Question 10.7: Find the range of gain for stability and instability, and th...

Since the open-loop poles are only in the left half-plane, the Nyquist criterion tells us that we want no encirclements of −1 for stability. Hence, a gain less than unity at ±180° is required. Begin by letting K = 1 and draw the portion of the contour along the positive imaginary axis as shown in Figure 10.34(a). In Figure 10.34(b), the intersection with the negative real axis is found by letting s = jω in G(s)H(s). Set the imaginary part equal to zero to find the frequency and then substitute the frequency into the real part of G(jω)H(jω). Thus, for any point on the positive imaginary axis,

G( jω) H( jω) = \frac{1}{(s² + 2s + 2)  (s + 2)} \mid_{s \rightarrow jω }      (10.46)

=\frac{4 (1  −  ω²)  −  jω (6  −  ω²)}{16 (1  −  ω²)²  +  ω²(6  −  ω²)²}

Setting the imaginary part equal to zero, we find ω = \sqrt{6} . Substituting this value back into Eq. (10.46) yields the real part, −(1/20) =(1/20)∠180°.

This closed-loop system is stable if the magnitude of the frequency response is less than unity at 180°. Hence, the system is stable for K < 20, unstable for K > 20, and marginally stable for K = 20. When the system is marginally stable, the radian frequency of oscillation is \sqrt{6} .