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Both <math>Y_k</math> yield the same output; therefore, the system is time invariant. | Both <math>Y_k</math> yield the same output; therefore, the system is time invariant. | ||
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| + | == 6B - What input X[n] would yield the output Y[n]=u[n-1]? == | ||
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| + | The system detailed above would output <math>\delta[n-1]</math> given the input of <math>\delta[n]</math>. Therefore, to obtain the output of <math>u[n-1]</math>, the input of <math>u[n]</math> should be sent through the system. You could also input the definition of <math>u[n]</math>, which is just a sum of many shifted delta functions. | ||
Latest revision as of 06:37, 12 September 2008
6A - Is the system time invariant?
System: $ Y_k[n]=(k+1)^2\delta[n-(k+1)] $
Time-delay: $ n-n_0 $
Time-delay, then system:
$ T_k[n]=\delta[(n-n_0)-(k+1)] $
$ Y_k[n]=(k+1)^2\delta[(n-n_0)-(k+1)] $
System, then time-delay:
$ T_k[n]=(k+1)^2\delta[n-k] $
$ Y_k[n]=(k+1)^2\delta[(n-n_0)-(k+1)] $
Both $ Y_k $ yield the same output; therefore, the system is time invariant.
6B - What input X[n] would yield the output Y[n]=u[n-1]?
The system detailed above would output $ \delta[n-1] $ given the input of $ \delta[n] $. Therefore, to obtain the output of $ u[n-1] $, the input of $ u[n] $ should be sent through the system. You could also input the definition of $ u[n] $, which is just a sum of many shifted delta functions.
