Update for spring 2020

A light update, mostly to take care of minor-to-moderate errata. Renumbered Matlab lab exercises so that their numbering matches the chapter numbers in the textbook. Closes #8

Matlab Labs/Lab Book.tex

@@ -115,7 +115,7 @@
 
 \mbox{}\hfill
 \parbox{1.5in}{\mbox{}}
-\parbox{3in}{Fall 2016}\\[0.25in]
+\parbox{3in}{Spring 2020}\\[0.25in]
 \mbox{}\hfill
 \parbox{3in}{\raggedright
 Michael Stiber\\
@@ -148,6 +148,7 @@
 
 \tableofcontents
 
+\setcounter{section}{-1}
 \include{lab1/lab1}
 \include{lab2/lab2}
 \include{lab3/lab3}

Matlab Labs/lab1/lab1.tex

@@ -188,14 +188,14 @@ \subsection{Matlab in a (Very Small) Nutshell}
 output.
 
 Add three lines of code to your script, so that it will plot a cosine
-on top of the sine in a different color. Use the \texttt{hold}
-function to add a plot of
+using the same axes as the sine (i.e., ``on top of the sine''). Use
+the \texttt{hold} function to add a plot of
 \begin{lstlisting}[style=Matlab-editor,basicstyle=\mlttfamily\small]
 0.75*cos(2*pi*f*t)
 \end{lstlisting}
-
-to the plot. Save the plot using the MATLAB \texttt{print} command as
-a PNG file named \texttt{step14.png} by typing:
+to the plot. So, your final graph will have two functions
+plotted. Save the plot using the MATLAB \texttt{print} command as a
+PNG file named \texttt{step14.png} by typing:
 \begin{lstlisting}[style=Matlab-editor,basicstyle=\mlttfamily\small]
 print -dpng step14
 \end{lstlisting}
@@ -274,7 +274,7 @@ \subsection{Trigonometric Functions and Complex Mathematics in Matlab}
 
 
 Include your code in your writeup. Additionally, include a plot of
-\texttt{x = sumcos(20, [0 pi/4 pi/2 3*pi/2], 200, 0.25);} versus
+\texttt{x = sumcos(20, [0 pi/4 pi/2 3*pi/2], [1 2 3 4], 200, 0.25);} versus
 time.
 
 Hint: the MATLAB \verb|length| function is useful in determining the

Matlab Labs/lab2/lab2.tex

@@ -12,8 +12,9 @@ \section{Let's Get Physical}
 \subsection{Beating}
 
 In this section, you will use the Matlab \texttt{AnalogSignal} class
-to simulate an analog signal generator. The constructor takes the
-following arguments:
+(described previously in lab~1 and its
+figure~\ref{fg:analogsignal-help}) to simulate an analog signal
+generator. The constructor takes the following arguments:
 \begin{lstlisting}[style=Matlab-editor,basicstyle=\mlttfamily\small]
 % AnalogSignal(type, amplitude, frequency, dur)
 %    Where:
@@ -32,7 +33,7 @@ \subsection{Beating}
 
 \paragraph{Step 1.1} Verify that you can get a triangle wave. What is
 the code to generate and plot a triangle wave ranging from -1 to 1
-Volts with a frequency of 10Hz and a duration of 1sec?
+Volts with a frequency of 10Hz and a duration of 1 second?
 
 \paragraph{Step 1.2} Next, verify that you can generate a sine wave of
 1 second duration at a frequency of 440Hz, ranging from -1 to +1. What
@@ -43,9 +44,10 @@ \subsection{Beating}
   \itshape{Xmax}])} function call to set the X axis limits so that the
 waveform is apparent (i.e., you're not just plotting a solid blob).
 
-\paragraph{Step 1.3} Generate sine waves of identical range and
-duration, but with frequencies of 442, 444, and 448 Hz. Now, generate
-the sums of 440Hz and each of these new signals to generate beating
+\paragraph{Step 1.3} Generate three sine waves of identical range and
+duration, but with frequencies of 442, 444, and 448 Hz. Next, generate
+the three sums of 440Hz and each of these new signals separately (so,
+440Hz + 442Hz, 440Hz + 444Hz, and 440Hz + 448Hz) to generate beating
 akin to tuning an instrument against the 440Hz standard. Play each sum
 signal; can you hear the beating? Plot each sum signal for its full 1s
 duration. The beating ``envelope'' should be obvious. What is the beat
@@ -57,8 +59,8 @@ \subsection{Fourier series representation of a physical signal}
 
 
 \paragraph{Step 2.1} Recall that any periodic signal can be
-represented as a sum of harmonic sinusoids.  The amplitude of each
-harmonic is known as the Fourier Series. It may at first seem like
+represented as a sum of harmonic sinusoids.  The amplitudes of these
+harmonics is collectively known as the Fourier Series. It may at first seem like
 sums of sinusoids would be poor approximations of real periodic
 signals, but this is not the case. We can illustrate this using a
 triangle wave. The formula for synthesis of a triangle wave with
@@ -75,7 +77,7 @@ \subsection{Fourier series representation of a physical signal}
   in Hz, and so $\omega_0 = 2\pi f_0$. Notice that the Fourier Series
   of the triangle wave only uses odd harmonics (i.e., the only
   non-zero frequencies are $(2k+1)\omega_0=\omega_0, 3\omega_0,
-  5\omega_0 \cdots$). Also notice that resulting wave will have zero
+  5\omega_0 \cdots$). Also notice that the resulting wave will have zero
   mean because there is no ``DC'' term (i.e., $2k+1 \neq 0$ for any
   integer k).
 

Matlab Labs/lab3/lab3.tex

@@ -32,13 +32,22 @@ \subsection{Sampling}
 sampled versions of this signal at 300Hz, 500Hz, 1000Hz, and
 2000Hz. Use the Matlab \texttt{subplot} command, and the
 \texttt{discreteplot} function provided with this class's Matlab code,
-to plot the original and all four sampled signals together. Clearly,
-the results are not the same, and none look identical to the original
-sine wave. What are the two essential pieces of information about a
-sine wave that need to be preserved when sampling it? Does it appear
-that all sampled versions are equally useful in achieving this? Why or
-why not (in other words, your answer to this question should not be
-just ``yes'' or ``no'')?
+to plot the original and all four sampled signals together. One of the
+things that you will note from these plots is that the X axes of these
+plots do not have units of continuous time; their units are \emph{not}
+seconds. Instead, the X axis values are sample count, starting with
+sample 1. You will want these plots to show their signals over the
+same time duration, and so you'll need to choose that duration,
+then figure out, for each sampled signal, how many samples correspond
+to that duration, and then set the X axis limits for each figure so
+that the plots show equal durations.
+
+Clearly, the results are not the same, and none look identical to the
+original sine wave. What are the two essential pieces of information
+about a sine wave that need to be preserved when sampling it? Does it
+appear that all sampled versions are equally useful in achieving this?
+Why or why not (in other words, your answer to this question should
+not be just ``yes'' or ``no'')?
 
 \paragraph{Step 1.3} Let's look at aliasing in a little more detail
 and with a lot more numerical precision. You'll recall from the text
@@ -121,13 +130,13 @@ \subsection{Analog to Digital Conversion}
   2, 4, 8, 12, and 16 bits quantization, and plot SNR on the Y-axis
   versus number of quantization bits on the X-axis.
 
-  \paragraph{Step 2.3} Repeat Step 3.2 using a square waveform with
+  \paragraph{Step 2.3} Repeat Step 2.2 using a square waveform with
   the same parameters.
 
-  \paragraph{Step 2.4} Repeat Step 3.2 using a triangle waveform with
+  \paragraph{Step 2.4} Repeat Step 2.2 using a triangle waveform with
   the same parameters.
 
   \paragraph{Step 2.5} As you double the number of bits used in
   quantization, how does the SNR change? How does this compare to what
   your learned from the textbook? Refer to specific features of your
-  plots from Steps~3.2--3.4 to justify your answer.
+  plots from Steps~2.2--2.4 to justify your answer.

Matlab Labs/lab4/lab4.tex

@@ -79,15 +79,18 @@ \subsection{Overview of Filtering and Matlab}
 
 You can visualize the zero locations with the following code:
 \begin{lstlisting}[style=Matlab-editor,basicstyle=\mlttfamily\small]
-plot(r, 'o')
+plot(complex(r), 'o')
 rectangle('Position', [-1 -1 2 2], 'Curvature', [1 1])
 line([-1 1], [0 0], 'Color', [0 0 0])
 line([0 0], [-1 1], 'Color', [0 0 0])
 axis equal
 \end{lstlisting}
-Here, we use the unfortunately named \verb|rectangle| function
-to draw the unit circle and the \verb|line| function to draw the real
-and imaginary axes.
+Here, we use the unfortunately named \verb|rectangle| function to draw
+the unit circle and the \verb|line| function to draw the real and
+imaginary axes. Note also that we have to make sure that the value of
+$r$ that we pass to the \verb|plot()| function is complex (since we
+want to plot it on the complex plane) using the \verb|complex()|
+function, because the roots of a polynomial can be real.
 
 Finally, you can compute and plot the magnitude of the filter's
 frequency response pretty directly in Matlab:

Matlab Labs/lab6/lab6.tex

@@ -94,7 +94,7 @@ \subsection{Feedback Filters as Recurrence Relations}
 \paragraph{Step 1.4} Let's do something similar with the recurrence
 relation for computing the series $y[n] = 1/3^n$ in the text (as
 always, remember that Matlab indices start at 1). Set the
-coefficients for a feedback filter to implement equation~(5-42) in the
+coefficients for a feedback filter to implement equation~(5-43) in the
 text, $y[n] = 1/3 y[n-1] + x[n]$. What are the filter coefficients?
 
 

Matlab Labs/lab7/lab7.tex

@@ -133,7 +133,7 @@ \subsection{Using the DFT}
 
 \paragraph{Step 2.4} Replace the sum of two sinusoids with your
 \verb|DTMFCoder| function from lab 6. Input a selection of button
-values. Does the FFT block show the separate frequencies for each
+values. Do the FFT function and plot show the separate frequencies for each
 button?
 
 

Matlab Labs/lab8/lab8.tex

@@ -240,22 +240,22 @@ \subsection{Lossy image coding: JPEG}
 
 \paragraph{Step 3.3} Let's quantize the image's spectral
 content. First, find the number of zero elements in the FFT, using
-something like \verb|origZero=length(find(abs(a)==0));|, where
-\verb|a| is the FFT. \emph{Remember to exclude the DC value in the
+something like \verb|origZero=length(find(abs(origX)==0));|, where
+\verb|origX| is the FFT. \emph{Remember to exclude the DC value in the
   \texttt{fft2} output in figuring out this range.} Then, zero out
 additional frequency components by zeroing out all with magnitudes
 below some threshold. You'll want to set the threshold somewhere
-between the min and max magnitudes of \verb|a|, which you can get as
-\verb|mn=min(min(abs(a)));| and \verb|mx=max(max(abs(a)));|. Let's
+between the min and max magnitudes of \verb|origX|, which you can get as
+\verb|mn=min(min(abs(origX)));| and \verb|mx=max(max(abs(origX)));|. Let's
 make four tests, with thresholds 5\%, 10\%, 20\%, and 50\% of the way
 between the min and max, i.e., \verb|th=0.05*(mx-mn)+mn;|. Zero out
 all FFT values below the threshold using something like:
 \begin{lstlisting}[style=Matlab-editor,basicstyle=\mlttfamily\small]
-b = a;
-b(abs(a)<th) = 0;   % Uses logical array indexing
+compX = origX;
+compX(abs(origX)<th) = 0;   % Uses logical array indexing
 \end{lstlisting}
 You can count the number of elements thresholded by finding the number
-of zero elements in \verb|b| at this point and subtracting the number
+of zero elements in \verb|compX| at this point and subtracting the number
 that were originally zero (i.e., \verb|origZero|). This is an estimate
 of the amount the image could be compressed with an entropy
 coder. Express the number of thresholded elements as a fraction of the
@@ -270,17 +270,20 @@ \subsection{Lossy image coding: JPEG}
 you suggest an approach that will take this into account? How does the
 JPEG algorithm deal with or avoid this problem?
 
+\begin{sloppypar}
 \paragraph{Step 3.4} Now we will see the effect of this thresholding
 on image quality. Convert the thresholded FFT back to an image using
-something like \verb|c = abs(ifft2(b));|. For each type of image and
-each threshold value, plot the original image and the final processed
-image. Compute the mean squared error (MSE) between the original and
-reconstructed image (mean squared error for matrices can be computed
-as \verb|mean(mean((a-c).^2))|).  What can you say about the effects
-on the image and MSE?  Collect your code together as a script to
-automate the thresholding and reconstruction, so you can easily
-compute MSE for a number of thresholds. Plot MSE vs. threshold
+something like \verb|reconstructed = abs(ifft2(compX));|. For each
+type of image and each threshold value, plot the original image and
+the final processed image. Compute the mean squared error (MSE)
+between the original and reconstructed image (mean squared error for
+matrices can be computed as
+\verb|mean(mean((original-reconstructed).^2))|).  What can you say
+about the effects on the image and MSE?  Collect your code together as
+a script to automate the thresholding and reconstruction, so you can
+easily compute MSE for a number of thresholds. Plot MSE vs. threshold
 percentage (just as you plotted fraction of pixels thresholded
 vs. threshold in step 3.3).
+\end{sloppypar}
 
 % LocalWords:  WebQ MATLAB DSP

Matlab/AnalogSignal.m

@@ -215,6 +215,7 @@ function soundsc(sig)
 
         function y = square(f, t)
             % SQUARE Generate a square wave with values in range [0, 1]
+	    %        with 50% duty cycle
             %   Y = SQUARE(F, T)
             %   Where:
             %   F = number of positive/negative waves per second

Signal Computing.tex

@@ -251,7 +251,7 @@
 \mbox{}\hfill
 \parbox{1.5in}{\mbox{}}
 \parbox{3in}{\raggedright
-Winter 2018\\[0.25in]
+Spring 2020\\[0.25in]
 Michael Stiber\\
 Bilin Zhang Stiber\\
 University of Washington Bothell\\
@@ -275,7 +275,7 @@
 \newpage
 \mbox{}\vspace{2in}
 \begin{flushright}
-Copyright \copyright\ 2002--2018 by Michael and Bilin Stiber and Eric
+Copyright \copyright\ 2002--2020 by Michael and Bilin Stiber and Eric
 C. Larson\\[1in]
 This material is based upon work supported by the National Science
 Foundation under Grant No. 0443118.\\[1in]

ch-conv/convolution.tex

@@ -342,7 +342,7 @@ \subsection{Example: z-transform of exponential signal}
 \end{equation}
 or
 \begin{equation}
-u[n] = \{1, 1, 1, \ldots\}, \quad k \ge 0
+u[n] = \{1, 1, 1, \ldots\}, \quad n \ge 0
 \end{equation}
 
 \index{unit step}

ch-fft/fft.tex

@@ -682,7 +682,7 @@ \subsection{The Fast Fourier Transform Algorithm}
 \ENDIF
 \STATE Shuffle the $x[n]$ in bit-reversed order
 \STATE $\mathit{Size} = 2$ \COMMENT{$\mathit{Size}=1$ DFTs already done}
-\WHILE{$\mathit{Size} < N$}
+\WHILE{$\mathit{Size} \leq N$}
    \STATE Compute $N/\mathit{Size}$ DFTs from the existing ones as
           $X_{\mathit{Size}/2}[k]^{\mathit{even}} +
           e^{-jk2\pi/\mathit{Size}}X_{\mathit{Size}/2}[k]^{\mathit{odd}}$
@@ -714,36 +714,38 @@ \subsubsection{Example: 8-Point FFT}
 1\}$, $n = \{0, 1, 2, 3, 4, 5, 6, 7\}$. The first thing we'll do is
 write the signal out as a 1D array, with indices in binary:
 \begin{center}
-  \begin{tabular}{|c|c|c|c|c|c|c|c|} \hline
-    0   & 1   & 0   & 1   & 0   & 1   & 0   & 1   \\ \hline
-    000 & 001 & 010 & 011 & 100 & 101 & 110 & 111 \\ \hline
+  \begin{tabular}{r|c|c|c|c|c|c|c|c|} \hline
+    $x[n]$ & 0   & 1   & 0   & 1   & 0   & 1   & 0   & 1   \\ \hline
+    $n$ & 000 & 001 & 010 & 011 & 100 & 101 & 110 & 111 \\ \hline
   \end{tabular}
 \end{center}
 
 Next, we sort the elements according to their bit-reversed indices:
 \begin{center}
-  \begin{tabular}{|c|c|c|c|c|c|c|c|} \hline
-    0   & 0   & 0   & 0   & 1   & 1   & 1   & 1   \\ \hline
-    000 & 100 & 010 & 110 & 001 & 101 & 011 & 111 \\ \hline
+  \begin{tabular}{r|c|c|c|c|c|c|c|c|} \hline
+    $x[n]$ & 0   & 0   & 0   & 0   & 1   & 1   & 1   & 1   \\ \hline
+    $n$ 000 & 100 & 010 & 110 & 001 & 101 & 011 & 111 \\ \hline
   \end{tabular}
 \end{center}
 
 Each of these elements is already a 1-point FFT; we now begin merging
-them into 2-point ones ($X_2[0]$ and $X_2[1]$), as in
+them into 2-point ones ($X_2[0]= x[\mathit{even}] + x[\mathit{odd}]$
+and $X_2[1]= x[\mathit{even}] - x[\mathit{odd}]$), as in
 equations~(\ref{eq:2fft-even}) and~(\ref{eq:2fft-odd}):
 \begin{center}
-  \begin{tabular}{|c|c||c|c||c|c||c|c|} \hline
-    0   & 0   & 0   & 0   & 2   & 0   & 2   & 0   \\ \hline
-    0   & 1   & 0   & 1   & 0   & 1   & 0   & 1   \\ \hline
+  \begin{tabular}{r|c|c||c|c||c|c||c|c|} \hline
+    $X_2[k]$ & 0   & 0   & 0   & 0   & 2   & 0   & 2   & 0   \\ \hline
+    $k$        & 0   & 1   & 0   & 1   & 0   & 1   & 0   & 1   \\ \hline
   \end{tabular}
 \end{center}
 
 Now we have four 2-point FFTs; we now merge adjacent pairs of
 points. So, we will be computing $X_4[k]$ for $k=\{0, 1, 2, 3\}$. To
-compute elements 2 and 3, we need to remember
-equations~(\ref{eq:fft-repeat-even}) and~(\ref{eq:fft-repeat-odd}): we
-get $X_2[2] = X_2[0]$ and $X_2[3] = X_2[1]$. The resulting equations
-are:
+compute elements $k=2$ and $k=3$, we need to remember
+equations~(\ref{eq:fft-repeat-even}) and~(\ref{eq:fft-repeat-odd}) ---
+that, since the transform $X$ is periodic, we simply repeat the
+2-point FFTs to get the ``extra'' two points. So, $X_2[2] = X_2[0]$
+and $X_2[3] = X_2[1]$. The resulting equations are:
 \begin{align*}
   X_4[0] &= X_2[0]^{\mathit{even}} +   e^{-j(0)2\pi/4}X_2[0]^{\mathit{odd}} \\
   X_4[1] &= X_2[1]^{\mathit{even}} +   e^{-j(1)2\pi/4}X_2[1]^{\mathit{odd}} \\
@@ -767,9 +769,9 @@ \subsubsection{Example: 8-Point FFT}
 \end{align*}
 yielding:
 \begin{center}
-  \begin{tabular}{|c|c|c|c||c|c|c|c|} \hline
-    0   & 0   & 0   & 0   & 4   & 0   & 0   & 0   \\ \hline
-    0   & 1   & 2   & 3   & 0   & 1   & 2   & 3   \\ \hline
+  \begin{tabular}{r|c|c|c|c||c|c|c|c|} \hline
+    $X_4[k]$ & 0   & 0   & 0   & 0   & 4   & 0   & 0   & 0   \\ \hline
+    $k$        & 0   & 1   & 2   & 3   & 0   & 1   & 2   & 3   \\ \hline
   \end{tabular}
 \end{center}
 
@@ -798,9 +800,9 @@ \subsubsection{Example: 8-Point FFT}
 
 The final result is:
 \begin{center}
-  \begin{tabular}{|c|c|c|c|c|c|c|c|} \hline
-    4   & 0   & 0   & 0   & -4  & 0   & 0   & 0   \\ \hline
-    0   & 1   & 2   & 3   & 4   & 5   & 6   & 7   \\ \hline
+  \begin{tabular}{r|c|c|c|c|c|c|c|c|} \hline
+    $X_8[k]$ & 4   & 0   & 0   & 0   & -4  & 0   & 0   & 0   \\ \hline
+    $k$        & 0   & 1   & 2   & 3   & 4   & 5   & 6   & 7   \\ \hline
   \end{tabular}
 \end{center}
 
@@ -1269,7 +1271,7 @@ \subsubsection{Hamming Window}
 
 \begin{figure}
 \centerline{\includegraphics[width=4in]{ch-fft/ufft_hamming_w512}}
-\caption{A 512-point Hann window in the time domain.\label{fig:ufft-hmw}}
+\caption{A 512-point Hamming window in the time domain.\label{fig:ufft-hmw}}
 \end{figure}
 
 \index{FFT!Hamming window and}