System type, steady state tracking, & Bode plot
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Transcript of System type, steady state tracking, & Bode plot
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System type, steady state tracking, & Bode plot
11
11)()(
21
sTsTs
sTsTKsGsCsG
Nba
p
C(s) Gp(s)R(s) Y(s)
At very low frequency:gain plot slope = –20N dB/dec.phase plot value = –90N deg
Type = N
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Type 0: gain plot flat at very low frequency phase plot approached 0 deg
Kv = 0
Ka = 0
Low freq phase = 0o
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Type 1: gain plot -20dB/dec at very low frequency phase plot approached 90 deg
Kp = ∞
Ka = 0
Low freq phase = -90o
=Kv
Low frequency tangent line
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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
Back to general theory
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log20log20log20 NKjG
0
0
Asymptotic straight line:
20 log 20log
At =1: 20log 20log
When y=1, straight line cross hor. axis.
The crossing frequency is:
0 dB 20log 20log
20 log 2 20log
a
s dB
a s dB
y K N
y K K
K N
K
20
20
20 log 20log
a s dB
a s dB
K
K
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Type 1: gain plot -40dB/dec at very low frequency phase plot approached 180 deg
Kp = ∞
Kv = ∞
Low freq phase = -180o
Low frequency tangent line
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Example
Ka
s0dB=Sqrt(Ka)
How should the phase plot look like?
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2 :freq lowAt
j
KjG
2
:linestraight
KjG
16,142
KK
4at dB 0 isit
1||
vK pK
16 aK
Example continued
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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 =
Example continued
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System type, steady state tracking, & Bode plot
At very low frequency:gain plot slope = –20N dB/dec.phase plot value = –90N deg
If LF gain is flat, N=0, Kp = DC gain, Kv=Ka=0
If LF gain is -20dB/dec, N=1, Kp=inf, Kv=LFg_tan_c , Ka=0
If LF gain is -40dB/dec, N=2, Kp=Kv=inf, Ka=(LFg_tan_c)2
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System type, steady state tracking, & Nyquist plot
11
11
21
jTjTj
jTjTKjG N
ba
C(s) Gp(s)
Nj
KjG
As ω → 0
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Type 0 system, N=0
Kp=lims0 G(s) =G(0)=K
0+
Kp
G(j)
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Type 1 system, N=1Kv=lims0 sG(s) cannot be determined easily from Nyquist plot
0+
infinity
G(j) -j∞
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Type 2 system, N=2
Ka=lims0 s2G(s) cannot be determined easily from Nyquist plot
0+
infinity
G(j) -∞
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System type on Nyquist plot
Kp
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System relative order
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Examples
System type =
Relative order =
System type =
Relative order =
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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)
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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
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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
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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
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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
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1. Where does cross the –180° lineAnswer: __________
at ωpc, how much is
2. Closed-loop stability: __________
_______jG
________ pc
jG
________GM
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1
120
2
40 :Example
21
ssss
sG
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1. crosses 0 dB at __________
at this freq,
2. Does cross –180° line? ________
3. Closed-loop stability: __________
_______ jG
________ gc jG
________GM
________PM
jG
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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
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-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
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-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