# Representations Transfer Functions Poles and Zeros by rma97348

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```									MEM 351 – Dynamic Systems Lab

Representations: Transfer Functions
Poles and Zeros
Recall          Damped Compound Pendulum Equations of Motion

&& + c θ + mL gd θ = 0
θ       &
J      J
d            Linearized 2nd order differential
θ
equation assumes small angles
c dθ
dt

L     Bar length [m]
L

m g
d     Pivot to CG distance [m]
L
mL    Mass of pendulum [kg]

Moment of Inertia [kg ⋅ m ]
2
θ=0                      J
Nms
c     Viscous damping coefficient [
]

General          2nd   order form
&          &
θ& + 2ςω nθ + ω n θ = 0
2
Tedious Math: Time domain differential equation

2nd order damped system             &          &
θ& + 2ζω nθ + ω n θ = 0
2

Yields complex roots               s1, 2 = −ζω n ± j ⋅ ω n 1 − ζ 2
Time domain solution

{        (
θ c (t ) = e −ζ ωnt A1 cos ω n t 1 − ζ   2
)+ A sin (ω t
2      n    1−ζ 2   )}   (1)

Small real root will yield
long settling times

Can be shown:
Time constant            Tc = ω nζ
4                          (2)
2% settling time t s =
Tc
Easier Math I: Laplace Domain

• Voltage V (s ) applied to motor

r
• Propeller spins, creating lift force F (s)
d
F
θ
c dθ
• Lift on lever arm r creates torque Τ(s)
dt                                    T = Fr
• Pendulum rotates angle Θ(s )
motor-propeller

mg

Motorized
Propeller
1/J
T(s) Torque                           Θ (s) = θ
V(s)             K
m                         2  c   m gd
[Volts]                        [Nm]            s +   s+ L         [ rad/sec ]
[Nm/Volts]                          J     J

Compound Pendulum
Calculating Constants
Motorized
Propeller
1/J
T(s) Torque                       Θ (s) = θ
V(s)             K
m                     2  c   m gd
[Volts]                        [Nm]        s +   s+ L         [ rad/sec ]
[Nm/Volts]                      J     J

Compound Pendulum

Km     Theory: can calculate lift force if have propeller pitch and radius
dimensions, air density and motor angular velocity.
Experimentally, apply known voltage V and pendulum will

&     &
Jθ& + cθ + mL gd sin θ = T

At steady-state angular acceleration and velocity are zero. The
torque at this known voltage is calculated by:
T
T      ss
= m L gd sin θ ss               And hence            Km =       ss

V
Open-Loop Transfer Function
Motorized
Propeller
1/J
T(s) Torque                       Θ (s) = θ
V(s)         K
m                     2  c   m gd
[Volts]                    [Nm]        s +   s+ L         [ rad/sec ]
[Nm/Volts]                      J     J

Compound Pendulum

Θ ( s)         Km / J
OLTF:                        =                  = Gol ( s)
V ( s)         c    mL gd
s + s+
2

J      J
Given

Km = 0.017 Nm/V
d = 0.023 m                           Θ( s)         1.89
= 2                 = Gol ( s)       (3)
J = 0.0090 kgm 2                      V ( s) s + 0.039s + 10.77
m L = 0.43     kg
Laplace domain OL Transfer function
OLTF
Simulations

Simulation reveals long settling time. This is consistent with the low
viscous damping coefficient. Poles of the characteristic equation
reveal the large oscillations. Recall from (1)

Θ( s)         1.89
= 2                 = Gol ( s)
V ( s) s + 0.039s + 10.77
Roots of the denominator (i.e. the poles) are:

s1 = −0.0019 + j 3.28
s 2 = −0.0019 − j3.28

Small real root will yield
long settling times

In other words, the system is bordering on the margin of stability.
Recall Equations (1) and (2).
System Poles and Zeros – What are they?

d
θ
dθ
c
dt

L

m g
L

θ=0

&          &
θ& + 2ςω nθ + ω n θ = 0
2                    (1)
Control Designer’s Goal:
Create compensators that
a   ς 2π    1   X
ln     =        = ln 1               (2A)
yield desired damping and
b   1 − ς 2 N X N +1                   rise time.
2π                                   In other words, place poles
= ωn 1 − ς 2               (2B)
where one wants them
T
Calculated the following:

&         &
θ& + 2ςω nθ + ω n2θ = 0
ζ = 0.0059                                s + 0.0295 s + 6.25 = 0
2

s1,2 = −0.01475 ± 2.5 j    poles

s = a + jb
1
X                      ωn     1- ζ
2

ζ = sin θ            ωn        θ

- ζ ωn

X                       - ωn    1- ζ 2
s = a - jb
2
s = -0.01475 +j 2.50
1
X                    ωn     1- ζ
2

ζ = sin θ            ωn     θ

- ζ ωn

X                - ωn    1- ζ 2
s = -0.01475 - j 2.50

θ = arctan
0.01475
= 0.0059
ζ = sin θ = sin 0.0059 = 0.0059
2.50                       ω n = a 2 + b 2 = 0. 01475 2 + 2.50 2 = 2.50

Matches experimental data!
Where are we going with this?

It’s called the characteristic equation because it
connotes system properties
Poles are the roots of the characteristic equation. As
such, the describe stability through ωn and ζ

Question: can we alter the locations of the poles? If we
can, then we change the characteristic of the system…

Answer: This is exactly what the control engineer does.
One popular method is called “pole placement” control

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