강의노트 1차앞선요소

보드선도 일차앞선요소

$$G(s) = s \tau + 1$$

$$\begin{align*} \vert G (j \omega ) \vert_{dB} &= 20 \log_{10} \vert \sqrt{1^2+ (\omega \tau)^2} \vert \end{align*}$$

$$\begin{align*} \angle G (j \omega ) &= \angle (1 + j \omega \tau) \end{align*}$$

이득

$$\omega \ll \dfrac1\tau \quad \Rightarrow \quad \vert j \omega \tau + 1 \vert_{dB} = 20 \log_{10} \vert j \omega \tau + 1 \vert \approx 20 \log_{10} 1 = 0$$

$$\omega = \dfrac1\tau \quad \Rightarrow \quad \vert j \omega \tau + 1 \vert_{dB} = 20 \log_{10} \vert j + 1 \vert = 20 \log_{10} \sqrt2 = 10 \log_{10} 2 \approx 3dB$$

$$\omega \gg \dfrac1\tau \quad \Rightarrow \quad \vert j \omega \tau + 1 \vert_{dB} = 20 \log_{10} \vert \omega \tau +1\vert \approx 20 \log_{10} \vert \omega \tau \vert = 20 \log_{10} \vert \omega \vert +20 \log_{10} \vert \tau \vert$$

위상각

$$\omega \ll \dfrac1\tau \quad \Rightarrow \quad \angle (j \omega \tau + 1) \approx \angle 1 = 0^\circ$$

$$\omega = \dfrac1\tau \quad \Rightarrow \quad \angle (j \omega \tau + 1) = \angle (j+1) = 45^\circ$$

$$\omega \gg \dfrac1\tau \quad \Rightarrow \quad \angle (j \omega \tau + 1) \approx \angle (j \omega \tau ) = 90^\circ$$

$\tau$ = 1일때의 보드선도