Bode Plots by Hand: Complex Poles or Zeros

This is a continuation of the Control Systems Lectures. This video describes the benefit of being able to approximate a Bode plot by hand and explains what a Bode plot looks like for a transfer function with either a pair of complex poles or zeros. This is the fourth and last of several videos where I describe step by step how to estimate a Bode plot from any transfer function.

I will be loading a new video each week and welcome suggestions for new topics. Please leave a comment or question below and I will do my best to address it. Thanks for watching!

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[TwoToedSloth]

10 years later, still useful. Thank you

[CodParadox]

Just killed my systems and control exam today thanks to you. This lecture is so information dense yet easy to understand! :)

[darjiaethera]

Our circuits professor introduced Fourier series and bode plots before teaching us about AC circuits or phasors.You're basically my hero for these videos, since I would probably fail without them.

[jingjie]

You are amazing! You should teach professors around the world how to teach control theory! I bet a even high school kid can understand your lectures!

[BrianBDouglas]

So far I only have these 10 videos. But I've been making new ones each week so stay tuned for more in the coming months.

[angrycloudofdogfood]

couldnt do my homework and where did that lead me? back to you
thank you sir

[alviur]

Thanks a lot man, my control professor must learn form you how to teach!!

[Peter_]

Why do I always find these videos like an hour before my own exams?
lol

[pritvarmora]

the way you explains everything i din't found all over the you tube till now at least for control theory lectures
great work it up for other subjects also.... i don't know why only this much views you are gating....

[GeletomWAbebe]

12 years later and still very useful video.

[BrianBDouglas]

You caught me while I was on so I'll reply right away! You solve for the roots of the denominator like you did, roots = -13.82, -1.59 + 4.85i, -1.59 - 4.85i]. Then you turn this into a first order times a second order equation. 1/(s+13.82) * 1/(s^2 + 3.18*s + 26.05). Then you need to pull the gain out so that the function is in the correct form: 1/(13.82 * 26.05) * 13.82/(s+13.82) * 26.05/(s^2 + 3.18*s + 26.05). Finally just draw the three Bode plots (gain, 1st and 2nd) and sum them together!

[animeshjain]

You just saved my semester course man. You're truly amazing!

[LureAngler]

Thank you very much for your videos. I discovered your videos just in time for my exams and it did help with my disillusionment with control theory.

[AirAdventurer]

You are truly a master of your art; thank you so much for sharing your knowledge

[renandebritoleme]

Great class! You are the best teacher that I have seen!

[GoingInFamoUs]

Yes thank you, Great videos! They really save me when I miss a lecture.

[DannyKng]

Making my degree so much easier, thank you so much!

[ricojia]

Thanks Brian!! I am reviewing my control systems course (going into control engineering in the future) and still find this video helpful.

[BrianBDouglas]

Yes there is something you don't see (or at least were too nice to point out), and that's my error in the Imaginary equation. It should be -2*zeta / (w/w0)^3. Then this will converge to zero faster than the real component and you'll get -180 of phase. I made too many errors in this last half of the video for it to be useful I think :( Thanks for working this out and letting me know. I think I'll redo the last bit and point to the correct video so it's not confusing.

[mahdo]

thank you for your GREAT videos ! i cant express how much helpfull your videos were ...