(New page: ==part a== This system can not be time invariant. Pf: Delay then system yields the constant <math>(K+1+n_o)^2</math>, but the system then delay yields a constant <math>(K+1)^2</math> =...) |
(→part b) |
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Since <math>X[n]=\delta[n]</math> yields <math>Y[n]=\delta[n-1]</math>, we can replace the unit impulses with unit steps because the system is linear, | Since <math>X[n]=\delta[n]</math> yields <math>Y[n]=\delta[n-1]</math>, we can replace the unit impulses with unit steps because the system is linear, | ||
| − | so <math> | + | so <math>X[n]=u[n]</math> yields <math>Y[n]=u[n-1]</math> |
Latest revision as of 15:02, 12 September 2008
part a
This system can not be time invariant.
Pf:
Delay then system yields the constant $ (K+1+n_o)^2 $, but the system then delay yields a constant $ (K+1)^2 $
part b
Since $ X[n]=\delta[n] $ yields $ Y[n]=\delta[n-1] $, we can replace the unit impulses with unit steps because the system is linear,
so $ X[n]=u[n] $ yields $ Y[n]=u[n-1] $
