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Continuous Time Signals & Syst

by: Ms. Isobel Rau

Continuous Time Signals & Syst EE 341

Ms. Isobel Rau

GPA 3.67


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This 1 page Class Notes was uploaded by Ms. Isobel Rau on Thursday October 15, 2015. The Class Notes belongs to EE 341 at New Mexico Institute of Mining and Technology taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/223652/ee-341-new-mexico-institute-of-mining-and-technology in Electrical Engineering at New Mexico Institute of Mining and Technology.

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Date Created: 10/15/15
EE 341 Fall 2004 EE 341 Solution of Difference Equations in the Time Domain The difference equation N M yin Zakyln kl waln kl k1 k0 has the solution if M g N yin A1ANKzypnl 1 V 2mm yssin where the Ak s are the roots of the characteristic polynomial of the system AN LlN l aN1 aN 0 For repeated roots use A 71A 712 A etc For example if 1 A2 you would use A1 A AgnVf instead of AM AZVf yp is the particular steadystate solution which depends on the input an yam 06n K6n Cu K Cu K Canuh K04 C cosw0n D sinw0n K1 cosmon 2 siann If the characteristic polynomial has a root at the value of an input exponential eg 1 and an you would use KnAZ for ypn eg yp Kn Solve for the unknowns by nding MO y1 3A2 from the initial conditions until you get as many equations as you have unknowns Solve these equations for the unknowns The transient response of the system is y Any ANV l 39ylz n 39yMN6n 7 M 7 N 7 it is the part of the response which dies out in time The steaajz state response of the system is yss yp 7 it stays around as long as the input drives it The total response is the sum of the two yt39rini 959 You can solve the difference equation for the case 0 subject to the initial conditions of the system This is the natural response or zero input response gymn of the system You can further solve for the case 0 for n lt 0 for the actual input This is the forced response or zero state response 3415M ofthe system The total response is the sum of the two yzi y


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