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┃        ┃
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┃      - ┃
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┃  Im d  ┃ L              ┃
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┃(       ┃                ┃
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┃        ┃                ┃
┗━━━━┻━━━━━━━━┛
c) F(s) : s - zi
Fig.5.17   nlustration ofcontour mapprng.
 corresponding points in the F-plane and obtain the contour B. Then the contour A
in the s-plane is said to be mapped to the contour B in the F -plane. We assume
 that the mapping is one to one, i.e., for every point in the s-plane, there is one and
 only one corresponding point in the F -plane and vice versa.
       To understand the concept ofmapping further,let F(s)  - s - Zi  and let the point
 s  - zi, which is the zero of F (s), lie outside the contour A as shown in Fig. 5.17a.
Instead of using the coordinates of point P,let us use the vector approach. Every
point P on the contour A is associated with a vector V. Let V' be the image
vector in the F-plane. For this case, IV'I = IVI and Z V' - / V. Now as we move
clockwise along the contour A, the magnitude and phase of the vector V varjr.
The phase oscillates betwcen.the two limiting values ~i and @z. In this case, a
clockwise movement along the contour A corresponds to a clockwise movement
along the image contour B in the F-plane.
     Now let F(s) = l/(s - pi) and let the pole s = Pi lie outside the contour A as
shown in Fig. 5.17b. For this case,  IV'I  =  1/I VI  and  L V'  =  -Z V. As a result, the
contour A in the first quadrant maps to contour B in the fourth quadrant. Observe
LINEAR SYSTEMS, THEORY, AND DESIGN: A BRIEF REVIEW       475
                 
  j(D d   /---   
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┃0     ┃~/y - ┃
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┃  ┃0r ~-   ┃
┃\ ┃~-,) ', ┃
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┃  ┃        ┃
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d) F(s) - l/(s - pi)
                       
        J(D J          
┏━━━━━━┳━━━┓
┃\           ┃~-y ' ┃
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  Im .                     
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┃0     ┃~,-, )  ┃    - ┃
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┗━━━┻━━━━┻━━━┛
e) F(s) - (s - zj)(s - zz)l(s - pl)(s - p,)
Fig. 5.17    nlustration of contour mapping, continued.
that a clockwise movement along the contour A in the s-plane corresponds to a
　
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