EGME 511 (Advanced Mechanical Vibration) Final Project Stabilization of an inverted pendulum on moving cart using feedback control

by Nasser Abbasi

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Introduction

Given the following system

Need to find control law $u\left( t\right)$ to stabilize the inverted pendulum. First we need to obtain the equations of motions.

Analysis

Let the Lagrangian coordinates be $\theta$ and as shown. Let be the Lagrangian. Let be the kinetic energy of the system and let be the potential energy. Hence MATH and MATH Where is the linear velocity of the blob relative to the inertial system.

Hence, since $v=l\dot{\theta}$ , we obtain MATH Hence becomes MATH And since the blob is losing potential energy as it move downwards, we obtain as (assuming zero potential energy is the ground level) MATH Therefore the Lagrangian is MATH To obtain the equation of motions, we need to evaluate for each Lagrangian coordinate $q_{i}$ and $Q_{i}$ is the generalized force for that coordinate. Hence for $\theta$ we obtain MATH Hence EQM for $\theta$ is MATH Now we need to obtain for the coordinate $\theta$ . Apply a virtual displacement $\delta\theta$ and determine the work done by $u\left( t\right)$

Hence the work done by is making virtual displacement $\delta\theta$ is zero, since is not in the line of force along this displacement. Therefore, the EQM for $\theta$ is from (1) above MATH Now we find EQM for coordinate

MATH Hence EQM for is MATH Now we need to find for . Apply virtual displacement in the direction, and find work done by

MATH But , hence we see that , therefore, the EQM becomes MATH

Conclusion

The 2 EQM found are MATH

Assume

small angle approximation

, we obtain MATH

Now, solve for $\ddot{x}$ and $\ddot{\theta}$ from (4) and (5). From the second equation above, eq (5) we find MATH Substitute the above into eq (4) we obtain MATH

Now, using result for $\ddot{\theta}$ found in (6), substitute into (5) we obtain MATH

Hence, to

summarize what we have so far

. We have obtained 2 linearized equations of motion for $\theta$ and

and they are the following

MATH

Now convert to state space. Let , hence MATH MATH Writing the above in the form $\dot{X}=AX+Bu$ we obtain MATH

Stability of open loop system

To determine the stability of the above system (it is a linear system now, since we have already linearized it), we first find the equilibrium point. This is found by setting , and this results in , i.e. and $\dot{\theta}=0$ . Notice that the value of is not important for the equilibrium point. Now we need to determine if this point is stable or not.

MATH

Hence

MATH

Since are all positive, we see that one root will be in the RHS of the complex plane. Therefore the open loop system is

unstable

To stabilize it, we need to supply a control law to force the roots of the new matrix to be all in the LHS of the complex plane.

Let MATH

Hence (7) becomes

MATH

Therefore

MATH

Hence

MATH

We $\allowbreak$ now need to determine $f_{1},f_{2},f_{3}$ and $f_{4}$ . Assume we require that the closed loop poles be located at

MATH

Hence, the characteristic polynomial is

MATH

Compare (8) and (9) we obtain the following

MATH

Hence MATH

Therefore, given we can now find which will generate force $u\left( t\right)$ which will keep the poles of the closed loop system in the LHS of the complex plane, and keep the inverted pendulum stable. For example, for , we obtain

MATH

Comparing solution with and without stabilizing control law

We will now generate the solution for some initial conditions and plot these solutions against time. In the first case, we assume $u\left( t\right)$ is zero. Hence we will observe that the system is unstable, i.e. will grow away from the marginally stable position which is $\theta=0^{0}$ and will not return back. Next, we will introduce $u\left( t\right)$ as determined in the previous section, and observe the new solution to see that it remains near or at the $\theta=0^{0}$ position.

First, we need to decide on some initial conditions. These must be such that close to zero and for $x\left( 0\right)$ we can use . Hence, let MATH

To determine $\QTR{bf}{y}$ , which is the solution of the system, we first must solve equation (7) and (8) for the above IC.

The solution to (7) is given by solution to MATH which is MATH

Where MATH and MATH

Taking Laplace transform of (10) we obtain

MATH

Hence

MATH

Therefore, the solution to

MATH

Therefore, the solution to which is is given by

MATH

Let , we plot the above solution for up to seconds

Now, we plot the solution to (8), which is the state space equation with the stabilizing control law derived above,which is the following

MATH

where

MATH

Where the above values determined to cause the closed loop poles to be located at MATH

Hence

MATH

To make the computation easier, we now substitute numerical values for all the above parameters $\allowbreak$ , which are , and we obtain

MATH

and

MATH

Hence

MATH

Using and solving for we obtain

MATH

Using CAS system to matrix inverse the above and obtain the inverse Laplace transform, and pick the solution and plot it, we observe that now the system becomes stable as expected.

Conclusion

We observe from the above plots and the plots shown in the computation section below, that with the control law derived to force the poles of the closed loop to be stable, the inverted pendulum has been stabilized and the final angle $\theta$ that the inverted pendulum makes with the vertical does go to zero. From the position plot, we see that the cart move to the right away from the position, then it return back to position, while in the same time, the pendulum has swung back and forth about the $\theta=0$ position before it finally settles down the at the stable position. This shows the using pole placement will result in an effective control law which stabilized the system. We have assumed small angle approximation here, and the initial angle used was small.