Lecture 1차지연요소

Lecture • Views 458 • Comments 0 • Last Updated at 2 months ago  
  • 요소별 보드선도
  • 주파수응답

보드선도 일차지연요소

G(s)=1sτ+1 G(s) = \dfrac{1}{s \tau + 1}

G(jω)dB=20log1012+(ωτ)21 \begin{align*} \vert G (j \omega ) \vert_{dB} &= 20 \log_{10} \vert \sqrt{1^2+ (\omega \tau)^2} \vert^{-1} \end{align*}

G(jω)=0(1+jωτ) \begin{align*} \angle G (j \omega ) &= 0^\circ - \angle (1 + j \omega \tau) \end{align*}

이득

ω1τ1jωτ+1dB=20log10jωτ+120log101=0 \omega \ll \dfrac1\tau \quad \Rightarrow \quad \vert \dfrac{1}{j \omega \tau + 1} \vert_{dB} = -20 \log_{10} \vert j \omega \tau + 1 \vert \approx -20 \log_{10} 1 = 0

ω=1τ1jωτ+1dB=20log10j+1=20log102=10log1023dB \omega = \dfrac1\tau \quad \Rightarrow \quad \vert \dfrac{1}{j \omega \tau + 1} \vert_{dB} = -20 \log_{10} \vert j + 1 \vert = -20 \log_{10} \sqrt2 = -10 \log_{10} 2 \approx -3dB

ω1τ1jωτ+1dB=20log10ωτ+120log10ωτ=20log10ω20log10τ \omega \gg \dfrac1\tau \quad \Rightarrow \quad \vert \dfrac{1}{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

ω1τ(1jωτ+1)1=0 \omega \ll \dfrac1\tau \quad \Rightarrow \quad \angle (\dfrac{1}{j \omega \tau + 1}) \approx -\angle 1 = 0^\circ

ω=1τ(1jωτ+1)=(j+1)=45 \omega = \dfrac1\tau \quad \Rightarrow \quad \angle (\dfrac{1}{j \omega \tau + 1}) = -\angle (j+1) = -45^\circ

ω1τ(1jωτ+1)(jωτ)=90 \omega \gg \dfrac1\tau \quad \Rightarrow \quad \angle (\dfrac{1}{j \omega \tau + 1}) \approx -\angle (j \omega \tau ) = -90^\circ

τ\tau = 1 일때의 보드 선도

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