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==Time Invariance== | ==Time Invariance== | ||
| − | For a fixed value of <math>k</math> input to the system <math>Y_k[n]=(k+1)^2\delta[n-(k+1)]</math>, the system is time invariant. As <math>n</math> goes through a time shift, the output will shift through time an equal amount. Since <math>k</math> is assumed to be fixed, the output amplitude does not change with time. | + | For a fixed value of <math>k</math> input to the system <math>Y_k[n]=(k+1)^2\delta[n-(k+1)]</math>, the system is time invariant. As <math>n</math> goes through a time shift, the output will shift through time an equal amount. Since <math>k</math> is assumed to be fixed, the output amplitude does not change with time except as prescribed by the <math>\delta</math> functional. |
==Linearity== | ==Linearity== | ||
Since the system is linear, the required input will be <math>X[n]=u[n]</math>. An input of a <math>\delta</math> functional produces an output of a time-shifted <math>\delta</math> functional, so an input of a unit step function will produce an output of a time-shifted unit-step function. | Since the system is linear, the required input will be <math>X[n]=u[n]</math>. An input of a <math>\delta</math> functional produces an output of a time-shifted <math>\delta</math> functional, so an input of a unit step function will produce an output of a time-shifted unit-step function. | ||
Revision as of 08:53, 9 September 2008
Time Invariance
For a fixed value of $ k $ input to the system $ Y_k[n]=(k+1)^2\delta[n-(k+1)] $, the system is time invariant. As $ n $ goes through a time shift, the output will shift through time an equal amount. Since $ k $ is assumed to be fixed, the output amplitude does not change with time except as prescribed by the $ \delta $ functional.
Linearity
Since the system is linear, the required input will be $ X[n]=u[n] $. An input of a $ \delta $ functional produces an output of a time-shifted $ \delta $ functional, so an input of a unit step function will produce an output of a time-shifted unit-step function.
