Lecture 2차앞선요소

2차 앞선요소

$$G(s) = (\dfrac{s}{\omega_n})^2 + (2a\dfrac{s}{\omega_n}) + 1 = \dfrac{s^2 + 2a\omega_ns+\omega^2_n}{\omega^2_n}$$

$$G(j\omega) = \dfrac{(j\omega)^2 + 2a\omega_n(j\omega)+\omega^2_n}{\omega^2_n} = 1-(\dfrac{\omega}{\omega_n})^2+j2a(\dfrac{\omega}{\omega_n})$$

$$\vert G(j\omega) \vert _{dB}= 20 \log _{10} \sqrt{(1-(\frac{\omega}{\omega_n})^2)^2+4a^2(\dfrac{\omega}{\omega_n})^2}$$

이득

$$\omega \ll \omega_n \quad \Rightarrow \quad \vert G(j\omega) \vert _{dB} \approx 20 \log _{10} \sqrt1 = 0$$

$$\omega = \omega_n \quad \Rightarrow \quad \vert G(j\omega) \vert _{dB} = 20 \log _{10} \sqrt{4a^2} = 20 \log _{10}(2a) = 20 \log _{10} 2 + 20 \log _{10} a = 6 +20 \log _{10} a$$

$$\omega \gg \omega_n \quad \Rightarrow \quad \vert G(j\omega) \vert _{dB} \approx 20 \log _{10} \sqrt{(\frac{\omega}{\omega_n})^4} = 40 \log _{10} \omega -40 \log _{10} (\omega_n) \approx 40 \log _{10} \omega$$

위상

$$\omega \ll \omega_n \quad \Rightarrow \quad \angle G(j\omega) = \tan ^{-1} (\frac{2 a (\frac{\omega}{\omega_n})}{1-(\frac{\omega}{\omega_n})^2}) = 0^\circ$$

$$\omega = \omega_n \quad \Rightarrow \quad \angle G(j\omega) = \tan ^{-1} (\frac{2 a (\frac{\omega}{\omega_n})}{1-(\frac{\omega}{\omega_n})^2}) = 90^\circ$$

$$\omega \gg \omega_n \quad \Rightarrow \quad \angle G(j\omega) = \tan ^{-1} (\frac{2 a (\frac{\omega}{\omega_n})}{1-(\frac{\omega}{\omega_n})^2}) = 180^\circ$$

• $\omega_n = 1$ 그리고 $a$는 0.1에서 0.9까지 0.1식 증가한 시스템의 보드선도이다.