4.1 HW2

4.1.1 Problem 1

Compute an appropriate sampling rate and DFT size \(N=2^{v}\) to analyze a signal with no significant frequency content above \(10khz\) and with a minimum resolution of 100 hz

4.1.1.1 Solution

From Nyquist sampling theory we obtain that sampling frequency is

\[ f_{s}=20000\ hz \]

Now, the frequency resolution is given by

\[ \Delta f=\frac {f_{s}}{N}\]

where N is the number of FFT samples. Now since the minimum \(\Delta f\) is 100 hz then we write

\[ \frac {f_{s}}{N}=\Delta f\geq 100 \]

or

\[ \frac {f_{s}}{N}\geq 100 \]

Hence

\begin{align*} N & \leq \frac {20,000}{100}\\ & \leq 200\text { samples}\end{align*}

Therefore, we need the closest N below 200 which is power of 2, and hence

\[ \fbox {$N=128$}\]

4.1.2 Problem 2

sketch the locus of points obtained using Chirp Z Transform in the Z plane for \(M=8,W_{0}=2,\phi _{0}=\frac {\pi }{16},A_{0}=2,\theta _{0}=\frac {\pi }{4}\)

Answer:

Chirp Z transform is defined as

\begin{equation} X\left ( z_{k}\right ) ={\displaystyle \sum \limits _{n=0}^{N-1}} x\left [ n\right ] z_{k}^{-n} \qquad k=0,1,\cdots ,M-1 \tag {1}\end{equation}

Where

\[ z_{k}=AW^{-k}\]

and \(A=A_{0}e^{j\theta _{0}}\) and \(W=W_{0}e^{-j\phi _{0}}\)

Hence

\begin{align*} z_{k} & =\left ( A_{0}e^{j\theta _{0}}\right ) \left ( W_{0}e^{-j\phi _{0}}\right ) ^{-k}\\ & =\frac {A_{0}}{W_{0}^{k}}e^{j\left ( \theta _{0}+k\phi _{0}\right ) }\end{align*}

Hence

\begin{align*} \left \vert z_{k}\right \vert & =\frac {A_{0}}{W_{0}^{k}}\\ & =\frac {2}{2^{k}}\end{align*}

and

\begin{align*} phase\ of\ z_{k} & =\theta _{0}+k\phi _{0}\\ & =\frac {\pi }{4}+k\frac {\pi }{16}\end{align*}

Hence

\(k\)	\(\left \vert z_{k}\right \vert =\frac {2}{2^{k}}\)	\(phase\ of\ z_{k}=\frac {\pi }{4}+k\frac {\pi }{16}\)	\(phase\ of\ z_{k\ }in\) degrees
\(0\)	\(\frac {2}{1}=2\)	\(\frac {\pi }{4}+0\times \frac {\pi }{16}=\frac {\pi }{4}\)	\(45\)
\(1\)	\(\frac {2}{2}=1\)	\(\frac {\pi }{4}+1\times \frac {\pi }{16}=\) \(\frac {5}{16}\pi \)	\(56.25\)
\(2\)	\(\frac {2}{4}=\frac {1}{2}\)	\(\frac {\pi }{4}+2\times \frac {\pi }{16}=\allowbreak \frac {3}{8}\pi \)	\(67.5\)
\(3\)	\(\frac {2}{8}=\frac {1}{4}\)	\(\frac {\pi }{4}+3\times \frac {\pi }{16}=\frac {7}{16}\pi \)	\(78.75\)
\(4\)	\(\frac {2}{16}=\frac {1}{8}\)	\(\frac {\pi }{4}+4\times \frac {\pi }{16}=\) \(\frac {1}{2}\pi \)	\(90\)
\(5\)	\(\frac {2}{32}=\frac {1}{16}\)	\(\frac {\pi }{4}+5\times \frac {\pi }{16}=\) \(\frac {9}{16}\pi \)	\(101.25\)
\(6\)	\(\frac {2}{64}=\frac {1}{32}\)	\(\frac {\pi }{4}+6\times \frac {\pi }{16}=\allowbreak \frac {5}{8}\pi \)	\(112.5\)
\(7\)	\(\frac {2}{128}=\frac {1}{64}\)	\(\frac {\pi }{4}+7\times \frac {\pi }{16}=\allowbreak \frac {11}{16}\pi \)	\(123.75\)

Figure 4.1: plot of the above contour

This is Mathematica notebook used to make plot of the Chirp Z transform contour.

4.1.3 graded HW2

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