Bode Plots and Interpretation PDF

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Summary

Building bode plots for magnitude and phase, reading them to estimate transfer functions
Converting to dB and deriving long and short-term behavior of the system with frequency...

Description

ESE 505 HW#05 Problem 1 (a) We expect to see an increase in magnitude when 𝜔 = 𝜔𝑛 (natural frequency). This magnitude peak occurs for 𝜔 = 4 𝑟𝑝𝑠. Hence, 𝜔𝑛 = 4 𝑟𝑝𝑠. We also know that for 𝜔 ≪ 𝜔𝑛 , 𝐺 ≈ 𝐴. Hence, observing the 𝐴 magnitude bode plot at low frequencies, we see it flatlining at 𝐴󰆹 = 0.5𝑑𝐵. We also notice that 𝑅 = for 2ζ

𝜔 = 𝜔𝑛 . From the graph we see a value of about 2.1dB. Hence, 𝜁 = function should look like 𝐺 =

0.5∗16 (𝑠2 +2∗0.12∗4∗𝑠+16)

=

8 𝑠2 +0.96𝑠+16

0.5 2∗2.1

= 0.12. Therefore, our transfer

.

Using num = [8], den = [1 0.96 16], and wdata = [0.2, 0.5, 2, 4, 8], and a time delay of 0, I obtained the following Bode plots.

1

1

= 10 2 − 10−2 ⇒ 𝜏 = 0.317. Hence, a good 𝜏 approximation for the transfer function would be num = [60] and den = [0.32, 1] with an input delay of (b) Based on OH and piazza, I obtained the expression

27s s.t. 𝐺 =

Problem 2

60𝑒 −27𝑠 0.32𝑠+1

. Using these and wdata = [2,4,6,8,10], I get the following bode plots:

The first few values of magnitude are 𝑅 = 6𝑑𝐵. The given transfer function looks like first order as it does 𝐴 not show a resonance peak at a natural frequency. Hence, 𝐺 = 𝜏𝑠+1. We know as 𝜔 → 0, 𝐺 → 𝐴 = 𝑅. Hence, 𝐴󰆹 = 6𝑑𝐵, implying A = 2. As 𝜔 ≫ 0, 𝐺 ≈ −53.7𝑑𝐵. Hence, R = 484.17. This gives us 𝜏 =

𝐴 −𝑗𝜋 𝑒 2, 𝜔𝜏

2 484.17∗314.16

where 𝑅 =

= 1.3 ∗ 10−4 .

𝐴 𝜔𝜏

. At 𝜔 = 314.16 𝑟𝑝𝑠, 𝑅 =...