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enter image description here


Refer to diagram above, in Ogata's text on discrete time control, he showed that you can map a curve in the S plane, namely curve 1,2,3,4,5 onto a circle in the Z plane through the complex mapping $e^{Ts}$

What I do not understand is why are we missing the negative branch? This is curious because we must have poles in the z-plane located on the negative branch. So if that branch is missing, how are we suppose to map those poles back to the S plane?

Can someone explain why the negative axis is missing in this mapping?

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If we denote $s=\sigma+j\omega$, then for the mapping $z=e^{sT}$, the negative real axis in the $z$-plane in the range $z\in[0,-1]$ corresponds to $\omega=\pm\omega_s/2$ (i.e. Nyquist) and $\sigma\in [-\infty,0]$, where $\omega_s$ is the sampling frequency in radians. So if the curve in the $s$-plane moved along the lines $s=\pm j\omega_s/2$ then the curve in the $z$-plane would move on the negative real axis of the $z$-plane. If it doesn't (as suggested in the figure) then this range of the $z$-plane will be excluded. So if with this mapping you want poles on the negative real axis of the $z$-plane, you need corresponding poles with imaginary parts $\pm \omega_s/2$ in the $s$-plane.

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  • $\begingroup$ Hi can you comment on the reasoning as to why the author may have deliberately chose to excuse the line $\omega = \omega_s/2$? $\endgroup$ – Carlos - the Mongoose - Danger Feb 15 '15 at 12:05
  • $\begingroup$ @IllegalImmigrant: The lines $\pm j\omega_s/2$ are the absolute limit for this transform to work. You can get arbitrarily close, but in the figure they probably wanted to show the two different paths in the $z$-plane (between 2 and 3, and 4 and 5), which would look like one single path otherwise. $\endgroup$ – Matt L. Feb 15 '15 at 20:06

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