nth_order

nth_order - n th Order Linear Systems Do not distribute...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

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

Unformatted text preview: n th Order Linear Systems Do not distribute these notes 63 Prof. R.T. MCloskey, UCLA Introduction to n th order ODEs Consider the n th order linear time-invariant ODE: d n dt n ( y ( t ))+ a 1 d n- 1 dt n- 1 ( y ( t )) + + a n- 1 d dt ( y ( t )) + a n y ( t ) = b 1 d n- 1 dt n- 1 ( u ( t )) + + b n- 1 d dt ( u ( t )) + b n u ( t ) , or more compactly, y ( n ) + a 1 y ( n- 1) + + a n- 1 y + a n y = b 1 u ( n- 1) + + b n- 1 u + b n u. (25) Note that this ODE is uniquely specified by the following sets of coefficients: { a 1 , a 2 , . . . , a n } , { b 1 , b 2 , . . . , b n } . (26) Any of these coefficients can be equal to zero, too. We assume constant coefficients (time-invariance) so we may take the starting time" of any initial value problem to be t = 0 (we havent demonstrated this fact for anything other than first order ODEs but it actually extends to ODEs of any order). We also assume that the input u has as many derivatives as required to make the right-hand-side of (25) well defined. The initial value problem is stated as: Initial Value Problem. Given the initial conditions { y (0) , y (0) , y (0) , . . . , y ( n- 1) (0) } , and the input u defined for t , find a function defined for t that satisfies (25) and the initial conditions. As in the first order case, the IVP may be solved in an ad hoc manner by considering homogeneous and particular solutions: 1. Homogeneous differential equation. The homogeneous differential equation is: y ( n ) + a 1 y ( n- 1) + + a n- 1 y + a n y = 0 , and a solution of the homogeneous ODE, denoted y h , satisfies y ( n ) h + a 1 y ( n- 1) h + + a n- 1 y h + a n y h = 0 for all t Do not distribute these notes 64 Prof. R.T. MCloskey, UCLA FACTS: It is possible to find n linearly independent functions of t , denoted y h, 1 , y h, 2 , etc., that satisfy the homoge- neous ODE. Linear independence is established from the non-zero Wronskian: W ( t ) = det y h, 1 y h, 2 y h,n y h, 1 y h, 2 y h,n . . . . . . . . . y ( n- 1) h, 1 y ( n- 1) h, 2 y ( n- 1) h,n 6 = 0 for all t An easy way to generate the linear independent set of functions is to first consider the characteristic equation : n + a 1 n- 1 + + a n- 1 + a n = 0 . The roots of characteristic equation are { 1 , 2 , . . . , n } , and may be distinct, repeated and/or complex-valued. If a root is distinct (it will be real) then we can define one of the functions in the linearly independent set to be e t . If a root is repeated m times then y h,k ( t ) = t k- 1 e t , k = 1 , 2 , . . . , m are m linearly independent functions. If a root is complex valued, say = r + j i , where j = - 1 and r and i represent the real and imaginary parts of the root, respectively, then the complex conjugate of this root is also a root of the characteristic equation due to the fact that we...
View Full Document

Page1 / 20

nth_order - n th Order Linear Systems Do not distribute...

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