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INC 341 PT & BP
INC341Frequency Response Method
Lecture 11
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Design controller to decrease peak time to 2/3 and steady-state error to 0
System has 20% overshoot
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3 expressions of sinusoidal signal
Starts from a sinusoidal signal, , which can be
rewritten as
• Polar form (showing magnitude and phase shift):
)sin()cos( tBtA
)/(tan 1
22
AB
BAM
i
i
)/(tancos 122 ABtBA
iiM
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2 expressions of sinusoidal signal (cont.)
• Rectangular form (complex number):
• Euler’s formula (exponential):
jBA
)sin()sin()cos()cos()cos(
)sin()sin()cos()cos()cos(
tMtMtM
ttt
B
ii
A
iiii
ijieM
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Frequency response of system
• Magnitude response:– ratio of output mag. To input mag.
• Phase response:– difference in output phase angle and input phase
angle
• Frequency response:
)(M
)(
)()( M
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Question
What is the output from a known system fed by a sinusoidal command?
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Basic property of frequency Response‘mechanical system’
input = forceoutput = distance
sinusoidal input gives sinusoidal output with same damped frequencyshifted by ,mag. expanded by
)(
)(M
Answer:
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The HP 35670A Dynamic Signal
Analyzer obtainsfrequency responsedata from a physical
system.
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Finding frequency response from differential equation
• Get transfer function• Set• Write
• Then the output is composed of
js
)()()(
)()()(
io
io MMM
)(sT
)()(
)()(
sT
sTM
)]()([)()()()( iioo MMM
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Finding frequency response from transfer function
s
)2(
1
)2(
1)(
)2(
1)(
jjjG
ssG
ω = 0, G = 0.5 0.5∟0ω = 2, G = 0.25 – j0.25 0.35 ∟-45ω = 5, G = 0.07 - 0.17i 0.19 ∟-68.2ω = 10, G =0.019 - j0.096 0.01∟-78.7ω = ∞, G = 0 0 ∟-90
Substitute with j
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What’s next?
After getting magnitude and phase of the system, we need to plot them but how???
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Types of frequency response plots
• Polar plot (Nyquist plot): real and imaginary part of open-loop system.
• Bode plot: magnitude and phase of open-loop system (begin with this one!!).
• Nichols chart: magnitude and phase of open-loop system in a different manner (not covered in the class).
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)2(
1)(
s
sGPolar plot of
so called ‘Nyquist plot’
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Bode plot
Note: log frequency and log magnitude
Magnitude
Phase
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Bode plot
• 1st order or higher terms that can be written as a product of 1st order terms– 4 cases:
• 2nd order terms– 2 cases:
ss
asas
1,,
)(
1),(
2222
2
1,2
nnnn
ssss
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)()( assG
)1()()( ajaajjG
ω = 0aM
ajG
log20log20
)(
ω >> a
log20log20
90)()(
M
jajajG
phase = 0
phase = 90
ω = a3log202log20log20
)()(
aaM
ajajG phase = 45
First order termsCase I: one zero at -a
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Asymptotes (approximation)
Break frequency
= freq. at which mag. has changed by 3 db
12 10
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3 dB at break frequency
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)(
1)(
assG
)1(
1
)(
1)(
aj
aajjG
ω = 0
)/1log(20log20
/1)(
aM
ajG
ω >> a
log20log20
9011
)(
1)(
M
jaja
jG
phase = 0
phase = -90
ω = a 3)/1log(20)2/1log(20log20 aaMphase = -45
First order termsCase II: one pole at -a
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G(s) = 1/s
Magnitude depends directly on jω (straight line down passing through zero dB at ω=1)Phase = - 90 (constant)
Case III: one zero at 0
G(s) = s
Magnitude depends directly on jω (straight line up passing through zero dB at ω=1)Phase = 90 (constant)
Case IV: one pole at 0
First order terms
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G(s) = s G(s) = 1/s
G(s) = s+a G(s) = 1/(s+a)
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)3)(2(
1)(
sssG
It’s convenient for calculation to plot magnitude in log scale!!!
What about ???
plot each term separately and sum them up
• log magnitude (s+2) added with log magnitude (s+3)
• phase (s+2) added with phase (s+3)
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Bode PlotsFind magnitude and phase of each term and sum them up!!!
)()()()()(
)(log20)(log20log20
)(log20)(log20log20)(log20
)()(
)()()(
))((
))(()(
2121
21
21
21
21
21
21
pspsszszsKsG
pspss
zszsKsG
pspss
zszsKsG
pspss
zszsKsG
m
m
m
m
mag(num)-mag(den)phase(num)-phase(den)
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Example
sketch bode plot of)2)(1(
)3()(
sss
ssG
break frequency at 1,2,3
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Frequency small 1 2 3
s -20 -20 -20 -20
1/(s+1) 0 -20 -20 -20
1/(s+2) 0 0 -20 -20
(s+3) 0 0 0 20
Total Slope -20 -40 -60 -40
Slope at each break frequency for magnitude plot
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Magnitude Plot
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Frequency small 0.1 0.2 0.3 10 20 30
s 0 0 0 0 0 0 0
1/(s+1) 0 -45 -45 -45 0 0 0
1/(s+2) 0 0 -45 -45 -45 0 0
(s+3) 0 0 0 45 45 45 0
Total Slope 0 -45 -90 -45 0 45 0
Slope at each point for phase plot
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Phase Plot
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Case I: 2 zeros22 2)( nnsssG
)2()()(2)()( 2222 nnnn jjjjG
Small ω = 0
large ω = ∞
log magnitude:
0)( 22 nnsG
180)()( 222 jsG
log40log20 2
set s = jω
2nd order terms
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Second-order bode plot
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22 2)( nnsssG Magnitude plot of
)2()()( 22 nn jjG
n 22)( njG
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22 2)( nnsssG Phase plot of
)2()()( 22 nn jjG
n 900
)2(tan)(
21 njG
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Magnitude plot of
)2()(
1)(
22
nn jjG
n 22
1)(
n
jG
22 2
1)(
nnsssG
Case II: 2 poles
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22 2
1)(
nnsssG
Phase plot of
)2()(
1)(
22
nn jjG
n 900
)2(tan)(
21 njG
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Example
sketch bode plot of
• Set then
• At DC, set s=0,
• Break frequency at 2, 3, (or 5)
js )252)(2(
)3()(
2
sss
ssG
)25)(2))((2)((
)3()(
2
jjj
jjG
50
3)0( G
25
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Magnitude Plot
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Phase plot
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Phase plot
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ConclusionsDrawing Bode plot
• Get transfer function• Set• Evaluate the break frequency• Approximate mag. and phase at low and high
frequencies, and also at the break frequency– Mag. plot: slope changes for 1st order,
for 2nd order (at break frequency)– Phase plot: slope changes for 1st order,
for 2nd order
js )(sT
decdB /20decdB /40
dec/90dec/180
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