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Problem 9.30PP
Consider the system
Find all values of a and for which the input u{t) = ay(t) + fS will achieve the goal of maintaining 
the output y(t) near 1.
Step-by-step solution
step 1 of 6
Step 1 of 6
Consider the following system.
Output equation is,
The input of the system is. 
a(/)=a;>(f)+P (2)
Substitute X| for in equation (2).
a (/) = cu ,+ P
Substitute oX| for in equation (1).
The nonlinear, closed loop system equations are,
i , =x,-t-x,{ax,+fi) (3)
i j = x 2 { x j+ a x ,+ f i ) (4)
Step 2 of 6
To find the equilibrium points of for the desired output of ^ substitute X | s l , i* |sO 
and sO in equations (3) and (4).
From equation (3).
0 = l + * , ( a ( l ) + / 9 )
I + j( ^ ( a + ^ ) = 0 
I
(5)
From equation (4). 
0 = J i( jC j+ a ( l)+ j9 ) 
j ^ ( * , + a + ^ ) = 0 
]4 + x ,(a + f i ) = 0 
Substitute _ ] for x , (a + /J )
j ^ - l = 0
jtJ -1
Step 3 of 6 ^
Case 1:
Consider the equilibrium point, s 1 and X2 — 1 •
Let.
...... (6)
y i = x 2 - i ...... (7)
0+ P = - l ......(8)
Differentiate the equation (6).
Differentiate the equation (7).
Substitute for ^ + l f o r aT| and y 2 -\-\^ox In equation (3).
>■1 - >>1 + i+ ( > ! +>)(“ ( :> 'i+> )+^)
= ; ' ! + i+ (> ’2 + i) ( “ .*'i+ “ +/?) 
= y , + \ + a y j / t + y t ( a + p ) * a y , * a + p 
My, + l+ay,y, +;>,(-l)+aj>, -1 
y ,= y ,( l+ a ) -y ,+ a y ,y j (9)
Substitute for ^ + l f o r aT| and y 2 -\-\^ox in equation (4).
>■2= ( ; '2 + O U 2 + 0 + ^ )
= { y i+ ^ ) ( y ,+ ^ + a y i+ a * P )
- + ̂ 2 + ‘*>’1^2+^2 ( « + ^ ) + ;>2+ 1 ^
= ; ^ + ;"2+ « J ’i>’2+ ; ’2( - 0+^2 + i+ a j> , - 1 
y 2 = a y ,+ y i+ a y , y i+ } i °)
Step 4 of 6 ^
The characteristic equation of the linearized system is,
j ' - ( a + 2 ) j+ ( 2 a + l ) = 0
There are no values of a which produce stable roots. So it is concluding that, s 1 and 
s i Is an unstable equilibrium point.
Step 5 of 6
Case 2:
Consider the equilibrium point, s 1 and X2- —I 
Let.
...... (11)
y i = x i + l ...... (12)
0 + P = l ...... (13)
Differentiate the equation (11).
>1=^1
Differentiate the equation (12).
Step 6 of 6
Substitute jĉ for ^ + 1 for x^ and y 2 —\ for X2 in equation (3).
- J ’l + l + ( j ’2 - l ) ( “ U + 0 + /®)
= ; ' i+ i+ ( > ’2 - i ) ( “ J’i + ‘* + ^ )
=>>,+ l+ « y ,y , + y ^ { a + P ) - a y , - ( a + p )
My, + l+ a y ,y j + y , ( l ) - a > , -1
y, = y i ( ^ - a ) * y i * a y ^ i ( I'l)
Substitute i |2 for ^ * i> lfo r JC| and I fo r X2 in equation (4).
.̂ 2 =(.>'2- l ) ( . ) ’2 - '+ a ( .> ’i+ > )+ t» )
=(^2 - >)(.> ’2 - 1+a>’i + «+>9 )
= A - y t +0W 2+>’2(«+/9)->’2 ~{a+P)
= A - y 2 + W 2+^ '2 (> )-J ’2 + > -o '; 'i - •
y , M - a y , - y , + a y , y , + ) i (15)
The characteristic equation of the linearized system is,
s ^ + a s + { 2 a - i) = 0
The system is stable for small signals near the equilibrium point if.
a > — and a+fi=\
Therefore, the values of or and are. o r > — a n d a + > 9 = l