1.27 (a,b,c,f) Old Kiwi - Rhea

Determine if the following are:

1. Memoryless
2. Time Invariant
3. Linear
4. Causal
5. Stable

A)

1. y(t)=x(t-2) + x(2-t)
   1. Let x3=ax1(t) + bx2(t)
   2. then the output of the function is y(t)=x3(t-2) + x(2-t) = (ax1(t-2) + bx2(t-s)) + (ax1(2-t) + bx2(2-t))
   3. Therefore the signal is Linear.

1. 1. x(t-T)-->S-->y(t)=x(t-T-2) + x(2-t-T) = x(t-T)
   2. Therefore the signal is Time Invariant because the output will be shifted by the same amount that the input was shifted by.

1. 1. Assuming that x(t) is bounded then the output y(t) is also bound because it is the sum of two bound functions.

1. 1. Since there is a time shift in both the positive and negative direction, the function is neither memoryless or causal.

1. 1. This function is Linear, Time Invariant, and Stable.

B)

1. y(t)=[cos(3t)]x(t)
   1. Since there is no time shift in the output function it is both memoryless and causal.

1. 1. Let x3=ax1(t) + bx2(t)-->S-->y(t)=cos(3t)x3(t)=cos(3t)[ax1(t) + bx2(t)]=ay(t) + by(t)
   2. Therefore the function is linear.

1. 1. Let x2(t)=x1(t-T)-->S-->y(t)=cos(3t)x2(t)=cos(3t)x1(t-T)
   2. This is not equal to y(t-T), therefore the function is not Time Invariant.

1. 1. Assuming that x(t) is bound, the function y(t) is also bound since it is the multiple of two bound functions.

1. 1. The function is Memoryless, Causal, Linear, Stable.

C)

1. y(t)=$ /int-/infty^2T x(/tau)d/tau $
   1. Let x3(t)=ax1(t) + bx2(t)-->S-->$ /int-/infty^2T x3(/tau)d/tau $=$ /int-/infty^2T ax1(/tau)d/tau $ + $ /int-/infty^2T bx2(/tau)d/tau $
   2. Therefore the function is linear.