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(a) Let x(n) = u(n) Find ya (n) by first convolving x(n) with h1 (n) and then convolving the result with h2 (n) ie ya(n) = x(n) * h1 (n) * h2 (n) (b) Again let x(n) = u(n) Find yb(n) by convolving x(n) with the result of the convolution of h1 (n) and h2 (n) ie yb(n) = x(n) * h1(n) * h2(n) Your results for parts (a) and (b) should be identical, illustrating.

Nunx o h. #title #points 687 #rows 1097 #sense 1 #xorigin 739 #yorigin #rotation 0 #ptseparation 005 #rwseparation 005 #transform #unit_length km,1000 #map_projection "nad27 / *lcc90" nad27,,,0. EC52 3 The zTransform definition (cont) So we may write hnr−n = 1 2π Z π −π H(rejΩ)ejΩndΩ •z = rejΩ →dz = jrejΩdΩ dΩ = 1 j z −1dz •As Ω goes from −π to π, z traverses a circle of radius r in a. See the answer Show transcribed image text Expert Answer 100% (1 rating).

The answer to this (and your other similar question) is most likely found in the somewhat unusual notation used for the system In ordinary terms, if you just take your system as a map between input and output signal, then you're absolutely right, it would not be time invariant.

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