"Show that |(1−α)(1+(ejw)−1)2(1−α(ejw)−1)|2=(1−α)2(1+cos⁡(w))2(1+α2−2αcos⁡(w)}So far, I have taken |HLP(ejω)|=1−α21+cos⁡(ω)1−αcos⁡(ω)Squaring..." was answered by our experts

Slade Higgins

Answered question

2022-05-01

Show that

{| \frac{(1 - α) (1 + {(e^{j w})}^{- 1})}{2 (1 - α {(e^{j w})}^{- 1})} |}^{2} = \frac{{(1 - α)}^{2} (1 + \cos (w))}{2 (1 + α^{2} - 2 α \cos (w)}}

So far, I have taken

| H_{L P} (e^{j ω}) | = \frac{1 - α}{2} \frac{1 + \cos (ω)}{1 - α \cos (ω)}

Squaring it subsequently gives

\frac{{(1 - α)}^{2}}{4} \frac{{(1 + \cos (ω))}^{2}}{1 - 2 α \cos (ω) + α^{2} \cos^{2} (ω)}

Simone Ali

Beginner2022-05-02Added 18 answers

You need to use

{| z |}^{2} = z \overset{―}{z}

. To warm up, consider

{| 1 + z^{- 1} |}^{2}

where

z = e^{i t}

with t real. Then

{| 1 + z^{- 1} |}^{2} = {| 1 + e^{- i t} |}^{2} = (1 + e^{- i t}) \overset{―}{(1 + e^{- i t})} = (1 + e^{- i t}) (1 + e^{i t}) = 2 + e^{i t} + e^{- i t}

etc.

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