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<math> (1) X_{k}[n-a] = d[n-k-a] </math><br><br> | <math> (1) X_{k}[n-a] = d[n-k-a] </math><br><br> | ||
<math> (2) Y_{k}[n-a] = (k+1)^2 d[n-(k+1)-a] </math><br><br> | <math> (2) Y_{k}[n-a] = (k+1)^2 d[n-(k+1)-a] </math><br><br> | ||
| − | As shown in (2), <math> (k+1)^2 </math> does not get shifted. Thus, | + | As shown in (2), <math> (k+1)^2 </math> does not get shifted. Thus, a shift made in <math> X_{k}[n] </math> does not accordingly shift <math> Y_{k}[n] </math> <br><br> |
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| + | == Part (b) == | ||
Revision as of 16:20, 11 September 2008
Part (a)
No. This system is not time-invariant. The general equation of the system is as follows.
$ X_{k}[n] = d[n-k] $
$ Y_{k}[n] = (k+1)^2 d[n-(k+1)] $
Shifting $ X_{k}[n] $ by a constant "a" yields $ X_{k}[n-a] $
Shifting $ Y_{k}[n] $ by a constant "a" yields $ Y_{k}[n-a] $
$ (1) X_{k}[n-a] = d[n-k-a] $
$ (2) Y_{k}[n-a] = (k+1)^2 d[n-(k+1)-a] $
As shown in (2), $ (k+1)^2 $ does not get shifted. Thus, a shift made in $ X_{k}[n] $ does not accordingly shift $ Y_{k}[n] $
