Nunx O H
(a) Since H is an integrator, H1 must be a differentiator H~' y(t) dx(t) = d dt G y(t) = x(2t) G 1 y(t) = x(t/2) (b) Solutions to Optional Problems S38 (a) x 2(t) = xi(t) xi(t 2) 2(t) = 1(t) y)(t.
Nunx o h. (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. 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. 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.
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