1. System type, steady state tracking, & Bode plot
R(s)
C(s)
Gp(s)
Y(s)
Type = N
At very low frequency:
gain plot slope = –20N dB/dec.
phase plot value = –90N deg

2. Kv = 0
Ka = 0
Low freq phase = 0o

3. Kp = ∞
Ka = 0
Low freq phase = -90o
=Kv

4. Kp = ∞
Kv = ∞
Low freq phase = -180o

5. System type, steady state tracking, & Nyquist plot
C(s)
Gp(s)
As ω → 0

6. Type 0 system, N=0
Kp=lims0 G(s)
=G(0)=K Kp
w0+
G(jw)

7. Type 1 system, N=1
Kv=lims0 sG(s) cannot be determined easily from Nyquist plot
winfinity
w0+
G(jw)  -j∞

8. Type 2 system, N=2
Ka=lims0 s2G(s) cannot be determined easily from Nyquist plot
winfinity
w0+
G(jw)  -∞

9. System type on Nyquist plot

10. System relative order

11. Examples
System type =
Relative order =
System type =
Relative order =

12. In most cases, stability of this closed-loop
Margins on Bode plots
In most cases, stability of this closed-loop
can be determined from the Bode plot of G:
Phase margin > 0
Gain margin > 0
G(s)

15. If never cross 0 dB line (always below 0 dB line), then PM = ∞.
If never cross –180° line (always above –180°), then GM = ∞.
If cross –180° several times, then there are several GM’s.
If cross 0 dB several times, then there are several PM’s.

16. Example:
Bode plot on next page.

18. Example:
Bode plot on next page.

20. Where does cross the –180° line Answer: __________ at ωpc, how much is
Closed-loop stability: __________

22. crosses 0 dB at __________ at this freq,
Does cross –180° line? ________
Closed-loop stability: __________

23. Margins on Nyquist plot
Suppose:
Draw Nyquist plot G(jω) & unit circle
They intersect at point A
Nyquist plot cross neg. real axis at –k

26. Stability from Nyquist plot
The complete Nyquist plot:
Plot G(jω) for ω = to +∞
Get complex conjugate of plot, that’s G(jω) for ω = 0– to –∞
If G(s) has pole on jω-axis, treat separately
Mark direction of ω increasing
Locate point: –1

28. Encirclement of the -1 point
As you follow along the G(jω) curve for one complete cycle, you may “encircle” the –1 point
Going around in c.w. once is +1 encirclement
c.c.w. once is –1 encirclement

32. Nyquist Criterion Theorem
# (unstable poles of closed-loop) Z = # (unstable poles of open-loop) P + # encirclement N or: Z = P + N To have closed-loop stable: need Z = 0, i.e. N = –P

33. That is: G(jω) needs to encircle the “–1” point c.c.w. P times.
If open loop is stable to begin with, G(jω) cannot encircle the “–1” point for closed-loop stability
In previous example:
No encirclement, N = 0.
Open-loop stable, P = 0
Z = P + N = 0, no unstable poles in closed-loop, stable

34. Example:

35. As you move around
from ω = –∞ to 0–,
to 0+, to +∞, you go
around “–1” c.c.w.
once.
# encirclement N = – 1.
# unstable pole P = 1

36. Check: c.l. pole at s = –3, stable.
i.e. # unstable poles of closed-loop = 0
closed-loop system is stable.
Check:
c.l. pole at s = –3, stable.

37. Example:
Get G(jω) for ω = 0+ to +∞
Use conjugate to get G(jω) for ω = –∞ to 0–
How to go from ω = 0– to ω = 0+? At ω ≈ 0 :

40. # encirclement N = _____
# open-loop unstable poles P = _____
Z = P + N = ________
= # closed-loop unstable poles.
closed-loop stability: _______

41. Q: 1. Find stability margins
Example:
Given:
G(s) is stable
With K = 1, performed open-loop sinusoidal tests, and G(jω) is on next page
Q: 1. Find stability margins
2. Find Nyquist criterion to determine closed-loop stability

43. Solution:
Where does G(jω) cross the unit circle? ________ Phase margin ≈ ________ Where does G(jω) cross the negative real axis? ________ Gain margin ≈ ________ Is closed-loop system stable with K = 1? ________

44. Note that the total loop T.F. is KG(s).
If K is not = 1, Nyquist plot of KG(s) is a scaling of G(jω).
e.g. If K = 2, scale G(jω) by a factor of 2 in all directions.
Q: How much can K increase before GM becomes lost? ________
How much can K decrease? ______

45. Some people say the gain margin is 0 to 5 in this example
Q: As K is increased from 1 to 5, GM is lost, what happens to PM?
What’s the max PM as K is reduced to 0 and GM becomes ∞?

46. To use Nyquist criterion, need complete Nyquist plot.
Get complex conjugate
Connect ω = 0– to ω = 0+ through an infinite circle
Count # encirclement N
Apply: Z = P + N
o.l. stable, P = _______
Z = _______
c.l. stability: _______

47. Correct
Incorrect

48. Example:
G(s) stable, P = 0
G(jω) for ω > 0 as given.
Get G(jω) for ω < 0 by conjugate
Connect ω = 0– to ω = 0+. But how?

49. Incorrect Choice a) : Where’s “–1” ? # encirclement N = _______
Z = P + N = _______
Make sense? _______
Incorrect

50. Correct Choice b) : Where is “–1” ? # encir. N = _____
Z = P + N = _______
closed-loop stability _______
Correct

51. Note: If G(jω) is along –Re axis to ∞ as ω→0+, it means G(s) has in it.
when s makes a half circle near ω = 0, G(s) makes a full circle near ∞.
choice a) is impossible, but choice b) is possible.

52. Incorrect

53. Example: G(s) stable, P = 0
Get conjugate for ω < 0
Connect ω = 0– to ω = 0+. Needs to go one full circle with radius ∞. Two choices.

54. Choice a) :
N = 0
Z = P + N = 0
closed-loop stable
Incorrect!

55. Correct! Choice b) : N = 2 Z = P + N = 2
Closed loop has two unstable poles
Correct!

56. Which way is correct?
For stable & non-minimum phase systems,

57. Example: G(s) has one unstable pole
P = 1, no unstable zeros
Get conjugate
Connect ω = 0– to ω = 0+. How? One unstable pole/zero If connect in c.c.w.

58. # encirclement N = ?
If “–1” is to the left of A
i.e. A > –1
then N = 0
Z = P + N = = 1
but if a gain is increased, “–1” could be inside, N = –2
Z = P + N = –1
c.c.w. is impossible

59. If connect c.w.:
For A > –1 N = ______
Z = P + N
= ______
For A < –1 N = ______
Z = ______
No contradiction This is the correct way.

60. Example: G(s) stable, minimum phase
G(jω) as given:
get conjugate.
Connect ω = 0–
to ω = 0+,

61. If A < –1 < 0 :. N = ______. Z = P + N = ______. stability of c
If A < –1 < 0 : N = ______ Z = P + N = ______ stability of c.l. : ______
If B < –1 < A : A=-0.2, B=-4, C=-20 N = ______ Z = P + N = ______ closed-loop stability: ______
Gain margin: gain can be varied between (-1)/(-0.2) and (-1)/(-4),
or can be less than (-1)/(-20)

62. If C < –1 < B :. N = ______. Z = P + N = ______
If C < –1 < B : N = ______ Z = P + N = ______ closed-loop stability: ______
If –1 < C : N = ______ Z = P + N = ______ closed-loop stability: ______