Rotary Inverted Pendulum Model

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Rotary Inverted Pendulum Model
C&DSP Lab. EE@MJU
which implies that 1 ¨ ˙ θ= −bc sin(α)α2 − cGθ ˙ ac − b2 cos2 (α) ηm ηg Kt Kg + bd sin(α) cos(α) + c V Rm 1 ˙ b2 sin(α) cos(α)α2 − b cos(α)Gθ ˙ α= ¨ 2 cos2 (α) ac − b η m η g Kt Kg + ad sin(α) + b cos(α) V Rm where a = J + mr2 , b = mLr, c = 4 mL2 , d = mgL, and 3 Fig. 1. A schematic diagram of the rotary inverted pendulum

2 ηm ηg Kt Km Kg . Rm

G=B+ Using the Lagrangian method, the equation of motion of rotary inverted pendulum can be derived as follows [1]: ¨ (J + mr )θ + mLr sin(α)α2 − mLr cos(α)¨ ˙ α ˙ = T − Bθ 4 2 ¨ ¨ 3 mL α − mLr cos(α)θ − mgL sin α = 0 2

Defining x= θ α ˙ θ α ˙
T

, y= θ

α

T

, u=V

(1)

where T is the torque on the load from the motor, α is the pendulum angle, θ is the horizontal arm angle and other system parameters are given in Table I. In addition, the torque T is generated by DC motor such that [2] T = ηm ηg Kt Kg ˙ V − Kg Km θ Rm (2)

and linearizing about the upright position i.e. x = 0, yields     0 0 0 1 0  0 0  0 0 1  x +  ηm ηg Kt Kg  u (4) x= ˙  bd cG  c R E  0 E −E 0 m η η Kt K 0 ad − bG 0 b mRg E g E E m

1 0 0 0 y= x 0 1 0 0

(5)

where E = ac − b2 . Using the system parameters of Table I, (4) becomes x = Ao x + Bo u ˙ y = Co x where  0 0 Ao =  0 0 1 Co = 0 0 0 41.68 84.05 0 0 1 0 1 0 −15.47 −14.89 0 . 0    0 0   1  , Bo =  0  , 27.13 0 0 26.12 (6)

where V is an applied armature voltage to DC motor.
TABLE I PARAMETERS OF ROTARY P ENDULUM Parameter J m r L g B Kt Kg Km Rm ηm ηg Description Moment of inertia at the load Mass of pendulum arm Rotating arm length Length to pendulum’s center of mass Gravitational constant Viscous damping coefficient Motor torque constant SRV02 system gear ratio Back-EMF constant Armature resistance Motor efficiency Gearbox efficiency Value (SI) 0.0033 0.1250 0.2150 0.1675 9.81 0.0040 0.0077 70 0.0077 2.6 0.69 0.90

Now, we will design the proposed output feedback controller. Using Q =...
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