EE3TP4_3_DifferentialEquationsReview_Lecture 8

EE3TP4_3_DifferentialEquationsReview_Lecture 8 -...

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Unformatted text preview: Differential Equations Review Differential Equations like the following are Linear and Time Invariant: a n d n y t dt n a n 1 d n 1 y t dt n 1 ... a y t = b m d m f t dt m ... b 1 df t dt b f t - coefficients are constants TI- No nonlinear terms Linear Can always ensure that a n = 1 f(t) y(t) LTI System These overheads were originally developed by Mark Fowler at Binghamton University, State University of New York. In the following we review the basics of solving Linear, Constant Coefficient Differential Equations under the Homogeneous Condition. Homogeneous means the forcing function is zero. Solutions are called complementary functions. (In the general case the full solution is the sum of the homogeneous solution and the particular solution found using either the method of undetermined coefficients or by variation of parameters method.) That means we are finding the zero-input response that occurs due to the effect of the initial conditions. Write D.E. like this: D n a n 1 D n 1 ... a 1 D a = Q D y t = b m D m ... b 1 D b = P D f t Diff . Eq . Q D y t = P D f t m is the highest-order derivative on the input side n is the highest-order derivative on the output side We will assume: m n d k y t dt k D k y t Use operational notation: Due to linearity: Total Response = Zero-Input Response + Zero-State Response Z-I Response : found assuming the input f ( t ) = 0 but with given ICs Z-S Response : found assuming ICs = 0 but with given f ( t ) applied D . E .: Q D y zi t = D n a n 1 D n 1 ... a 1 D a y zi t = t () Consider y t = ce t c and are possibly complex numbers linear combination of y zi ( t ) & its derivatives must be = 0 Can we find c and such that y 0 ( t ) qualifies as a homogeneous solution?...
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EE3TP4_3_DifferentialEquationsReview_Lecture 8 -...

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