Stability via the Nyquist diagram

Range of gain for stability

Problem: For the unity feedback system be- low, where K G(s)= , s(s + 3)(s + 5) find the range of gain, K, for stability, insta- bility 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 sys- tem. 1 Solution: K G(jω)= s jω s(s + 3)(s + 5)| →

8Kω j K(15 ω2) = − − · − 64ω3 + ω(15 ω2)2 − When K = 1, 8ω j (15 ω2) G(jω)= − − · − 64ω3 + ω(15 ω2)2 −

Important points:

Starting point: ω = 0, G(jω)= 0.0356 j − − ∞

Ending point: ω = , G(jω) = 0 270 ∞ 6 − ◦

Real axis crossing: found by setting the imag- inary part of G(jω) as zero, K ω = √15, , j0 {−120 }

2 When K = 1, P = 0, from the Nyquist plot, N is zero, so the system is stable. The real axis crossing K does not encircle [ 1, j0)] −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

0.03 1.5

0.02 System: G 1 Real: −0.00824 0.01 Imag: 1e−005 Frequency (rad/sec): −3.91 0.5 0 0

Imaginary Axis −0.01 Imaginary Axis −0.5 −0.02 −1 −0.03

−0.04 −1.5

−0.05 −2 −0.1 −0.08 −0.06 −0.04 −0.02 0 0.02 0.04 0.06 0.08 0.1 −3 −2.5 −2 −1.5 −1 −0.5 0 Real Axis Real Axis (a) (b)

Figure above; Nyquist plots of ( )= K ; 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 unsta- ble.

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 un- stable.

4 Figure above; Nyquist diagram showing gain and phase margins

Problem: Find the gain and phase margin for the unity feedback system with 6 G(s)= . (s2 + 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. 6 G(jω)= s jω (s2 + 2s + 2)(s + 2)| →

6[4(1 ω2) jω(6 ω2)] = − − − 16(1 ω2)2 + ω2(6 ω2)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

1

0.5

0 Imaginary Axis

−0.5

−1

−1.5 −1 −0.5 0 0.5 1 1.5 Real Axis

Figure above; Nyquist diagram for 6 G(s)= . (s2 + 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 com- puting 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 diagram for ( )= 40 G s (s+2)(s+4)(s+5).

The Nyquist criterion tells us that we want zero encirclement of 1, j0 . Thus the Bode log- {− } magnitude 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 mag- nitude 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 − ◦ Φ = 165 ( 180) = 15 . M − ◦ − − ◦ ◦ 10