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Bode plots for ratio of first/second order factors
Problem: Draw the Bode plots for
G(s) =s + 3
(s + 2)(s2 + 2s + 25)
Solution: We first convert G(s) showing eachterm normalized to a low-frequency gain ofunity. The second order term is normalizedby factoring ω2
n, forming
s2
ω2n
+ 2ζs
ωn+ 1
Thus
G(s) =3
2 × 25
(s3 + 1)
(s2 + 1)( s2
25 + 225s + 1)
The Bode log-magnitude diagram is the sumof the individual first and second terms of G(s).
The low frequency value for G(s) is 350, or
−24.44dB, which is found by letting s → 0.We see that the break frequencies are at 2,3 and 5, and set the extent of our plot fromω = 0.01rad/s to ω = 100rad/s.
1
Table below: Bode magnitude diagram slopes
Start: Start: Start:pole zero atat -2 at -3 ωn = 5
Frequency 0.01 2 3 5(rad/s)
pole at -2 0 -20 -20 -20
zero at -3 0 0 20 20
ωn = 5 0 0 0 -40
Total slope 0 -20 0 -40(dB/dec)
The Bode plot starts at −24.44dB and con-
tinue until the first break frequency at 2rad/s,
yielding -20dB/decade slope downwards un-
til the next break frequency at 3rad/s, which
causes +20dB/decade slope upwards, which
when added to the previous -20dB, gives a net
slope of 0. At ωn = 5, the second order term
initiates a -40dB downward slope which con-
tinues to infinity.
2
The correction to the log-magnitude curve due
to the underdamped second order term is found
by plotting a point −20 log2ζ above the asymp-
tote at the natural frequency ωn = 5. Note
ζ = 0.2, the correction is 7.9dB.
Figure above: Bode log-magnitude plot for
G(s) = s+3(s+2)(s2+2s+25)
; (a) components; (b)
composite
3
We now turn to the phase plot. The first or-
der pole at −2 yields a phase angle that starts
at 0 and end at −90 via a −45◦ starting a
decade below the break frequency and ending
at a decade above the break frequency.
The first order zero at −3 yields a phase angle
that starts at 0 and end at +90 via a +45◦
starting a decade below the break frequency
and ending at a decade above the break fre-
quency.
Table below: Bode phase diagram slopes.
Start: Start: Start: End: End: End:pole zero pole zeroat -2 at -3 ωn = 5 at -2 at 3 ωn = 5
Frequency 0.2 0.3 0.5 20 30 50(rad/s)
pole at -2 -45 -45 -45 0
zero at -3 45 45 45 0
ωn = 5 -90 -90 -90 0
Total slope -45 0 -90 -45 -90 0(deg/dec)
4
The second order poles yield a phase angle that
starts at 0 and end at −180 via a −90◦ start-
ing a decade below the break frequency (i,e.
natural frequency at ωn = 5) and ending at a
decade above the break frequency.
Figure above: Bode phase plot for G(s) =s+3
(s+2)(s2+2s+25); (a) components; (b) com-
posite
5
Frequency response using Matlab
We can use Matlab to make Bode plots using
bode(G), where G(s) = numgdeng
, and G is an LTI
transfer function object.
Problem: Draw the Bode plots for
G(s) =s + 3
(s + 2)(s2 + 2s + 25)
using Matlab.
≫ numg=[1 3];
≫ deng=conv([1 2],[1 2 25]);
≫ G=tf(numg,deng);
≫ bode(G);
≫ grid on;
≫ title(’open loop frequency response’)
≫ [mag,phase,w] = bode(G);
≫ points=[20*log10(mag(:,:))’, phase(:,:)’,w]
6
−80
−60
−40
−20
0
Mag
nitu
de (
dB)
10−1
100
101
102
−180
−135
−90
−45
0
Pha
se (
deg)
Bode Diagram
Frequency (rad/sec)
Figure above; The Bode plots for
G(s) =s + 3
(s + 2)(s2 + 2s + 25).
7
We can also use Matlab to make polar plots
using nyquist(G), where G(s) = numgdeng
, and G is
an LTI transfer function object.
Problem: Draw the polar plots for
G(s) =s + 3
(s + 2)(s2 + 2s + 25)
using Matlab.
≫ numg=[1 3];
≫ deng=conv([1 2],[1 2 25]);
≫ G=tf(numg,deng);
≫ nyquist(G);
≫ grid on;
≫ title(’polar plots’)
8
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4−0.2
−0.15
−0.1
−0.05
0
0.05
0.1
0.15
0.20 dB
−20 dB
−10 dB−6 dB−4 dB−2 dB
20 dB
10 dB 6 dB4 dB 2 dB
polar plots
Real Axis
Imag
inar
y A
xis
Figure above; The polar plots for
G(s) =s + 3
(s + 2)(s2 + 2s + 25).
9
Problem: Sketch the Bode plots and polar
plots by hand and verify the results using Mat-
lab for
G(s) =s + 20
(s + 2)(s + 7)(s + 50).
≫ numg=[1 20];
≫ deng=conv([1 2],[1 7]);
≫ deng=conv(deng,[1 50]);
≫ G=tf(numg,deng);
≫ bode(G);
≫ grid on;
≫ title(’open loop frequency response’)
≫ [mag,phase,w] = bode(G);
≫ points=[20*log10(mag(:,:))’, phase(:,:)’,w]
10
−120
−100
−80
−60
−40
−20
Mag
nitu
de (
dB)
10−1
100
101
102
103
−180
−135
−90
−45
0
Pha
se (
deg)
open loop frequency response
Frequency (rad/sec)
Figure above; The Bode plots for
G(s) =s + 20
(s + 2)(s + 7)(s + 50).
11
Problem: Sketch the Bode plots by hand and
verify the results using Matlab for
G(s) =s + 20
(s + 2)(s + 7)(s + 50).
≫ numg=[1 20];
≫ deng=conv([1 2],[1 7]);
≫ deng=conv(deng,[1 50]);
≫ G=tf(numg,deng);
≫ nyquist(G);
≫ grid on;
≫ title(’polar plots’)
12
−1 −0.8 −0.6 −0.4 −0.2 0 0.2−0.02
−0.015
−0.01
−0.005
0
0.005
0.01
0.015
0.020 dB −20 dB−10 dB−6 dB−4 dB−2 dB20 dB 10 dB 6 dB4 dB2 dB
polar plots
Real Axis
Imag
inar
y A
xis
Figure above; The polar plots for
G(s) =s + 20
(s + 2)(s + 7)(s + 50).
13