哈工大惯性技术(导航原理)大作业

Assignment of Inertial Technology 《惯性技术》作业

Autumn 2015

Assignment 1: 2-DOF response simulation

A 2-DOF gyro is subjected to a sinusoidal torque with amplitude of 4 g.cm and frequency of 10 Hz along its outer ring axis. The angular moment of its rotor is 10000 g.cm/s , and the angular inertias in its equatorial plane are both 4 g.cm/s 2. Please simulate the response of the gyro within 1 second, and present whatever you can discover or confirm from the result.

In this simulation, we are going to discuss the response of a 2-DOF gyro to sinusoidal torque input. According to the transfer function of the 2-DOF gyro, the outputs can be expressed as:

122

22

()()(

)()

y

x y x y x y J H

s M s M s J J s H s J J s H α=

-

++ 12222()()()()

x y

x x y x y J H

s M s M s J J s H s J J s H β=

+++ The original system transfer function is a 2-input, 2-output coupling system. But the given input only exists one input, we can treat the system as 2 separate FIFO system

As a consequence, we can establish the block diagram of the system in simulink in Fig 1.1. Substitute the parameters into the system and input, then we have the input signals as follow: 0,4sin(20)y ox M M t π==

Then the inverse Laplace transform of the output equals the response of the gyro in time

domain as follows:

02222

0002

02222

00()sin sin ()()

()cos cos ()()

ox a ox

a x a x a ox ox a ox a a a a a M M t t t J J M M M t t t H H H ωαωωωωωωωωω

βωωωωωωωω=-+--=

+---

Fig 1.1 The block diagram of the system in simulink

And the simulation results in time domain within 1 second are shown in the follwing pictures. Fig1.2 is the the output of outer ring ()t α. Fig1.3 is the output of inner ring ()t β. Fig1.4 is the trajectory of 2-DOF gyro ’s response to sinusoidal input. As we can see from the Fig1.3,there are obvious sawtooth wave in the output of the inner ring. It ’s a unexpected phenomenon in my original theoretical analysis.

Fig1.2 The output of inner ring ()t β

Fig 1.3 The output of outer ring ()t α

I believe the sawtooth wave is caused by the nutation. For the frequency of the nutation is obtained as 010*******/3974

e H rad s Hz J ω=

==≈, which is far higher than the frequency

of the applied sinusoidal torque , namely 0a ωω.

Fig 1.4 Trajectory of 2-DOF gyro ’s response to sinusoidal input

The trajectory of 2-DOF gyro ’s response to sinusoidal input are shown in Fig1.4. As we can see, it ’s coupling of X and Y channel scope output. The overall shape is an ellipse, which is not perfect for there are so many sawteeth on the top of it.

Note that the major axis of ellipse is in the direction of the forced procession, amplitude of which is 0

ox M H ω, whereas the minor axis is in the direction of the torsion spring effects,

with amplitude ox a

M H ω.

The nutation components are much smaller than that of the forced vibration, which can be eliminated to get the clear static response.

2200

2

22

0()sin sin ()cos (1cos )()ox ox

a

a x a ox ox ox

a a a a a a

M M t t t J H M M M t t t H H H αωωωωωωβωωωωωωω≈

≈-≈

-=--

To prove it, we eliminate the effects of the nutation namely the quadratic term in the denominator and get Fig 1.5, which is a perfect ellipse. We can conclude that when input to the 2-DOF gyro is sinusoidal torque, the gyro will do an ellipse conical pendulum as a static response, including procession and the torsion spring effects, together with a high-frequency vibration as the dynamic response.