Timeline for Understanding log-log axes
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 27, 2018 at 1:10 | comment | added | AgentS | This is of form $$\color{blue}{y} = \dfrac{1}{2}\color{red}{x}$$ shouldn't we be letting \$\color{red}{x=\log((\omega/2)^2+1)}\$ instead ? | |
| Jun 27, 2018 at 1:08 | comment | added | AgentS | I have a small question, why are they using \$\log \dfrac{\omega}{\|z_1\|}\$ on x axis ? Because, for the magnitude we have $$\|H(j\omega)\| = \sqrt{\omega^2+4}=2\sqrt{(\omega/2)^2+1}$$ Taking log both sides and rearranging gives $$\color{blue}{\log \dfrac{\|H(j\omega)\|}{2}} = \dfrac{1}{2}\color{red}{\log((\omega/2)^2+1)}$$ | |
| Jun 26, 2018 at 16:43 | comment | added | AgentS | Ohk... right side graph is only for \$\omega \gt 0\$, then I guess it makes sense... when \$\omega = |z_1|10^{-2}\$ etc the x values get negative, but the y value stays around \$0\$ because \$\log \dfrac{\|H(j\omega)\|}{\|z_1\|} \approx \log \dfrac{\|z_1\|}{\|z_1\|} = 0\$. I think I get this. Thank you so much :) | |
| Jun 26, 2018 at 16:12 | history | answered | The Photon | CC BY-SA 4.0 |