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7/31/2019 Ch-8 Bode Plot Slides
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Chapter 8
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Input is sinusoidal
Steady state output will be sinusoidal with
same frequency for linear systems
amplitude and phase will vary with
frequency
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Example :
1s
1
(s)G ++++====
r (t) = sin
t
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In steady state
)-t(sin
1
1(t)C2
+
=
where = tan 1
tsin1tcos-e)(1
t-2
++
C (t) =
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Examine
G (s) with s = j
1j
1)(jG
++++====
tan
1
1 -1
2
++++
====
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Hence when input isSin
t
)t(sin)j(G
statesteadyinputout
++++====)j(Gwhere =
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Hence we plotmagnitude and phase
angle of as a
function of frequency
G ( j )
- frequency response
- sinusoidal transfer function.
put s = j in the transfer function
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Bode plot
magnitude as a
function of frequency phase angle as a
function of frequency
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Polar plot
For a particularfrequency, mark ( r, )on the polar graph,
Where r is magnitude & is phase angle.
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Example
1j
1
)(jG ++++====
1
1
)(M 2++++==== ( ) = - tan
1
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= 0 M = 1 = 0
= M = 0 = - 90
= 1 M = 21
= - 45
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Polar plot of 1s
1
(s)G ++++====
. .
==== 0====
6K-
Polar plot of
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Bode plot
)(je)(jG)j(G ====
)(j434.0)(jGlog +=
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graphsemilogonlogversus)(and)(jGlog20plot
scalelinearmagnitude
scalelog
.dbordecibelin)j(Glog20
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Ex:
Tj1
1)j(G
++++====
tan-
1
1
1-
22
++++====
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(((( )))) 21-22 1log20)j(Glog20
++++====
)1(log10-22
++++====
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for frequencies
,1
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For high frequencies
,1 >>>>>>>>
)T(gol01-
)j(Ggol0222
====
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Tgol02-====Tgol02-gol02- =
Value is zero db
,T1At ====
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. .
.
0
db
-20
0 db line
T1====
T10====
log
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The db magnitude plot for ,1 >>>>>>>>
is a straight line with -20 per unit change in log
This slop is - 20 db/decade or -6 db / octave
The straight line approximation is
called asymptotic plot Two asymptotesmeet at the corner frequency 1 / T.
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2log10-error
,T
1at
====
====
= - 3 db
1log10)411(log10-
T21aterror
++=
=
= - 1 db
To get actual plot,
T10for
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1log20(2)log10-error
,T
1at
+=
=
= - 3 db
2log204)1(log10-error
,T2at
++=
=
= - 1 db
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Ex:Bode plot for
)1s(s
)2s(5(s)G ++++
++++====
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1)(jj
2)(j5)j(G
++++++++====
)1j(1j
)2j(110
++++++++
====
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10 20 log 10
= 20 db
constant for allfrequencies
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log20-
1
log20j
1
====
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It is a straight line
with - 20 db/decade
slope ; passes
through 0 db line
for= 1
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rad/sec2frequencycorner)2j1(
====++++
decade/db20,2
db0,2
++++>>>>>>>>
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rad/sec1frequencycorner
)j(1
1
====
++++
decade/db20-,1
db0,1
>>>>>>>>
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db
20
log
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db
.log
1
- 20 db/decade
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db
. .
log
2
+ 20 db/decade
0 db
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db
.log
1
- 20 db/decade
0 db
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20
log
db
.
.
12
- 20 db/decade
- 20 db/decade
- 40 db/decade
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0
log
32 db.
.
1
+12
5
6
. . .
.
0.5
-6
Slopes in db/octave
db
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Corner frequencies :0.5 , 1, 5
+ 12 db / octave : s2
)0.5/s(1
1
:octave/db6 ++++++++
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)s1(
1:db0
++++
)5/s1(
1
:db6- ++++
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..
0.5 1
26 3212
6
38
log
db
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20 log K = 38K = 79.8
)s0.21()s1()2s1(
s8.97
(s)G
2
++++++++++++
====
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Gain margin & phasemargin :
..
>
. - a
- 1
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Gain margin =
a
1
for stable system a
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Phase margin()
=G(j)H(j) + 180
phase margin is + ve for
stable system
= 1
G i C O F
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log180-
0
db
PM
GH
log
GM
Gain Cross Over Freq.
Phase Cross Over Freq.
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Put s = j to get
the sinusoidal transfer function
)s2s()s(R
)s(C
2nn
2
2
n
++
=
Frequency Response of Second Order
Under damped Systems
2
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u
n
=
Putnormalised driving
Signal frequency
2
nn
2
2
n
)(j2)(j
)R(j)C(j ++=
u2ju1
12
+
=
2/122
])u2()u1[(
1M
+
=
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]u1
u2[tan2
1
=
u = 0 M = 1 = 0
u = 1 M = 1/2 = -/2u = M = 0 = -
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( )
uuu
u
for
u
u
log40log201
01
4)u-(1log20
)j(Glog20
2
212222
=>>
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Correlation between time &
frequency response:
Steady state output
for r(t) = Sin t
c(t) = M Sin(t+)
0dM
will give resonant Frequency where M is max
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2r
2nr
2r
121M
21
21u
====
====
====
Whereur : Normalised resonant freq
r : resonant freqM : resonant peak value
0
du
= will give resonant Frequency where M is max.
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M
Mr
1.0
0.707
0 r c
Bandwidth
Typical magnification curve of a feedback control system.
where
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r resonant frequencyc cut off frequency
Bandwidth
where
The range of frequencies over which M 1/2
2
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0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
0.2 0.6 1 1.4 1.8
Normalized
bandwidth
ub
Bandwidth versus damping factor
0.707
Time domain
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2nd
d
p
1
p
1
t
eM
2
====
====
====
Frequency domain
2
nr
2
r
21
12
1M
====
====