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\newcommand{\mytitle}[1]{\begin{center} {\Large 6.003 Problem Set \psnum --
Juliet Bourne -- #1 -- TA R. Kim} \\ \rule{6in}{.5mm} \end{center}}
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{\Large 6.003 Problem Set \psnum -- Juliet Bourne -- {\bf Page \thepg}}
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\newcommand{\seepage}{\noindent See page \rule{.2in}{.1mm} for the
computer output.}

\begin{document}

\mytitle{4/6/90}

\prob{6.8 (PE)}

\subprob{a} The missing code is:

\begin{verbatim}
(convolve (delay-stream x-stream) (cdr h-list))
\end{verbatim}

\subprob{b} i)

\beq
y[n] & = & \sum_{m = 0}^N h[m]x[n-m] \\
x[n] & = & (1/2)u[n] \\
h[n] & = & 0, \, except \, h[0] = 1, h[1] = -.5, h[2] = 2, h[3] = -1.5 
\eeq

\vspace{6in} \seepage

\mynewpage

\noindent ii)

\beq
y[n] & = & \sum_{m = 0}^N x[m]h[n-m] \\
x[n] & = & 0, \, \, except \, \, x[n] = 1.5 \, \, for -2 < n < 2 \\
h[n] & = & (1/2)^n u[n]
\eeq

\vspace{7in} \seepage

\mynewpage

\noindent iii)

\beq
y[n] & = & \sum_{m = 0}^N h[m]x[n-m] \\
x[n] & = & n(u[n] - u[n-6]) \\
h[n] & = & 2(u[n-2] - u[n-5])
\eeq

\vspace{7in} \seepage

\mynewpage

\prob{6.9 (PE)}

\subprob{a}

\mynewpage

\subprob{b}

This produces the required matched system since the system function
corresponding to \verb+hx+ is $\tilde{H} = -1 + z^{-1} - z^{-2} -
z^{-3} + z^{-4} + z^{-5} + z^{-6}$. This implies that the input stream
\verb+x-stream = (1 1 1 -1 -1 1 -1)+ has been reversed and that the
elements of this reversed stream are advanced successively (multiplied
by $(z^{-1})^n$). This equivalent to the reversal and shifting done to
$x[n]$ to achieve $h[n]$.

\seepage

\subprob{c} 1)

\begin{verbatim}
a  =  ( 1  1 -1  -1 -1  -1  1  1  1  1  1  1  -1  -1  1  1) 
b  =  (-1 -1  1  -1  1  -1  1 -1  1  1  1  1  -1  -1  1  1) 
c  =  (-1  1  1  -1 -1  -1  1 -1 -1 -1  1 -1  -1  -1  1 -1)
\end{verbatim}

\seepage

\vspace{.1in}\noindent 2) message = a b a c a b a c a

\seepage

\vspace{.1in}\noindent 3) \seepage

The output of the correct filter the of the three outputs at the
correct times. I believe that I could deduce the correct message
sequence without knowing the correct times since the outputs of all
the filters at all times other than the correct times are small
compared to the the outputs at the correct times. 

\prob{6.10 Revised}

\subprob{a} \seepage

\subprob{b} \seepage

\subprob{c} \seepage

\prob{6.11 (PE)}

\subprob{a}

\beq
\tilde{H}_a(z) & = & \frac{(1 - a)^2}{(1 - az^{-1})} \\
\mbox{} & = & {\left( \frac{1 - a}{1 - az^{-1}} \right)}^2 \\
\mbox{} & = & \frac{1 - a}{1 - az^{-1}} \cdot \frac{1 - a}{1 -
az^{-1}} \\
h_a[n] & = & (a^n + a^{n + 1}) * (a^n + a^{n + 1}) \\
\eeq

\mynewpage

\beq
h_a[0] & = & (1 + a)^2 \\
h_a[1] & = & (1 + a)(a + a^2) + (1 + a)(a + a^2) \\
\mbox{} & = &  2 a (1 + a)^2 \\
h_a[2] & = & (1 + a)(a^2 + a^3) + (a + a^2)(a + a^2) \\
\mbox{} & = & 3 a^2 (1 + a)^2 \\
h_a[3] & = & (1 + a)(a^3 + a^4) + (a^2 + a^3)(a + a^2) + (a^2 + a^3) +
(1 + a)(a^3 + a^4) \\
\mbox{} & = & 4 a^3 (1 + a)^2 \\
h_a[n] & = & (n + 1) (1 + a)^2 a^n
\eeq

\subprob{b}

\vspace{5in}

\beq
x * h_a & = & \sum_{m = 0}^{N} h_a[m] x[n - m] \\
N \rightarrow \infty & \Rightarrow & h_a[m] = 1
\eeq

\noindent Thus, when a slowly-changing waveform is convolved with $h_a$,
it remains relatively unaltered since at long time periods $h_a
\rightarrow  1$. A quickly-changing waveform will be considerably
smoothed out by $h_a$, however.

\mynewpage

\subprob{c} \seepage

\subprob{d} \seepage

\vspace{1in}

\prob{6.12}

\vspace{.2in}
\seepage

\vspace{1in}

\prob{6.13}

\vspace{.2in}
\seepage

\vspace{1in}

\prob{6.14}

\vspace{.2in}
\seepage

\end{document}


