control engineering project submission

19
F  = k θ θ

Upload: surajkumar-harikumar

Post on 04-Jun-2018

219 views

Category:

Documents


0 download

TRANSCRIPT

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 1/19

   

   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  F   = kθθ  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 2/19

 

 

G(s)  

 

GA(s) =  (s + 10)

s(s + 90)

℄  

℄  

 

 

 

 

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 3/19

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 4/19

 

1(s+100+i10)(s+100−i10)

(s+300)(s+40)

(s+10)s(s+20+i100)(s+20−i100)

(s+10)(s+150)

(s+8+i60)(s+8−i60)(s+1+i10)(s+1−i10)

 

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 5/19

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 6/19

  GA(s)  

 

   jω  

  ζ  ∼ 0.1    ωn ∼  10    M  p  = e−ζΠ√ 1−ζ2 = 73.04%  

  Lims→0sG(s)1s

 = 0.01  

 

 

1(s+1−i10)(s+1+i10)

s+300s+25

 

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 7/19

 

s−10(s+20+i20)(s+20−i20)

1(s+5)(s+2+i20)(s+2−i20)

(s+40)(s+300)

s(s+1+i10)(s+1−i10)

 

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 8/19

 

 

Gc(s)G p(s) =  H 1(s) =  (s + α)

s3 + (1 + α)s2 + (α− 1)s + (1− α)

 

  α    < ±10%  

  K  p  

Steady StateE rror   =  1

1 + K  p≤ 0.1

=⇒   K  p   ≥ 9

  K  p = lims→0 H 1(s) = lims→0(s+α)

s3+(1+α)s2+(α−1)s+(1−α)   =   α1−α

 ≥ 9 =⇒   α ≥ 0.9

 

  α  

 H 1(s)    s3+(1+α)s2+(α−1)s+(1−α) = (s3+s2−s+1)+α(s2+s−1)

 H 2(s) = 1 +   α(s2+s−1)(s3+s2−s+1)    H 1(s)  

  H 2(s)

 H 2(s)      

  α

 

  α  

 

 

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 9/19

  H (s) =   H 1(s)1+H 1(s) .  

  1+H 1(s) = 0 =⇒   1+  (s+α)

s3+(1+α)s2+(α−1)s+(1−α)  = 0 =⇒   s3+(1+α)s2+αs+1 = 0  

  α    (s3+s2+1)+α(s2+s) = 0

  H 3(s) =   s2+ss3+s2+1    1 + H 3(s) = 0  

 

 

−0.469    α  = 4.48  

  α  = 4.48  

 

 

  2nd   2    =   4

ξωn  

 

ξωn   α = 4.48

 

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 10/19

  α = 4.48  

 

 

℄  

 

℄ ℄  

 

 

  α

 

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 11/19

  α >   4.48  

  αjust < 4.48  

 

  3.8  

  11.3s  

α   = 3.82  

  α <   3.82  

  α  

 

 

  G(s) =   1(s+1)3  

 

 

  H (s) =  KG(s)1+KG(s)  =   1

(s+1)3+K   

 

Characteristic Equation =  s3 + 3s2 + 3s + (K  + 1)

 

 

 

s3  

s2  

s   8−K 3  

1  

 

  0 < K < 8

 

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 12/19

  0 < K < 8   K  = 8

 

 

 

 −0.124  

 

G( jω) =  1

( jω  + 1)3  =

 (1− jω)3

(1 + ω2)3  =

 (1− 3ω2) + j(−3ω + ω3)

(1 + ω2)3

  ω =√ 

3, 0    ω =√ 

3  

  G( jω)  

G( jω) = (1− 3 ∗ 3)

(1 + 3)3  = −8

64  = −0.125

  −a,a = 0.125

  N   =   P  − Z   

  s = −1  

  G(s)    P   = 0  

  Z  = 0    N  = 0    s = −1

  s   =−Ka, a   = 0.125    Ka   = 1  

Ka = 1    K  =   1a

 = 8

  K  = 8    18.06 dB

  K  = 8  

 

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 13/19

 

 

 

   

 

 

  18.1 dB    20 log(K ) = 18.1 dB    K  = 8.035

 

℄  

 

 

 

 

 

  K   = 8.0011

 

Range of P roportional Gain K 

 ∈[0, 8]

 

 

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 14/19

 

 

 

 

 

  kx    F   =  kxx + kθθ  

 

  θ, ω    α  

  θ

 

Iα  =  F L− bvω − mgLsin(θ)

2

  sin(θ)    θ  

 

Iα  =  F L− bvω − mgLθ

2  

mω2L

2   = T  − mgcos(θ)

 

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 15/19

 

 

2kx =  Tsin(θ)

 

  θ    ω  

X    =

θ

ω

X ′

=

  0 1−mgL2I 

  − bvI 

X  +

0LI 

 

  θ

 

F    = kθθ

 

kθ   0

 

  ω  

X ′

=

  0 1−mgL2I 

  − bvI 

X  +

0LI 

kθ   0

X   

 

X ′

=   0 1

−mgL

2I    +  kθL

I    −bv

I X   

  A =

  0 1

−mgL2I 

  +   kθLI   − bv

 

 

s2 + sbv

I   +

 mg L

2I   − kθL

I   = 0  

 

 

 

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 16/19

b2vI 2 ≥ 4(

mgL

2I   − kθL

I   )

  kθ  

kθ ≤mg

2  − b2v

4IL

 

kθ  =  mg

2  − b2v

4IL

 

kθ  >  mg

2  − b2v

2IL

  F   = kxx + kθθ    x  

θ  

  x    θ  

 

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 17/19

 

 

 

 

 

 

  G1(s) =  (1−s)(1+s)(1+ s

2)    G1(s) =

  (1−s)0.5s2+1,5s+1  

  ℄ ℄     s  

   

 

 

 

  G2(s) =  (1−s)(1− s

2)

(1+s)(1+ s2)(1+ s

3)    G2(s) =   3s2−9s+6

s3+6s2+11s+6  

  ℄ ℄        

   

 

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 18/19

 

  G1(s)  

 

  G2(s)  

 

           

        limt→0f (0) =   lims→∞sF (s)      

 

H (s)s

 

 

H (s)s

     

limt→00f ′(0) = lims→∞sH (s)

 

 

 

 

 

 

  ∞  

 

 

 

 

 

  ∞  

(1−s)0.5s2+1,5s+1

 

 

 

 

 

 

 

 

 

 

∞  

 

 

 

 

 

 

∞  

3s2−9s+6s3+6s2+11s+6

 

 

 

 

 

 

 

 

 

 

 

H (s)s

 

 

 

 

H (s)s

 

 

 

 

 

  srU (s)  

  U (s) =  H (s)

s  

 

 

  sr−1U (s)  

 

  ∞srH (s)

 

 

 

 

 

8/13/2019 Control Engineering Project Submission

http://slidepdf.com/reader/full/control-engineering-project-submission 19/19

 

H (s) =  K 

(s− zi)(s− zi∗)

(s− zj)

(s + zk)

(s− pm)(s− pm∗)

(s + pn)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

K (s−zi)(s−zi∗)

(s−zj)

(s+zk)

(s− pm)(s− pm∗)(s+ pn)

 

  S  =  K ( |zi|2)(−1)b(

zj)(

zk)( | pm|2)(

 pn)

 

 

 

  ∞sr

H (s)  

D =  lims−>∞sr.K 

(s− zi)(s− zi∗)

(s− zj)

(s + zk)

(s− pm)(s− pm∗)

(s + pn)

 

 

 

 

 

 

 

 

 

 

  K 2( |zi|2)(−1)b(

zj)(

zk)( | pm|2)(

 pn)

  <  0