Difference between revisions of "HW2.E Jeff Kubascik ECE301Fall2008mboutin" - Rhea

Latest revision as of 20:22, 11 September 2008

Part A: Can the system be time invariant?

The system cannot be time invariant.

For instance, the input

$ \,X_0[n]=\delta [n]\, $

yields the output

$ \,Y_0[n]=\delta [n-1]\, $

Thus,

$ \,Y_0[n-1]=\delta [n-2]\, $

However, the input

$ \,X_0[n-1]=\delta [n-1]=X_1[n]\, $

yields the output

$ \,Y_1[n]=4\delta[n-2]\, $

Since these two are not equal

$ \,\delta [n-2]\not= 4\delta[n-2]\, $

the system is time variant (by not fitting the definition of time invariance).

Part B: Find input given output

The given output is:

$ \,Y[n]=u[n-1]\, $

This can be re-written as:

$ \,Y[n]=\sum_{k=0}^{\infty}\delta [n-(k+1)]\, $

$ \,Y[n]=\delta [n-1]+\delta [n-2]+\delta [n-3]+\ldots +\delta [n-(k+1)]\, $

$ \,Y[n]=Y_0[n]+\frac{1}{4}Y_1[n]+\frac{1}{9}Y_2[n]+\ldots +\frac{1}{(k+1)^2}Y_k\, $

Because the system is assumed to be linear, we can write the input as

$ \,X[n]=X_0[n]+\frac{1}{4}X_1[n]+\frac{1}{9}X_2[n]+\ldots +\frac{1}{(k+1)^2}X_k\, $

$ \,X[n]=\delta [n]+\frac{1}{4}\delta [n-1]+\frac{1}{9}\delta [n-2]+\ldots +\frac{1}{(k+1)^2}\delta [n-k]\, $

$ \,X[n]=\sum_{k=0}^{\infty}\frac{1}{(k+1)^2}\delta[n-k]\, $

Therefore, the input is:

$ \,X[n]=\frac{1}{(n+1)^2}u[n]\, $ when $ \,n\not= -1\, $

$ \,X[n]=0\, $ otherwise