Control Systems Bode plots

Transcription

Control Systems
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Bode plots
L. Lanari
Monday, November 3, 2014
Outline
•
•
•
•
•
Bode’s canonical form for the frequency response
Magnitude and phase in the complex plane
The decibels (dB)
Logarithmic scale for the abscissa
Bode’s plots for the different contributions
Lanari: CS - Bode plots
2
Frequency response
The steady-state response of an asymptotically stable system P(s)
to a sinusoidal input u(t) = sin ω̄t is given by
yss (t) = |P (j ω̄)| sin(ω̄t + ∠P (j ω̄))
P(j!) is the restriction of the transfer function to the imaginary axis P(j!)|s = j!
Frequency response
P (jω)
�
|P (jω)|
∠P (jω)
gain curve
(or magnitude)
phase curve
for ω ∈ R+ = [0, +∞)
3
Bode canonical form
pole/zero representation of the transfer function
with m such that
•
•
•
�
� 2
2
(s
−
z
)
(s
+
2ζ
ω
s
+
ω
)
1
k
� n�
�
n�
k
�
� 2
F (s) = K m �
2 )
s
(s
−
p
)
(s
+
2ζ
ω
s
+
ω
i
z nz
nz
i
z
m = 0 if no pole or zero in s = 0
m < 0 if m zeros in s = 0
m > 0 if m poles in s = 0
remarks
•
•
•
numerator and denominator are by hypothesis coprime
denominator is monic
K’ is not the system gain
4
Bode canonical form
•
the terms (s - zk) and (s - pi) are relative to
‣
‣
•
real zeros (in s = zk)
real poles (in s = pi)
2
2
) and (s2 + 2ζz ωnz s + ωnz
) are relative to
the terms (s2 + 2ζ� ωn� s + ωn�
‣ complex conjugate zeros (in s = α� ± jβ� )
‣
complex conjugate poles (in s = αz ± jβz )
with
‣
‣
natural frequency
ωn∗ =
�
α∗2 + β∗2
damping coefficient ζ∗ = −α∗ /ωn∗
�
= −α∗ / α∗2 + β∗2
5
Bode canonical form
factoring out the constant terms
s − zk = −zk (1 − 1/zk s) = −zk (1 + τk s)
s − pi = −pi (1 − 1/pi s) = −pi (1 + τi s)
with
with
τk = −1/zk
τi = −1/pi
with ¿i and ¿k being time constants
�
� 2 �
�
2
2
(−z
)
(ω
)
(1
+
τ
s)
(1
+
2ζ
/ω
s
+
s
/ω
)
1
k
k
�
n�
�
n�
n�
k
�
k
�
�
�
�
F (s) = K m �
2
2
2
s
i (−pi )
z (ωnz )
i (1 + τi s)
z (1 + 2ζz /ωnz s + s /ωnz )
defining
�
� 2
� �k (−zk )� � (ωn� )
K=K
2 )
(−p
)
(ω
i
nz
i
z
K = [sm F (s)]s=0
for any
m�0
how to compute K
�
�
2
(1 + τk s) � (1 + 2ζ� /ωn� s + s2 /ωn�
)
1
k
�
�
F (s) = K m
2 /ω 2 )
s
(1
+
τ
s)
(1
+
2ζ
/ω
s
+
s
i
z
nz
nz
i
z
Bode canonical form
6
Gains
�
�
2
(1 + τk s) � (1 + 2ζ� /ωn� s + s2 /ωn�
)
1
k
�
�
F (s) = K m
2 /ω 2 )
s
(1
+
τ
s)
(1
+
2ζ
/ω
s
+
s
i
z
nz
nz
i
z
generalized gain
K = [sm F (s)]s=0
for any
= 1 for s = 0
m�0
Note that
•
for a system with no poles (i.e. m negative or zero) we have defined as dc-gain (or static gain)
�
�
Ks = F (s)�
= F (0)
s=0
if m < 0 (zeros in s = 0) we have F(0) = 0
•
static and generalized gain coincide only when m = 0
K = Ks
•
⇔
m=0
for an asymptotically stable system, the step response tends to the static gain Ks = F(0)
7
Bode canonical form
Examples
F (s)
=
K
=
F (s)
=
K
=
F (s)
=
K
=
s−1
s−1
1
1−s
=
=−
2
2s + 6s + 4
2(s + 1)(s + 2)
2 (1 + s)(1 + s/2)
1
Ks = −
2
s(s − 1)
1
s(1 − s)
=
=−
2
2(s + 1) (s + 2)
2 (1 + s)2 (1 + s/2)
1
−
Ks = 0
2
s−1
1
1−s
=−
2s(s + 1)(s + 2)
2 s(1 + s)(1 + s/2)
1
−
� Ks
2
8
Bode canonical form
frequency response
1
F (jω) = K
(jω)m
�
�
2
2
(1
+
jωτ
)
(1
+
2ζ
jω/ω
+
(jω)
/ω
)
k
�
n�
n�
k
�
�
�
2
2
i (1 + jωτi )
z (1 + 2ζz jω/ωnz + (jω) /ωnz )
has 4 elementary factors
1. constant K (generalized gain)
2. monomial j! (zero or pole in s = 0)
3. binomial 1 + j!¿ (non-zero real zero or pole)
4. trinomial 1 + 2³(j!)/!n + (j!)2/!n2 (complex conjugate pairs of zeros or poles)
so first check which kind of root you have and then factor it out
9
Bode diagrams
for any real value of the angular frequency ! the frequency response F(j!) is a complex number
|F (jω)|
magnitude of the frequency response as a function of the angular frequency !
∠F (jω) angle or phase of the frequency response as a function of the angular frequency !
F (jω) = Re[F (jω)] + jIm[F (jω)]
F (jω) = |F (jω)|e
j∠F (jω)
F(j!)
Im[F(j!)]
|F
)|
ω
(j
∠F (jω)
Re[F(j!)]
|F (jω)| =
�
Re[F (jω)]2 + Im[F (jω)]2
∠F (jω) = atan2(Im[F (jω)], Re[F (jω)])
10
Phase
Phase[F.G] = Phase[F ] + Phase[G]
the phase of a product is the sum of the phases
and therefore
�
�
F
= Phase[F ] − Phase[G]
Phase
G
the phase of a ratio is the difference of the phases
since
Phase
�1�
G
= −Phase[G]
very useful since we can find the contribution to the
phase of each term and then just do an algebraic sum
11
Phase
F(j!)
Im[F(j!)]
|F
)|
ω
(j
principal argument takes on values
in (- ¼, ¼] and is implemented by
the function with two arguments
∠F (jω)
Re[F(j!)]
for
P = ® + j¯
atan2(β, α)
=

� �
β

arctan

α





� �



β

arctan

α +π




� �
β
arctan

α −π






π



2 sign(β)






undefined
atan2
if
α>0
(I & IV quadrant)
if
β ≥ 0 and α < 0 (II quadrant)
if
β < 0 and α < 0 (III quadrant)
if
α = 0 and β �= 0
if
α = 0 and β = 0
II
I
III
IV
12
Magnitude
in order to have the same property we need to go through some logarithmic function
|F (jω)|dB = 20 log10 |F (jω)|
same nice properties
|0.1|dB = −20 dB
|10|dB = 20 dB
|F.G|dB = |F |dB + |G|dB
� �
�1�
� �
= − |F |dB
�F �
dB
as phase
|1|dB = 0 dB
decibels (dB)
�√ �
� 2�
dB ≈ 3 dB
|100|dB = 40 dB
|F |dB
� +∞ if
|F | � ∞
|F |dB
� −∞ if
|F | � 0
13
Logarithmic scale
we use a logarithmic (log10) scale for the abscissa (angular frequency !)
log10(!) becomes a straight line
if ! is in a logarithmic scale
20 log10(|x|)
a decade corresponds to multiplication by 10
20
0
−20
−40
−60
−3
10
−2
−1
10
0
10
x in logarithmic scale
1
10
10
very useful when we add
|x|
10
different contributions
5
20 log10(|x|)
0
−3
10
−1
0
10
x in logarithmic scale
1
10
10
20
0
−20
−40
−60
−2
10
1
2
3
4
5
6
x in linear scale
7
8
9
10
14
Logarithmic scale
advantages
•
•
quantities can vary in large range (both ! and magnitude)
easy to build the magnitude plot in dB of a frequency response given in its Bode canonical
•
form from the magnitudes of the single terms
easy to represent series of systems
|F(jω)|
300
200
100
0
2
4
6
8
10
linear scale
same data
60
|F(jω)|dB
different scales
for abscissa and
0
40
20
0
−20
ordinates
1
2
3
4
5
6
linear scale
7
8
9
10
|F(jω)|dB
60
40
20
0
−20
−1
10
0
10
logarithmic scale
1
10
15
Bode diagrams
|F (jω)|dB
∠F (jω)
magnitude in dB of the frequency response as a function of the angular
frequency ! with logarithmic scale for !
angle or phase of the frequency response as a function of the angular
frequency ! with logarithmic scale for !
we need to find the magnitude (in dB) and phase for the 4 elementary factors
1. constant K (generalized gain)
2. monomial j! (zero or pole in s = 0)
3. binomial 1 + j!¿ (non-zero real zero or pole)
4. trinomial 1 + 2³(j!)/!n + (j!)2/!n2 (complex conjugate pairs of zeros or poles)
for ω ∈ R+ = [0, +∞)
16
Constant
Im
√
K
K3 = − 10
K1 =
√
10
Re
√
| 10|dB = 10 dB
√
| − 10|dB = 10 dB
|0.5|dB ≈ −6 dB
magnitude
20 log10 |K|
Magnitude (dB)
K2 = 0.5
20
|K 1 | dB , |K 3 | dB
10
0
|K 2 | dB
−6
−10
−2
10
√
∠0.5 = 0◦
0
10
frequency (rad/s)
1
10
2
10
0
K1, ! K2
!
phase
∠K
Phase (deg)
∠ 10 = 0◦
√
∠ − 10 = −180◦ = −π
−1
10
−50
−100
−150
K3
!
−180
−2
10
−1
10
0
10
frequency (rad/s)
10
1
10
2
17
Monomial - Numerator
Im
j!
j!
90°
|jω|dB = 20 log10 ω
Re
log scale
|jω|dB = 20x
magnitude
Magnitude (dB)
40
20 dB/dec
20
0
−20
phase
Phase (deg)
−40
−2
10
−1
10
0
10
frequency (rad/s)
1
10
2
10
90
80
60
40
20
0
−2
10
−1
10
0
10
frequency (rad/s)
10
1
2
10
18
Monomial - Denominator
from properties of log and phase
magnitude
Magnitude (dB)
40
20
0
-20 dB/dec
−20
−40
−2
10
−1
10
0
10
frequency (rad/s)
1
10
2
10
phase
Phase (deg)
0
−20
−40
−60
−80
−90
−2
10
−1
10
0
10
frequency (rad/s)
10
1
2
10
19
Binomial - Numerator
magnitude
1 + j!¿
|1 + jωτ |dB = 20 log10
�
1 + ω2 τ 2
approximation wrt the cutoff frequency 1/|τ |
(corner frequency)

1
if ω � 1/|τ |

�
1 + ω2 τ 2 ≈
 √ 2 2
ω τ
if ω � 1/|τ |
and therefore
|1 + jωτ |dB
at the cutoff frequency
∗
≈
ω = 1/|τ |



0 dB
if
ω � 1/|τ |
20 log10 ω + 20 log10 |τ |
if
ω � 1/|τ |
|1 + jτ /|τ | |dB = 20 log10
√
2 ≈ 3 dB
two half-lines approximation: 0 dB until the cutoff frequency, + 20dB/decade after
20
1 + j!¿
phase depends on the sign of ¿
Im
¿<0
j!¿
Im
Re
1 Re
1
¿>0
j!¿
see how the phase changes as ! increases
21
case ¿ > 0
case ¿ < 0
∠(1 + jωτ )
∠(1 + jωτ )
≈
≈

 0




phase depends on the sign of ¿
1 + j!¿
ω � 1/|τ |
if
π
2
ω � 1/|τ |
if
0
if
− π2
if
and
τ >0
ω � 1/|τ |
ω � 1/|τ |
and
τ <0
the two asymptotes are connected by a segment starting a decade before (0.1/| ¿ | ) the cutoff
frequency and ending a decade after (10/| ¿ |). The approximation is a broken line.
¼/2
1/| ¿ |
0
-¼/2
at the cutoff frequency
ω ∗ = 1/|τ |
∠(1 + jτ /|τ |) =



π
4
if
τ >0
− π4
if
τ <0
22
Binomial - numerator
1/| ¿ |
1 + j!¿
magnitude
Magnitude (dB)
40
30
20
10
3
0
−10
frequency (rad/s)
0.1/| ¿ |
10/| ¿ |
¿>0
phase
Phase (deg)
90
45
0
frequency (rad/s)
0.1/| ¿ |
1/| ¿ |
10/| ¿ |
¿<0
phase
Phase (deg)
0
−45
−90
frequency (rad/s)
23
Binomial - denominator
1/| ¿ |
1 /(1 + j!¿)
magnitude
Magnitude (dB)
10
0
−3
−10
−20
−30
−40
frequency (rad/s)
10/| ¿ |
0.1/| ¿ |
¿>0
phase
Phase (deg)
0
−45
−90
frequency (rad/s)
0.1/| ¿ |
1/| ¿ |
10/| ¿ |
¿<0
phase
Phase (deg)
90
45
0
frequency (rad/s)
24
Trinomial
magnitude
�
�
2
�
�
�1 + 2 ζ (jω) + (jω) �
�
ωn
ωn2 �
=
=
�
�
2
�
�
�1 − ω + j2ζ ω �
�
ωn2
ωn �
��
1−
ω2
ωn2
�2
�
+ 4ζ 2
ω2
ωn2
�
approximation wrt !n
|TRINOMIAL|
|TRINOMIAL|dB
≈
≈




1
��
2

2

ω

=
ω2



n
ω2
2
ωn
if
ω � ωn
if
ω � ωn
0 dB
if
ω � ωn
40 log10 ω − 20 log10 ωn2
if
ω � ωn
25
Trinomial
in ! = !n the magnitude | TRINOMIAL | is equal to 2 | ³ |
|ζ|
0
0.5
√
1/ 2 ≈ 0.707
|T RIN |dB in ωn
−∞
0 dB
3 dB
1
6 dB
large variation of the magnitude in ! = !n depending upon the value of the damping coefficient ³
no approximation around the natural frequency !n
26
Trinomial
")
How does a generic complex root
"(
varies in the plane as a function of !
&'
" !!
" ) ! " (! " #
"
#
"#
!"
$#
Phase
�
ζ
(jω)
∠ 1 + 2 (jω) +
ωn
ωn2
2
�
=

0







π







−π
%
#
ω � ωn
if
if
ω � ωn
and
ζ≥0
if
ω � ωn
and
ζ<0
transition between - ¼ and ¼ is symmetric wrt !n and becomes more abrupt as | ³ |
becomes smaller. When ³ = 0 the phase has a discontinuity in !n
27
Trinomial - numerator
magnitude
Magnitude (dB)
60
40
0.7
20
1
0
0.5
−20
0.3
0.1
0
−40
frequency (rad/s)
0.1 !n
10 !n
!n
phase
ζ≥0
Phase (deg)
180
0.1
135
0.3
0.7
90
1
45
0
0.5
0
frequency (rad/s)
0.1 !n
!n
10 !n
phase
ζ<0
Phase (deg)
0
−45
0.01
1
−90
−135
−180
frequency (rad/s)
28
Trinomial - denominator
magnitude
Magnitude (dB)
40
0
20
0
1
−20
−40
−60
frequency (rad/s)
0.1 !n
!n
10 !n
phase
ζ≥0
Phase (deg)
0
0
−45
1
−90
−135
−180
frequency (rad/s)
0.1 !n
!n
10 !n
phase
ζ<0
Phase (deg)
180
135
90
45
1
0.01
0
frequency (rad/s)
29
Trinomial
When | ³ | = 1 the trinomial reduces to a product of identical binomials (real roots)
roots
�
�
=
2
ζ
s
1+2 s+ 2
ωn
ωn
−ωn
if
ζ=1
ωn
if
ζ = −1
�
=
ζ=±1
�
s
1±
ωn
�2
and therefore the magnitude and phase coincides with that of a double binomial with corner
frequency
1
= ωn
|τ |
that is in ! = !n when | ³ | = 1
2 × (3 dB)
=
6 dB
2 × (−3 dB)
=
−6 dB
(numerator)
(denominator)
example: MSD system with critical value for the damping (µ2 = 4km)
30
Trinomial
√
if |ζ| < 1/ 2 ≈ 0.707 the magnitude of a trinomial factor at the denominator has a peak
at the resonance frequency
1
�
|F (jωr )| =
2|ζ| 1 − ζ 2
resonance
peak
�
ωr = ωn 1 − 2ζ 2
Magnitude (dB)
20
0
0.5
0.3
−20
0.1
0.01
34
0.01
14
5
1
−20
−40
frequency (rad/s)
anti-resonance
peak
Magnitude (dB)
(similarly for the anti-resonance peak)
frequency (rad/s)
resonance
peak
31

Control Systems Bode plots

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