강의노트 2차지연요소

강의노트 • 조회수 436 • 댓글 0 • 수정 2개월 전  
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2차 지연 요소

G(s)=ω2s2+2aωns+ωn2=1(sωn)2+(2asωn)+1 G(s) = \frac{\omega^2}{s^2 + 2a \omega_n s + \omega^2_n } = \frac{1}{ (\frac{s}{\omega_n})^2 + (2a\frac{s}{\omega_n}) + 1}

G(jω)=11(ωωn)2+j2a(ωωn) G(j\omega) = \dfrac{1}{1-(\dfrac{\omega}{\omega_n})^2+j2a(\dfrac{\omega}{\omega_n})}

G(jω)dB=20log101(1(ωωn)2)2+4a2(ωωn)2=20log10(1(ωωn)2)2+4a2(ωωn)2 \vert G(j\omega) \vert_{dB}= 20 \log _{10} \vert \dfrac{1}{\sqrt{(1-(\dfrac{\omega}{\omega_n})^2)^2 +4a^2(\dfrac{\omega}{\omega_n})^2}} \vert \\ = - 20 \log _{10} \sqrt{(1-(\dfrac{\omega}{\omega_n})^2)^2 +4a^2(\dfrac{\omega}{\omega_n})^2}

이득

ωωnG(jω)dB20log101=0 \omega \ll \omega_n \quad \Rightarrow \quad \vert G(j\omega) \vert _{dB} \approx -20 \log _{10} \sqrt1 = 0

ω=ωnG(jω)dB=20log104a2=20log10(2a)=20log10220log10a=620log10a \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

ωωnG(jω)dB20log10(ωωn)4=40log10ω+40log10(ωn)40log10ω \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

ωωnG(jω)=tan1(2a(ωωn)1(ωωn)2)=0 \omega \ll \omega_n \quad \Rightarrow \quad \angle G(j\omega) = - \tan ^{-1} (\dfrac{2 a (\dfrac{\omega}{\omega_n})}{1-(\dfrac{\omega}{\omega_n})^2}) = 0^\circ

ω=ωnG(jω)=tan1(2a(ωωn)1(ωωn)2)=90 \omega = \omega_n \quad \Rightarrow \quad \angle G(j\omega) =- \tan ^{-1} (\dfrac{2 a (\dfrac{\omega}{\omega_n})}{1-(\dfrac{\omega}{\omega_n})^2}) = -90^\circ

ωωnG(jω)=tan1(2a(ωωn)1(ωωn)2)=180 \omega \gg \omega_n \quad \Rightarrow \quad \angle G(j\omega) = - \tan ^{-1} (\dfrac{2 a (\dfrac{\omega}{\omega_n})}{1-(\dfrac{\omega}{\omega_n})^2}) = -180^\circ

  • ωn=1 \omega_n = 1 그리고 aa는 0.1에서 0.9까지 0.1식 증가한 시스템의 보드선도이다.

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