Bode Plots by Hand: Complex Poles or Zeros

[@TwoToedSloth]

10 years later, still useful. Thank you

[@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.

[@Cod4Paradox]

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

[@Peter_1986]

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

[@74-KA]

Why do I always find these videos AFTER my own exams? lol

[@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.

[@abuelc]

I was going crazy on this part! Thanks for confirming the solution. :)

[@aniket9806]

@@abuelc me too..i was going mad over it

[@alviur]

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

[@jingjie3000]

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!

[@SuperCaptain4]

I am writing a paper about control theory in my mathematics class in high school so yeah, this is definitely the best explanation i've come across so far.

[@sadidrahman4608]

@@SuperCaptain4 stfu nerd

[@SuperCaptain4]

@@sadidrahman4608 mad cuz bad? ;)

[@sadidrahman4608]

@@SuperCaptain4 lol nah bro, just a compliment 😎. I am kinda jealous tbh.

[@SuperCaptain4]

@@sadidrahman4608 alrighty ^^ Funny thing is that I gave up on math and physics after writing that shit and now I'm doing economics in uni 😅😅😅

[@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.

[@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!

[@animeshjain5712]

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

[@pritvarmora2036]

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

[@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.

[@BrianBDouglas]

Yes to your first question. If the pair of poles are complex then you'll have -40 roll off and if zeta is approximately less than 0.5 then you won't have that spike before the roll off. If zeta is > or equal 1 and both poles are on the real line then really what you have is two real poles. Then you can just approximate each real pole separately and add them together. If they are right on top of each other, zeta is 1, then the first half adds to 0 and the second half adds to -40 roll off.

[@BrianBDouglas]

It would be difficult to get an exact value of zeta from the Bode plot but you could certainly get an estimate, but only then for a second order system. The main reason is that the bode plot sketching method is only an estimate. -20log(2*zeta) is an exact value for the gain at the natural frequency however it isn't necessarily the highest peak on the graph. Try plotting 1/(s^2 + s + 1) in Matlab. This has zeta=0.5 and wo=1. So you'd expect the peak to be 0 at the natural frequency 1 rad/s.

[@BrianBDouglas]

But the peak is really closer to 1 dB around 0.65 rad/s. But at 1 rad/s the gain is exactly 0 like we predicted. But if you looked only at the Bode plot you might be tempted to say the natural frequency is ~0.65 rad/s and zeta is ~0.45. That wouldn't be far off, but not exact. Of course if you are intimately familiar with the exact shape of various bode plots then you could very accurately measure what zeta would be from them. Hope that helps.

[@angrycloudofdogfood7128]

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

[@BrianBDouglas]

The gain points up at that point only when the damping ration, zeta, is less than 0.5 (which is why you only draw a peak when you have less damping then that). For zeta less than 0.5, 2 times zeta is then less than 1. The log of any number less than 1 is negative. Then -20 times a negative number is the positive number that causes it to point up. For zeta greater than 0.5 the damping is high enough that you don't get any positive gain at that frequency, hence, no upward pointing peak.

[@AirAdventurer194]

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

[@BrianBDouglas]

You're fast! You beat me to my second comment. I'll put out a correction this week (I did see the other mistake in the real part of the equation also). I tried to skip too many steps and not write out the work ahead of time before making the video :( I think the entire second half of the video needs to be re-done correctly and *hopefully* without any errors!

[@renandebritoleme3097]

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.

[@BrianBDouglas]

Hi Miles, hopefully I can clear it up here. The transfer function 1/(s+5-5j) only has a single complex pole (s = -5 + 5j). The video only covered transfer functions that have two complex poles that are complex conjugates of each other. In this case you could add a second pole (s = -5 - 5j) so you'd have a two pole system and the video will cover that. However, you can solve for the Bode plot of your system in the exact same way. See next comment for help.

[@BrianBDouglas]

If you substitue s=jw in your transfer function you'll get H(jw) = 1/(jw+5-5j). If you separate the real and imaginary parts of the solution you'll get, real = 5/(25+(w-5)^2) and imaginary = -(w-5)/(25+(w-5)^2). if you let w approach 0 you'll see that for this transfer function the phase approaches 45 degrees. If you let w approach infinity you'll see that the phase approaches -90 degrees. Does this help?

[@ricojia7322]

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

[@DannyKng]

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

[@BrianBDouglas]

Hi Jason, the number of poles and zeros makes no difference in the method for finding gain and phase ... just the complexity. You could do it two ways, 1) substitute jw everywhere there is an 's', then solve for the real and imaginary parts and find phase and gain for each omega as the angle and magnitude of the point in the imaginary plane (like I did in the video). Or 2) approximate the gain by summing the 3 individual bode plots, the two poles and the 1 zero. Did that answer your question?

[@BrianBDouglas]

Then as the two real poles drift further from each other you'll add the response of each separately still and you'll start to get no slope at the first part of the graph, -20 slope between the two poles, and -40 after the second pole. Of course this is approximate near the pole because there is some smoothing in the line. It's not these perfectly straight lines we draw when sketching by hand.

[@arthurgomes6124]

Very good video, explained clearly and objectively. I Liked.

[@mahdo7]

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

[@BrianBDouglas]

Another way of thinking of this is that you have a negative imaginary part and a negative real part. When you take atan you're not really dividing. You are just using trigonometry to find the angle between the positive real line and your point in the imaginary plane. So by dividing and canceling out the minuses you're wiping 180 degrees off the phase and moving the point into the +,+ quadrant instead of the -,-. Did that make sense? Sorry for all of the confusion.

[@vanityngreed]

This is unbelievably good

[@covingtonkua9404]

binge watch these series at 2am

[@aman_]

You are just awesome... Great teaching... You just made super simple...

[@BrianBDouglas]

Hello Ahmed, that is true, but I'll revisit that when I start the series on Nyquist Stability Criterion. This video was really just supposed to be on how to draw the Bode plot and not interpret it just yet. Stay tuned!

[@BrianBDouglas]

Timofte my old friend! What made you choose zeta = -50:1:1? Negative damping ratio? That would be adding energy into the system instead of removing it. I wouldn't go lower than a damping ratio of 0. Also, a critically damped system has a zeta of 1 but you can have an over damped system with zeta>1, however that creates two real poles and not complex like this video is about. Try this and see if it still contradicts what I stated. zeta = 0.01:0.01:1; f=-20*log10(2*zeta);p­lot(zeta,f). Cheers.

[@BananaManZang]

Legend.. My lecture obviously thought this didn't need explaining in his notes

[@MohamedNourElDin5]

Thank you so much for this awesome tutorial !

[@74-KA]

Thanks for your video a lot. I love them.

[@michlgilbertclements5378]

In the mechanical example, k represents the oscillating spring, b represents the viscus dampener, then the equation should be 1/(ks+b).

[@cnstantinoschronakis7743]

You are the real MVP.

[@pb48711]

Brian, at 8:07 of the video your approximation for the {Im} component seems wrong as it must assume that in the reduced denominator, (2*zeta*(w/wo) >> ((1/2*zeta)*(w/wo)^3. How did you arrive at this conclusion or is this a plug and chug error? I am totally lost. Please assist. Thanks.

[@shinjones7490]

Same here

[@AxlTheallmighty]

First of all, I can't thank you enough for what you're doing. It really is because of you I've started to like control system theory. But to my question: Let's say you have a transfer function that contains exponentials, how do you handle that using your methods? I assume you can rewrite it to a real pole somehow, but I'm not sure how. Again, many thanks! Nyqvist plots are by the way something I'm requesting as well for your future videos.

[@sekomiko]

Sir, you are the best!!

[@ArlindoFernandes]

You have done a great material! Gongrats!

[@fhcaglayan]

Allright thank you. I think now we understand each other hahaha! You thought that I did not see where the angle came from. I do see where it comes from IF indeed both the real and imaginary part are negative, but I have calculated them myself and I don't end up with negative real and imaginary parts. Probably, I took one of the mistakes in the video into my calculations without being aware of it. So the mistake must be in my early steps. I'll have a look myself as well agian!

[@BrianBDouglas]

Hello again, you have to use the function atan2 instead of atan in this case (I didn't use it in the video because the error I had made tricked me into thinking atan would have worked, in hind sight I should have used atan2 throughout. I'll correct that in my follow up video). Atan2 works the same way at atan but it keeps track of the quadrant too. Try atan2(-.00001, -1) in Matlab and see that it approaches -180 degrees.

[@tcharlesmackinson5899]

Good work man... Keep it up

[@burningsilicon149]

At 8:10 should it not follow the imaginary component since it is larger? I mean the (w) goes to infinity so the 1/(w/w0) would be larger than the (1/(w/w0))^2 so should the imaginary component dominate?

[@fhcaglayan]

Hi Brian, Thank you for your amazing videos. I have my re-exam in August for the course control systems and thanks to your videos I have even chosen for a master in control engineering! I also have a question about the video at 8:00. I don't see how you get to the imaginary part. The weird thing is that I do see how you get to the real part, so I think there is something small I don't see!

[@Mezor]

thank you again!!! l'll try that now.

[@20hawkar10]

Hey Brian, thank you for the video, I have one question regarding the mistake you made on the phase of your second case. What does the phase have been approximated by, because when I simplify down I seem to get, w/(2*w0*zeta) which approaches 90 or -90 as w>>w0.

[@ryanmccormack6403]

Hi Brian. Thanks alot for all these videos. In relation to Bode plots, I was wondering if you were planning on or had done any Quantitive Feedback Videos, using Nichols and Inverse Nichols plots.

[@matheusramos8791]

Thanks! Amazing.

[@jessstuart7495]

1/(s+j) looks like a Hilbert transformer plus low-pass filter to me. It will phase shift large (>1) positive frequencies +90deg, and phase shift negative frequencies (

[@NXTangl]

5:03 - so what you're saying is, if we have a LTI system acting on quantum wave packets, it could have unbalanced poles? (Since quantum waveforms are inherently, physically complex.)

[@SaeEDKh313]

Great! Big thanks

[@austinfritzke9305]

Thank you thank you thank you

[@abuye2011]

Brian, please try to give some more practical examples about Bode,Nyquist and root locus. It confuses me all the time... thx.

[@hl8807]

Thank you!!

[@mhayek4588]

Hello Brian, thank you for your lectures, really nice! I have a question about the imaginary part when w>>w0, how did you end up with that value? I couldn't find it. I have 1/((w/w0)^3 + (4*zeta^2*w/w0)). Please let me know, thanks!

[@benjamincastro541]

for the case were w>>w_0, do you draw the asymptote at the point of interest or one decade higher?

[@peterandersson6846]

The math/algebra in this video is quite hairy! I'm having a bit of trouble following all the steps. I'm following when you say 4ac > b^2, then divide both sides with 4ac, you get b^2/(4ac) < 1. But that is not what you write at 03:50. Why is there's suddenly a square root appearing in this equation? How is this possible? Also, at 05:16, how do you get to this transfer function from what was written earlier? Can you please show the steps in between, that leads from what was written previously to this transfer function?

[@flygerian1231]

Thank You so much.

[@praneethnanduri6220]

Hi Brian, Amazing video series. Thank you very much! A question about the cut-off frequency, is there any definition for cut-off frequency in general? I see that it is the frequency at which the power becomes half because it doesn't make sense in this video because "wo" is the cutoff frequency you took in this video but it is not clearly -3dB point.

[@RaedMohsen]

Thanks for These Tutorials, will you explain state space representation?

[@r410a8]

8:41 Why in order to find the |H(jω)| and arg(H(jω)) for ω>>ω0 you didn't use the same formula as used to find them for ω

[@ShariefSaleh]

Great Videos

[@planauts_111]

Please update the playlist with the correction video.

[@GoingInFamoUs]

What happens when you have a 2 real poles and 1 real zero. How does the zero on top effect finding magnitude and phase. For example G(s) = (s+5)/(s+2)(s+4) when i need to find gain and phase?

[@kshatreshmishra5608]

Sir, please tell what would be the gain plot when we have a positive real pole as according to me the argument of log would become negative for frequencies above the break frequency.

[@sukyun0403]

Thank you:)

[@abuye2011]

Hi Brain, I found your video unexpectedly and find it very useful. would you mind to explain what's the w0(break frequency thing) and how i'm supposed to get it. is that the point where the GM is zero? i'm an elec. eng. student.... thx and keep up the good work.

[@mothman000]

this video is proof that you don't need a good school for a great education!

[@martincho8002]

Hi Brian! Thanks for all your videos: I have a newbie question. Looking at eigenvalues on a x-real/y-imag plot, changes along the x-axis (real part changes) modify the natural frequency, but do changes along the y-axis (imaginary part changes) modify the natural frequency ?

[@dunnasuryanarayana5918]

i'm bit confused..what is the behavior of system if there are poles only at the orizon..

[@fhcaglayan]

I know that if both the real and imaginary part are negative then this will happen. You don't need atan2 if you ask me. You can visualize in your head what will happen when one approaches zero faster than the other one. My point is that there is probably another mistake in the video, because when I calculate the real and imaginary parts, they are not negative. This is also the case in the video.

[@hussam05652]

Thanks

[@Muhammad_Saleh754]

yo wussap brain i wanna ask you about the real part in case #2 how did u calculate the real part and the imaginary part ??!!

[@sanjeevang9581]

There is a error in finding the phase at w>>w0. Correct it

[@ricnzambanzamba8235]

hello sir you didnt do block diagram reduction please do it

[@ronpearson1912]

So if complex poles are pairs of real and complex parts wouldnt there be 4 parts (2 real and 2 imaginary) in the bode plots? Or rather 3 since 2 of the reals are the same ...

[@fhcaglayan]

I had noticed that one as well, but still the real part is positive. Then the argument should be 0 degrees instead of -180? And the imaginary part is positive as well. The minusses cancel out?

[@haileymarsh7559]

Around 6:20 or so when you took the complex conjugate why didn't you get 4zeta^2 in your term on the bottom? you had 2zeta(w^2/wo^2)

[@MsHShuaib]

You're awesome!

[@munibowais]

which software are you using for drawing?

[@theducksneezes4987]

I second that question

[@ShajeeJurangpathy]

He put up a video on how he makes his videos. Just check his channel, hope it helps :)

[@megatronzm]

This is his video: kzbin.info/www/bejne/o16wZpKjft57m5Y

[@borockin4004]

I don't understand the value of the real and imag when omega is much larger than omega 0

[@tayenjamjeneetaa494]

anybody got the answer?

[@josepaul2000]

So.. since the complex poles come in pairs, the attenuation is not -20dB/dec, it is -40dB/dec? And also, the effect of the complex pole on the magnitude Bode plot is the (lack of) damping of the spike before the -40dB/dec drop? I mean, if I understand this correctly, if the poles were real, the damping factor zeta > 1, and this would mean better damping, no? And this would mean that the spike at the pole (in the Bode plot) would be lesser?

[@chaimachalouche5013]

Thank you, you saved my life hahah xD

[@Ghewar123]

plot the bode diAGRAM FOR tranfer function (10-s)

[@mosesjames360]

why us 0.1 w and 10 w in sketching phase diagram ?thank you!

[@abdullahfaqihalmubarok8232]

great!

[@lcable99]

What a homie

[@camisa666]

Thanks for your videos! they are really helping me understand. One question: I could not find a video for bode plots of functions with RHP poles/zeros. ej: 1/(s² -9) Are you planning to make one :-) ?

[@carultch]

Usually, if the poles are on the right half of the plane, it is a pretty useless control system, since all it does is exacerbate the output out of control. Take the inverse Laplace of your example, and it is 1/3*sinh(3*t). As t gets large, the output grows exponentially, rather than approaching a steady state value with an exponential decay.

[@LucasRato89]

First of all : Thank you very much!! Your videos are amazing, very intuitive and useful!! I have a question. How can I interpret the damping ratio and the break frequency mathematically? I mean, i understand what they do and how to find them, but while looking to the pure formula, i would like to understand their mathematical influence in the transfer function. Any insight?

[@carloseduardocarneirodemel2551]

Dear Bian, I am from Brazil. I have to thank you for the lectures bode plot videos. Regards to the Complex ploles. Could you send me a tuturial about the mathematic manipulation to reach the final formule.

[@OnurKaratalay]

So, I can find the systems' damping ratio from bode plot by equating X dB = -20log (2*zeta) ?

[@sprobablycancr4457]

This!

[@benjohnson7974]

how to take 20db , 40db,60 db slope angle in exam copy?

[@philipsoben8108]

Could you please explain "Nyquist Diagram"?

[@thitlwin4257]

Hi Brian. I am trying to solve k/ [(s+1)^3 (s+4)]. I managed to get the starting, ending points and the break-out points. I have seen the sketch of the locus (from modal solutions) but am not able to understand why the poles will converge at the breaking point(-3.25, -1 and -1) only to curl away to infinity symmetrically. I thought that poles which converges on the real axis should break out 90degree to the it. Can u help?