Page MenuHomec4science
Files F104874576

2014final.tex
No OneTemporary
Actions

File Metadata

Created
Thu, Mar 13, 01:33

2014final.tex
View Options

\documentclass[12pt,a4paper]{article}
\usepackage{styles/SPhw}
\includecomment{sol}
\begin{document}
\begin{exercise}{Infinite sums}
Consider the discrete-time signal $x[n] = \sinc(an)$ with $0 < a < 1$; compute the following sums:
\begin{enumerate}
\item $\displaystyle \sum_{n = -\infty}^{\infty} x[n]$
\item $\displaystyle \sum_{n = -\infty}^{\infty} x^2[n]$
\end{enumerate}
\begin{sol}
\Solution
{\em the impulse response of an ideal lowpass filter with cutoff frequency $\omega_c$ is }
\[
h[n] = (\omega_c/\pi)\sinc((\omega_c/\pi)n)
\]
{\em therefore $x[n]$ is the impulse response of an ideal lowpass filter with cutoff frequency $a\pi$, scaled by $1/a$ so that} $X(e^{j\omega}) = (1/a)\rect(\omega/(2a\pi))$. {\em From this:}
\begin{enumerate}
\item $\displaystyle \sum_{n = -\infty}^{\infty} x[n] = X(e^{j\omega})|_{\omega = 0} = 1/a$
\item$\displaystyle \sum_{n = -\infty}^{\infty} x^2[n] = \frac{1}{2\pi}\int_{-\pi}^{\pi}|X(e^{j\omega})|^2 = \frac{1}{2\pi}\int_{-a\pi}^{a\pi}a^{-2} = 1/a$ {\em (by using Parseval's theorem)}
\end{enumerate}
\end{sol}
\end{exercise}
\begin{exercise}{DFT}
Compute the DFT of the $\mathbb{C}^4$ vector $\mathbf{x} = [ 1 \ \ 1 \ -1 \ -1]^T$ \\ \\
\begin{sol}
\Solution
the DFT matrix for $\mathbb{C}^4$ is
\[
\mathbf{W} = \begin{bmatrix}
1 & 1 & 1 & 1 \\
1 & -j & -1 & j \\
1 & -1 & 1 & -1 \\
1 & j & -1 & -j
\end{bmatrix}
\]
By computing the matrix-vector product $\mathbf{X} = \mathbf{Wx}$ it is easy to obtain $\mathbf{X} = [0 \ \ (2-2j) \ \ 0 \ \ (2+2j)]^T$
\end{sol}
\end{exercise}
\begin{exercise}{Pole-zero plot}
Compute the impulse response of the causal filter with the following pole-zero plot:
\begin{center}
\begin{dspPZPlot}[width=5cm,xticks=none,yticks=none]{1.3}
\dspPZ[label={$\alpha$}]{0.6,0}
\dspPZ[type=zero,label=none]{1,0}
\end{dspPZPlot}
\end{center}
\begin{sol}
\Solution
The system has a pole in $z = \alpha$ and a zero in $z = 1$. We can write the transfer function of the system as
\[
H(z) = \frac{1 - z^{-1}}{1 - \alpha z^{-1}} = \frac{1}{1 - \alpha z^{-1}} - z^{-1}\frac{1}{1 - \alpha z^{-1}}
\]
A first order section with a pole in $z = \alpha$ has a transfer function $G(z) = 1/(1 - \alpha z^{-1})$ and impulse response $g[n] = \alpha^n u[n]$. Therefore the impulse response of the above system is
\[
h[n] = g[n] - g[n-1] = \alpha^n u[n] - \alpha^{n-1} u[n-1] = \begin{cases}
0 & n < 0 \\
1 & n = 0 \\
\alpha^{n-1}(\alpha - 1) & n > 0
\end{cases}
\]
\end{sol}
\end{exercise}
\begin{exercise}{System analysis}
Sketch the magnitude response of the following causal system:
\begin{center}
\begin{dspBlocks}{1.1}{0.5}
$x[n]~~$ & \BDadd & & \BDsplit & \BDadd & & \BDsplit & $~~y[n]$ \\
& & \BDdelay & & & \BDdelay &
\end{dspBlocks}
\ncline{1,4}{2,4}\ncline{1,7}{2,7}
\ncline{2,2}{2,3}\ncline{2,6}{2,5}
\psset{arrows=->}
\ncline{1,1}{1,2}\ncline{1,2}{1,5}\ncline{1,5}{1,8}
\ncline{4,2}{4,3}\ncline{4,3}{4,4}\ncline{4,4}{4,5}
\ncline{2,2}{1,2}\ncline{2,4}{2,3}\tbput{$\qquad\qquad (1+j)$}
\ncline{2,5}{1,5}\ncline{2,7}{2,6}\tbput{$\qquad\qquad (1-j)$}
\end{center}
\begin{sol}
\Solution
The transfer function of the system is $H(z) = H_1(z)H_2(z)$ where
\[
H_{1,2}(z) = \frac{1}{1 - (1\pm j)z^{-1}}.
\]
The system has therefore no zeros and two poles at $z = (1\pm j)$ or, in polar coordinates, at $z = e^{\pm j\frac{\pi}{4}}$ (note that the filter is not stable); its frequency response in magnitude follows the classic resonator pattern:
\begin{center}
\begin{dspPlot}[xtype=freq,yticks=none,xticks=4]{-1,1}{0,3}
\dspFunc{x \dspTFM{1}{1 -2 2}}
\end{dspPlot}
\end{center}
\end{sol}
\end{exercise}
\end{document}

Event Timeline

c4science · Help