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Catégorie :Category: mViewer GX Creator App HP-Prime
Auteur Author: antoniodagama
Type : Application
Page(s) : 71
Taille Size: 3.72 Mo MB
Mis en ligne Uploaded: 14/08/2019 - 17:56:16
Uploadeur Uploader: antoniodagama (Profil)
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Shortlink : http://ti-pla.net/a2295060

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Solutions to Problems 11-7


8. Not only is the magnitude curve increased at higher frequencies, but so is the phase curve. Thus the

180o point moves up in frequency with the increase in gain.
9. To correct for the negative phase angle of the uncompensated system
10. When designing the lag portion of a lag-lead compensator, we do not worry about the transient design.
The transient response will be considered when designing the lead portion of a lag-lead compensator.



SOLUTIONS TO PROBLEMS
1.
a. Plot Bode plots for K = 1; angle is 180o at = 15.8 rad/s where the magnitude is –76.5 dB.
Therefore a 66.5 dB ( or K = 2113) increase will yield a 10 dB gain margin.




b. Plot Bode plots for K = 1; angle is 180o at = 6.32 rad/s where the magnitude is –55 dB.
Therefore a 45 dB ( or K = 177.8) increase will yield a 10 dB gain margin.
11-8 Chapter 11: Design via Frequency Response




c. Plot Bode plots for K = 1; angle is 180o at = 9.45 rad/s where the magnitude is –63.8 dB.
Therefore a 53.8 dB ( or K = 489.8) increase will yield a 10 dB gain margin.
Solutions to Problems 11-9



2.
a. For a 40o phase margin, the phase must be -140o when the magnitude plot is zero dB. The phase is
-140o at = 9.12 rad/s. At this frequency, the magnitude curve is –67.48 dB. Thus a 67.48 dB
increase (K = 2365) will yield a 40o phase margin.
b. For a 40o phase margin, the phase must be -140o when the magnitude plot is zero dB. The phase is
-140o at = 2.76 rad/s. At this frequency, the magnitude curve is – 42.86 dB. Thus a 42.86 dB
increase (K = 139) will yield a 40o phase margin.
c. For a 40o phase margin, the phase must be -140o when the magnitude plot is zero dB. The phase is
-140o at = 5.04 rad/s. At this frequency, the magnitude curve is – 54.4 dB. Thus a 54.4 dB increase
(K = 525) will yield a 40o phase margin.
3.
20% overshoot => = 0.456 => M = 48.15o.

a. Looking at the phase diagram, where M = 48.15o (i.e. -131.85o), the phase margin frequency =
4.11 rad/s. At this frequency, the magnitude curve is -55.2 dB. Thus the magnitude curve has to be
raised by 55.2 dB (K = 575).
b. Looking at the phase diagram, where M = 48.15o (i.e. = -131.85o), the phase margin frequency =
7.14 rad/s. At this frequency, the magnitude curve is – 65.6 dB. Thus the magnitude curve has to be
raised by 65.6 dB (K = 1905).
c. Looking at the phase diagram, where M = 48.15o (i.e. = -131.85o), the phase margin frequency =
8.2 rad/s. At this frequency, the magnitude curve is – 67.3 dB. Thus the magnitude curve has to be
raised by 67.3 dB (K = 2317).
11-10 Chapter 11: Design via Frequency Response



4.
a. Bode plots for K = 1:




Using Eqs. (4.39) and (10.73) a percent overshoot = 15 is equivalent to a = 0.517 and M = 53.170.
The phase-margin frequency = 2.61 rad/s where the phase is 53.170 – 1800 = -126.830. The
magnitude = -13 dB, or 0.0.2239. Hence K = 1/ 0.2239 = 4.466.
b.
Program:
G=zpk([-20 -25],[0 -6 -9 -14],1)
K=4.466
T=feedback(K*G,1);
step(T)

Computer response:
Zero/pole/gain:
(s+20) (s+25)
--------------------
s (s+6) (s+9) (s+14)

K =

4.466
Solutions to Problems 11-11




5.
For Kv = 50, K = 350. Plot the Bode plots for this gain.




Also, since %OS = 15%, = 0.517. Using Eq. (10.73), M = 53.17o. Increasing M by 10o we will

design for a phase margin of 63.17o. The phase margin frequency is where the phase angle is
63.17 - 180o = -116.83o, or M = 3.54 rad/s. At this frequency, the magnitude is 22 dB. Start the

magnitude of the compensator at - 22 dB and draw it to 1 decade below M.
11-12 Chapter 11: Design via Frequency Response



30

20

10
dB
0

-10

-20

-30
.001 .01 v .1 1

Then begin +20 dB/dec until zero dB is reached. Read the break frequencies as 0.028 rad/s and 0.354

rad/s from the Bode plot and form a lag transfer function that has unity dc gain:

s 0.354
Gc(s) =0.0791
s 0.028
The compensated forward path is

350 * 0.0791(s 0.354) 27.69(s 0.354)
G(s) =
s(s 7)(s 0.028) s(s 7)(s 0.028)
6.
a. For Kv = 1000, K = 1473. Plotting the Bode for this value of K:




Using Eqs. (4.39) and (10.73) a percent overshoot = 15 is equivalent to a = 0.517 and M = 53.17.

Using an extra 10o, the phase margin is 63.17o. The phase-margin frequency = 1.21 rad/s. At this
Solutions to Problems 11-13


frequency, the magnitude = 57.55 dB = 754.2. Hence the lag compensator K = 1/754.2 = 0.001326.
Following Steps 3 and 4 of the lag compensator design procedure in Section 11.3,

s + 0.121
Glag(s) = 0.001326 s + 0.0001604
b.
Program:
%Input system
numg=1473*poly([-10 -11]);
deng=poly([0 -3 -6 -9]);
G=tf(numg,deng);
numc=0.001326*[1 0.121];
denc=[1 0.0001604];
Gc=tf(numc,denc);
Ge=G*Gc;
T=feedback(Ge,1);
step(T)

Computer response:




7.
Uncompensated system:
Searching along the 121.1o line (15% overshoot), find the dominant pole at -2.15 ± j3.56 with K =
97.7
97.7. Therefore, the uncompensated static error constant is Kvo = = 1.396. On the frequency
70
11-14 Chapter 11: Design via Frequency Response


response curves, plotted for K = 97.7, unity gain occurs at = 1.64 rad/s with a phase angle of -71o.
Therefore the uncompensated phase margin is 180o - 71o = 109o.
Compensated system:
1 1
The old steady-state error, estep ( ) 0.4174 . For a 5 times improvement
1 K po 97.7
1
70
1 K
in steady-state error, estep ( ) 0.0835 , yielding , K pn 10.98 . Thus
1 K pn 70
K 768.6 . Plotting the Bode plots at this gain,




Adding 5o, the desired phase margin for 15% overshoot is 58.17o, or a phase angle of -121.83o. This
phase angle occurs at = 3.505 rad/s. At this frequency the magnitude plot is +12 dB. Start the
magnitude of the compensator at - 12 dB and draw it to 1 decade below M.
Solutions to Problems 11-15




Then, begin +20 dB/dec until zero dB is reached. Read the break frequencies as 0.08797 rad/s and

0.3505 rad/s from the Bode plot and form a lag transfer function that has unity dc gain,


s 0.3505
Gc ( s) 0.251
s 0.08797
The compensated forward path is

( s 0.3505) 192.91(s 0.3505)
G ( s ) 0.251*768.6 .
( s 2)( s 5)( s 7)( s 0.08797) ( s 2)( s 5)( s 7)( s 0.08797)

8.
K (4)
For K p 100 , K = 2400. Plotting the Bode plot for this gain,
(2)(6)(8)
11-16 Chapter 11: Design via Frequency Response


Uncompensated system


40

30

20




20 log M
10
0

-10

-20
0.1 1 10 100
Frequency (rad/s)

0


-50
Phase (degrees)




...

Archive contentsContenu de l'archive

Action(s) SizeTaille FileFichier
3.06 Ko KB readme.txt
3.65 Ko KB lisezmoi.txt
1.09 Ko KB SOLCAP11.hpprgm
2.03 Mo MB SOLCAP11 01_36.hpappdir.zip
1.70 Mo MB SOLCAP11 37_71.hpappdir.zip
168 octets bytes appslist.txt

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