How to Build Bode Plots for Complex Systems | Understanding Bode Plots, Part 4
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- เผยแพร่เมื่อ 21 เม.ย. 2013
- Learn how to build Bode plots for first-order systems in this MATLAB® Tech Talk by Carlos Osorio. A Bode plot describes the frequency response of a dynamic system and displays the magnitude and phase of the system response as a function of frequency in a logarithmic scale. You will learn how to interactively construct transfer functions of second and higher-order systems and study the frequency response of these systems using Bode plots.
From the series - Understanding Bode Plots: bit.ly/3HQ4Xog
Check out these resources on Bode plots:
- Frequency Response of a SISO System: bit.ly/323lGVV
- Bode Diagram Design: bit.ly/3GQ62vQ
- Tips and tricks for using Bode plots in MATLAB: bit.ly/3K4srbt
- Bode function to create Bode plots of dynamic systems: bit.ly/3V5LFoy
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Learn how to build Bode plots for second- and higher-order systems in this MATLAB® Tech Talk by Carlos Osorio. Watch other MATLAB Tech Talk videos here: www.mathworks.com/videos/tech-... - วิทยาศาสตร์และเทคโนโลยี
Really good presentations, thank you!
Great explanation. Thanks!
Great explanation.
Why do we look at the behaviour of the system before/after the natural frequency of the system? Is this because we can assume that "(k/Wn)*j" is similar to a pole time constant in simple systems?
Hello sir , first of all many thanks for your videos they are really helpfull.
I have only one question it might be silly but its vital for me understanding Bode,am a student from Greece , Automation engineer,
Q: why the slope is 20db/dec in either zero/pole ?
here the default slope we are using is 6db/oct am having really bad time finding an answer so plz can you ? anyone else of course is welcome to answer my question.
thanks a lot.
MrProzaki Watch the 3rd video, Simple Systems. In that video Carlos explains why the slope for a single pole or zero is 20 db/dec. For a single pole, at high frequencies, the magnitude is inversely proportional to frequency. So if we consider w2=10*w1, the magnitude at w2 will be 10 times smaller than magnitude at w1. Now take a log of that and multiply by 20: 20*log10(1/10)= - 20 db/dec. Octave is by definition the doubling of frequency. So, w3=2*w1, and the magnitude at w3 will be 1/2 of magnitude at w1. So take the ratio of magnitude at w3 to magnitude at w1, and compute a log of that ratio, and multiply by 20 and you get: 20*log10(1/2)= - 6 db/octave.
Is critical damping acheived by a damping ratio of 1 or 0.707?
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