Linearity and Time Variance

From the given equation:

$ \,\!Y_k[n]=(k+1)^2\delta [n-(k+1)] $ ,

It is clear that this is a non-time-invariant, or more simply, a time variant function because the amplitude changing $ (k+1)^2 $ term.

The inputs to this system are all shifted $ \delta $ functions, this means that if the $ X[n] $ inputs were changed to unit step functions $ u[n] $ , then the output will be a time shifted step function. For this specific example we want an output of $ Y[n]=u[n-1] $ , so we can notice that when :

$ \,\!X_0[n]=\delta [n] $ yields $ \,\!Y_0[n]=\delta [n-1] $ , then


$ \,\!X_0[n]=u[n] $ will yield $ \,\!Y_0[n]=u[n-1] $

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett