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Problem 9.02PP
The circuit shown in Fig has a nonlinear conductance G such that iG = g(vG) = vG(vG - 1 ){vG - 
4). The state difTerential equations are
di
- = - . + v.
Figure Nonlinear circuit
R = l
dv
= - i + g ( i i - v ) .
where / and v are the state variables and u is the input.
(a) One equilibrium state occurs when i/ = 1. yielding /1 = v1 = 0. Find the other two pairs of v 
and /that will produce equilibrium.
(b) Find the linearized model of the system about the equilibrium point u = 1. /1 = v1 = 0.
(c) Find the linearized models about the other two equilibrium points.
Step-by-step solution
step 1 of 5
Refer to Figure 9.57 in the textbook.
The current flowing through the resistor is,
'o -s C v o )
The differential equations are,
^ = - / + g ( a - v )
Where,
i and v are the state variables 
If is the input
Step 2 of 5
(a)
At equilibrium state,
4 - 0
( 1)- i + v = 0
^ = 0
dt
- i + g ( t t - v ) = 0 (2)
Substitute ( v « - l ) ( v o - 4 ) for ig in equation (1).
From Figure 9.7 in the text book,
V c - l f - V
Substitute i f - v for vg in equation (3).
- ( i f - v ) ( i f - v - l ) ( i f - v - 4 ) + v = 0 
Substitute 1 for If 
- ( l - v ) ( l - v - l ) ( l - v - 4 ) + v = 0 
- ( l - v ) ( - v ) ( - 3 - v ) + v = 0
v(v* + 2v - 2 ) = 0 ...... (4)
Solve the equation (4).
Therefore, 
v = 0 , - l± > / 3 
From equation (1),
- i+ v = 0 
i= v
1= o , - \ ± S
Therefore, the pairs of y and i ^^at will produce equilibrium is,
v = 0 , - l ± S 
i = 0 ,- l± y /3
Step 3 of 5
leplace u, v, and / by 1 îf, Svy and Si ■
Tom the differential equations.
Si = -S i + ......(5)
= -S i + g (l + Su - 5 v ) ......(6)
Simplify the expression s -S i + ^ ( l + ^ if - ^ v ) further.
= - g i + - ^ v ) [ ( l + Su - ^ v ) - l ] [ ( l + Su - tfv) - 4 ]
= -S i + (1 + 5if - Sv)[Su - Sv){Su - - 3)
^ v s - A + 3^ if + 3 ^ v ...... (7)
Vrite the equations (5) and (7) in matrix form.
'hus, the linearized model of the system at equilibrium point is
'F '-
-1
Su
Step 4 of 5
(c)
Write the general linearized model equation.
Su
-1 I ■[ s n O'
-1 ^ 
dv. W * " .dv.
When u = 1,
« (" .v ) = « ( l.v )
= v^+2v-2
Apply partial differentiation.
^ = v’ + 2 v - 2 + v(2v + 2) 
dv
= 3v*+4v-2
Substitute -1 ± for v .
^^3(-i±S)\A{-X±S)-2
- 5 t 2 - ^
Step 5 of 5
It is know that,
gg (« - v) -g '{u - v) (8)
And,
( " - *') g '{u -v ) (9)
Similarly,
^ = 5 w 2 S
dv
And,
du dv 
m -5±2 S
The linearized model about the other two equilibrium points is.
4* 1= ■' i M i .
m u
Su
-1 5 T 2 â JL ^ vJ ’' [ - 5 ± 2>5J
Thus, the linearized model about the other two equilibrium points is.
d [ S n ■-1 1 r ^ / 1 0
-1 5 T 2 , ^ U J" -5 ± 2 ^ 3
^If