MEM 640 Lecture 2: Bode Plots

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MEM 640 Lecture 2: Bode Plots 1 Bode Plots: Why Study Them? Recall: RC Low-pass filter time response plot Input: Vi (t) =1.0 Output (dots):Vo (t) • Output reaches 63.6% of steady-state at 0.47 ms • Note: 1/0.00047 = 2127 radians/sec 2 Low Pass Filter Bode Plot Definition: A Bode diagram consists of 2 plots. The first plots the output/input ratio [dB] versus frequency. The second plots the phase angle versus frequency. Typically a semi-log plot for frequency is used Low Pass Filter Bode Plot Diagram: -3 dB 2127 radians/sec Thus the -3 dB point represents the frequency corresponding to about 1 time constant 3 Phase Shift Significance Bode phase plot on previous slide says 45-degree lag at 2127 radians/sec [338 Hz] Observe: Apply a sine voltage input (338 Hz) into a low-pass RC filter Period T = 0.0049 − 0.0019 = 0.003sec 0.003 sec = 333.3 Hz Δt = 0.0019 − 0.00148 = 0.00042 sec Δt φ = 2π = 50 deg T Hence see that this is approximately the 45 degree lag shown on Bode plot 4 Bode Plot: Step-by-Step Instructions Problem: Below is data collected by applying sine voltage inputs into a low pass RC filter. Sketch the Bode diagram on semi-log paper 5 Solution Step 1: Label axis on semi-log paper. Choose units and be consistent! 6 Step 2: Plot your points 7 Step 3: Draw asymptotes, mark -3 dB point, cutoff frequency and label slopes 8 Step 4: On new semi-log paper, plot the phase angle versus frequencies 9 System ID Given the Bode Plot Problem: Given the following Bode Plot, calculate the transfer function 10 Step 1: Note that the magnitude is not zero at start and slope is -20 dB/dec K TF1 = s Step 2: Slope decreases -20 dB/dec, hence have another pole. The cutoff frequency ωc = 0.2 with time constant 1 τ = = 5 Hence 0.2 1 TF2 = 5s +1 Step 3: Slope increase +20 dB/dec, hence a zero has been added. The cutoff frequency ωc = 0.8 The time constant is 1 τ = =1.25 0.8 Hence TF3 =1.25s +1 11 Step 4: Slope decreases another 20 dB/dec. This means a simple pole was added The cutoff frequency is ωc =10 The time constant 1 τ = = 0.1 10 Hence 1 TF4 = 0.1s +1 Step 5: Take product of all sub transfer functions K(1.25s +1) TF = G(s) = s(5s +1)(0.1s +1) Step 6: Determine the value of K Take decibel log of each side and replace s with jω 20 2 2 20 2 2 20 2 2 20logG(s) = 20log K + log(1+1.25 ω )− 20logω − log(5 ω +1)− log(0.1 ω +1) 2 2 2 (1) 12 From Bode magnitude plot, see that at ω = 0.1 have 20logG( jω) = 60 dB Thus substituting this frequency into the (1) 60 dB = 20log K + 0.0673 + 20 − 0.969 − 4.3×10−4 20log K = 40.90 K =110.9 Hence 110.9(1.25s +1) G(s) = s(5s +1)(0.1s +1) 13.

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