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Unformatted text preview: n th Order Linear Systems Do not distribute these notes 63 Prof. R.T. M’Closkey, UCLA Introduction to n th order ODEs Consider the n th order linear timeinvariant 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 (timeinvariance) so we may take the “starting time" of any initial value problem to be t = 0 (we haven’t 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 righthandside 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. M’Closkey, 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 nonzero 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 : λ n + a 1 λ n 1 + ··· + a n 1 λ + a n = 0 . The roots of characteristic equation are { λ 1 , λ 2 , . . . , λ n } , and may be distinct, repeated and/or complexvalued. If a root ˜ λ is distinct (it will be real) then we can define one of the functions in the linearly independent set to be e ˜ λt . If a root ˜ λ is repeated m times then y h,k ( t ) = t k 1 e ˜ λt , k = 1 , 2 , . . . , m are m linearly independent functions. If a root is complex valued, say ˜ λ = λ r + jλ i , where j = √ 1 and λ r and λ 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...
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 Spring '06
 TSAO
 Prof. R.T. M’Closkey

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