INC 341 Feedback Control Systems:
Lecture 4 Linearization
Asst. Prof. Dr.-Ing. Sudchai Boonto
Department of Control Systems and Instrumentation Engineering
King Mongkut’s University of Technology Thonburi
Linear system vs Nonlinear system
Output f (x)
Output f (x)
• Linear systems possesses two properties: Superposition and homogeneity.
• Nonlinear systems do not.
2
1
1
2
3
Input x
4
• Linear function f (αx) = αf (x)
2
1
1
2
3
Input x
4
• Nonlinear function f (αx) ̸= αf (x)
INC 341 Feedback Control Systems:, Lecture 4 Linearization
J 2/10 I }
Physical nonlinearities
Some examples of physical nonlinearities:
• An electronic amplifier is linear over a specific range by exhibits the nonlinearity called
saturation at high input voltages.
• A motor that does not respond at very low input voltages due to frictional forces
exhibits a nonlinearity called dead zone.
• Gears that do not fit tightly exhibit a nonlinearity called backlash.
Dead Zone
Input x
Backlash in Gears
Output f (x)
Output f (x)
Output f (x)
Saturation
Input x
INC 341 Feedback Control Systems:, Lecture 4 Linearization
Input x
J 3/10 I }
Linearization
The systems that we consider so far assumed to be linear. However, if there are some
nonlinear components are presented, we have to linearize the systems before finding the
transfer functions. The linearization steps are:
• Check where is the nonlinear component and write the nonlinear differential equation
of the system.
• Linearize the nonlinear part using a first order Taylor series approximation at the
equilibrium point.
Output f (x)
f (x) = f (x0 ) +
df (x − x0 ) + high order term .
dx x0
f (x)
f (x) − f (x0 ) ≈ ma (x − x0 )
f (x0 )
⇒
δf (x) ≈ ma δx
and
f (x) ≈ f (x0 ) + ma (x − x0 ) ≈ f (x0 ) + ma δx
x0
x
Input x
where ma is a slope at the tangent point.
INC 341 Feedback Control Systems:, Lecture 4 Linearization
J 4/10 I }
Linearization
Example
Problem: Linearize f (x) = 5 cos x about x = π/2.
Solution: Start with find the slope at x = x0 = π/2.
m=
df
= −5 sin x
= −5
dx
x= π
2
Since f (x0 ) = 5 cos
π
2
= 0, then
f (x) = f (x0 ) + mδx = −5δx.
This system can be represented as
f (x) = −5δx,
for small excursion of x.
INC 341 Feedback Control Systems:, Lecture 4 Linearization
J 5/10 I }
Linearization
Example
Problem: Linearize the system below for small excursions about x = π/4.
d2 x
dx
+2
+ cos x = 0.
dt2
dt
Solution: the nonlinear term cos x can be linearized as follow:
cos x = cos
√
√
π
d cos x 2
2
+
δx
=
−
δx
4
dx x= π
2
2
4
Substituting x = x0 + δx and the linearized of cos x to the system, we have
d2
(π
4
+ δx
dt2
)
+2
d
(π
4
)
√
2
2
−
δx = 0
dt
2
2
√
√
dδx
2
2
d2 δx
+
2
−
δx
=
−
dt2
dt
2
2
+ δx
√
+
INC 341 Feedback Control Systems:, Lecture 4 Linearization
J 6/10 I }
Linearization of a simple pendulum
Example
τ
The motion equation:
J
l
θ
d2 θ
M gl
+
sin θ = τ (t),
dt2
2
where τ (t) is a small-signal force.
Mg
Since θ0 = 0, the nonlinear term sin θ can be linearized as follow:
sin θ = sin θ0 +
d sin θ (θ − θ0 ) = cos θ0 δθ = δθ
dθ θ=0
Then,
J
(
)
M gl
M gl
d2 (θ + δθ)
2
+
δθ
=
τ
(t)
⇒
Js
+
δΘ(s) = τ̂ (s)
dt2
2
2
δΘ(s)
1/J
=
gl
τ̂ (s)
s2 + M
2J
INC 341 Feedback Control Systems:, Lecture 4 Linearization
J 7/10 I }
Transfer Function – Nonlinear Electrical Network
+ vL (t) −
1H
RNL
v(t)
+
−
i(t)
20 V
Find the transfer function, VL (s)/V (s),
for the electrical network shown in Fig.,
which contains a nonlinear resistor whose
voltage-current relationship is defined by
ir = 2e0.1vr , where ir and vr are the
resistor current and voltage, respectively.
Also, v(t) in Fig is a small-signal source.
Applying Kirchhoff’s voltage law around the loop, where ir = i(t) yields
L
di
1
+ 10 ln i(t) − 20 = v(t)
dt
2
Set the small-signal source, v(t), equal to zero. We have di/dt = 0 and 10 ln 12 i = 20. Then
we have i0 = ir = 14.78 A.
d(ln 12 i) 1
1
ln i(t) = ln i0 +
2
2
di i=i0
(i − i0 ) = ln
1
1
1
1
i0 + δi = ln i0 + δi
2
i i=i0
2
i0
INC 341 Feedback Control Systems:, Lecture 4 Linearization
J 8/10 I }
Transfer Function – Nonlinear Electrical Network
Substituting the result into the system equation, the linearized equation becomes
L
(
)
dδi
1
1
+ 10 ln i0 + δi − 20 = v(t)
dt
2
i0
dδi
+ 10 (2 + 0.0677δi) − 20 = v(t)
dt
dδi
+ 0.677δi = v(t).
dt
Taking Laplace transform, yield
δI(s) =
V (s)
s + 0.677
Since vL (t) = Ldi/dt, we have vL (t) = dδi/dt and VL (s) = sδI(s). Finally, we obtain
VL (s)
s
=
,
V (s)
s + 0.667
for small excursions about i = 14.78 A or, equivalently, about v(t) = 0.
INC 341 Feedback Control Systems:, Lecture 4 Linearization
J 9/10 I }
Reference
1. Norman S. Nise, ”Control Systems Engineering,6th edition, Wiley, 2011
2. Gene F. Franklin, J. David Powell, and Abbas Emami-Naeini, ”Feedback Control of
Dyanmic Systems”,4th edition, Prentice Hall, 2002
INC 341 Feedback Control Systems:, Lecture 4 Linearization
J 10/10 I }