Projeto de Controlador e Erro Estacionário

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Problem 4.22PP
You are given the system shown in Fig., where the feedback gain jS is subject to variations. You 
are to design a controller for this system so that the output y(t) accurately tracks the reference 
input r(t).
(a) Let f3 =1. You are given the following three options for the controller Dci(s):
Choose the controller (including particular values for the controller constants) that will result in e 
Type 1 system with a steady-state error to a unit reference ramp of less than .
(b) Next, suppose that there is some attenuation in the feedback path that is modeled by p = 0.9. 
Find the steady-state error due to a ramp input for your choice of Dei = (s) in part (a).
(d; iNexi, suppose inai iiieie is some uuenuauun in uie leeuuauK paui inai is iiiuueieu oy p = u.s. 
Find the steady-state error due to a ramp input for your choice of Dei = (s) in part (a).
(c) If = 0.9, what is the system type for part (b)? What are the values of the appropriate error 
constant?
Figure Control system 
X
S tep -by-s tep s o lu tio n
step 1 of 8
Refer to Figure 4.35 in the textbook.
(a)
In order to choose a controller that will result in Type 1 system, first evaluate the transfer function 
and then determine the steady-state error.
First choose as an integrator in the loop and evaluate the transfer function.
10____ Y * » ^ * * ' j
( ( i + l) ( i+ 1 0 )
[(.-m)U io)A
' * ^ ( ( , + l)(“ + 1 0 ) ) ( ^ ) 
1 0 ( V + * , )
_ 5(s + l ) ( ^ + IO)
10(A ^ + Jfc,)
1 + A j ( 5 + !)(« +10)
Step 2 of 8
Now evaluate the system error.
io(*^+*,)
1 —
5 ( j + l ) ( j + I0 )
10(*,5 + * , )
j ( 5 + l) (« + 10)^
1 -
j ( j + l ) ( « + 10)
j ( j + l ) ( i+ 1 0 )+ 1 0 ( * , j+ * , ) /? 
j ( j + l) (5 + 10)
_ r r i o ( v + * , ) ¥ 4 ^ ^ i ) ( » - M o ) ] ] 1
( ( i ( * + l) ( j + 10 )J(j(j+ l)(j+ 1 0 )+ 1 0 (* ,s+ * ,)/jJJ« ’
f i 1 0 ( V + * , ) ) 1
[ i(s+ l) (5 + 10)+10(t,^+t,))»
p(^+i)(»+io)+io(V+*,)/>-io(V+*,)~| 1
[ j ( i+ l ) ( s + 10)+ 10(*^+*,)^
(1)
step 3 of 8
Substitute ^ = t in equation (1).
[ i ( i + l) (s + 1 0 ) + 1 0 ( * ^ + t , ) ( l ) J s '
■ t
s(5 + l ) ( j + 10) ^ I[̂ j(5 + l)(i + 10)+10(*^+*,) Js*
Now apply final value theorem to find the steady state emor.
= ---------] j .
^^j(i + l) (s+ 1 0 )+ 1 0 (t^ + * ,)J i
f ( i + l ) ( i + 10) )
[ i ( i + l)(i+ 10 )+ I0 (* ,i+ * ,) J
s lim
 ̂ 10
‘ lOJfe,
...... (2)
Step 4 of 8
From equation (2), it is clear that 2 1 0 which meets the steady state specifications. The next 
step is to determine the stability, and see if all the poles are on the open left half plane, use the 
Routh criterion
The characteristic equation of the closed-loop poles is,
t i l s ’ 4 - ia s 4 - lo ( i^ + 1^) = 0
Step 5 of 8
To determine the Routh array, we first arrange the coefficients of the characteristic polynomial in 
two rows, beginning with first and second coefficients and followed by the even numbered and 
odd-numbered coefficients.
Write the Routh array,
s’ : 1
f ’ : 11 KM ,
1 1 0 (ltl:,)-l(l^
11
s*: lo t.
From the Routh array, it is clear that > 0 and l l ( l + * , ) > 0 for the system to be stable.
Step 6 of 8
(b)
The value of feedback gain, f i = 0.9 
The input is,
Substitute f i = 0.9 in equation (1).
f ̂ (^ + i)(» * io )-^ io (V + * ,)^ -io (V + *,)] j . 
^ ( s ( i + l) (s + 1 0 ) + 1 0 ( * , f+ t , ) ^ J '■
i ( s + l ) ( s + l0 ) + 1 0 ( * , s + t , ) ( 0 .9 ) - 1 0 ( t , s + t , ) j^ 1 j
i(i + l)(j + 10)+10(*,s+t,)(0.9)
s (j +1)(s + 10)+9(*,s +*,)-10(1 ,s + 1 ,)Y i ^ 
i(s + l)(s + 10)+9(*,s+*,) JU’J
l)(j + 10)+9t,s+9*,-10*,j 
s(s+l)(s+10)+9(t,s+*,)
JTfil f V 1 ) ̂ (s{s + l)(s + 10)+9(*,s+i,)Jl.s’J
s(s + l ) ( j + 1 0 )+ 9 t ,s + 9 * ,-1 0 * ,s -1 0 * , Y I j
Step 7 of 8
Apply final value theorem to find the steady state error. 
e . = l im s [£ ( j ) ]
|^5(5+ l)(j+10)+9(A :^5+ ifc ,)JL5*y
Therefore, the steady-state emor due to ramp input is 0
Since the steady-state error due to ramp input is infinite, the system is no longer Type 1.
Step 8 of 8
(c)
The value of feedback gain, f i = 0.9
The input is, —s
Calculate the steady state error. 
e . - t o s [ £ ( j ) ]
^4 (4+ l)(j + 10) + 9(jfc^J+*;) j\s)
■ lim
j{ 5 + l ) ( j + 10 )-ik^ j-ifc ,
j ( j + 1 )( j + 10)+9(A:,j + * ,)
9k,
= - i
9
From error formula we see the system is Type 0. For the system Type 0 the error for step position
1
l+ K ,
9 l + K^
\ + K , ^ - 9
a: = -1 0
Therefore, the appropriate value for error constant. is |» |q| .