# Chapter8 - CHAPTER 8 8.1 THE NYQUIST STABILITY CRITERION...

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CHAPTER 8 THE NYQUIST STABILITY CRITERION AND STABILITY MARGINS 8.1 (a) Revisiting Example 8.1 will be helpful. (i) 1 (ii) 0 – 180º (iv) t t t t 1 2 1 2 90 + — - º (b) Revisiting Example 8.2 will be helpful. (i) – t j ¥ (ii) 0 – 180º (c) Revisiting Example 8.3 will be helpful. (i) ¥— – 180º (ii) 0 – 270º (d) (i) – ( t 1 + t 2 ) – j ¥ (ii) 0 – 270º (iii) tt 12 180 + —- º (e) (i) ¥— - 180º (ii) 0 0º (iv) tt tt 12 12 90 () º + 8.2 (a) Number of poles of G ( s ) in right half s -plane, P = 1 Number of clockwise encirclements of the critical point, N = 1 Z = number of zeros of 1 + G ( s ) in right half s -plane = N + P = 2 The closed-loop system is unstable. (b) P = 2 Number of counterclockwise encirclements of the critical point = 2 Therefore N = number of clockwise encirclements = – 2 Z = N + P = – 2 + 2 = 0 The closed-loop system is stable. (c) P = 0 N = (Number of clockwise encirclements of the critical point – number of counterclockwise encirclements of the critical point) = 0 Z = N + P = 0; the closed-loop system is stable. 8.3 The polar plot of G ( j w ) for w = 0 + to w = + ¥ is given in Fig. P8.3b. Plot of G ( j w ) for w = – ¥ to w = 0 is the reflection of the given polar plot with respect to the real axis. Since Gj w w = 0 = – 180º, G ( s ) has double pole at the origin. The map of Nyquist contour semicircle s = r e j f , r 0, f varying from – 90º at w = 0 through 0º to + 90º at w = 0 + , into the Nyquist plot is given by – 2 f (an infinite semicircle traversed clockwise).

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74 CONTROL SYSTEMS: PRINCIPLES AND DESIGN With this information, Nyquist plot for the given system can easily be drawn. (i) P = 0, N = 0; Z = N + P = 0 The closed-loop system is stable (ii) P = 1, N = 0; Z = N + P = 1 The closed-loop system is unstable (iii) P = 0, N = 0; Z = N + P = 0 The closed-loop system is stable. 8.4 (a) The key points to the polar plot are: Gj Hj () ww w = 18 w = 0 – 270º The intersections of the polar plot with the axes of G ( s ) H ( s )-plane can easily be ascertained by identifying the real and imaginary parts of G ( j w ) H ( j w ). When we set Im [ G ( j w ) H ( j w )] to zero, we get w = 4.123 and . w = 4 123 = – 1.428 Similarly, setting Re [ G ( j w ) H ( j w )] to zero, we get intersection with the imaginary axis. Based on this information, a rough sketch of the Nyquist plot can easily be made. From the Nyquist plot, we find that N = 2. Since P = 0, we have Z = N + P = 2, i.e., two closed-loop poles in right half s -plane. (b) The key points of the polar plot are (refer Problem 8.1d): w = 0 = – 3 – j ¥ w = 0 – 270º Intersection with the real axis at w = 1/ 2 ; w = 1 2 = - 4 3 The map of Nyquist contour semicircle s = r e j f , r 0, f varying from – 90º at w = 0 through 0º to + 90º at w = 0 + , into the Nyquist plot is given by ¥— f (an infinite semicircle traversed clockwise). Based on this information, a rough sketch of Nyquist plot can easily be made. P = 0, N = 2. Therefore Z = N + P = 2
SOLUTION MANUAL 75 (c) Gj Hj () ww w = 0 = 2 – 180º w = 0 – 90º No more intersections with real/imaginary axis. From the Nyquist plot, we find that N = – 1. Since P = 1, we have Z = N + P = 0; the closed- loop system is stable.

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Chapter8 - CHAPTER 8 8.1 THE NYQUIST STABILITY CRITERION...

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