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Homework 2 Ben Horst: A  :: B  :: C  :: D  :: E


Given the system

Input      Output
X0[n]=δ[n]  ->   Y0[n]=δ[n-1]
X1[n]=δ[n-1]  ->   Y1[n]=4δ[n-2]
X2[n]=δ[n-2]  ->   Y2[n]=9 δ[n-3]
X3[n]=δ[n-3]  ->   Y3[n]=16 δ[n-4]
...       ...
Xk[n]=δ[n-k]  ->   Yk[n]=(k+1)2 δ[n-(k+1)]

For any non-negative integer k

Test for Time Invariance

Start with X2[n]=δ[n-2]

Delay the signal by 2 => X2[n] = δ[n-4]

Then run the signal through the system:

X2[n] = δ[n-4] -> system -> Y2 = 25δ[n-5]


Repeat in reverse...

Run the signal through the system:

X2[n]=δ[n-2] -> system -> Y2 = 9δ[n-3]

Then delay the signal by 2 => Y2 = 9δ[n-5]


Since 9δ[n-5] ≠ 25δ[n-5], the system is not Time Invariant.


Second Part

Question: Assuming that this system is linear, what input X[n] would yield the output Y[n]=u[n-1]?

X[n] = u[n] would yield Y[n]=u[n-1] since u[n] is simply the sum of shifted delta functions, and linearity dictates that they could be sent through the system (producing δ[n-1]'s) and then summed after the fact to give Y[n]=u[n-1].

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