Linearization (introduction)
Consider the following example
m θ l
g
mg
T θ
[ ∑ MO = JOθ&&]
θ
θ 2 &&
sin ml
mgl =
−
0
sin =
⎟⎠
⎜ ⎞
⎝
+ ⎛ θ
θ l
&& g
Non-linear equation
• Difficult or cannot solve for the exact solution
• Superposition of solutions does not allow
Linear and non-linear equations
d dx cy
b dy dx
y
a d
2 2+ + =
• a, b, c, d = constant, or f(x) Linear equation
• a, b, c, d = f(y)
• There are transcendental functions in the equation
Non-linear equation
Function that cannot be defined directly by algebraic formulas Ex. Exponential, logarithmic, trigonometric functions
K
K
− + +
′′ + + −
− ′ +
=
)
! ( ) (
)
! ( 2
) ) (
( ) (
) ( )
(
0 ) 0 (
0 2
0 0
0 0
x n f
x x
x x f
x x f x x x
f x
f
n n
Linearization (1)
To obtain a linear model of non-linear mechanical systems or components based on Taylor series.
Non-linear model: y = f(x)
Taylor series
This approximation is valid only for x near x0 (operating point)
Linearized model
(first-order approximation)
) ( ) (
) ( )
(x f x0 x x0 f x0
f ≅ + − ′
Linearization (2)
Example: y = sin(x)
0
)) (sin(
) (
) sin(
)
sin( 0 0
x
dx x
x x d
x x
x
=
− +
≅
) ( ) (
) ( )
(x f x0 x x0 f x0
f ≅ + − ′
) cos(
) (
) sin(
)
sin(x ≅ x0 + x − x0 x0
From
0 = 0
x sin(x) ≅ 0+ (x −0)(1) = x
0 2
= π
x )(0) 1
( 2 1 )
sin( ≅ + − π =
x x
0
sin =
⎟⎠
⎜ ⎞
⎝
+⎛ θ
θ l
&& g ⎟ = 0
⎠
⎜ ⎞
⎝
+⎛ θ θ l
&& g for small θ
) 0
( →
for x near 0 for x near π/2
θ
Linearized EOM (1)
Given non-linear EOM: M (x)x&&+C(x, x&) + K(x) = F(t)
Procedures
(1) Select an “operating point” t0, x0, x&0, x&&0, F(t0) (2) At operating point, we have
) ( )
( )
, ( )
(x0 x0 C x0 x0 K x0 F t0
M && + & + =
(3) Near the operating point, we have
) ( )
(
) (
) ,
( ]
)[
(
0 0 0
0 0
0
t F t
F
x x
K x
x x x
C x
x x x
M
Δ +
=
Δ + +
Δ + Δ
+ +
Δ + Δ
+ && && & &
x x
x x
x x
x x
x
t F t
F t
F t
t t
&&
&&
&&
&
&
& = + Δ = + Δ
Δ +
=
Δ +
= Δ
+
=
0 0
0
0 0
,
,
) ( )
( )
( ,
Linearized EOM (2)
) ( )
(
) (
) ,
( ]
)[
(
0 0 0
0 0
0
t F t
F
x x
K x
x x x
C x
x x x
M
Δ +
=
Δ + +
Δ + Δ
+ +
Δ + Δ
+ && && & &
(4) Using Taylor series for non-linear terms
x x x M
M x
x M
x
∂ Δ + ∂
≅ Δ
+
0
) ( )
( 0 0
x x x C
x x C
x C x
x x x
C
x x x
x
∂ Δ + ∂
∂ Δ + ∂
≅ Δ
+ Δ
+
0 0 0
0, ,
0 0 0
0 , ) ( , )
(
&
&
&
&
&
&
&
x x x K
K x
x K
x
∂ Δ + ∂
≅ Δ
+
0
) ( )
( 0 0
Linearized EOM (3)
(5) Substituting (4) into (3), using relation in (2) and neglecting higher-order terms,
)
~ (
~
~ x C x K x F t
M Δ && + Δ & + Δ = Δ
)
~ (
x0
M M =
0 0,
~
x
x x
C C
&
&
∂
= ∂
0 0
0
0 ,
0
~
x x
x
x x
K x
x C x
K M
∂ + ∂
∂ + ∂
∂ Δ
= ∂
&
&&
Where
This “linearized EOM” is valid only near the operating point x0 )
( ,
,
, x x F t
x Δ Δ Δ
Δ & && are small
Example (1)
Derive the EOM of the system in the figure. You may have to make some approximation of the cosine. Assume the bearings provide a viscous damping force only in the vertical direction.
[Inman/1.46]
Example (2)
Derive the EOM of an airplane’s steering-gear mechanism for the nose wheel of its landing gear. The mechanism is modeled as the single-degree-of-freedom system illustrated in the figure.
[Inman/1.49]
Example (3)
A control pedal of an aircraft can be modeled as the single- degree-of-freedom system shown in the figure. Consider the level as a massless shaft and the pedal as a lumped mass at the end of the shaft. Determine the EOM in θ. Assume the spring to be unstretched at θ = 0. [Inman/1.50]
Example (4)
Consider the disk of the figure connected to two springs.
Derive EOM for small angle θ (t). [Inman/1.82]
Example (5)
Consider the pendulum mechanism shown in the figure, which is pivoted at point O. Derive EOM for small angles. Assume that the masses of the rod, spring, and damper are negligible.
[Inman/2.16]
Example (6)
A foot pedal for a musical instrument is modeled by the sketch in the figure. Derive EOM of the system. Also, use the small angle approximation. [Inman/2.25]
