Here, we will obtain the relationship of jΩ axis in s-plane to the unit circle in the z-plane (r = 1)

$Ω=\frac{2}{T_s} ×\frac{(2r sin⁡ω)}{(r^2+2r cos⁡ω+1 )}$ ………………(1)

For the unit circle,r=1. Thus putting r=1in the equation 1, we get

$Ω=\frac{2}{T_s} ×\frac{(2 sin⁡ω)}{(1+2 cos⁡ω+1 )}$

$∴Ω=\frac{2}{T_s} ×\frac{(2 sin⁡ω)}{(2+2 cos⁡ω )}$

$∴Ω=\frac{2}{T_s} ×\frac{sin⁡ω}{(1+cos⁡ω )}$ …………….….(2)

We have trigonometric identities

$sin⁡ω=2sin⁡\frac{ω}{2}.cos⁡\frac{ω}{2} and 2 cos^2⁡\frac{ω}{2}$

$1+ cos⁡ω$

Thus equation (1) becomes,

$Ω=\frac{2}{T_s} ×\frac{2 sin⁡\frac{ω}{2} cos⁡\frac{ω}{2}}{2 cos^2⁡\frac{ω}{2}}$

$Ω=\frac{2}{T_s} ×\frac{2 sin⁡\frac{ω}{2}}{2 cos⁡\frac{ω}{2}}$

$Ω=\frac{2}{T_s} tan\frac{⁡ω}{2}$

$ω=2tan^(-1)⁡\frac{ΩT_s}{2}$ …..(2)

Now for different values of $ΩT_s$ ; the graph of $ΩT_s$ versus ω is as shown in below Fig.