# discrete time signal(particular and homogenous solution

Discussion in 'Homework Help' started by TAKYMOUNIR, Apr 14, 2013.

1. ### TAKYMOUNIR Thread Starter Active Member

Jun 23, 2008
351
1
can someone help me to know how to find particular solution and homogenous solution for discrete time signal ,if you know links for that subject ,i did search on google but i can not find something
thanks

2. ### blah2222 Well-Known Member

May 3, 2010
565
33
Are you referring to difference equations, which are analogous to differential equations for continuous-time signals?

Tons of stuff if you know how to use Google.

Is there a particular example that you could work through and post on here? That might be more useful.

JP

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3. ### TAKYMOUNIR Thread Starter Active Member

Jun 23, 2008
351
1
like y[n]=y[n-1]+y[n-2]+x[n]
where x[n]=n^2*u[n]
u[n] is unit step function

4. ### blah2222 Well-Known Member

May 3, 2010
565
33
Okay, are you given any initial conditions of y[n] for n < 0? If not, assume y[-2] = 0 and y[-1] = 0 then go through each value of y[n] for n.

y[0] = y[-1] + y[-2] + x[0] = 0 + 0 + 0^2*u[0] = 0 + 0 + 0*1 = 0
y[1] = ...

Do this until the values converge and stop changing.

In the example that you have given, if there are no initial conditions your function will be a zero function for all n.

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5. ### TAKYMOUNIR Thread Starter Active Member

Jun 23, 2008
351
1
y[n]=yP[n]+yh[n]
where yp[n] is the particular solution and yh[n] is the homogenous solution so i find this answer yp and yh
yh=k^n
yp=m*u[n]