1. Frequency Resp. method Given:
G(jω) as a function of ω is called the freq. resp.
For each ω, G(jω) = x(ω) + jy(ω) is a point in the complex plane
As ω varies from 0 to ∞, the plot of G(jω) is called the Nyquist plot

2. Can rewrite in Polar Form:
|G(jω)| as a function of ω is called the amplitude resp.
as a function of ω is called the phase resp.
The two plots:
With log scale-ω, are Bode plot

3. To obtain freq. Resp from G(s):
Select
Evaluate G(jω) at those to get
Plot Imag(G) vs Real(G): Nyquist
or plot
With log scale ω
Matlab command to explore: nyquist, bode

4. To obtain freq. resp. experimentally:
Select
Given input to system as:
Adjust A1 so that the output is not saturated or distorted.
Measure amp B1 and phase φ1 of output:

5. Then is the freq. resp. of the system at freq ω
Repeat for all ωK
Either plot
or plot

6. Product of T.F.

8. System type, steady state tracking, & Bode plot
R(s)
C(s)
Gp(s)
Y(s)

9. As ω → 0
Therefore: gain plot slope = –20N dB/dec.
phase plot value = –90N deg

10. If Bode gain plot is flat at low freq, system is “type zero”
Confirmed by phase plot flat and  0°
at low freq
Then: Kv = 0, Ka = 0
Kp = Bode gain as ω→0
= DC gain
(convert dB to values)

11. Example

12. Steady state tracking error
Suppose the closed-loop system is stable:
If the input signal is a step,
ess would be =
If the input signal is a ramp,
If the input signal is a unit acceleration,

13. N = 1, type = 1
Bode mag. plot has –20 dB/dec slope at low freq. (ω→0)
(straight line with slope = –20 as ω→0)
Bode phase plot becomes flat at –90° when ω→0
Kp = DC gain → ∞
Kv = K = value of asymptotic straight line at ω = 1
=ws0dB =asymptotic straight line’s 0 dB crossing frequency
Ka = 0

14. Example
Asymptotic straight line
ws0dB
~14

15. The matching phase plot at low freq. must be → –90°
type = 1
Kp = ∞ ← position error const.
Kv = value of low freq. straight line
at ω = 1
= 23 dB
≈ 14 ← velocity error const.
Ka = 0 ← acc. error const.

16. Steady state tracking error
Suppose the closed-loop system is stable:
If the input signal is a step,
ess would be =
If the input signal is a ramp,
If the input signal is a unit acceleration,

17. N = 2, type = 2
Bode gain plot has –40 dB/dec slope at low freq.
Bode phase plot becomes flat at –180° at low freq.
Kp = DC gain → ∞
Kv = ∞ also
Ka = value of straight line at ω = 1
= ws0dB^2

18. Example Ka
ws0dB=Sqrt(Ka)
How should the phase plot look like?

20. Steady state tracking error
Suppose the closed-loop system is stable:
If the input signal is a step,
ess would be =
If the input signal is a ramp,
If the input signal is a unit acceleration,

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

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

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

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

25. 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)

28. 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.

29. Example:
Bode plot on next page.

31. Example:
Bode plot on next page.

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

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

36. 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