Question 10.5: Sketch the Nyquist diagram of the unity-feedback system of F...
Sketch the Nyquist diagram of the unity-feedback system of Figure 10.10, where G(s) = (s + 2)/s².
The system’s two poles at the origin are on the contour and must be bypassed, as shown in Figure 10.29(a). The mapping starts at point A and continues in a clockwise direction. Points A, B, C, D, E, and F of Figure 10.29(a) map, respectively, into points A′, B′, C′, D′, E′, and F′ of Figure 10.29(b).
At point A, the two open-loop poles at the origin contribute 2 × 90° = 180°, and the zero contributes 0°. The total angle at point A is thus −180°. Close to the origin, the function is infinite in magnitude because of the close proximity to the two open-loop poles. Thus, point A maps into point A′, located at infinity at an angle of −180°.
Moving from point A to point B along the contour yields a net change in angle of +90° from the zero alone. The angles of the poles remain the same. Thus, the mapping changes by +90° in the counterclockwise direction. The mapped vector goes from −180° at A′ to −90° at B′. At the same time, the magnitude changes from infinity to zero, since at point B there is one infinite length from the zero divided by two infinite lengths from the poles.
Alternately, the frequency response can be determined analytically from G(jω) = (2 + jω)/(− ω²), considering ω going from 0 to ∞. At low frequencies, G(jω) ≈ 2/(− ω² ), or ∞∠180°. At high frequencies, G(jω) ≈ j/(− ω), or 0∠ −90°. Also, the real and imaginary parts are always negative.
As we travel along the contour BCD, the function magnitude stays at zero (one infinite zero length divided by two infinite pole lengths). As the vectors move through BCD, the zero’s vector and the two poles’ vectors undergo changes of −180° each. Thus, the mapped vector undergoes a net change of +180°, which is the angular change of the zero minus the sum of the angular changes of the poles {− 180 − [2(− 180)] = + 180}. The mapping is shown as B′ C′ D′, where the resultant vector changes by +180° with a magnitude of ε that approaches zero.
From the analytical point of view,
G(s) = \frac{R_{-2} ∠ θ_{-2} }{(R_{0} ∠ θ_{0} ) (R_{0} ∠ θ_{0} )} (10.44)
anywhere on the s-plane where R_{-2} ∠ θ_{-2} is the vector from the zero at −2 to any point on the s-plane, and R_{0} ∠ θ_{0} is the vector from a pole at the origin to any point on the s-plane. Around the infinite semicircle, all R_{-i} = ∞, and all angles can be approximated as if the vectors originated at the origin. Thus at point B, G(s) = 0∠ − 90°, since all θ_{-i} = 90° in Eq. (10.44). At point C, all R_{-i} = ∞, and all θ_{-i} = 0° in Eq. (10.44). Thus, G(s) = 0∠0°. At point D, all R_{-i} = ∞, and all θ_{-i} = −90° in Eq. (10.44). Thus, G(s) = 0∠90°.
The mapping of the section of the contour from D to E is a mirror image of the mapping of A to B. The result is D′ to E′.
Finally, over the section EFA, the resultant magnitude approaches infinity. The angle of the zero does not change, but each pole changes by +180°. This change yields a change in the function of −2 × 180° = −360°. Thus, the mapping from E′ to A′ is shown as infinite in length and rotating −360°. Analytically, we can use Eq. (10.44) for the points along the contour EFA. At E, G(s) = (2∠ 0°)/[(ε∠ − 90°) (ε∠ − 90°)] = ∞∠180°. At F, G(s) = (2∠0°)/[(ε∠ 0°)(ε∠ 0°)] = ∞∠ 0°. At A, G(s) = (2∠ 0°)/[(ε ∠90°) (ε∠90°)] = ∞∠ − 180°.
The Nyquist diagram is now complete, and a test radius drawn from −1 in Figure 10.29(b) shows one counterclockwise revolution, and one clockwise revolution, yielding zero encirclements.
Students who are using MATLAB should now run ch10apB2 in Appendix B. You will learn how to use MATLAB to make a Nyquist plot and list the points on the plot. You will also learn how to specify a range for frequency. This exercise solves Example 10.5 using MATLAB.
