1. Frequency Resp. method Given a system: G(s)
G(jω) as a function of ω is called the frequency response of the system, or of G(s)
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 of G
y(s)
u(s)
G(s)

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

3. Relationship between bode and nyquist
length
vector

4. 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: Bode
Matlab command to explore: nyquist, bode

5. To obtain freq. resp. experimentally: only if system is stable
Select
Give input to system:
Adjust A1 so that the output is not saturated or distorted.
Measure amp B1 and phase φ1 of output:
u(s)
y(s)
System

6. Then is the freq. resp. of the system at freq ω1
Repeat the steps for all ωK
Either plot
or plot

7. y(s)
u(s)
G1(s)
G2(s)
Product of T.F.
G(s)

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

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

11. 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)

13. Example

14. Example continued 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,

15. 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 evaluated at ω = 1
=ws0dB =asymptotic straight line’s 0 dB crossing frequency
Ka = 0

18. Example
Asymptotic straight line
ws0dB
~14

19. Example continued 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.

20. Example continued 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,
Notice: dashed line intersect 0dB line at w=14

21. Back to general theory 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

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

25. Example continued

26. Example continued 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,

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

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

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

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

31. System type on Nyquist plot

32. System relative order

33. Examples
System type =
Relative order =
System type =
Relative order =