homoonous solution(discret time signal)

Discussion in 'Homework Help' started by TAKYMOUNIR, Apr 16, 2013.

  1. TAKYMOUNIR

    Thread Starter Active Member

    Jun 23, 2008
    351
    1
    y[n]=y[n-1]+u[n]
    where u[n] is step function
    y[n]=yp[n]+yh[n]
    wher yp[n] is the particular solution
    and yh[n] is the homogenous solution
    in this example yh[n]=k^n so why ? how this come from
    i know in homogenous solution we put input to be zero so u[n]=zero but how yh[n]=k^n
    thanks
     
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