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<math> Y_{k}[n] = (k+1)^2 d[n-(k+1)] </math><br><br> | <math> Y_{k}[n] = (k+1)^2 d[n-(k+1)] </math><br><br> | ||
Shifting <math> X_{k}[n] </math> by a constant "a" yields <math> X_{k}[n-a] </math><br><br> | Shifting <math> X_{k}[n] </math> by a constant "a" yields <math> X_{k}[n-a] </math><br><br> | ||
| + | Shifting <math> Y_{k}[n] </math> by a constant "a" yields <math> Y_{k}[n-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> | ||
| + | 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><br> | ||
| + | == Part (b) == | ||
| + | <br><br> | ||
| + | <math> X[n] = u[n] </math><br> This would yield the expected result. <br> | ||
| + | Since the system is linear, the input u[n] would result in u[n-1] as d[n] resulted in d[n-1] | ||
Latest revision as of 16:51, 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] $
Part (b)
$ X[n] = u[n] $
This would yield the expected result.
Since the system is linear, the input u[n] would result in u[n-1] as d[n] resulted in d[n-1]
