Question 10.9: Use Bode plots to determine the range of K within which the ...
Use Bode plots to determine the range of K within which the unity-feedback system shown in Figure 10.10 is stable. Let G(s) = K/[(s + 2)(s + 4)(s + 5)].
Since this system has all of its open-loop poles in the left half-plane, the open-loop system is stable. Hence, from the discussion of Section 10.5, the closed-loop system will be stable if the frequency response has a gain less than unity when the phase is 180°.
Begin by sketching the Bode magnitude and phase diagrams shown in Figure 10.36. In Section 10.2, we summed normalized plots of each factor of G(s) to create the Bode plot. We saw that at each break frequency, the slope of the resultant Bode plot changed by an amount equal to the new slope that was added. Table 10.6 demonstrates this observation. In this example, we use this fact to draw the Bode plots faster by avoiding the sketching of the response of each term.
TABLE 10.6 Magnitude diagram slopes for Example 10.9
| Frequency (rad/s) | ||||
| Description | 0.01 (Start: Plot) |
2 (Start: Pole at −2) |
3 (Start: Zero at −3) |
5 (Start: ω_{n} = 5) |
| Pole at −2 | 0 | −20 | −20 | −20 |
| Zero at −3 | 0 | 0 | 20 | 20 |
| ω_{n} = 5 | 0 | 0 | 0 | −40 |
| Total slope (dB/dec) |
0 | −20 | 0 | −40 |
The low-frequency gain of G(s)H(s) is found by setting s to zero. Thus, the Bode magnitude plot starts at K/40. For convenience, let K = 40 so that the log-magnitude plot starts at 0 dB. At each break frequency, 2, 4, and 5, a 20-dB/decade increase in negative slope is drawn, yielding the log-magnitude plot shown in Figure 10.36.
The phase diagram begins at 0° until a decade below the first break frequency of 2 rad/s. At 0.2 rad/s, the curve decreases at a rate of −45°/decade, decreasing an additional 45°/decade at each subsequent frequency (0.4 and 0.5 rad/s) a decade below each break. At a decade above each break frequency, the slopes are reduced by 45°/decade at each frequency.
The Nyquist criterion for this example tells us that we want zero encirclements of −1 for stability. Thus, we recognize that the Bode log-magnitude plot must be less than unity when the Bode phase plot is 180°. Accordingly, we see that at a frequency of 7 rad/s, when the phase plot is −180°, the magnitude plot is −20 dB. Therefore, an increase in gain of +20 dB is possible before the system becomes unstable. Since the gain plot was scaled for a gain of 40, +20 dB (a gain of 10) represents the required increase in gain above 40. Hence, the gain for instability is 40 × 10 = 400. The final result is 0 < K < 400 for stability.
This result, obtained by approximating the frequency response by Bode asymptotes, can be compared to the result obtained from the actual frequency response, which yields a gain of 378 at a frequency of 6.16 rad/s.
Students who are using MATLAB should now run ch10apB4 in Appendix B. You will learn how to use MATLAB to find the range of gain for stability via frequency response methods. This exercise solves Example 10.10 using MATLAB.
