Compute an appropriate sampling rate and DFT size \(N=2^{v}\) to analyze a single with no significant frequency content above \(10khz\) and with a minimum resolution of 100 hz
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\[ \fbox{$N=128$}\]
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 |
| \(0\) | \(\frac{2}{1}=2\) | \(\frac{\pi }{4}+0\times \frac{\pi }{16}=\frac{\pi }{4}\) | \(45\) |
| \(1\) | \(\frac{2}{2}=1\) | \(\frac{\pi }{4}+1\times \frac{\pi }{16}=\) \(\frac{5}{16}\pi \) | \(56.25\) |
| \(2\) | \(\frac{2}{4}=\frac{1}{2}\) | \(\frac{\pi }{4}+2\times \frac{\pi }{16}=\allowbreak \frac{3}{8}\pi \) | \(67.5\) |
| \(3\) | \(\frac{2}{8}=\frac{1}{4}\) | \(\frac{\pi }{4}+3\times \frac{\pi }{16}=\frac{7}{16}\pi \) | \(78.75\) |
| \(4\) | \(\frac{2}{16}=\frac{1}{8}\) | \(\frac{\pi }{4}+4\times \frac{\pi }{16}=\) \(\frac{1}{2}\pi \) | \(90\) |
| \(5\) | \(\frac{2}{32}=\frac{1}{16}\) | \(\frac{\pi }{4}+5\times \frac{\pi }{16}=\) \(\frac{9}{16}\pi \) | \(101.25\) |
| \(6\) | \(\frac{2}{64}=\frac{1}{32}\) | \(\frac{\pi }{4}+6\times \frac{\pi }{16}=\allowbreak \frac{5}{8}\pi \) | \(112.5\) |
| \(7\) | \(\frac{2}{128}=\frac{1}{64}\) | \(\frac{\pi }{4}+7\times \frac{\pi }{16}=\allowbreak \frac{11}{16}\pi \) | \(123.75\) |
This is Mathematica notebook used to make plot of the Chirp Z transform contour. This is my graded HW2