**1 Answer**

written 4.6 years ago by |

In Impulse Invariance Transformation, the mapping of points from the s-plane to the z-plane is given by the relation $z=e^{p_{i} T}=e^{s T},$ where $p_{i}$ are poles of analog filter and T is sampling frequency.

If $s=\sigma+j \Omega$ then $z=e^{(\sigma+j \Omega) T}$

Let $s=\sigma+j\left(\Omega+\frac{2 \pi k}{T}\right)$

$\therefore z=e^{\left[\sigma+j\left(\Omega+\frac{2 \pi k}{T}\right)\right] T}$

$=e^{[\sigma T+j \Omega T+j 2 \pi k]}$

$=e^{(\sigma+j \Omega) T} \cdot e^{j 2 \pi k}$

$=e^{(\sigma+j \Omega) T}$

Thus, in general, the mapping of any pole $s=\sigma+j\left(\Omega+\frac{2 \pi k}{T}\right)$ in the s-plane is same as that of the pole $s=\sigma+j \Omega$ for all integer values of $k.$

A strip of width $\frac{2 \pi}{T}$ in the s-plane for $\frac{-\pi}{T} \leq \Omega \leq \frac{\pi}{T}$ is mapped into entire z-plane.

Similarly, another strip of width $\frac{2 \pi}{T}$ in the s-plane for $\frac{\pi}{T} \leq \Omega \leq \frac{3 \pi}{T}$ is mapped into entire z-plane.

Thus, any strip of width $\frac{2 \pi}{T}$ in the s-plane for $(2 k-1) \frac{\pi}{T} \leq \Omega \leq(2 k+1) \frac{\pi}{T}$ is mapped into entire z-plane.

For each strip, left half portion in s-plane is mapped inside the unit circle while the right half portion in s- plane is mapped outside the unit circle in z-plane. The $j \Omega$ -axis is mapped on the unit circle.

Hence, in IIT there is many to one mapping of poles from s-plane to z-plane.