Find the error in phase angle and dB gain Function Transfer

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]

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

EDIT: Sorry picture this question:

What I did wrong?

• 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. – TimWescott Nov 21 '19 at 20:02

• i got this question finally! I do wrong the first place due to monolog scale on frequency graph. Look again the problem I compute: $\dfrac{\log a-\log 0.1a}{45} = \dfrac{\log 0.5a - \log 0.1a}{X}\Rightarrow X = 31.45$ – miguel747 Nov 22 '19 at 2:24