What is the particular solution of the difference equation y(n)= 5/6y(n-1)- 1/6y(n-2) x(n) when the forcing function x(n)=2n, n≥0 and zero elsewhere - Study24x7
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30 Mar 2019 11:04 AM study24x7 study24x7

What is the particular solution of the difference equation y(n)= 5/6y(n-1)- 1/6y(n-2)+x(n) when the forcing function x(n)=2n, n≥0 and zero elsewhere

A

(1/5) 2n

B

(5/8) 2n

C

 (8/5) 2n

D

(5/8) 2-n

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  • geet sharma
  • The assumed solution of the difference equation to the forcing equation x(n), called the particular solution of the difference equation is
    yp(n)=Kx(n)=K2nu(n) (where K is a scale factor)
    Upon substituting yp(n) into the difference equation, we obtain
    K2nu(n)=5/6K2n-1u(n-1)-1/6 K2n-2u(n-2)+2nu(n)
    To determine K we must evaluate the above equation for any n>=2, so that no term vanishes.
    => 4K= 5/6(2K)-1/6 (K)+4
    => K= 8/5
    => yp(n)= (8/5) 2n.

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