- #1

Larrytsai

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## Homework Statement

y[n] - (2/3)y[n-1] = x[n]

what is y[n] if x[n] = diracdelta[n]

## The Attempt at a Solution

for some reason, i argued that y[n-1] = diracdelta[n-1]

so

y[n] = diracdelta[n] + (2/3)diracdelta[n-1]

Im pretty sure this is wrong, anybody can help?