
%\centering
%\includegraphics{images/kronecker.ps}
%\caption{ The Unit Impulse, a.k.a Kronecker Delta }
%\label{ fig:kronecker }
%\end{figure}

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\rput(3.428,-0.3){ $n$ }
\rput(2.928,1.857){ $\delta(n)$ }
	
\end{pspicture}
\end{center}

The Kronecker Delta function is used in the linear combination:
$$ \sum_{m \in M} x(m) \delta(n-m)  $$

Another signal that we use to decompose signals is the \emph{Complex Exponential}.

This signal is given by

$$ e^{i \theta t};  t \in \R$$ 
$$ e^{i \theta n};  n \in \Z$$ 


Signals to Decompose Signals

Week 2

$$z = \alpha + i\beta,$$ where $\alpha,\beta \in \R$. $\alpha$ is the real part, denoted by Re$(z)$. $\beta$ is the complex part, denoted as Im$(z)$.

Cartesian Representation

$$z = Re^{i\theta},$$ where $R$ is the \emph{magnitude}, or norm, and $\theta$ is the \emph{phase}, or angle.

$$R = \left| z \right| \geq 0$$

$$\theta = \angle z$$

Note that $\theta = $undefined if $R=0$. In other words, $\left| z \right| = 0 \Rightarrow \theta $ is undefined.

Generally keep in these in mind: 
\begin{enumerate}
\item $R \in \R_{\oplus}$
\item $\left| z \right| = 0 \Rightarrow \theta $ is undefined.
\item Calculators use the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
\end{enumerate}


Polar Representation

\begin{definition}
Let $z = x + iy = Re^{i \theta}$, then the \emph{complex conjugate} of $z$ is defined:
$$z^* = x-iy=Re^{-i \theta}$$
\end{definition}

Some properties of the complex conjugate:

\begin{claim}
The product of a complex number and its conjugate is the magnitude squared.
$$ z \cdot z^* = r^2 = \left| z \right|^2 $$
\end{claim}

\begin{proof}
\begin{align*}
z \cdot z^* &= (a + ib)(a - ib) \quad \mbox{Using rectangular representation}\\
z \cdot z^* &= a^2 +aib - aib -i^2b^2 \\
z \cdot z^* &= a^2 +b^2 \\
z \cdot z^* &= \sqrt{a^2 +b^2}^2 \\
z \cdot z^* &= R^2
\end{align*}
\end{proof}

\begin{proof}
\begin{align*}
z \cdot z^* &= Re^{i \theta} \cdot Re^{-i \theta} \quad \mbox{using Polar representation} \\
z \cdot z^* &= R^2e^{i(\theta -\theta)} \\
z \cdot z^* &= R^2e^0 \\
z \cdot z^* &= R^2
\end{align*}
\end{proof}

\begin{claim}
The sum of a complex number and its conjugate is twice the complex number's real part.
$$ z + z^* = 2\mbox{Re}(z) $$
\end{claim}

\begin{proof}
\begin{align*}
z + z^* &= (a + ib) + (a - ib) \\
z + z^* &= 2a
\end{align*}
\end{proof}

\begin{claim}
The difference of a complex number and its conjugate is twice the complex number's imaginary part.
$$ z - z^* = 2i\mbox{Im}(z) $$
\end{claim}

\begin{proof}
\begin{align*}
z - z^* &= (a + ib) - (a - ib) \\
z - z^* &= 2ib 
\end{align*}
\end{proof}

\begin{claim}
The division of a complex number by its conjugate results in a doubling of the angle $\theta$
$$ z / z^* = e^{i2 \theta}  $$
\end{claim}

\begin{proof}
\begin{align*}
z / z^* &=   e^{i\theta} / e^{-i\theta} \\
z / z^* &=   e^{i\theta} \cdot e^{i\theta} \\
z / z^* &=   e^{i(\theta + \theta)} = e^{i2 \theta} 
\end{align*}
\end{proof}

Complex Conjugate

$z_1 + z_2$ is easy in Caresian coordinates.
$$ z_1 + z_2 = (\alpha_1 + \alpha_2) + i(\beta_1 + \beta_2) $$

$z_1 \cdot z_2$ is easy in Polar coordinates.
$$z_1 \cdot z_2 = R_1R_2e^{i(\theta_1 + \theta_2)} $$

\begin{problem}
Solve $z^3 - 1= 0$ in terms of the magnitude and phase.
\end{problem}

\begin{solution}
To solve for the magnitude is simple. $z^3 - 1= 0 \Rightarrow z^3 = 1 \Rightarrow z = 1 \Rightarrow \left| z \right| = 1$. The phase requires a little more work:
\begin{align*}
z^3 - 1&= 0 \\
(e^{i\theta})^3 - 1&= 0 \quad \mbox{(Polar Representation)} \\
e^{i3\theta} - 1&= 0 \\
e^{i3\theta}  &= 1  \\
\cos(3\theta) + i\sin(3 \theta)  &= 1  \quad \mbox{(Euler's Formula)} \\
3\theta &= 0 + 2 \pi k \mbox{ where } k \in \Z \\
\theta &= 0 + \frac{2 \pi}{3} k
\end{align*}

So the solution set gives us $ \theta \in \{ 0, \frac{2\pi}{3}, \frac{4\pi}{3} \}$.
\end{solution}

Choosing a Representation

A brief note for when $z$ is real:

If $z \in \R_{\oplus} \Rightarrow \angle z = 2\pi k $, where $k\in\Z$. The standard is to use $\angle z = 0$. Also when $z$ is a positive real number $\left| z \right| = z$. If the notation $z \gt 0 $ or $z \lt 0$ is used, this means that $z$ is \emph{not} imaginary. For example, if $z<0 \Rightarrow \angle z = \pi $ and $ \left| z \right| = -z$. For instance, $-2 = -2e^{i\pi}$, and $2=2e^{i2\pi}$

Some useful formulae:


$$ \frac{z_1}{z_2} = \frac{R_1}{R_2}e^{i(\theta_1 -\theta_2)}$$

$$ \left| z_1 \cdot z_2 \right| = \left| z_1 \right| \cdot \left| z_2 \right| $$

$$ \left| \frac{z_1}{z_2} \right| = \frac{\left| z_1 \right|}{\left| z_2 \right|} $$

$$ \angle \left( z_1 \cdot z_2 \right) = \angle z_1 + \angle z_2 $$

$$ \angle \frac{z_1}{z_2} = \angle z_1 - \angle z_2 $$


Useful Information

Polar representations have $2 \pi k$ ambiquiity in phase, where $k \in \Z$. This means that $e^{i(\theta - 2\pi)} = e^{i\theta} = e^{i(\theta+2\pi)} = \cdots$


Usually we will use the range [$ 0, 2\pi)$ or [$-\pi,\pi)$.


\begin{problem}
Given the complex number $z = e^{i\frac{2\pi}{3}}$, determine $\sqrt[4]{z}$ . Be mindful of how many fourth roots $z$ has and identify each 
of them.
\end{problem}

\begin{solution}
\begin{align*}
z &= e^{i\frac{2\pi}{3}} \\
z &= e^{i(\frac{2\pi}{3} + 2\pi k) } \mbox{ for some } k \in \Z \\
\sqrt[4]{z} &= e^{i(\frac{2\pi}{3} + 2\pi k) \frac{1}{4}} \\
\sqrt[4]{z} &= e^{i(\frac{\pi}{6} + \frac{\pi}{2} k) \frac{1}{4}} 
\end{align*}
So we can determine that within the range [$0,2\pi$), we find that $\sqrt[4]{z} = e^{i\frac{\pi}{6}}, e^{i\frac{2\pi}{3}}, e^{i\frac{7\pi}{6}}, e^{i\frac{5\pi}{3}}$
\end{solution}


Polar Representation Ambiquity

\begin{align*}
i &= \sqrt{-1} \\
i &= 0 + 1i \\
i &= 1e^{i\frac{\pi}{2}} \\
i^0 &= 1 \\
i^1 &= i \\
i^2 &= -1 \\
i^3 &= -i \\
i^4 &= 1 \\
\frac{1}{i} &= \frac{1}{i} \cdot \frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i \\
i^i &= (e^{i\frac{\pi}{2}})^i = e^{i^2\frac{\pi}{2}} = e^{-\frac{\pi}{2}}
\end{align*}


%\includegraphics{images/imag.ps}

\begin{center}
\begin{pspicture}(-3,-3)(3,3)
		
% y-axis
\rput(0.3,2.8){ $Im$ }
\psline{->}(0,-2.5)(0,2.5)

% x-axis
\rput(2.8,0.3){ $Re$ }
\psline{->}(-2.5,0)(2.5,0)
			
% the circle
\pscircle(0,0){ 2 }
			
% quadrants
\rput(0.3,2.3){ $i^1$ }
\rput(0.3,-2.3){ $i^2$ }
\rput(-2.3,0.3){ $i^3$ }
\rput(2.3,0.3){ $i^0$ }
		
\end{pspicture}
\end{center}



The Imaginary Unit

$$\frac{d}{dt}e^{it}$$

When taking the derivative of a \emph{complex-valued exponential}, we insist on the same property as a \emph{real-valued exponential}:
$$\frac{d}{dt}e^{\omega t} = \omega e^{\omega t} $$
$$\frac{d}{dt}e^{it} = i \cdot e^{it} e^{i\frac{\pi}{2}} \cdot e^{it} = e^{i(t + \frac{\pi}{2})}$$

\begin{nicebox}
$z \cdot i$ is rotation by $90^o$
\end{nicebox}

$$ z \cdot i = e^{i\theta} \cdot i = e^{i\theta} \cdot e^{i\frac{\pi}{2}} = e^{i(\theta + \frac{\pi}{2})}$$ 
$$ (\alpha + i\beta)i = i\alpha + i^2\beta = -\beta + i\alpha $$


% \includegraphics{images/euler/multrot.ps}
% \caption{ Multiplication and Rotation }

\begin{center}
\begin{pspicture}(-4,-4)(4,4)
		
% y-axis
\rput(0.3,3.75){ $Im$ }
\psline{->}(0,-3.75)(0,3.75)

% x-axis
\rput(3.75,0.3){ $Re$ }
\psline{->}(-3.75,0)(3.75,0)
			
% the circle
\pscircle(0,0){ 3 }
			
% quadrants
\rput(0.3,3.3){ $i^1$ }
\rput(0.3,-3.3){ $i^3$ }
\rput(-3.3,0.3){ $i^2$ }
\rput(3.3,0.3){ $i^0$ }
	
 % new vector

\psline[linewidth=1.5 pt]{->}(0,0)(2.598,1.500)
\rput(2,1.500){$\alpha+i\beta$}

%\pcline[linewidth=1.5 pt]{->}(0,0)(2.598,1.500)
%\Aput{\small $\alpha+i\beta$}

 % new vector
\psline[linewidth=1.5 pt]{->}(0,0)(-1.500,2.598)
\rput(-1.500,1.5){$-\beta+i\alpha$}

%\pcline[linewidth=1.5 pt]{->}(0,0)(-1.500,2.598)
%\Aput{\small $-\beta+i\alpha$}
	
\end{pspicture}
\end{center}

Derivatives of Complex Exponentials

Converting from Cartesian to Polar:
$$ \alpha + i\beta = \left| \sqrt{ \alpha^2 + \beta^2 } \right| e^{i\tan^{-1}(\frac{\beta}{\alpha})} $$

Converting from Polar to Cartesian:
$$ Re^{i\theta} = R\left( \cos \theta + i \sin \theta \right) $$


\begin{problem}
Express $z = 1 + i\sqrt{3}$ in rectangular coordinates.
\end{problem}

\begin{solution}
\begin{align*}
z &= 2\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)   \quad \mbox{If you see a way to pull out typical trig values, do it} \\
z &= 2e^{i\frac{\pi}{3}}   \quad \mbox{Since we know $\theta = \frac{\pi}{3}$ when $\sin \theta = \frac{\sqrt{3}}{2}$ and $\cos \theta = \frac{1}{2}$}
\end{align*}
\end{solution}


2.1 Complex Numbers