Stability via the Nyquist diagram Range of gain for stability Problem: For the unity feedback system below, where K G(s) = , s(s + 3)(s + 5) find the range of gain, K, for stability, instability and the value of K for marginal stability. For marginal stability, also find the frequency of oscillation. Use the Nyquist criterion.

Figure above; Closed-loop unity feedback system. 1

Solution: G(jω) =

K |s→jω s(s + 3)(s + 5)

−8Kω − j · K(15 − ω 2) = 64ω 3 + ω(15 − ω 2)2

When K = 1,

−8ω − j · (15 − ω 2) G(jω) = 64ω 3 + ω(15 − ω 2)2 Important points: Starting point: ω = 0, G(jω) = −0.0356 − j∞ Ending point: ω = ∞, G(jω) = 06 − 270◦ Real axis crossing: found by setting the imaginary part of G(jω) as zero, ω=

√

15,

K {− , j0} 120 2

When K = 1, P = 0, from the Nyquist plot, N is zero, so the system is stable. The real K does not encircle [−1, j0)] axis crossing − 120 until K = 120. At that point, the system is marginally stable, and the frequency of oscilla√ tion is ω = 15 rad/s. Nyquist Diagram Nyquist Diagram

0.05 2

0.04 1.5

0.03 System: G Real: −0.00824 Imag: 1e−005 Frequency (rad/sec): −3.91

0.01

1 Imaginary Axis

Imaginary Axis

0.02

0 −0.01

0.5 0 −0.5

−0.02 −1

−0.03

−1.5

−0.04 −0.05 −0.1

−0.08 −0.06 −0.04 −0.02 0 0.02 Real Axis

(a)

0.04

0.06

0.08

0.1

−2 −3

−2.5

−2

−1.5 Real Axis

−1

−0.5

0

(b)

K ; Figure above; Nyquist plots of G(s) = s(s+3)(s+5) (a) K = 1; (b)K = 120.

3

Gain/phase margin via the Nyquist diagram We use the Nyquist diagram to define two quantitative measures of how stable a system is. These are called gain margin and phase margin. Systems with greater gain margin and phase margins can withstand greater changes in system parameters before becoming unstable. Gain margin, GM , The gain margin is the change in open-loop gain, expressed in decibels (dB), required at 180◦ of phase shift to make the closed-loop system unstable. Phase margin, ΦM , The phase margin is the change in open-loop phase shift, required at unity gain to make the closed-loop system unstable. 4

Figure above; Nyquist diagram showing gain and phase margins Problem: Find the gain and phase margin for the unity feedback system with G(s) =

6 . 2 (s + 2s + 2)(s + 2)

5

Solution: From G(s), we see P = 0. The Nyquist diagram shows that N is zero, so the closed-loop system is stable. G(jω) =

6 |s→jω 2 (s + 2s + 2)(s + 2)

6[4(1 − ω 2) − jω(6 − ω 2)] = 16(1 − ω 2)2 + ω 2(6 − ω 2)2

The Nyquist diagram √ crosses the real axis at a frequency of ω = 6. The real part is found to be −0.3. Nyquist Diagram 1.5

Imaginary Axis

1

0.5

0

−0.5

−1

−1.5 −1

−0.5

0

0.5

1

1.5

Real Axis

Figure above; Nyquist diagram for G(s) =

6 . 2 (s + 2s + 2)(s + 2) 6

The gain margin is GM = 20 log(1/0.3) = 10.45dB. To find the phase margin find the frequency for which G(jω) has a unit gain. Using a computing tool, we can find G(jω) has a unit gain at a frequency of 1.251, at this frequency the phase angle is −112.3◦. The difference of this angle with −180◦ is 67.7◦, which is the phase margin ΦM . Gain/phase margin via the Bode plots

Figure above; Gain and phase margins on the Bode plots. 7

Problem: Let a unit feedback system have K G(s) = . (s + 2)(s + 4)(s + 5) Use Bode plots to determine the range of gain within which the system is stable. If K = 200 find the gain margin and the phase margin. The low frequency gain is found by setting s to zero. Thus the Bode magnitude plots starts at K/40. For convenience set K = 40, so that the log-magnitude plots starts at 0dB. At each break frequency, 2, 4 and 5, a slope of -20dB/decade is added. The phase diagram starts at 0◦ until 0.2rad/s (a decade below the break frequency of 2), the curve decreases at a slope of 45◦/decade at each subsequent frequency at 0.4rad/s and 0.5rad/s (a decade below the break frequency of 4 and 5 respectively). Finally at 20rad/s, 40rad/s and 50rad/s (a decade above the break frequencies of 2,4,5), a slope of +20dB/decade is added, until the curve levels out. 8

Figure above; Bode log-magnitude and phase 40 diagram for G(s) = (s+2)(s+4)(s+5) . The Nyquist criterion tells us that we want zero encirclement of {−1, j0}. Thus the Bode logmagnitude plot must be less than unity when the Bode phase plot is −180◦. Accordingly we see that at frequency 7 rad/s, when the phase plot is −180◦. The magnitude is -20dB. 9

Thus an increase of 20dB is possible before the system becomes unstable, which is a gain of 10, so the gain for instability is K > 10 × 40 = 400. If K = 200 (five times greater than K = 40), the magnitude plot would be 20 log 5 = 13.98dB higher, as the Bodes plots was scaled to a gain of 40. At −180◦, the gain is −20 + 13.98 = −6.02dB, so GM = 6.02dB. To find phase margin, we look on the magnitude plot for the frequency where the gain is 0dB. As the plot should be 13.98dB higher, so we look at −13.98dB crossing to find the frequency is 5.5rad/s. At this frequency, the phase angle is −165◦. Thus ΦM = −165◦ − (−180)◦ = 15◦. 10