|
Bode Magnitude and Phase Plots |
System Gain and Phase Margins &
Bandwidths |
Transfer Function from Bode Plots |
Bode Plots of Open Loop and Closed
Loop Systems |
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Problem 5.1 Bode Magnitude and Phase
Plots
Given
the transfer function
G(s)
= 300/[s(s+4)(s+8)]
(a)
sketch the Bode
magnitude and phase plots
(b)
sketch the polar
plot
(c)
sketch the log-magnitude/phase diagram.
(d)
Why does using logarithmic co-ordinates reduce the drudgery in
plotting the Bode plots considerably?
(e) How are the gain and phase
margins determined by using the Bode plots?
Solution:
G(s) = 300/[s(s+4) (s+8)] = (75/8)/[s (1+ s) (1+s)
4
8
Slope = -18 dB /octave
0.5 1 2 4 8
1 2 4 8 w (log scale)
(-130.80)
-2700
(b)
(c)
dB
(d) Log scale is used for the w axis. The simplification in the Bode plot results due to the basic advantage of logarithmic representation that multiplication and division are replaced by addition and subtraction respectively. There are several other advantages of Bode plots.
(e)
Theoretical.
Problem 5.2 System Gain and Phase Margins & Bandwidths
(a)
Obtain the phase and
gain margins of a unity feedback system
whose open loop transfer function is
G(s)
= K/[s(s+1)(s+5)]
for the two
cases where K=10 and K =100 respectively.
(b)
For the following two systems:
System I:
C(s)/R(s)= 1/(s+1)
System II:
C(s)/R(s) = 1/(3s+1)
Solution:
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K=10 |
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30
20
10
0
│G│in dB -10
-20
-30
0
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0
-90
ÐG in deg.
-180
-270
0.2 0.4 0.6 0.8 1 2 4 6 8 10
-12 dB gain margin
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K=100 |
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50
40
30
20
│G│in dB 10
0
-10
0
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0
-90
ÐG in deg.
-180
-270
0.2 0.4 0.6 0.8 1 2 4 6 8 10
-20
The above figure shows the closed loop response curves for the two systems (Asymptotes are shown by dotted lines)
(ii) System I: Bandwidth is 0≤ w≤1 rad/s
System II: Bandwidth is 0≤ w≤0.33 rad/s
I II
t
The above Figure shows the step response curves of the two systems
1
The above Figure shows the ramp response curves of the two systems.
(iv) System I whose bandwidth is three times wider than that of system II has a faster speed of response and can follow the input much better.
Problem 5.3 Polar Plot and Bode Diagrams
(a)
Sketch the polar
plot of the frequency response for the following transfer function at w = 0, 1, 5…infinity:
(1+
0.5 s) (1+ 2s)
(b)
Draw the Bode
diagram representation of the frequency response for the transfer function
given in Part (a). Show the points on the diagram at w =0.5, 1, 2, 8, and so on.
Solution:
(a)
|
w |
0 |
1 |
5 |
µ |
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1 |
0.4 |
0.037 |
0 |
|
f |
00 |
-900 |
-1530 |
-1800 |
(b)
|
w |
0.5 |
1 |
2 |
8 |
|
dB |
-3.27 |
-8 |
-15.3 |
-36.4 |
|
f |
-590 |
-900 |
-1210 |
-1620 |
The polar plot and Bode diagrams can be plotted from the above calculations.
Problem 5.4 Transfer Function from Bode
Plots
The frequency response of a control system is given
in Table 1. Sketch the Bode plots and hence estimate the transfer function.
|
w |
M (dB) |
f |
w |
M (dB) |
f |
|
0.1 |
20.0 |
-90.30 |
10 |
-14 |
-1800 |
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0.2 |
14.0 |
-90.30 |
15 |
-26.8 |
-2390 |
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0.4 |
8.0 |
-90.30 |
20 |
-36 |
-2520 |
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0.6 |
4.46 |
-90.30 |
30 |
-47.8 |
-2590 |
|
1 |
0.08 |
-90.30 |
40 |
-55.6 |
-2620 |
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2 |
-5.71 |
-90.30 |
50 |
-61.6 |
-2640 |
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4 |
-10.77 |
-90.30 |
60 |
-66.5 |
-2650 |
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6 |
-12.55 |
-90.30 |
80 |
-74.1 |
-2660 |
|
8 |
-12.7 |
-90.30 |
100 |
-79.9 |
-2670 |
Solution:
Draw the magnitude and phase plots.
As w tends to zero, f tends to infinity. Hence, we have one pole at the origin. For large w, f tends to –270 deg. Hence the excess of poles over zeroes is three.
A rapid change in f near w = 10 indicates a pair of complex conjugate poles with wn= 10. Since f = -180deg. For w= 10, we have wn= 10. Thus
G(s)= K/[s (s2+2xwn+wn2)]
As the gain approaches 0 dB foe small values of w, we have K=wn2= 100.
For w= 10, s=j10, hence
G (j10) = 0.1/2xÐ180deg. The gain for w= 10 is –14 db or 10-14/20=0.2
Or, x= 0.25
Hence G(s)= 100/[(s (s2+5s+100)]
Problem 5.5 Bode Plots of Open Loop and
Closed
(a)
The block diagram of a speed-control system is shown in Fig.1.
Motor Load
80s2(s+7) (s+2)2(s2+6s+25) 2.5 4s+2
Fig.1
Draw and label the Bode magnitude and phase plots of the
open-loop and the closed- loop systems.
(b)
State the practical use of the Bode plots.
(i)
The desirable
frequency- domain specifications for a control system are:
(a)
relatively large
resonant magnitude, Mpw
(b)
relatively large
bandwidths so that the system time constant, t = 1/xwn , is sufficiently small
(c)
relatively small
resonant magnitude, Mpw
(d)
relatively large
bandwidths so that the system time constant, t = 1/xwn, is sufficiently large
Ans:(
b )