Frequency Resp. method

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Frequency Resp. method

• Given
• G(j?) as a function of ? is called the freq.
  resp.
• For each ?, G(j?) x(?) jy(?) is a point in
  the complex plane
• As ? varies from 0 to 8, the plot of G(j?) is
  called the Nyquist plot

y(s)
G(s)
u(s)
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• Can rewrite in Polar Form
• G(j?) as a function of ? is called the
  magnitude resp.
• as a function of ? is called
  the phase resp.
• The two plots
• with log scale-?, are the Bode plot

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Relationship between bode and nyquist
length
vector
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• To obtain freq. Resp from G(s)
• Select
• Evaluate G(j?) at those to get
• Plot Imag(G) vs Real(G) Nyquist
• Or plot
• with log scale ? Bode
• Matlab command to explore nyquist, bode

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• 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 f1 ofoutput

u(s)
y(s)
System
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• Then is the freq. resp.
  of the system at freq ?1
• Repeat the steps for all ?K
• Either plot
• or plot

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y(s)
G2(s)
G1(s)
u(s)
Product of T.F.
G(s)
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System type, steady state tracking, Bode plot
C(s)
Gp(s)
R(s)
Y(s)
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As ? ? 0 Therefore gain plot slope 20N
dB/dec. phase plot value 90N deg
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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 ??0 DC gain (convert dB to
values)
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Example
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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
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N 1, type 1 Bode mag. plot has 20 dB/dec
slope at low freq. (??0) (straight line with
slope 20 as ??0) Bode phase plot becomes
flat at 90 when ??0 Kp DC gain ? 8 Kv K
value of asymptotic straight line evaluated at
? 1 ws0dB asymptotic straight lines 0 dB
crossing frequency Ka 0
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Example
Asymptotic straight line
ws0dB
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The matching phase plot at lowfreq. must be ?
90 type 1 Kp 8 ? position error
const. Kv value of low freq. straight line
at ? 1 23 dB 14 ? velocity error
const. Ka 0 ? acc. error const.
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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
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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 ? 8 Kv 8 also Ka value of straight
line at ? 1 ws0dB2
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Example
Ka
ws0dBSqrt(Ka)
How should the phase plot look like?
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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
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System type, steady state tracking, Nyquist plot
C(s)
Gp(s)
As ? ? 0
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Type 0 system, N0
Kplims?0 G(s) G(0)K
Kp
w?0
G(jw)
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Type 1 system, N1
Kvlims?0 sG(s) cannot be determined easily from
Nyquist plot
w?infinity
w?0
G(jw) ? -j8
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Type 2 system, N2
Kalims?0 s2G(s) cannot be determined easily from
Nyquist plot
w?infinity
w?0
G(jw) ? -8
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System type on Nyquist plot
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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 gt 0
• Gain margin gt 0

G(s)
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If never cross 0 dB line (always
below 0 dB line), then PM 8. If
never cross 180 line (always above 180), then
GM 8. If cross 180 several
times, then there are several GMs. If
cross 0 dB several times, then there are
several PMs.
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Example Bode plot on next page.
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Example Bode plot on next page.
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1. Where does cross the 180
   lineAnswer __________at ?pc, how much is
2. Closed-loop stability __________

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1. crosses 0 dB at __________at
   this freq,
2. Does cross 180 line? ________
3. Closed-loop stability __________

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Margins on Nyquist plot

• Suppose
• Draw Nyquist plot G(j?) unit circle
• They intersect at point A
• Nyquist plot cross neg. real axis at k

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