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The system does a phase shift to the right by 1 unit and then multiplies the amplitude of the function by the square of total shift. | The system does a phase shift to the right by 1 unit and then multiplies the amplitude of the function by the square of total shift. | ||
| − | if Y[n] = u[n-1] | + | if Y[n] = u[n-1] |
| − | <math>delta[n-(k+1)]</math> | + | then in the function |
| + | <math>\delta[n-(k+1)]</math> | ||
| + | k must be equal to 0 | ||
| + | |||
| + | we then multiply the amplitude by the square of the total shift (-1), which has no effect because it is the same as multiplying by 1. | ||
| + | |||
| + | We can finally solve and say that in order to produce Y[n], the input must be <math>x(t) = u(t)</math> | ||
Revision as of 14:04, 10 September 2008
Part A
Can the system be time invariant?
Part B
The system does a phase shift to the right by 1 unit and then multiplies the amplitude of the function by the square of total shift.
if Y[n] = u[n-1]
then in the function $ \delta[n-(k+1)] $ k must be equal to 0
we then multiply the amplitude by the square of the total shift (-1), which has no effect because it is the same as multiplying by 1.
We can finally solve and say that in order to produce Y[n], the input must be $ x(t) = u(t) $
