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How to Build Bode Plots for Complex Systems | Understanding Bode Plots, Part 4

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MATLAB

MATLAB

Күн бұрын

Пікірлер: 8
@beevees1636
@beevees1636 6 жыл бұрын
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?
@orsmas7997
@orsmas7997 7 жыл бұрын
Really good presentations, thank you!
@alexandrpetrov1110
@alexandrpetrov1110 4 жыл бұрын
Great explanation. Thanks!
@sayednaseel
@sayednaseel 3 жыл бұрын
Is critical damping acheived by a damping ratio of 1 or 0.707?
@matthewjames7513
@matthewjames7513 2 жыл бұрын
1
@MrProzaki
@MrProzaki 10 жыл бұрын
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.
@arturevskiy
@arturevskiy 9 жыл бұрын
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.
@elfaber
@elfaber 10 жыл бұрын
The individuals with hearing loss disabilities would apreciate ClosedCaptions...
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