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This is a question paper that I've tried to solved but seems I am missing something.

My attempt:

$$ |G|_{w=0} = K$$

So,

$\require{cancel}|G|_{w=0.5a} = \dfrac{K}{\left|1+\dfrac{j0.5\cdot \cancel{a}}{\cancel{a}}\right|} = \dfrac{K}{\underbrace{|1+j0.5|}_{Gain}}$

$Gain = 20 \log(\sqrt{1,25}) = 0.96910013008$ [OK]

But about the phase error?

$\angle(1+j0.5) = \tan^{-1}(0.5/1) = 26.56º$

EDIT: Sorry picture this question:

w

What I did wrong?

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  • $\begingroup$ I get the same numbers you do. The question seems ill-posed, in that "error" is not defined (error from what?), and the gain is relative to K, not absolute. Best to ask your prof, find out what the thinking is. $\endgroup$ – TimWescott Nov 21 '19 at 20:02
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The "Error" would be the actual phase and amplitude relative to the asymptotic plot shown. You computed the gain and phase, so now compare this to the asymptotic lines given by the plot to determine the error.

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  • $\begingroup$ Oww. Nice hint man. But if u do right the phase on a is 45o (1 decade to 0.1a) at 0.5a the phase is 22.5o. So the error is 26.56 - 22.5 = 4.06. It is still away the answers options $\endgroup$ – miguel747 Nov 21 '19 at 23:04
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    $\begingroup$ i got this question finally! I do wrong the first place due to monolog scale on frequency graph. Look again the problem I compute: $\dfrac{\log a-\log 0.1a}{45} = \dfrac{\log 0.5a - \log 0.1a}{X}\Rightarrow X = 31.45$ $\endgroup$ – miguel747 Nov 22 '19 at 2:24
  • $\begingroup$ Good work! Glad you figured it out. $\endgroup$ – Dan Boschen Nov 22 '19 at 15:35

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