Projeto e Discretização PID

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Problem 8.08PP
For the system shown in Fig, find values for K, TD, and 7/ so that the closed-loop poles satisfy ^ 
> 0.5 and wn > 1 rad/sec. Discretize the PID controller using
(a) Tustin’s method
(b) The matched pole-zero method
Use Matlab to simulate the step response of each of these digital implementations for sample 
times of 7 = 1.0.1. and 0.01 sec.
Figure Control system
Step-by-step solution
step 1 of 10
Write the plant transfer function. 
1i— >0.5
sec
Select the dominant closed loop poles to exceed the specifications. 
Consider s = -0 .8 ± J .
Calculate the values of ^ .
= 1 .2$2 !t
sec
Compare » = -0 .8 ± y with 
(ta , = 0. i
^ 1.28
*0.625
Therefore, the required specifications are met with s ~ -O .S±J.
step 3 of 10
Evaluate the characteristic equation at » = -0 .8 ± y
^1.888 - 0.36(1 + ATj,) - 0.8A:+ + ( 0 . 9 2 -1.6(1 + CTj ) + 8C)y = 0
Equate the real and imaginary parts to zero.
1.888 - 0.36(1+ATo ) -0 .8 8 :+ — = 0 
2|
1.528-(0.367„+ 0 .8 )8 :+ ^ = 0
‘ t
1.528 = ̂ 0 .3 6 r „ + 0 .8 - ^ j 8 : ...... (1)
o .9 2 - i .6 ( i+ 8 :r„ )+ 8 := o 
0.92 + 8:
1 + 8 T „
1.6
(2 )
Arbitrarily choose the value, 7} ■ 10.
Solve equations (1) and (2) by substituting 7} s 10.
Therefore,
a: = 1.817,71, = 0.3912
Therefore, the transfer function of continuous PID controller is.
D (5 ) = 1 .8 1 7 ^ 1 + 0 .3 9 1 2 s + ^ j
Step 4 of 10
(a)
Discretize the PID controller using Tustin’s method.
Substitute —[-i— for every occurrence of s in Z)(5).
r ( ^ i+ 2 " 'J ' ' '
= 1.817 1+ 0.'■ 3 9 1 2 [|( ) ) h
10
C ( r ) =
T ( . l+ z " 'J l - z " '
1 . 8 1 7 ( l- z - " ) + l : ^ ( l - 2 z - '+ z - " ) + 0 .0 5 r ( l+ 2 z - ' + z-*)
1 .1 7 - jz - '+ ^ i ; ^ + 0 .0 5 r - 1 .8 1 7 jr
l - z - '
1 .8 1 7 + i^ ^ + 0 .0 5 7 -+ f- ^ ® ^ ^ ^. 1 : ^ 6 .
T
+ 0.1
Step 5 of 10
Calculate the expression D (z ) at r = i , r = o . i , r = o .o i sec. 
3.2886 - 2.7432Z-' - 0 .3 4 5 4 z ‘'
l - z - "
16.042 - 28.422z-' +12.404z-“
143.97 - 284.32z-' + 140.345Z-*
1-Z-"
Step 6 of 10
(b)
Observe the PID controller transfer function
G ( z ) = 8 : ( i + v + ^ )
There is one more zero than pole.
Therefore, for the matched pole zero approximation add a pole at 2 s -1 . 
Map poles and zeros according to the relation z —
( z + l ) ( z - l )
There is no DC gain for this transfer function D (z ) .Therefore, match the gain of D (z ) at 
5 = y ,r )+ e '^ * *^ }s in (2 ff l, r )+
{sm (2 ^ )- l}
Step 7 of 10
Equate the DC gains.
D (z ) = K , ^ -------------^ --------------->-
CalculatiCalculate the expression Z )(z) at r = l, r = 0.1, 7 = 0.01 sec. 
3.2886 - 2.74325-' -0.34542-*
l - z - '
16.042 - 28.422z-' + 12.404z''
l - z
143.97 - 284.32z-' +140.345Z-'
1 - z ^
Step 8 of 10
Sketch the step response for each of the digital implementations at 7 = 1, 7 = 0.1, 7 = 0.01 sec 
using MATLAB.
Write a MATLAB code for step response at 7 = 1 sec • 
num=[3.2886 -2.7432 -0.3454];den=[1 0 -1];sys=tf(num,den,1) 
sysd=feedback(sys,1) 
step(sysd)
Step response for j* = 1 is shown in Figure 1.
Step 9 of 10
Write a MATLAB code for step response at 7 = 0.1 sec •
num=[16.042 -28.422 12.404];den=[1 0 -1];sys=tf(num,den,0.1)sysd=feedback(sys,1)step(sysd) 
Step response for 7 = 0.1 >s shown in Figure 2.
Figure 2
Step 10 of 10
Write a MATLAB code for step response at 7 = 0.01 sec. 
num=[143.97-284.32 140.345];den=[1 0 
-1];sys=tf(num,den,0.01)sysd=feedback(sys,1)step{sysd)
Step response for 7 = 0.01 is shown in Figure 3.
S t^ Response
Figures