Question 10.6: For the unity-feedback system of Figure 10.10, where G(s) = ...

First set K = 1 and sketch the Nyquist diagram for the system, using the contour shown in Figure 10.31(a). For all points on the imaginary axis,

G( jω) H( jω) = \frac{K}{s (s  +  3)  (s  +  5)} \mid_{\underset{s=jω}{K= 1}}  =  \frac{−8ω²  −  j  (15ω  −  ω³)}{64ω^{4}  +  ω²  (15  −  ω²)²}      (10.45)

At ω = 0, G(jω)H(jω) = −0.0356 − j ∞.

Next find the point where the Nyquist diagram intersects the negative real axis. Setting the imaginary part of Eq. (10.45) equal to zero, we find ω = \sqrt{15} . Substituting this value of ω back into Eq. (10.45) yields the real part of −0.0083. Finally, at ω = ∞, G(jω)H(jω) = G(s)H(s)\mid _{s \rightarrow j\infty } = 1/(j∞)³ = 0∠ − 270°.

From the contour of Figure 10.31(a), P = 0; for stability N must then be equal to zero. From Figure 10.31(b), the system is stable if the critical point lies outside the contour (N = 0), so that Z = P − N = 0. Thus, K can be increased by 1/0.0083 = 120.5 before the Nyquist diagram encircles −1. Hence, for stability, K < 120.5. For marginal stability K = 120.5. At this gain, the Nyquist diagram intersects −1, and the frequency of oscillation is \sqrt{15} rad/s.