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385
Frequency (rad/sec)
Phas
e (d
eg);
Mag
nitu
de (d
B)
Bode Diagrams
-60
-40
-20
0
20
40
60
80Gm = Inf, Pm=59.479 deg. (at 62.445 rad/sec)
10-1 100 101 102 103-180
-160
-140
-120
-100
-80
and using the step routine on the closed loop system shows the step re-sponse to be less than the maximum allowed 20%.
51. The open-loop transfer function of a unity feedback system is
G(s) =K
s(1 + s/5)(1 + s/20).
(a) Sketch the system block diagram including input reference commandsand sensor noise.
(b) Design a compensator for G(s) using Bode plot sketches so that theclosed-loop system satisÞes the following speciÞcations:
i. The steady-state error to a unit ramp input is less than 0.01.
ii. PM ≥45◦iii. The steady-state error for sinusoidal inputs with ω < 0.2 rad/sec
is less than 1/250.
iv. Noise components introduced with the sensor signal at frequen-cies greater than 200 rad/sec are to be attenuated at the outputby at least a factor of 100,.
386 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
(c) Verify and/or reÞne your design using MATLAB including a compu-tation of the closed-loop frequency response to verify (iv).
Solution :
a. The block diagram shows the noise, v, entering where the sensor wouldbe:
b. The Þrst speciÞcation implies Kv ≥ 100 and thus K ≥ 100. The bodeplot with K = 1 and D = 1 below shows that there is a negative PM butall the other specs are met. The easiest way to see this is to hand plotthe asymptotes and mark the constraints that the gain must be ≥ 250 atω ≤ 0.2 rad/sec and the gain must be ≤ 0.01 for ω ≥ 200 rad/sec.
10-1 100 101 102 10310-5
100
105
Frequency (rad/sec)
Mag
nitu
de
uncompensated Bode Plot
10-1 100 101 102 103-300
-250
-200
-150
-100
-50
Frequency (rad/sec)
Phas
e (d
eg)
387
In fact, the specs are exceeded at the low frequency side, and slightlyexceeded on the high frequency side. But it will be difficult to increasethe phase at crossover without violating the specs. From a hand plotof the asymptotes, we see that a combination of lead and lag will do thetrick. Placing the lag according to
Dlag(s) =(s/2 + 1)
(s/0.2 + 1)
will lower the gain curve at frequencies just prior to crossover so that a-1 slope is more easily achieved at crossover without violating the highfrequency constraint. In addition, in order to obtain as much phase atcrossover as possible, a lead according to
Dlead(s) =(s/5 + 1)
(s/50 + 1)
will preserve the -1 slope from ω = 5 rad/sec to ω = 20 rad/sec whichwill bracket the crossover frequency and should result in a healthy PM. Alook at the Bode plot shows that all specs are met except the PM = 44.Perhaps close enough, but a slight increase in lead should do the trick. Soour Þnal compensation is
D(s) =(s/2 + 1)
(s/0.2 + 1)
(s/4 + 1)
(s/50 + 1)
with K = 100. This does meet all specs with PM = 45o exactly, as canbe seen by examining the Bode plot below.
388 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
10-1 100 101 102 10310-5
100
105
Frequency (rad/sec)
Mag
nitu
de
compensated Bode Plot
10-1 100 101 102 103-300
-250
-200
-150
-100
-50
Frequency (rad/sec)
Phas
e (d
eg)
52. Consider a type I unity feedback system with
G(s) =K
s(s+ 1).
Design a lead compensator using Bode plot sketches so thatKv = 20 sec−1
and PM > 40◦. Use MATLAB to verify and/or reÞne your design so thatit meets the speciÞcations.
Solution :
Use a lead compensation :
D(s) =Ts+ 1
αTs+ 1, α > 1
From the speciÞcation, Kv = 20 sec−1,
=⇒ Kv = lims→0
sD(s)G(s) = K = 20
=⇒ K = 20
391
Figure 6.105: Control system for Problem 54
The Matlab margin routine shows a GM = 6.3 db and PM = 48◦ thusmeeting all specs.
54. In one mode of operation the autopilot of a jet transport is used to con-trol altitude. For the purpose of designing the altitude portion of the au-topilot loop, only the long-period airplane dynamics are important. Thelinearized relationship between altitude and elevator angle for the long-period dynamics is
G(s) =h(s)
δ(s)=
20(s+ 0.01)
s(s2 + 0.01s+ 0.0025)
ft
deg.
The autopilot receives from the altimeter an electrical signal proportionalto altitude. This signal is compared with a command signal (proportionalto the altitude selected by the pilot), and the difference provides an errorsignal. The error signal is processed through compensation, and the resultis used to command the elevator actuators. A block diagram of this systemis shown in Fig. 6.105. You have been given the task of designing thecompensation. Begin by considering a proportional control law D(s) = K.
(a) Use MATLAB to draw a Bode plot of the open-loop system forD(s) = K = 1.
(b) What value of K would provide a crossover frequency (i.e., where|G| = 1) of 0.16 rad/sec?
(c) For this value of K, would the system be stable if the loop wereclosed?
(d) What is the PM for this value of K?
(e) Sketch the Nyquist plot of the system, and locate carefully any pointswhere the phase angle is 180◦ or the magnitude is unity.
(f) Use MATLAB to plot the root locus with respect to K, and locatethe roots for your value of K from part (b).
392 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
(g) What steady-state error would result if the command was a stepchange in altitude of 1000 ft?
For parts (h)and (i), assume a compensator of the form
D(s) = KTs+ 1
αTs+ 1.
(h) Choose the parameters K, T , and α so that the crossover frequencyis 0.16 rad/sec and the PM is greater that 50◦. Verify your design bysuperimposing a Bode plot of D(s)G(s)/K on top of the Bode plotyou obtained for part (a), and measure the PM directly.
(i) Use MATLAB to plot the root locus with respect to K for the systemincluding the compensator you designed in part (h). Locate the rootsfor your value of K from part (h).
(j) Altitude autopilots also have a mode where the rate of climb is senseddirectly and commanded by the pilot.
i. Sketch the block diagram for this mode,
ii. deÞne the pertinent G(s),
iii. design D(s) so that the system has the same crossover frequencyas the altitude hold mode and the PM is greater than 50◦
Solution :
The plant transfer function :
h(s)
δ(s)=
80³ s
0.01+ 1´
s
½³ s
0.05
´2
+ 20.1
0.05s+ 1
¾
393
(a) See the Bode plot :
10-4
10-3
10-2
10-1
100
102
104
106
Frequency (rad/sec)
Magnitude
B ode D iagram s
10-4
10-3
10-2
10-1
100
-200
-150
-100
-50
0
Frequency (rad/sec)
Phase (deg)
(b) Since |G| = 865 at ω = 0.16,
K =1
|G| |ω=0.16 = 0.0012
(c) The system would be stable, but poorly damped.
(d) PM = 0.39◦
(e) The Nyquist plot for D(jω)G(jω) :
The phase angle never quite reaches −180◦.
394 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
(f) See the Root locus :
The closed-loop roots for K = 0.0012 are :
s = −0.009, −0.005± j0.16
(g) The steady-state error e∞ :
ess = lims→0
s1
1 +Kh(s)
δ(s)
1000
s
= 0
as it should be for this Type 1 system.
(h) Phase margin of the plant :
PM = 0.39◦ (ωc = 0.16 rad/sec)
Necessary phase lead and1
α:
necessary phase lead = 50◦ − 0.39◦ ' 50◦
From Fig. 6.53 :
=⇒ 1
α= 8
Set the maximum phase lead frequency at ωc :
ω =1√αT
= ωc = 0.16 =⇒ T = 18
so the compensation is
D(s) = K18s+ 1
2.2s+ 1
395
For a gain K, we want |D(jωc)G(jωc)| = 1 at ω = ωc = 0.16. Soevaluate via Matlab
¯̄̄̄D(jωc)G(jωc)
K
¯̄̄̄ωc=0.16
and Þnd it = 2.5× 103
=⇒ K =1
2.5× 103= 4.0× 10−4
Therefore the compensation is :
D(s) = 4.0× 10−4 18s+ 1
2.2s+ 1
which results in the Phase margin :
PM = 52◦ (ωc = 0.16 rad/sec)
10-3
10-2
10-1
100
102
104
106
Frequency (rad/sec)
Magnitude
B ode D iagram s
10-3
10-2
10-1
100
-200
-150
-100
-50
0
Frequency (rad/sec)
Phase (deg)
51.4 deg -180 deg
G D G /K D G /K
K=4.09x10-4
w =0.16
G
D G /K
396 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
(i) See the Root locus :
The closed-loop roots for K = 4.0× 10−4 are :
s = −0.27, −0.0074, −0.095± j0.099
(j) In this case, the referance input and the feedback parameter are therate of climb.
i. The block diagram for this mode is :
ii. DeÞne G(s) as :
G(s) =úh(s)
δ(s)=
80³ s
0.01+ 1´
s³ s
0.05
´2
+ 20.1
0.05s+ 1
iii. By evaluating the gain of G(s) at ω = ωc = 0.16, and setting Kequal to its inverse, we see that proportional feedback :
D(s) = K = 0.0072
satisÞes the given speciÞcations by providing:
PM = 90◦ (ωc = 0.16 rad/sec)
397
The Bode plot of the compensated system is :
Frequency (rad/sec)
Phase (deg); M
agnitude (dB)
B ode D iagram s
-20
-10
0
10
20
30G m = Inf, Pm =90.349 deg. (at 0.16509 rad/sec)
10-3
10-2
10-1
100
-200
-150
-100
-50
0
50
100
55. For a system with open-loop transfer function c
G(s) =10
s[(s/1.4) + 1][(s/3) + 1],
design a lag compensator with unity DC gain so that PM ≥40◦. What isthe approximate bandwidth of this system?
Solution :
Lag compensation design :
Use
D(s) =Ts+ 1
αTs+ 1
K=1 so that DC gain of D(s) = 1.
(a) Find the stability margins of the plant without compensation byplotting the Bode, Þnd that:
PM = −20◦ (ωc = 3.0 rad/sec)GM = 0.44 (ω = 2.05 rad/sec)
(b) The lag compensation needs to lower the crossover frequency so thata PM ' 40◦ will result, so we see from the uncompensated Bodethat we need the crossover at about
=⇒ ωc,new = 0.81
402 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
(c) For a low frequency gain increase of 3.16, and the pole at 3.16 rad/sec,the zero needs to be at 10 in order to maintain the crossover atωc = 31.6 rad/sec. So the lag compensator is
D2(s) = 3.16
s
10+ 1
s
3.16+ 1
and
D1(s)D2(s) = 100s
20 + 1s
100 + 1
s
10+ 1
s
3.16+ 1
The Bode plots of the system before and after adding the lag com-pensation are
100 101 102 10310-2
10-1
100
101
102
Frequency (rad/sec)
Mag
nitu
de
Bode Diagrams
Lead and LagLead only
100 101 102 103-400
-300
-200
-100
0
Frequency (rad/sec)
Phas
e (d
eg)
(d) By using the margin routine from Matlab, we see that
PM = 49◦ (ωc = 3.16 deg/sec)ec)
58. Golden Nugget Airlines had great success with their free bar near the tailof the airplane. (See Problem 5.41) However, when they purchased a much
403
larger airplane to handle the passenger demand, they discovered that therewas some ßexibility in the fuselage that caused a lot of unpleasant yawingmotion at the rear of the airplane when in turbulence and was causing therevelers to spill their drinks. The approximate transfer function for thedutch roll mode (See Section 9.3.1) is
r(s)
δr(s)=
8.75(4s2 + 0.4s+ 1)
(s/0.01 + 1)(s2 + 0.24s+ 1)
where r is the airplane�s yaw rate and δr is the rudder angle. In performinga Finite Element Analysis (FEA) of the fuselage structure and addingthose dynamics to the dutch roll motion, they found that the transferfunction needed additional terms that reßected the fuselage lateral bendingthat occurred due to excitation from the rudder and turbulence. Therevised transfer function is
r(s)
δr(s)=
8.75(4s2 + 0.4s+ 1)
(s/0.01 + 1)(s2 + 0.24s+ 1)· 1
( s2
ω2b+ 2ζ s
ωb+ 1)
where ωb is the frequency of the bending mode (= 10 rad/sec) and ζ is thebending mode damping ratio (= 0.02). Most swept wing airplanes havea �yaw damper� which essentially feeds back yaw rate measured by a rategyro to the rudder with a simple proportional control law. For the newGolden Nugget airplane, the proportional feedback gain, K = 1, where
δr(s) = −Kr(s). (3)
(a) Make a Bode plot of the open-loop system, determine the PM andGM for the nominal design, and plot the step response and Bodemagnitude of the closed-loop system. What is the frequency of thelightly damped mode that is causing the difficulty?
(b) Investigate remedies to quiet down the oscillations, but maintain thesame low frequency gain in order not to affect the quality of thedutch roll damping provided by the yaw rate feedback. SpeciÞcally,investigate one at a time:
i. increasing the damping of the bending mode from ζ = 0.02 toζ = 0.04. (Would require adding energy absorbing material inthe fuselage structure)
ii. increasing the frequency of the bending mode from ωb = 10rad/sec to ωb = 20 rad/sec. (Would require stronger and heavierstructural elements)
iii. adding a low pass Þlter in the feedback, that is, replace K in Eq.(3) with KD(s) where
D(s) =1
s/τp + 1. (4)
Pick τp so that the objectionable features of the bending modeare reduced while maintaing the PM ≥ 60o.
404 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
iv. adding a notch Þlter as described in Section 5.5.3. Pick thefrequency of the notch zero to be at ωb with a damping of ζ =0.04 and pick the denominator poles to be (s/100 + 1)2 keepingthe DC gain of the Þlter = 1.
(c) Investigate the sensitivity of the two compensated designs above (iiiand iv) by determining the effect of a reduction in the bending modefrequency of -10%. SpeciÞcally, re-examine the two designs by tab-ulating the GM, PM, closed loop bending mode damping ratio andresonant peak amplitude, and qualitatively describe the differencesin the step response.
(d) What do you recommend to Golden Nugget to help their customersquit spilling their drinks? (Telling them to get back in their seats isnot an acceptable answer for this problem! Make the recommenda-tion in terms of improvements to the yaw damper.)
Solution :
(a) The Bode plot of the open-loop system is :
Frequency (rad/sec)
Phase (deg); M
agnitude (dB)
B ode D iagram s
-100
-80
-60
-40
-20
0
20G m =1.2833 dB (at 10.003 rad/sec), Pm =97.633 deg. (at 0.083342 rad/sec)
10-3
10-2
10-1
100
101
102
-300
-200
-100
0
100
PM = 97.6◦ (ω = 0.0833 rad/sec)GM = 1.28 (ω = 10.0 rad/sec)
The low GM is caused by the resonance being close to instability.
405
The closed-loop system unit step response is :
0 20 40 600
0.3
0.6
0.9
Tim e (sec)
Amplitude
U nit Step R esponse
The Bode plot of the closed-loop system is :
10-3
10-2
10-1
100
101
102
10-4
10-2
100
Frequency (rad/sec)
Magnitude
B ode D iagram s
10-3
10-2
10-1
100
101
102
-300
-200
-100
0
Frequency (rad/sec)
Phase (deg)
From the Bode plot of the closed -loop system, the frequency of thelightly damped mode is :
ω ' 10 rad/sec
406 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
and this is borne out by the step response that shows a lightly dampedoscillation at 1.6 Hz or 10 rad/sec.
i. The Bode plot of the system with the bending mode dampingincreased from ζ = 0.02 to ζ = 0.04 is :
Frequency (rad/sec)
Phase (deg); M
agnitude (dB)
B ode D iagram s
-100
-80
-60
-40
-20
0
20G m =7.3092 dB (at 10.006 rad/sec), Pm =97.614 deg. (at 0.083342 rad/sec)
10-3
10-2
10-1
100
101
102
-300
-200
-100
0
100
PM = 97.6◦ (ω = 0.0833 rad/sec)GM = 7.31 (ω = 10.0 rad/sec)
and we see that the GM has increased considerably because theresonant peak is well below magnitude 1; thus the system willbe much better behaved.
ii. The Bode plot of this system (ωb = 10 rad/sec =⇒ ωb = 20
407
rad/sec) is :
Frequency (rad/sec)
Phase (deg); M
agnitude (dB)
B ode D iagram s
-80
-60
-40
-20
0
20G m =7.3481 dB (at 20.003 rad/sec), Pm =97.643 deg. (at 0.083338 rad/sec)
10-3
10-2
10-1
100
101
102
-300
-200
-100
0
100
PM = 97.6◦ (ω = 0.0833 rad/sec)GM = 7.34 (ω = 20.0 rad/sec)
and again, the GM is much improved and the resonant peak issigniÞcantly reduced from magnitude 1.
iii. By picking up τp = 1, the Bode plot of the system with the low
408 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
pass Þlter is :
Frequency (rad/sec)
Phase (deg); M
agnitude (dB)
B ode D iagram s
-150
-100
-50
0
50G m =34.972 dB (at 8.6186 rad/sec), Pm =92.904 deg. (at 0.083064 rad/sec)
10-3
10-2
10-1
100
101
102
-400
-300
-200
-100
0
100
PM = 92.9◦ (ω = 0.0831 rad/sec)GM = 34.97 (ω = 8.62 rad/sec)
which are healthy margins and the resonant peak is, again, wellbelow magnitude 1.
409
iv. The Bode plot of the system with the given notch Þlter is :
Frequency (rad/sec)
Phase (deg); M
agnitude (dB)
B ode D iagram s
-120
-100
-80
-60
-40
-20
0
20G m =55.275 dB (at 99.746 rad/sec), Pm =97.576 deg. (at 0.083336 rad/sec)
10-3
10-2
10-1
100
101
102
103
-300
-200
-100
0
100
PM = 97.6◦ (ω = 0.0833 rad/sec)GM = 55.3 (ω = 99.7 rad/sec)
which are the healthiest margins of all the designs since the notchÞlter has essentially canceled the bending mode resonant peak.
(b) Generally, the notch Þlter is very sensitive to where to place the notchzeros in order to reduce the lightly damped resonant peak. So if youwant to use the notch Þlter, you must have a good estimation ofthe location of the bending mode poles and the poles must remainat that location for all aircraft conditions. On the other hand, thelow pass Þlter is relatively robust to where to place its break point.Evaluation of the margins with the bending mode frequency loweredby 10% will show a drastic reduction in the margins for the notchÞlter and very little reduction for the low pass Þlter.
Low Pass Filter Notch Filter
GM 34.97 (ω = 8.62 rad/sec) 55.3 (ω = 99.7 rad/sec)PM 92.9◦ (ω = 0.0831 rad/sec) 97.6◦ (ω = 0.0833 rad/sec)
Closed-loop bendingmode damping ratio
' 0.02 ' 0.04Resonant peak 0.087 0.068
410 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
The magnitude plots of the closed-loop systems are :
10-3
10-2
10-1
100
101
102
10-4
10-2
100
Frequency (rad/sec)
Magnitude (Low Pass Filter)
B ode D iagram s
10-3
10-2
10-1
100
101
102
10-4
10-2
100
Frequency (rad/sec)
Magnitude (Notch Filter)
The closed-loop step responses are :
0 20 40 600
0.3
0.6
0.9
Tim e (sec)
Amplitude
U nit Step R esponse
N otch Filter
Low Pass Filter
(c) While increasing the natural damping of the system would be thebest solution, it might be difficult and expensive to carry out. Like-wise, increasing the frequency typically is expensive and makes the
411
Figure 6.106: Control system for Problem 59
structure heavier, not a good idea in an aircraft. Of the remainingtwo options, it is a better design to use a low pass Þlter because ofits reduced sensivity to mismatches in the bending mode frequency.Therefore, the best recommendation would be to use the low passÞlter.
Problems and Solutions for Section 6.8
59. A feedback control system is shown in Fig.6.106. The closed-loop systemis speciÞed to have an overshoot of less than 30% to a step input.
(a) Determine the corresponding PM speciÞcation in the frequency do-main and the corresponding closed-loop resonant peak valueMr. (SeeFig. 6.37)
(b) From Bode plots of the system, determine the maximum value of Kthat satisÞes the PM speciÞcation.
(c) Plot the data from the Bode plots (adjusted by the K obtained inpart (b)) on a copy of the Nichols chart in Fig. 6.73 and determine theresonant peak magnitude Mr. Compare that with the approximatevalue obtained in part (a).
(d) Use the Nichols chart to determine the resonant peak frequency ωrand the closed-loop bandwidth.
Solution :
(a) From Fig. 6.37 :
Mp ≤ 0.3 =⇒ PM ≥ 40o =⇒Mr ≤ 1.5resonant peak : Mr ≤ 1.5
(b) A sketch of the asymptotes of the open loop Bode shows that a PMof ∼= 40o is obtained when K = 8. A Matlab plot of the Bode can beused to reÞne this and yields
K = 7.81
for PM = 40o.
426 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
(d) A lead compensator may provide a sufficient PM, but it increases thegain at high frequency so that it violates the speciÞcation above.
(e) A lag compensator could satisfy the PM speciÞcation by lowering thecrossover frequency, but it would violate the low frequency speciÞca-tion, W1.
(f) One possible lead-lag compensator is :
D(s) = 100
s
8.52+ 1
s
22.36+ 1
s
4.47+ 1
s
0.568+ 1
which meets all the speciÞcation :
Kv = 100
PM = 47.7◦ (at ωc = 12.9 rad/sec)|KG| = 50.45 (at ω = 1 rad/sec) > 49
|KG| = 0.032 (at ω = 100 rad/sec) < 0.0526
The Bode plot of the compensated open-loop system D(s)G(s) is :
10-1
100
101
102
103
10-2
100
102
Frequency (rad/sec)
Magnitude
B ode D iagram s
10-1
100
101
102
103
-200
-160
-120
-80
Frequency (rad/sec)
Phase (deg)
Problems and Solutions for Section 6.1067. Assume that the system
G(s) =e−Tds
s+ 10,
427
has a 0.2-sec time delay (Td = 0.2 sec). While maintaining a phase margin≥ 40◦, Þnd the maximum possible bandwidth using the following:
(a) One lead-compensator section
D(s) = Ks+ a
s+ b,
where b/a = 100;
(b) Two lead-compensator sections
D(s) = K
µs+ a
s+ b
¶2
,
where b/a = 10.
(c) Comment on the statement in the text about the limitations on thebandwidth imposed by a delay.
Solution :
(a) One lead section :
With b/a = 100, the lead compensator can add the maximum phaselead :
φmax = sin−1 1− ab
1 + ab
= 78.6 deg ( at ω = 10a rad/sec)
By trial and error, a good compensator is :
K = 1202, a = 15 =⇒ Da(s) = 1202s+ 15
s+ 1500PM = 40◦ (at ωc = 11.1 rad/sec)
The Bode plot is shown below. Note that the phase is adjusted forthe time delay by subtracting ωTd at each frequency point while thereis no effect on the magnitude. For reference, the Þgures also includethe case of proportional control, which results in :
K = 13.3, PM = 40◦ (at ωc = 8.6 rad/sec)
428 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD
100
101
102
100
Frequency (rad/sec)Magnitude
B ode D iagram s
100
101
102
-250
-200
-150
-100
-50
0
Frequency (rad/sec)
Phase (deg)
-180 deg
KG (s)
D a(s)G (s)
D a(s)G (s)
KG (s)
PM =40 deg PM =40 deg
w =11.1 w =8.6
(b) Two lead sections :
With b/a = 10, the lead compensator can add the maximum phaselead :
φmax = sin−1 1− ab
1 + ab
= 54.9 deg ( at ω =√10a rad/sec)
By trial and error, one of the possible compensators is :
K = 1359, a = 70 =⇒ Db(s) = 1359(s+ 70)2
(s+ 700)2
PM = 40◦ (at ωc = 9.6 rad/sec)
429
The Bode plot is shown below.
100
101
102
100
Frequency (rad/sec)
Magnitude
B ode D iagram s
100
101
102
-250
-200
-150
-100
-50
0
Frequency (rad/sec)
Phase (deg)
-180 deg
D b(s)G (s)
KG (s)
D b(s)G (s)
KG (s)
PM =40 deg
w =9.57 w =8.6
(c) The statement in the text is that it should be difficult to stabilizea system with time delay at crossover frequencies, ωc & 3/Td. Thisproblem conÞrms this limit, as the best crossover frequency achievedwas ωc = 9.6 rad/sec whereas 3/Td = 15 rad/sec. Since the band-width is approximately twice the crossover frequency, the limitationsimposed on the bandwidth by the time delay is veriÞed.
68. Determine the range of K for which the following systems are stable:
(a) G(s) = K e−4s
s
(b) G(s) = K e−ss(s+2)
Solution :
(a) ¯̄̄̄G(jω)
K
¯̄̄̄= 2.54, when ∠G(jω)
K= −180◦
range of stability : 0 < K <1
2.54
