What is the relation between digital and analog frequency in impulse invariant transformation? What is aliasing?
1 Answer


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}\\ $


$\omega \rightarrow$ 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.

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