Download - Frequency Resp. method
Frequency Resp. methodGiven:
• 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
• 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
)()G(j)( jGjejG
)( jG
~)(
~)(log20 10
jG
jG
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
N 21
NGGGG 21
vsGangle
vsGabs
)(
))((log20 10
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 ofoutput:
N 21)sin()( 11 tAtu
)sin()( 111 tBty
• Then is the freq. resp. of the system at freq ω
• Repeat for all ωK
• Either plot
or plot
1
1
1
11
1
jeA
Bor
andA
B
BodeA
B
KK
KK
K
~
~
)( Kj
K
K eA
Bimag )( Kj
K
K eA
Breal
NyquistA
B
A
BK
K
KK
K
K cos~sin
Product of T.F.
sGsGsG 21
jGjGjG 21
jGjGjG 21
21 G of Bode mag.
2
G of Bode mag.
1
G of Bode mag.
log20log20log20
jGjGjG
plots. Bode mag. add
21 G ofplot phase Bode
2
G ofplot phase Bode
1
product theG, ofplot Bode phase
jGjGjG
product theofplot phase the
plot. phase individual
product theofplot Bode :Overall
plot. Bode individual
System type, steady state tracking, & Bode plot
11
11)()(
21
sTsTs
sTsTKsGsCsG
Nba
p
sintegrator # w.r.t. typeSystem NsR
11
11
21
jTjTj
jTjTKjG N
ba
C(s) Gp(s)R(s) Y(s)
As ω → 0
Therefore: gain plot slope = –20N dB/dec.
phase plot value = –90N deg
Nj
KjG
N
KjG
log20log20log20 NKjG
90Nj
KjG N
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)
Example
typeSystem
vKpK
aK
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, ess would be =
If the input signal is a unit acceleration, ess would be =
N = 1, type = 1Bode mag. plot has –20 dB/dec slope at
low freq. (ω→0)(straight line with slope = –20 as
ω→0)
Bode phase plot becomes flatat –90° when ω→0
Kp = DC gain → ∞Kv = K = value of asymptotic straight line
at ω = 1 =s0dB =asymptotic straight line’s 0 dB
crossing frequencyKa = 0
Example
freq. low
20slope
Asymptotic straight line
s0dB ~14
The matching phase plot at lowfreq. 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.
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, ess would be =
If the input signal is a unit acceleration, ess would be =
N = 2, type = 2Bode 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
= s0dB^2
Example
Ka
s0dB=Sqrt(Ka)
How should the phase plot look like?
2 :freq lowAt
j
KjG
2
:linestraight
KjG
16,142
KK
4at dB 0 isit
1||
vK
pK
16 aK
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, ess would be =
If the input signal is a unit acceleration, ess would be =
System type, steady state tracking, & Nyquist plot
11
11
21
jTjTj
jTjTKjG N
ba
C(s) Gp(s)
Nj
KjG
As ω → 0
Type 0 system, N=0
Kp=lims0 G(s) =G(0)=K
0+
Kp
G(j)
Type 1 system, N=1Kv=lims0 sG(s) cannot be determined easily from Nyquist plot
0+
infinity
G(j) -j∞
Type 2 system, N=2
Ka=lims0 s2G(s) cannot be determined easily from Nyquist plot
0+
infinity
G(j) -∞
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)
freq.over -crossgain :gc dB0or 1 ,at jGgc
margin phase :PM
gcjG 180
freq.over -cross phase :pc
180pcjG
dB log20margingain : pcjGGM
in value 1 pcjG
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.
jG
jG
jG
jG
52
1100
ss
ssGExample:
Bode plot on next page. 11
110
51
21
ss
s
100near line dB 0 cross .1 jG100 gc
_______PM
______at 180 cross .2 pcωjG _______GM
254
252
sss
sGExample:
Bode plot on next page.
______near line dB 0 cross .1 jG
______ gc
______about is at gcωjG
_______PM
1
1
2542
251
sss
1. Where does cross the –180° lineAnswer: __________
at ωpc, how much is
2. Closed-loop stability: __________
_______jG
________ pc
jG
________GM
1
120
2
40 :Example
21
ssss
sG
1. crosses 0 dB at __________
at this freq,
2. Does cross –180° line? ________
3. Closed-loop stability: __________
_______ jG
________ gc jG
________GM
________PM
jG
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
in value1kGM
indicated angle :Then PM
-100 -50 0 50-200
-150
-100
-50
0
50
100
150
200 Nyquist Diagram
Real Axis
Imag
inary
Axis
-2 -1.5 -1 -0.5 0-2
-1.5
-1
-0.5
0
0.5
1
1.5
2 Nyquist Diagram
Real Axis
Imagin
ary Ax
is
-4 -2 0 2 4-10
-5
0
5
10 Nyquist Diagram
Real Axis
Imagin
ary Ax
is
-2 -1.5 -1 -0.5 0-2
-1.5
-1
-0.5
0
0.5
1
1.5
2 Nyquist Diagram
Real Axis
Imagin
ary Ax
is