(New page: ==Time Invariance== For a given input <math>X_i[n]=\delta[n-(i-1)]</math> it yields the output <math>Y_i[n]=(k+1)^2 \delta[n-(k+1)]</math> is NOT time invariant. Say we shifted the input...) |
(→Linearity) |
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==Linearity== | ==Linearity== | ||
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| + | Assuming the system is linear, <math>\mu[n]</math> should yield the output <math> \mu[n-1] </math> based on the response to <math>\delta[n]</math>. Since the input of a function involving <math>\delta[n]</math> yields a time shifted delta function, a step function input should yield a time shifted step function for output. | ||
Latest revision as of 16:11, 11 September 2008
Time Invariance
For a given input $ X_i[n]=\delta[n-(i-1)] $ it yields the output $ Y_i[n]=(k+1)^2 \delta[n-(k+1)] $ is NOT time invariant.
Say we shifted the input at $ X_3[n] $ by +3, it would yield the output of $ Y[n]=\delta[n-1] $
However if we look at the output of $ X_3[n] $ and then shift it by 3, we get $ Y[n] = 16*\delta[n-4+3] = 16*\delta[n-1] $
Which is clearly 16 * the other input and thus they are not equal, so the system is not time invariant.
Linearity
Assuming the system is linear, $ \mu[n] $ should yield the output $ \mu[n-1] $ based on the response to $ \delta[n] $. Since the input of a function involving $ \delta[n] $ yields a time shifted delta function, a step function input should yield a time shifted step function for output.
