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\author{Nada Amin\\
\texttt{namin@mit.edu}
}

\title{6.003 Final Cheat Sheet}

\begin{document}

\date{April 14, 2005}

%\maketitle
\section{Definitions}
\begin{description}
\item[DT convolution] $y[n] = x[n] * h[n] = \sum_{k=-\infty}^{+\infty}x[k]h[n-k]$
\item[CT convolution] $y(t) = x(t) * h(t) = \int_{-\infty}^{+\infty}x(\tau)h(t-\tau]d\tau$
\item[CFS synthesis] $x(t) = \sum_{k=-\infty}^{+\infty}a_ke^{jk\omega_0t} = \sum_{k=-\infty}^{+\infty}a_ke^{jk(2\pi/T)t}$
\item[CFS analysis] $a_k = \frac{1}{T} \int_{T}x(t)e^{-jk\omega_0t}dt = \frac{1}{T} \int_{T}x(t)e^{-j(2\pi/T)t}dt$
\item[CFT synthesis] $x(t) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}X(j\omega)e^{j\omega t} d\omega$
\item[DFT analysis] $X(j\omega) = \int_{-\infty}^{+\infty}x(t)e^{-jkt} dt$
\item[DFS synthesis] $x[n] = \sum_{k=<N>}a_ke^{jk\omega_0n} = \sum_{k=<N>}a_ke^{jk(2\pi/N)n}$
\item[DFS analysis] $a_k = \frac{1}{N} \sum_{n=<N>}x[n]e^{-jk\omega_0n} = \frac{1}{N} \sum_{n=<N>}x[n]e^{-j(2\pi/N)n}$
\item[DFT synthesis] $x[n] = \frac{1}{2\pi}\int_{2\pi}X(e^{j\omega})e^{j\omega n} d\omega$
\item[DFT analysis] $X(e^{j\omega}) = \sum_{n=-\infty}^{+\infty}x[n]e^{-jkn}$
\item[LCC differential equation] $\sum_{k=0}^{N}a_k\frac{d^ky(t)}{dt^k}=\sum_{k=0}^{M}b_k\frac{d^kx(t)}{dt^k} \Longleftrightarrow H(j\omega) = \frac{\sum_{k=0}^{M}b_k(j\omega)^k}{\sum_{k=0}^{N}a_k(j\omega)^k}$
\item[LCC difference equation] $\sum_{k=0}^{N}a_ky[n-k]=\sum_{k=0}^{M}b_kx[n-k] \Longleftrightarrow H(e^{j\omega}) = \frac{\sum_{k=0}^{M}b_ke^{-jk\omega}}{\sum_{k=0}^{N}a_ke^{-jk\omega}}$
\end{description}

\twocolumn

\section{Summary of Bode Plotting Rules}
First, normalize so that $H(s) = c \frac{(s+z_1)(s+z_2)\cdot\cdot\cdot(s+z_n)}{(s+p_1)(s+p_2)\cdot\cdot\cdot(s+p_n)}$.
\subsection{Plotting Magnitude $20\log|H(j\omega)|$}
\begin{enumerate}
\item Identify locations and order (how many) of all poles and zeros -
  these are the breakpoints.
\item Draw axes. Note that it is impossible to include $\omega=0$ on a
  log scale. Start with a small $\omega$, like $1$, $0.00001$, or
  whatever is appropriate. It is also useful to draw vertical dashed
  lines at breakpoints.
\item Starting at the left, the magnitude plot starts flat unless
  there is a pole at $s=0$ (start plot with slope of $-20$dB/dec for
  each pole at origin) or there is a zero at $s=0$ (start with a plot
  of $+20$dB/dec for each zero at origin).
\item Continue drawing asymptote in a straight line until you reach a
  breakpoint (pole/zero).
\item For \underline{each} pole, {\it decrease} slope of asymptote by
  $20$dB/dec. For \underline{each} zero, {\it increase} slope by
  $20$dB/dec. Go to step 4 unless there are no more breakpoints left.
\item Label one point on the $y$-axis by plugging in a value of
  $\omega$ into $H(j\omega)$ from any flat region of the $y$-axis
  using slopes of asymptotes as guides.
\item Round corners by $\pm 3$dB for a more accurate magnitude plot.
\end{enumerate}
\subsection{Plotting Phase $\angle H(j\omega)$}
\begin{enumerate}
\item Identify locations and order (how many) of all poles and zeros -
  these are the breakpoints.
\item Draw axes and vertical dashed lines at breakpoints.
\item Starting at the left, the phase plot starts at $\angle
  H(j\omega=0)$ (usually $0^o$). Plot starts at $+90^o$ for each zero
  at origin and $-90^o$ for each pole at origin. A leading minus sign
  will add $180^o$ to the phase. Plug in a very small imaginary number
  for $j\omega$ and evaluate the phase manually if you're
  confused. Also, remember, shifting the phase curve up or down by
  $360^o$ doesn't change anything.
\item The phase plot continues as a flat line until reaching
  $0.1\times$breakpoint.
\item \underline{Each} pole {\it substracts} $90^o$ from the phase,
  spread over a distance of $0.1\times$pole location to
  $10\times$pole location. At the pole location, the phase has
  dropped by $45^o$ (halfway there). The situation is the same for
  zeros, but this time phase is {\it added}. Watch out for multiple
  poles/zeros. Go back to step 4 unless there are no more breakpoints
  left.
\item Round all corners to ressemble an $\arctan$ curve (that's how
  phase is calculated) for more accurate plotting; the phase round by
  about $6^o$ at $0.1\times$breakpoint and at $10\times$breakpoint.
\end{enumerate}

\section{Partial Fraction Expansion}
\subsection{linear, non-repeated factors}
\begin{align*}
X(s) =& \frac{k_1}{s-p_1} + frac{k_2}{s-p_2} + \ldots\\
k_i =& X(s)(s - p_i) |_{s=p_i}
\end{align*}
\subsection{linear, repeating factors}
The formula is inhuman, but reduces to the one for non-repeated roots
for the residual of \underline{the highest power term}. So find that,
then substract that term from the equation and keep going.

\section{Quadratic Formula}
\begin{align*}
ax^2 + b x + c = 0\\
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\end{align*}

\section{Sampling}
\subsection{sampling period}
{\it The sampling rate ($\omega_s$) must be greater than twice the maximum frequency present in the input signal ($\omega_M$).}
\begin{align*}
\omega_s &> 2 \omega_M\\
\omega_s = \frac{2 \pi}{T}
\end{align*}
\subsection{low-pass filter}
\begin{description}
\item[gain] $T$
\item[cutoff] between $\omega_M$ and $\omega_s - \omega_M$
\end{description}
\subsection{c/d conversion}

If $x_p = x_c$ sampled with period $T$, then $X_p(j\omega)$ is
$X_c(j\omega)$ repeated every $\frac{2\pi}{T}$. If $x_d[n] = x_c(nT)$,
then $X_d(e^{jw})$ is $X_c(j\omega)$ repeated every $2\pi$. In short,
to convert a sampled CT signal into a DT signal, in the frequency
domain, multiply the $\omega$-axis by the sampling period.
\begin{align*}
\omega_{ct} = T \omega_{dt}
\end{align*}

\section{C Systems Properties}
\subsection{Causality}
\begin{itemize}
\item LTI system is causal $\leftrightarrow$ $h(t) = 0$ for $t<0$.
\item system is causal $rightarrow$ ROC is a right-half plane.
\item rational system is causal $\leftrightarrow$ ROC is a rightmost right-half plane.
\end{itemize}
\subsection{Stability}
\begin{itemize}
\item LTI system is stable $\leftrightarrow$ $\int_{-\infty}^{+\infty}|h(\tau)|d\tau < \infty$.
\item LTI system is stable $\leftrightarrow$ ROC includes $j\omega$-axis ($\Re e \{s\} = 0$).
\end{itemize}
\subsection{Causality and Stability}
Rational system is causal and stable $\leftrightarrow$ all poles lie in the left-half of the $s$-plane.

\section{D Systems Properties}
\subsection{Causality}
\begin{itemize}
\item LTI system is causal $\leftrightarrow$ $h[n] = 0$ for $n<0$.
\item system is causal $\rightarrow$ ROC is the exterior of a circle including infinity.
\item rational system is causal $\leftrightarrow$ ROC is the exterior of a circle outside the outermost pole.
\end{itemize}
\subsection{Stability}
\begin{itemize}
\item LTI system is stable $\leftrightarrow$ $\sum_{k = -\infty}^{+\infty}|h[k]| < \infty$.
\item LTI system is stable $\leftrightarrow$ ROC includes the unit circle, $|z| = 1$.
\end{itemize}
\subsection{Causality and Stability}
Rational system is causal and stable $\leftrightarrow$ all poles lie
inside the unit circle -- i.e., thez must all have magnitude smaller
than $1$.

\section{Checking if All Poles Are in LHP}
Calculate all the roots and see! or apply Routh-Hurwitz -- without
having to solve for roots.
\subsection{Routh-Hurwitz}
\begin{tabular}{ccc}
& Polynomial & Condition \\&&so that all roots \\&&are in the LHP \\
1st order & $s + a_0$ & $a_0 > 0$ \\
2nd order & $s^2 + a_1s + a_0$ & $a_1 > 0, a_0 > 0$ \\
3rd order & $s^3 + a_2s^2 + a_1s + a_0$ & $a_2 > 0, a_1 > 0, a_0 > 0$ \\&& \emph{and} $a_0 < a_1a_2$\\
\end{tabular}

\section{Bode Plots from Pole-Zero Plot}
For a rational system, 
\begin{align*}
X(s) = M \frac{\Pi_{i=1}^{R}(s - \beta_i)}{\Pi_{j=1}^{P}(s - \alpha_j)}
\end{align*}

The magnitude of $X(s_1)$ is then the magnitude of the scale factor
$M$, \emph{times} the product of the lengths of the zero vectors (i.e., the
vectors from the zeros to $s_1$) \emph{divded} by the product of the lengths
of the pole vectors (i.e., the vectors from the poles to $s_1$).

The angle of the complex number $X(s_1)$ is the \emph{sum} of the
angles of the zero vectors \emph{minus} the sum of the angles of the
pole vectors. If the scale factor $M$ is negative, an additional angle
of $\pi$ would be included.

\section{Root Locus}
$K = \frac{1}{|G(s_0)H(s_0)|}$
\begin{itemize}
\item For $K=0$, poles of $G(s)H(s)$.
\item For $K=\infty$, zeroes of $G(s)H(s)$.
\item For $K>0$, \emph{odd} number of real poles and zeros of $G(s)H(s)$.
\item For $K<0$, \emph{even} number of real poles and zeros of $G(s)H(s)$.
\item Branches of the root locus between two real poles must break off into complex plane for $|K|$ large enough.
\end{itemize}

\end{document}
