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Bode Plots: Understanding Sinusoidal Steady-State Behavior of Dynamical Systems, Study notes of Electrical and Electronics Engineering

An outline for a lecture on bode plots, which are graphical representations of the sinusoidal steady-state behavior of dynamical systems. The lecture covers the relationship between input and output phasors, the transfer function in s-domain analysis, and the significance of bode plots. Examples are included to illustrate the application of bode plots in determining the output of a system given a complex input.

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Pre 2010

Uploaded on 08/19/2009

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Download Bode Plots: Understanding Sinusoidal Steady-State Behavior of Dynamical Systems and more Study notes Electrical and Electronics Engineering in PDF only on Docsity! lecture 18 outline 18-1 Bode Plots Bode plots are widely used as graphical protrayals of the sinusoidal steady-state behavior of dynamical systems. Bode plots show the relationship, in the sinusoidal steady-state, between the input and output phasors. Recall that, in s-domain analysis the input and output are related via the system transfer function. To be specific, consider the input and output to both be voltages (of course, depending on the system, they could be temperatures, currents, or forces). Vo(s) = TF(s) V i(s) The sinusoidal steady-state response is given when ω→s j . Notice when s is replaced by jω, that Vo, TF, and V i all become complex functions of ω. That is, they become phasor quantities. Vo(ω)= TF (ω)Vi(ω) Think of these phasor quantities in polar form. ( ) θ θ θ θ θ θ θ θ ω θ θ θ θ ω ∠ ∠ ∠ ∠ ∠ ∠ ∠ o o TF i i o o o TF o i i i i o i TF o i i V = TF V V V TF = = - V V V TF = (TF is a function of ) V = - ( is a function of ) There are two Bode plots. 1)The Bode magnitude plot gives the magnitude (in decibels) of TF as a function of ω. 2)The Bode phase plot gives θTF as a function of ω. What good are Bode plots? What are they used for? lecture 18 outline 18-2 Example Suppose a system has the frequency response shown below. What is the ouput, if the input is V i = 10 cos (500t – 15°) volts? lecture 18 outline 18-5 Bode magnitude plot ( ) ( ) 10db 10 b 10db bz1 2 z1 2 nz1 nz1 bp1 2 p1 2 np1 np1 = 20 log j =20 log K 20 n-m log j + 20 log + 1 + j 2 j + 20 log + j + 1 + - 20 log + 1 - 2s - 20 log + j + 1 - ω ω ω ω ζ ω ω ω ω ω ζ ω ω ω + TF TF TF L L L L Bode phase plot ( ) ( ) ( ) b bz1 2 z1 2 nz1 nz1 bp1 2 p1 2 np1 np1 j = K n-m j + + 1 + j 2 j + + j + 1 + - + 1 - 2s - + j + 1 - ω θ ω ω ω ζ ω ω ω ω ω ζ ω ω ω   ∠ + ∠ ∠        ∠ ∠           ∠     L L L L Both the magnitude and phase plots are sums of individual terms. Let's consider each of these terms in turn. Kb s lecture 18 outline 18-6 1/s b s + 1 ω b 1 s + 1 ω lecture 18 outline 18-7 2 2 nn s 2 + s + 1ζ ωω 2 2 nn 1 s 2 + s + 1ζ ωω
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