Linearização por Realimentação

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Problem 9.05PP
Linearizing effect of feedback: We have seen that feedback can reduce the sensitivity of the 
input-output transfer function with respect to changes in the plant transfer function, and reduce 
the effects of a disturbance acting on the plant. In this problem we explore another beneficial 
property of feedback: It can make the input-output response more linear than the open-loop 
response of the plant alone. For simplicity, let us ignore all the dynamics of the plant and assume 
that the plant is described by the static nonlinearity
y «) =
« 1 .
(a)Suppose we use proportional feedback 
u(t) = r(t) + a(r(t)-y(t)),
where a > 0 is the feedback gain. Find an expression for y(t) as a function of r(t) for the closed-
loop system. {This function is called the nonlinear characteristic at the system.) Sketch the 
u(tf= r(t) + a(r(t) - y(t)).
where a > 0 is the feedback gain. Find an expression for y(t) as a function of r(t) for the closed- 
loop system. (This function is called the nonlinear characteristic ot the system.) Sketch the 
nonlinear transfer characteristic for a = 0 (which is really open loop), a = 1, and a = 2 .
(b) Suppose we use integral control: 
t
h( 0 = r ( t ) + J ( r { r ) - y ( T ) ) d r .
The closed-loop system is therefore nonlinear and dynamic. Show that if r(t) is a constant, say r, 
then lim y ( /) = r . Thus, the integral control makes the steady-state transfer characteristic of
/-►oo'
the closed-loop system exactly linear. Can the closed-loop system be described by a transfer 
function from rto y?
Step-by-step solution
>1 of11 ^
Given that the plant is described by the static non linearity as 
u, u £ 1 
u +1^ ( 0 = - . u > \
w
The given proportional feedback is « ( i) = r ( i ) + ” ,y (0 ] •
Here, a ^ 0 is the feedback gain.
Step 2 of 11 ^
The nonlinear system with saturation proportional control is sketched.
Find,y for u ^ 1. 
y = r + a { r - y ) 
y = { l - a ) y 
y = ( l - i i ) r
Thus, we get 
y - r - u
Obtain y for u > 1 
u + \
Simplify the egression further. 
2y s \ + r + a r - £^,
( 2 + a ) ^ = l + ( l + a ) r 
l + ( l + a ) r
y = -------i --------- i—
2 + a
Step 4 of 11
Sketch the response of the open-loop sjrstem v/s the closed-loop system
2h 3
Find the egression for,y when and 2.
Thus we get
I 1 2y\ , = - + —/ 
3 3
4 4
Step 6 Of 11 ^
Sketch the open-loop v/s closed loop gri^h for a s 0,1, and 2.
Open loop closed loop lor winous a
a = ^
y
z
X '
y '
zz
(b)
Sketch the non-linear system with saturation inte^'al control
) I
output.
V J input
step 8 of 11 ^
Given that
The closed loop system is non-linear and dynamic.
Thus, we getu = r -I- ,
where r is a constant 
Findj', w hena 
In stable condition,y stays bounded. Hence, y y^-
Hence, we get ^ = r + (r - y j)d t oo, if y ^ * r . 
The e^»ression infers that y^ = r .
Find 
y - ^ y 
y = r - y 
y - ^ y - r 
Find,y w hena> 1.
We know that =j^a).
Find X and y .
y - - \ i r - y )
Step 11 of 11
From the equation we get 2y y ^ r .
Hence, on further simplification we obtain the e^sression —^ —̂ ,
R (s) 2s + 1
Therefore, in steady-state condition.^ = r. 
Thus, we get lim = lim r ( t ) .
So, we infer that lim .y (t) = r , since, r is a constant