Lecture 보드선도 _ 1차앞선요소

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보드선도 일차앞선요소

G(s)=sτ+1 G(s) = s \tau + 1

이득

ω1τjωτ+1dB=20log10jωτ+120log101=0 \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

ω=1τjωτ+1dB=20log10j+1=20log102=10log1023dB \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

ω1τjωτ+1dB=20log10ωτ+120log10ωτ=20log10ω+20log10τ \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

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

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

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

  • τ\tau = 1일때의 보드선도

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