Post on 21-Dec-2015
Chapter 6
Frequency Response
• http://kunst.gymszbad.de/kunstgeschichte/motivgeschichte/altaere/frame-menue.htmHieronymus Bosch
Garden of Delights
• http://kunst.gymszbad.de/kunstgeschichte/motivgeschichte/altaere/frame-menue.htmMaster of Flemalle ( Robert Campin)
Mérode-Altar The central panel shows the Annunciation.
The child is already on his way on golden rays…
• http://kunst.gymszbad.de/kunstgeschichte/motivgeschichte/altaere/frame-menue.htmMaster of Flemalle ( Robert Campin)
Mérode-Altar The Cloisters New York, NY
Detail:The child is already on his way on golden rays,
carrying the cross of the passion with him.
Master of Flemalle ( Robert Campin)Mérode-Altar
The Metropolitan Museum of Art, The Cloisters New York, NY
Another Detail:St. Joseph the carpenter (right panel) has just completed a mousetrap (on the table),possibly to trap the devil.
Time domain signals: Square Wave and triangular wave.
An Example: Analysis of Sound Waves
Time domain signal analysis: Spectrum of Square Wave
Fundam
. freq
HARMONICS The second harmonic is twice the fundam-ental frequency, the third harmonic is three times the fundam. frequency, and so forth.
Analysis of Sound Waves
Fourier Transform:Let period T infinity
The interval betweenDiscrete frequencies 0
The Fourier series becomesthe Fourier Transform
dtBtAtf )sin)(cos)((2
1)(
0
The A() and B() terms of the Fourier Transform can be combined into the complex termC(j)
becomes
dejCtf tj
0
)(2
1)(
dtBtAtf )sin)(cos)((2
1)(
0
dejCtf tj
)(2
1)(
dtetfjC tj
)(
2
1)(
where
Compare with the definition of the Laplace Transform
A
SoundWave
andits
Spectrum
Question:
How do we recognize voices or musical instruments?
Answer:
Our brains perform a real time spectral analysis of the incoming sound signal. The spectrum, not the signal itself, informs us about the source.
Question:
How do we recognize color?
Bode Plots:•Same content as polar plot, just a different mode of presentation.
Bode Plots:•Logarithmic -axis.
Logarithmic |F| (magnitude axis) Why?
Phase values are entered directly Why?
jeFjF *||)(
)(2121
)(2121
21
21
*||/||/
*||*||*
j
j
eFFFF
eFFFF
Basic Bode Plot (First Order)
f = -45deg. at b
Break Frequency b
G(s)= K/(s+1)
2. b at -45 deg. And |F| = 0.707
1. Note K and b
2. Draw |F| from low freq to b
3. Draw |F| from b , slope -1/decade
Bode Magnitude Plot
K = 2
b =5
Bode Phase Plot1. Phase = -450 at b
2. Draw from 0 to b/10 , slope =0
3. Draw from b/10 freq to 10*b
4. Min Phase is -900 from 10*b
Decibels
• An alternate unit of Magnitude or Gain
• Definition: xdB = 20* lg(x)
• dB Notation is widely used in Filter theory and Acoustics
x lg(x) X(db)
10 1 20
100 2 40
0.1 -1 -20
Decibels
• An alternate unit of Magnitude or Gain
• Definition: xdB = 20* lg(x)
Bode Plot of Integrator
10-1
100
101
10-1
100
101
mag
nitu
de Bode plot of Integrator 1/s (a) magnitude
10-2
10-1
100
101
-91
-90.5
-90
-89.5
-89
(rad/sec)
phas
e
Integrator phase
G(s) = 1/(s) |F|= 1/ = -tan-1(/0) = -900
Memorize!
10-1
100
101
-180
-160
-140
-120
-100
-80
-60
-40
-20
0
/ n
phase
Fig. 6.3 (b) phase
Underdamped second order systems and Resonance
10-1
100
101
10-2
10-1
100
101
/ n
magnitude
Fig. 6.3 (a) Magnitude
Underdamped second order systems and Resonance
AsymptoteSlope = -2
Phase is -90 deg. at n
Bode Plot ConstructionG(s) = 2/(s)(s+1)
10-1
100
101
10-1
100
101
magnitude
Bode Plot (a) magnitude
10-1
100
101
-180
-160
-140
-120
-100
-80
(rad/sec)
phase (
deg)
(b) phase
1. Construct each Element plot
Integrator Slope = -1
Integrator Phase = -90 deg.
2. Graphical Summation
Gain = 2.
Slope = -2
Bode Plot of 1/(s(s+1)): Matlab Plot
10-1
100
101
10-2
100
102
mag
nitu
de Bode Plot (a) magnitude
10-1
100
101
-180
-160
-140
-120
-100
-80
(rad/sec)
phas
e (d
eg)
(b) phase
Bode Plot Construction
0.01 0.1 1 100.1
1
10
z
0.01 0.1 1 10100
50
0
50
100
G(s) = 5*(s+1)/(10s+1)(100s+1)
1. Construct each Element plot
2. Graphical Summation: Complete plot. Note beginning and final values
K = 5 Slope = -1
Slope = -2
Slope = -1
Phase Plot Construction
0.01 0.1 1 100.1
1
10
z
0.01 0.1 1 10100
50
0
50
100
G(s) = 5*(s+1)/(10s+1)(100s+1)
2. Graphical Summation of phase angles. Note beginning and final phase values. Here: = 0 at = 0, and = -90 final angle
K = 5
Initial Phase is zero to 0.001, follows the first Phase up to 0.01
- 90 deg./decade
0 deg./decade+45 deg./decade
Final phase:Constant - 90 deg
Bode Plot Construction: Matlab Plot
10-2
10-1
100
101
10-4
10-2
100
102
magnitude
Bode Plot (a) magnitude
10-2
10-1
100
101
-150
-100
-50
(rad/sec)
phase (
deg)
(b) phase
Nyquist Criterion:
Closed Loop Stability: Evaluate Frequency response at Phase of
-180 degrees
Nyquist Stability Criterion
Nyquist Criterion:
Stability in the Frequency
Domain
Nyquist Criterion in the Bode
Plot:
Gain Margin and Phase
Margin Phase Margin
Gain Margin
Nyquist Criterion in the Bode
Plot:
Gain Margin and Phase
Margin
Bode Lead Design
1. Select Lead zero such that the phase margin increases while keeping the gain crossover frequency as low as reasonable.
2. Adjust Gain to the desired phase margin.
0
5
10
15
20M
agni
tude
(dB
)
100
101
102
103
0
30
60
Pha
se (
deg)
Bode Example Lead = 10*(s+10)/(s+100)
Frequency (rad/sec)
Lead compensator |p| = 10*z G(s) =1. Construct each Element plot
Slope = 0
2. Graphical Summation
Gain = 1
Slope = +1
1*1
1*1
spole
szero
Slope = 0
Slope = 0
Phase = 0
Slope = 0
Slope = 0
Note Break Frequencies
-150
-100
-50
0
50
Magnitu
de (
dB
)
10-1
100
101
102
-270
-180
-90
0
Phase (
deg)
Bode Example of plant addition, Plant = 2/[(2s+1)(2s+1)(s/5+1)
Frequency (rad/sec)
Bode Lead DesignObjectives: Increase Loop Gain and damping by raising the phase margin at the 0dB crossover frequency. Phase Margin = 45 deg.
Phase Margin
Try: Lead Zero at 0.9, pole at 9Draw new Phase and Mag. Plots
Phase with Lead. The new crossover freq. is 3 rad/s.
Magn. with Lead.
Final Step: Adjust Gain. Here K is raised approx. 3-fold
Bode Lead Design
-120
-100
-80
-60
-40
-20
0
20
Mag
nitu
de (
dB)
10-1
100
101
102
-270
-225
-180
-135
-90
-45
0
45
90
Pha
se (
deg)
Bode Example of plant addition, Plant+ Lead
Frequency (rad/sec)
Bode plot with Lead Zero at 0.9, pole at 9 (in Red).
Phase Margin
Note phase crossing at =3 with -135 deg. phase margin
Adjust gain at =3 phase crossing. Here: raise gain by about 10 dB or by a factor of 3
Bode Lead Design
-120
-100
-80
-60
-40
-20
0
20
40
Mag
nitu
de (d
B)
10-1
100
101
102
-270
-225
-180
-135
-90
-45
0
45
90
Phas
e (d
eg)
Bode Example of plant addition, gain adjusted, Plant* Lead
Frequency (rad/sec)
Final Design: Raise Gain K = 3From Matlab: Phase Margin =
38.3356 degrees
Bode Lag Design
1. All other design should be complete. Gain K and phase margin are fixed
2. Select Lag zero such that the phase margin does not drop further. (Slow)
3. Steady State Gain should now be about 10 times larger than without Lag.
Lag compensator |p| = 0.1*zG(s) =Construct each Element plot
Slope = 0 Gain = 0.1
1*1
1*1
spole
szero
Slope = 0
Phase = 0
Slope = 0
Slope = 0
Slope = -1
Slope = 0
Note Break Frequencies
Bode Lag Design
-60
-40
-20
0
20
40
Magnitu
de (
dB
)
10-2
10-1
100
101
-270
-225
-180
-135
-90
Phase (
deg)
Bode plot of plant 1/[s(0.2s+1)(s+1)
Frequency (rad/sec)
Bode Lag Design
-80
-60
-40
-20
0
20
40
Magnitu
de (
dB
) Plant
LAG
PLant*LAG
10-2
10-1
100
101
-270
-225
-180
-135
-90
-45
0
Phase (
deg)
Bode Example of plant addition, Plant+ Lag
Frequency (rad/sec)
Bode Lag Design
-100
-80
-60
-40
-20
0
20
40
60
80
100
Magnitu
de (
dB
)
Plant
LAG
PLant*LAG
Final with adjusted Gain
10-2
10-1
100
101
-270
-225
-180
-135
-90
-45
0
Phase (
deg)
Bode Example of plant addition, gain adjusted, Plant* Lag
Frequency (rad/sec)
-margin = 39 deg. K = 10
Lead Design Example
• (a) P-control for phase margin of 45 degrees. Controller gain K = 0.95
)11.0(*)1(
2)(
2
sssG
• (b) Lead-control for phase margin of 45 degrees. Lead zero and pole in RED. Initial design: Lead is too slow
)11.0(*)1(
2)(
2
sssG
Lead is too slow. Lead Zero should be near the phase margin. Here: Place Lead zero around 3 rad/s.
• (b) Lead-control for phase margin of 45 degrees. Lead zero and pole in RED. Improved design: Lead zero at 3, pole at 30 rad/s
)11.0(*)1(
2)(
2
sssG
Lead zero at 3. Lead pole at 30.
New gain crossover at 5 rad/s
Final step: adjust gain K such that |F| = 0 dB at cr.
Result: The controller gain is now K = 3.4 (4 times better than P-
control)
Bode Lead and Lag Design:General placement rules
10-1
100
101
102
-270
-180
-90
0
Pha
se (
deg)
Bode Example of plant addition, Plant = 2/[s(0.1s+1)(s+1)2
Frequency (rad/sec)
-100
-80
-60
-40
-20
0
20
Mag
nitu
de (
dB)
Place Lead Zero near desired Gain Crossover Frequency
Phase Margin
Place Lag Zero at a decade belowGain Crossover Frequency