written 22 months ago by | • modified 22 months ago |
Solution:
The mapping of points from solans to z plane implied by the relation
$ z=e^{s t} $
If we substitute
$ \begin{aligned}\\ &S=\sigma+j \Omega \\\\ &z=\gamma e^{j \omega}\\ \end{aligned}\\ $
Where,
$\omega \rightarrow$ Digital frequency (rad/sample)
$\Omega \rightarrow$ Analog frequency (rad /sec).
$r \rightarrow$ Magnitude
$\sigma \rightarrow$ Attenuation
$T \rightarrow$ sampling perseid.
$ \begin{aligned}\\ &\gamma e^{j \omega}=e^{(\sigma+j \Omega) T} \\\\ &\gamma e^{j \omega}=e^{\sigma T} e^{j \Omega T} \\\\ &\gamma e^{j \omega}=e^{\sigma T} e^{j \Omega T} \\\\ &r=e^{\sigma T} \\\\ &\omega=\Omega T\\ \end{aligned}\\ $
As $\omega=\Omega T$, the mapping of $j \Omega$ axis into a unit circle is not one-to-one.
Since $\omega$ is unique over the range $(-\pi, \pi)$ the mapping $\omega=\Omega T$ implies that the interval $-\frac{\pi}{T} \leq \Omega \leq \frac{\pi}{T}$ maps into the corresponding value $-\pi \leq \omega \leq \pi$
Furthermore the frequency interval
$\frac{\pi}{T} \leq \Omega \leq \frac{3 \pi}{T}$ also maps into the interval
$ \longrightarrow \text { In general, the interval } \frac{(2 k-1) \pi}{\top} \leq \Omega \leq \frac{(2 k+1) \pi}{T}\\ $
$ \text { mane into the interval }-\pi \leq \omega \leq \pi \text {. }\\ $
Thus mapping from analog frequency $\Omega$ to the variable frequency $\omega$ in the digital domain is many to one which reflects the effect of. aliasing due to sampling.