Bode PlotsSinusoidal Frequency AnalysisSlide 3Slide 4Slide 5Slide 6Bode Plot SummarySingle Pole & Zero Bode PlotsBode Plot RefinementsBode Plots to Transfer FunctionClass ExamplesLect21 EEE 202 1Bode PlotsDr. HolbertApril 16, 2008Lect21 EEE 202 2Sinusoidal Frequency Analysis•The transfer function is composed of both magnitude and phase information as a function of frequencywhere |H(jω)| is the magnitude and (ω) is the phase angle•Plots of the magnitude and phase characteristics are used to fully describe the frequency response)()()(jejHj HLect21 EEE 202 3Bode Plots•A Bode plot is a (semilog) plot of the transfer function magnitude and phase angle as a function of frequency•The gain magnitude is many times expressed in terms of decibels (dB)dB = 20 log10 Awhere A is the amplitude or gain–a decade is defined as any 10-to-1 frequency range–an octave is any 2-to-1 frequency range20 dB/decade = 6 dB/octaveLect21 EEE 202 4Bode Plots•Straight-line approximations of the Bode plot may be drawn quickly from knowing the poles and zeros–response approaches a minimum near the zeros–response approaches a maximum near the poles•The overall effect of constant, zero and pole termsTerm Magnitude Break Asymptotic Magnitude Slope Asymptotic Phase Shift Constant (K) N/A 0 0 Zero upward +20 dB/decade + 90 Pole downward –20 dB/decade – 90Lect21 EEE 202 5Bode Plots•Express the transfer function in standard form•There are four different factors:–Constant gain term, K–Poles or zeros at the origin, (j)±N–Poles or zeros of the form (1+ j)–Quadratic poles or zeros of the form 1+2(j)+(j)2 222221)()(21)1()()(21)1()(bbbaNjjjjjjjKjHLect21 EEE 202 6Bode Plots•We can combine the constant gain term (K) and the N pole(s) or zero(s) at the origin such that the magnitude crosses 0 dB at•Define the break frequency to be at ω=1/ with magnitude at ±3 dB and phase at ±45°NdBNNdBNKjKZeroKjKPole/10/10)/1()(:)(:Lect21 EEE 202 7Bode Plot Summarywhere N is the number of roots of value τLect21 EEE 202 8Single Pole & Zero Bode PlotsωPole at ωp=1/GainPhaseω0°–45°–90°One Decade0 dB–20 dBωZero at ωz=1/GainPhaseω+90°+45°0°One Decade+20 dB0 dBωpωzAssume K=120 log10(K) = 0 dBLect21 EEE 202 9Bode Plot Refinements•Further refinement of the magnitude characteristic for first order poles and zeros is possible sinceMagnitude at half break frequency: |H(½b)| = ±1 dBMagnitude at break frequency: |H(b)| = ±3 dBMagnitude at twice break frequency: |H(2b)| = ±7 dB•Second order poles (and zeros) require that the damping ratio ( value) be taken into account; see Fig. 9-30 in textbookLect21 EEE 202 10Bode Plots to Transfer Function•We can also take the Bode plot and extract the transfer function from it (although in reality there will be error associated with our extracting information from the graph)•First, determine the constant gain factor, K•Next, move from lowest to highest frequency noting the appearance and order of the poles and zerosLect21 EEE 202 11Class Examples•Drill Problems P9-3, P9-4, P9-5, P9-6 (hand-drawn Bode plots)•Determine the system transfer function, given the Bode magnitude plot belowω (rad/sec)|H(ω)|6 dB–20 dB/decade+20 dB/decade–40 dB/decade0.1 0.7 2 11
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