(New page: == Part A == Yes the system is time invariant. <math>x[n] \to (system) \to y[n] = (k+1)^2 x[n] \to (time shift) \to z[n] = y[n-(k+1)] = (k+1)^2 x[n-(k+1)]</math> and then switch the syst...) |
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== Part A == | == Part A == | ||
Yes the system is time invariant. | Yes the system is time invariant. | ||
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<math>x[n] \to (system) \to y[n] = (k+1)^2 x[n] \to (time shift) \to z[n] = y[n-(k+1)] = (k+1)^2 x[n-(k+1)]</math> | <math>x[n] \to (system) \to y[n] = (k+1)^2 x[n] \to (time shift) \to z[n] = y[n-(k+1)] = (k+1)^2 x[n-(k+1)]</math> | ||
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| + | Then switch the system and time shift and compare to see if they are equal | ||
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<math>x[n] \to (time shift) \to y[n] = x[n-(k+1)] \to (system) \to w[n] = (k+1)^2 y[n] = (k+1)^2 x[n-(k+1)]</math> | <math>x[n] \to (time shift) \to y[n] = x[n-(k+1)] \to (system) \to w[n] = (k+1)^2 y[n] = (k+1)^2 x[n-(k+1)]</math> | ||
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== Part B == | == Part B == | ||
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| + | To get the output of Y[n]=u[n-1] from and input of X[n], X[n] = u[n]. | ||
Latest revision as of 06:42, 12 September 2008
Part A
Yes the system is time invariant.
$ x[n] \to (system) \to y[n] = (k+1)^2 x[n] \to (time shift) \to z[n] = y[n-(k+1)] = (k+1)^2 x[n-(k+1)] $
Then switch the system and time shift and compare to see if they are equal
$ x[n] \to (time shift) \to y[n] = x[n-(k+1)] \to (system) \to w[n] = (k+1)^2 y[n] = (k+1)^2 x[n-(k+1)] $
Since z[n] = w[n] the signal is time invariant.
Part B
To get the output of Y[n]=u[n-1] from and input of X[n], X[n] = u[n].
