π
<-
Chat plein-écran
[^]

SOLCAP90


Hierarchy of files

 Downloads
 Files created online(20005)
 HP-Prime(2888)

 mViewer GX Creator App(2419)

DownloadTélécharger


LicenceLicense : Non spécifiée / IncluseUnspecified / Included

 TéléchargerDownload

Actions



Vote :

ScreenshotAperçu


Informations

Catégorie :Category: mViewer GX Creator App HP-Prime
Auteur Author: antoniodagama
Type : Application
Page(s) : 76
Taille Size: 2.69 Mo MB
Mis en ligne Uploaded: 14/08/2019 - 17:55:05
Uploadeur Uploader: antoniodagama (Profil)
Téléchargements Downloads: 1
Visibilité Visibility: Archive publique
Shortlink : http://ti-pla.net/a2295057

Description 

Answers to Review Questions 9-3




UFSS Vehicle: Lead and Feedback Compensation
0.25K2(s+0.437)
Minor loop: Open-loop transfer function G(s)H(s) = (s+2)(s+1.29)(s+0.193) ; Closed-loop transfer


0.25K2 (s 0.437)
function: TML (s) 3 . Searching along the 126.87o line ( = 0.6), find the
s(s ...)
dominant second-order poles at -1.554 ± j2.072 with 0.25K2 = 4.7. Thus K2 = 18.8. Searching the
real axis segment of the root locus for a gain of 4.7 yields a 3rd pole at -0.379.
Major loop: The unity feedback, open-loop transfer function found by using the minor-loop closed-
-0.25K1(s+0.437)
loop poles is GML(s) = s(s+0.379)(s+1.554+j2.072)(s+1.554-j2.072) . Searching along the 120o line

( = 0.5), find the dominant second-order poles at -1.069±j1.85 with 0.25K1 = 4.55. Thus K1 = 18.2.
Searching the real axis segment of the root locus for a gain of 4.55 yields a 3rd pole at -0.53 and a 4th
pole at -0.815.



ANSWERS TO REVIEW QUESTIONS
1. Chapter 8: Design via gain adjustment. Chapter 9: Design via cascaded or feedback filters
2. A. Permits design for transient responses not on original root locus and unattainable through simple gain
adjustments. B. Transient response and steady-state error specifications can be met separately and
independently without the need for tradeoffs
3. PI or lag compensation
4. PD or lead compensation
5. PID or lag-lead compensation
6. A pole is placed on or near the origin to increase or nearly increase the system type, and the zero is
placed near the pole in order not to change the transient response.
7. The zero is placed closer to the imaginary axis than the pole. The total contribution of the pole and zero

along with the previous poles and zeros must yield 1800 at the design point. Placing the zero closer to the
imaginary axis tends to speed up a slow response.
8. A PD controller yields a single zero, while a lead network yields a zero and a pole. The zero is closer to
the imaginary axis.
9. Further out along the same radial line drawn from the origin to the uncompensated poles
10. The PI controller places a pole right at the origin, thus increasing the system type and driving the error
to zero. A lag network places the pole only close to the origin yielding improvement but no zero error.
11. The transient response is approximately the same as the uncompensated system, except after the
original settling time has passed. A slow movement toward the new final value is noticed.
9-4 Chapter 9: Design via Root Locus


12. 25 times; the improvement equals the ratio of the zero location to the pole location.
13. No; the feedback compensator's zero is not a zero of the closed-loop system.
14. A. Response of inner loops can be separately designed; B. Faster responses possible; C. Amplification
may not be necessary since signal goes from high amplitude to low.




SOLUTIONS TO PROBLEMS
1.
Uncompensated system: Search along the = 0.5 line and find the operating point is at -1.5356 ±

/ 1
2 4
j2.6598 with K = 73.09. Hence, %OS e x100 = 16.3%; Ts = = 2.6 seconds; Kp
1.5356
73.09
= =2.44. A higher-order pole is located at -10.9285.
30
Compensated: Add a pole at the origin and a zero at -0.1 to form a PI controller. Search along the =
0.5 line and find the operating point is at -1.5072 ± j2.6106 with K = 72.23. Hence, the estimated
2
/ 1
performance specifications for the compensated system are: %OS e x100 = 16.3%; Ts =
4
= 2.65 seconds; Kp = . Higher-order poles are located at -0.0728 and -10.9125. The
1.5072
compensated system should be simulated to ensure effective pole/zero cancellation.
2.
s 0.01
a. Insert a cascade compensator, such as Gc (s) .
s
b.
Program:
K=1
G1=zpk([],[0,-3,-6],K) %G1=1/s(s+3)(s+6)
Gc=zpk([-0.01],[0],1) %Gc=(s+0.01)/s
G=G1*Gc
rlocus(G)
T=feedback(G,1)
T1=tf(1,[1,0]) %Form 1/s to integrate step input
T2=T*T1
t=0:0.1:200;
step(T1,T2,t) %Show input ramp and ramp response


Computer response:
K =

1
Solutions to Problems 9-5


Zero/pole/gain:
1
-------------
s (s+3) (s+6)


Zero/pole/gain:
(s+0.01)
--------
s


Zero/pole/gain:
(s+0.01)
---------------
s^2 (s+3) (s+6)


Zero/pole/gain:
(s+0.01)
-------------------------------------------
(s+6.054) (s+2.889) (s+0.04384) (s+0.01304)


Transfer function:
1
-
s


Zero/pole/gain:
(s+0.01)
---------------------------------------------
s (s+6.054) (s+2.889) (s+0.04384) (s+0.01304)



Step Response
200

180

160

140

120

100

80

60

40

20

0
0 20 40 60 80 100 120 140 160 180 200
Time (sec)
9-6 Chapter 9: Design via Root Locus




3.
a. Searching along the 126.16o line (10% overshoot, = 0.59), find the operating point at
41.1905
-1.8731 + j2.5633 with K = 41.1905. Hence, K p 0.9807
2 * 3* 7
s 0.40787
b. A 4.0787 x improvement will yield Kp = 4. Use a lag compensator, Gc ( s ) .
s 0.1
c.

Uncompensated
0.7



0.6



0.5



0.4



0.3



0.2



0.1



0
0 0.5 1 1.5 2 2.5
Time (sec)




Compensated
0.8


0.7


0.6


0.5
Amplitude




0.4


0.3


0.2


0.1


0
0 5 10 15 20 25
Time (sec)
Solutions to Problems 9-7



4.
a. Searching along the 126.16o line (10% overshoot, = 0.59), find the operating point at
27.9948
-1.1207 + j1.5336 with K = 27.9948. Hence, Kv = 1.3331 .
3x7
s 0.3.0006
b. A 3.0006 x improvement will yield Kv = 4. Use a lag compensator, Gc(s) = .
s 0.1
c.
Program:
K=17.5
G=zpk([],[0,-3,-5],K)
Gc=zpk([-0.3429],[-0.1],1)
Ge=G*Gc;
T1=feedback(G,1);
T2=feedback(Ge,1);
T3=tf(1,[1,0]); %Form 1/s to integrate step input
T4=T1*T3;
T5=T2*T3;
t=0:0.1:20;
step(T3,T4,T5,t) %Show input ramp and ramp responses


Computer response:
K =

27.9948


Zero/pole/gain:
27.9948
-------------
s (s+3) (s+7)


Zero/pole/gain:
(s+0.3001)
----------
(s+0.1)
9-8 Chapter 9: Design via Root Locus




Step Response
25




20




15




...

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 SOLCAP90.hpprgm
1.44 Mo MB SOLCAP90 01_40.hpappdir.zip
1.25 Mo MB SOLCAP90 41_76.hpappdir.zip
168 octets bytes appslist.txt

Pub / Ads

-
Search
-
Featured topics
Offre TI-Planet/Jarrety pour avoir la TI-83 Premium CE avec son chargeur pour 79,79€ port inclus !
Offre TI-Planet/Jarrety pour avoir la TI-Nspire CX CAS à seulement 130€ TTC port inclus!
Jailbreake ta TI-Nspire avec Ndless et profite des meilleurs jeux et applications !
123
-
Donations / Premium
For more contests, prizes, reviews, helping us pay the server and domains...

Discover the the advantages of a donor account !
JoinRejoignez the donors and/or premium!les donateurs et/ou premium !


-
Featured files
Partner and ad
Notre partenaire Jarrety 
-
Stats.
352 utilisateurs:
>341 invités
>6 membres
>5 robots
Record simultané (sur 6 mois):
6892 utilisateurs (le 07/06/2017)
-
Other interesting websites
Texas Instruments Education
Global | France
 (English / Français)
Banque de programmes TI
ticalc.org
 (English)
La communauté TI-82
tout82.free.fr
 (Français)