(6B - What input X[n] would yield the output Y[n]=u[n-1]?)
 
Line 14: Line 14:
  
 
== 6B - What input X[n] would yield the output Y[n]=u[n-1]? ==
 
== 6B - What input X[n] would yield the output Y[n]=u[n-1]? ==
 +
 +
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 07:37, 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.

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett