Inverse Transforms-Digital Signal Processing-Assignment Solution, Exercises of Digital Signal Processing

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465 11.1. Using the fact that x,[n] is the inverse transform of Re we get Re{X(e”)} 2-ae™ — ae” u 28[n} — af{n + 1] — ad{n - 3) zeln} Since z[n} is causal, we can recover it from z,{n] = 2x4{nJufn] — z¢[0|é[n] = 24{n] — 2odjn ~ 1} This implies that Ze{n] = ael= ail = ad[n +1] - ad[n ~ 1] and since j7m{X(e)} is the transform of z,[n] we find Im{X (e**)} = asinw 11.2, Taking the inverse transform of Re{X(ei)} = 5/4 —cosw, we get zefn] = Fain) ~ fin +1) - Féln—1] Since 2{n] is causal, we can recover it from 2,{n] s(n] = 2re{nluln ~ 2-[olstn] = Safe] - ale — 1, 11.3. Note that Ix(eyP 8 cosu 4 (-Je*)(-H4) X (eM )X*(e) if X(e") = (1 ~ $e7™) we get af] = én] — 35f0 — 1) but this does not satisfy the conditions on z[n] given in the problem statement. However, if we let X(e%) = (1— fe7J¥)e“™ we get =4[n-1]- pain ~ 2] which satisfies all the constraints. ‘The idea behind this choice is that cascading a signal with an allpass system does not change the magnitude squared response. Another choice that works is X(e%) = $(1 — 2e7*)e7#* for which we get s[n] = dan ~I- gn — 2 The idea behind this choice was to flip the zero to its reciprocal location outside the unit circle. This has the same magnitude squared response up to 2 scaling factor; hence, the } term.