ECE108HW%233Sol_06

ECE108HW%233Sol_06 - ECE 108 HW#3 - Due 2/13/06 1. For a...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE 108 HW#3 --- Due 2/13/06 1. For a LTI system described by the differential equation () () () 2 2 35 2 43 2 dyt d yt y tx t dt dt ++= where ( ) ( ) xt tut = (a) Determine the Characteristic Equation 2 0 2 pp ++ = (b) Determine the Homogeneous Solution Roots of Characteristic Equation 1.2 31 9 2 0 31 821 63 2 88 pj =− ± Then the most general solution to the Homogeneous Equation is 12 31 88 121 2 j tj t pt bt y t Ae  −+ −−   =+= + (c) Using the Method Of Undetermined Coefficients, determine the Particular Solution Assume we want to find the solution for t > 0, then over that range of values and 1 dx t xt t dt == Further derivatives produce no new independent functions and so the assumed solution should be ( ) 01 p yt B B t =+ Substituting this assumed value for y ( t ) in the differential equation along with the value for x ( t ) we obtain the equation 10 1 2 2 BB B t t = Then 1 5 20 2 3 2 B t ++− = This is only possible for all values of t if 1 0 2 5 32 B += −=
Background image of page 1

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

View Full DocumentRight Arrow Icon
2 Solving these two equations for the constants 1 2 64 5 1536 25 B B = =− Then () 1536 64 25 5 p yt t + (d) Assuming the system is "initially at rest", determine the Complete Solution The complete solution is ( ) ( ) ( )
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/24/2009 for the course ECE 108 taught by Professor Li during the Spring '08 term at Lehigh University .

Page1 / 5

ECE108HW%233Sol_06 - ECE 108 HW#3 - Due 2/13/06 1. For a...

This preview shows document pages 1 - 3. Sign up to view the full document.

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