14diagrammi_risposta_in_frequenza_14

# Log10 j db 1 j 90 j 90 90 80 60 0 40 db gradi 20 1 0

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Unformatted text preview: &lt;0 -180 -80 0 10 10 1 ω 10 2 10 3 10 4 ( rad/s) -270 0 10 1 10 2 10 ω (rad/s) 3 10 4 10 G ( j ω ) = ( jω ) − g ω 1 = − jω dB = −20 log10 ω jω dB ω ω − 1 ∠ = −∠jω = −90° jω 90 90 80 60 0 40 dB gradi 20 −1 0 -90 -20 g=1 -40 -180 -60 -80 -2 10 -270 0 10 -1 10 0 10 ω (rad/s) 1 10 10 2 1 10 2 10 ω (rad/s) 3 10 4 10 G ( j ω ) = ( jω ) − g ω) = − ( jω ) 1 ( jω ) g g ∠ dB ω = −20 g log10 ω dB 1 ( jω ) g = − g∠jω = − g ⋅ 90° − (1 + Tk jω ) (1 + τ k jω ) τ 1 1 + jωτ ∠ = − 1 + jωτ dB = −20 log 1 + jωτ = −20 log 1 + ω 2τ 2 dB 1 = −atan (ωτ ) (1 + jωτ ) ω &lt;&lt; 1 → −20 log 1 + ω 2τ 2 ≅ 0 τ 1 ω &gt;&gt; → −20 log 1 + ω 2τ 2 τ ≅ −20 log ω 2τ 2 = −20 log ωτ = −20 log ω + 20 log 1 τ l’approssimazione asintotica del diagramma di Bode del modulo è pertanto costituita dalle due semirette (asintoti): dB 1 0 per ω ≤ τ = 1 − 20 log ω + 20 log τ per ω ≥ 1 τ ω τ τ|| Diagramma di Bode - Modulo 0 diagramma asintotico -5 diagramma effettivo ω= -10 -1 1 τ 20 log10 2 = 3dB -15 dB -20 ω= -25 -30 -35 -40 -45 -2 10 -1 10 0 10 pulsazione 1 10 2 10 1 τ ∠ 1 = −atan (ωτ ) (1 + jωτ ) ω &lt;&lt; ω &lt;&lt; 1 τ → −∠(1 + jωτ ) ≅ 0° − 90° τ &gt; 0 → −∠(1 + jωτ ) = τ + 90° τ &lt; 0 1 1 ω= τ − 45° τ &gt; 0 atan (ωT ) = + 45° τ &lt; 0 l’approssimazione asintotica del diagramma di Bode della fase è pertanto costituita dalle due semirette...
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