achapt 6 - TRACKING THE REFERENCE PATH 6 TRACKING THE...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
TRACKING THE REFERENCE PATH 6 TRACKING THE REFERENCE PATH As stated earlier, the problem of controlling a dynamical system is customarily divided into two basic problems – tracking and regulation. The former is associated with moving a component from one equilibrium point to another, and the latter is associated with keeping the component at a given equilibrium point. The simplest form of tracking is accomplished without feedback – by applying a force that's an explicit function of time. A tracking force that is an explicit function of time is also called an open-loop tracking force . The design of an open-loop tracking force can be divided into two steps. The first step is associated with specifying a desirable path to follow. The second step determines the tracking force. Rest-to- rest tracking and spin-up tracking are considered below. 1. Rest-to-Rest Tracking Consider a single degree of freedom point mass ( c = k = 0) subjected to a tracking force f . The MAE 461: DYNAMICS AND CONTROLS
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
TRACKING THE REFERENCE PATH response of the system will be assumed in the form of a polynomial: (6 – 1) n n t a t a a x f x m L & & + + = = 1 0 where a r are coefficients that need to be determined. The response x is the desired path that the point mass should follow and that is being specified by the designer. The coefficients in the polynomial are determined from the boundary conditions that are imposed on the desired path. The quadratic path (n = 2) In the case of a quadratic path, three coefficients need to be determined in Eq. (6 – 1). The number of coefficients, in turn, determines the number of boundary conditions that can be imposed. Let’s impose the boundary conditions (6 – 2) 1 ) ( 0 ) 0 ( 0 ) 0 ( x T x x x = = = & Equation (6 – 2) is substituted into Eq. (6 – 1) to produce a set of three linear algebraic equations in terms of the three unknown coefficients. The three equations and the solution are expressed below in matrix-vector form (6 – 3) = = 2 1 2 1 0 1 2 1 0 2 / 0 0 0 0 1 0 1 0 0 0 1 T x a a a x a a a T T Substituting Eq. (6 – 3) into Eq. (6 – 1), and dividing by the appropriate quantities, yields the response and the tracking force; both expressed in non-dimensional form. We get MAE 461: DYNAMICS AND CONTROLS
Background image of page 2
TRACKING THE REFERENCE PATH (6 – 4) 1 ) / ( ) ( 1 2 2 1 = = mx f T T t x t x The response given in Eq. (6 – 4) is not a true rest- to-rest solution in that the system is not brought to rest. However, this is the best solution that can be found using a quadratic form for the desired path (See Fig. 6 – 1). 0 0.5 1 0 1 0 1 1 T t mx f T T t x t x - versus and versus ) ( : 1 6 Fig. 1 2 1 = The cubic path (n = 3) Let’s now consider a cubic path. Four coefficients in Eq. (6 - 1) need to be determined. Thus, we select the boundary conditions (6 – 5) 1 1 ) ( ) ( 0 ) 0 ( 0 ) 0 ( x T x x T x x x = = = = & & Substituting Eq. (6 – 5) into Eq. (6 – 1) yields four linear algebraic equations and the solution given below: MAE 461: DYNAMICS AND CONTROLS
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/17/2009 for the course MAE 469 taught by Professor Silverberg during the Fall '08 term at N.C. State.

Page1 / 15

achapt 6 - TRACKING THE REFERENCE PATH 6 TRACKING THE...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online