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| − | <math>\delta[n - k] \to sys \to (k + 1)^2 \delta[n - (k + 1)] \to timedelay \to (k + 1)^2 | + | <math>\delta[n - k] \to sys \to (k + 1)^2 \delta[n - (k + 1)] \to timedelay \to (k + 1)^2 = (k + 1)^2 \delta[n -(k + 1 +t_0)] </math> |
| Line 23: | Line 23: | ||
'''6.b)''' | '''6.b)''' | ||
| + | |||
| + | Assumed that the system given is linear, then '''X[n]=u[n]''' is the input to yield the output''' Y[n] = u[n-1]. | ||
| + | ''' | ||
| + | since u[n] is simply the summation of shifted delta functions we can say that | ||
| + | |||
| + | <math> X_k[n]= \ u[n] = \delta[n]- \delta[n - N]</math> | ||
| + | |||
| + | substituting k = 0 and N = 1 in the above equation will give the output Y'''[n] =u[n − 1]''' after the input X[n] goes through the system. | ||
Latest revision as of 14:34, 12 September 2008
Linearity and Time Invariance
6.a)
the system is defined as
$ X_k[n] = \delta[n - k] \to sys \to Y_k[n] = (k + 1)^2 \delta[n - (k + 1)] $
let us check for time invariance
System followed by time delay
now,let us apply a time-delay of $ t_0 $ to the system.
$ \delta[n - k] \to sys \to (k + 1)^2 \delta[n - (k + 1)] \to timedelay \to (k + 1)^2 = (k + 1)^2 \delta[n -(k + 1 +t_0)] $
Time-delay followed by system:
$ \delta[n - k] \to timedelay \to \delta[n-(k + t_0)] \to sys \to (k + t_0 + 1)^2 \delta[n - (k + t_0 + 1)] $
For the system to be time invariant both the outputs should be same but they are not. so the system is not time variant it is rather a time variant system as the output varies with time.
6.b)
Assumed that the system given is linear, then X[n]=u[n] is the input to yield the output Y[n] = u[n-1]. since u[n] is simply the summation of shifted delta functions we can say that
$ X_k[n]= \ u[n] = \delta[n]- \delta[n - N] $
substituting k = 0 and N = 1 in the above equation will give the output Y[n] =u[n − 1] after the input X[n] goes through the system.
