CORRECTION: Bode Plots by Hand: Complex Poles or Zeros

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I explain how to determine the straight-line estimate of the Bode Plot for a second order transfer function with a pair of complex poles. This video is a repeat of the last half of the Bode Plot By Hand: Complex Poles or Zeros video - minus all of the mistakes! Hopefully you'll find this video easier to follow and less confusing than the other one.
If you haven't seen the first video you should start there:
• Bode Plots by Hand: Co...
Errata:
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[@arien7371]

Every time Brian says ‘Thanks for watching’ it’s like a saint saying ‘Thanks for letting me save your life’

[@simplyxem33]

You say thanks for watching but thanks for saving my life in ece3101 signals and systems. My professor can't teach at all and you just cleared up bode plots for me with your videos! Great work!

[@antorariful1387]

heiii...

[@theboxingbiker]

simplyxem33 Me too I dont understand why do teachers explain it so badly?

[@jimrummy1300]

Simple answer, most aren't meant to be teachers. Brian gets it.

[@aryangaurav7883]

@@antorariful1387 😂😂

[@yasmeenalsadi9557]

thank you to the moon and back! I can't even tell you how efficient these videos are!

[@astafzciba]

i entered an endless loop of videos, my system become undamped, please help.

[@chanakyasinha8046]

Keep your roots at lhf

[@aparnasadhukhan5567]

Try to add a zero in your system.It'll bend the root locus in the left half s plane and you may get back your stability.

[@kkaleeny]

It would be very helpful if you were to use examples using numbers in order to show us what an exam question would look like. Love your work. Thanks so much.

[@gustavoalbino8422]

@Zoe TheCat I think the generic treatment is very important to the correct understanding, but some examples at the end of the video could be nice too.

[@liuxoB]

an exam question at my course would just be like this style and not use real numbers lol

[@TurdBoi666]

​@@liuxoB same lol

[@_Junkers]

It'd be great if you could extend this example to an actual application e.g. active filters. Would help put this into context.

[@roey224]

dude its 2am and i just binged watched all your videos about this subject, thank you so so much! it was super helpful!

[@vivarun]

Dear Brian Douglas, Thank you for these wonderful video lectures. These video lectures are so good to understand the fundamental math and the concept.

[@888899991]

This is pure gold.

[@jouleSansLoi]

At Steven. Remember when he had the transfer function, placed unity on top, did s = jw, and then multiplied&divided the expression with the complex conjugate? Well, the numerator is just unity times the complex conjugate, and you can just simplify the denominator with a foil technique. Do it.. See how it beautifully simplifies to the expression we were using?

[@subratadey1464]

please give some more examples when there is a complex pole along with other poles

[@theFatMan5050]

As someone with ADHD I find it impossible to focus in class. Thank you for the lectures good sir you're saving lives.

[@RuffRoadDesigns]

I just completed a university level linear circuits class that was very confusing and I learned more in this video than I did all semester. #youtubeUniversity

[@anjishnu8643]

9:30 how did you arrive at the expressions w0/5^zeta and w0*5^zeta? Doubt aside, have found these videos extremely helpful. Thanks a lot.

[@carultch]

It probably comes from replacing omega with 10^x, to represent the logarithmic scale. Then you can take the derivative of the phase relative to x, and you'll get the straight line approximation, centered on the break frequency. Where this intersects the flat lines, will approximately occur around 10^x = omega0*5^(-zeta) and 10^x = omega0*5^zeta.

[@youngidealist]

You are amazing Brian. Studying for a final right now and really needed this review (feels like a better introduction than I got this semester).

[@gustavoalbino8422]

First of all, congratulations for all your control system lectures Brian! But what about the bode plot of transfer functions with delay, like [1/(s+1)]*e^(-2s). How can I do this bode plot?

[@Rene_Christensen]

Gustavo Albino A delay can be seen as a extra added linear phase.

[@jakethiem]

Thank you! I've subscribed, this will hopefully help a lot as we have been covering bode plots for weeks and I'm still so so lost.

[@Nerrvih]

Thank you for this series! It sure does explain things nicely

[@niyooshashahgaldi7065]

thank you thank you this series saved my life!!

[@Muz493]

Nice video series. Keep it up Brian!

[@HugoCaen1]

Nice work, dude! Helped me a lot! Keep up the videos, they're just amazing. Congrats from Brazil o/

[@ayadawsy7828]

Excellent, could you told me what is the name of software did you use for explanation.

[@MindfulFreedom]

I have a more general question: What textbook would you recommend for learning more about control systems?

[@stancosmin-bogdan1236]

Have you found a good one?

[@jyotiradityasatpathy3546]

IS Nagrath

[@hiteshchaudhary7964]

thanks a lot brian for all u r control system lectures.my many concept has been cleared.would like to watch videos on pid tuning and how to tune for any transfer function of plant.

[@thkdec1]

Hey Brian, what is the trick to plot bode with multiple poles and zeros? I've review this video a few times but still stuck on this.. Thanks :)

[@JZ1917]

Thank you for doing these videos. Very good work

[@alexeyorekhov3982]

Thank you. I really appreciate the work you do here!

[@shounakdas9393]

Good lesson.Nice work!

[@olafurbjarkibogason7148]

Thanks for the great videos Brian, they are lucid and easy to follow. I did catch one small error in this video though. On 8:58 you state that when xi < 0.5 (under-damped system), the peak of the gain is at omega_0 (the pole/natural frequency). That can happen (as in series/parallel RLC circuits) but generally is not the case. Instead the peak is somewhere "really close to" omega_0, more precisely at the damped natural frequency = omega_0 * sqrt(1 - xi^2) [0]. Keep on making these great videos, all the best! [0] en.wikipedia.org/wiki/Damping#Under-damping_.280_.E2.89.A4_.CE.B6_.3C_1.29

[@theboxingbiker]

You are the best. Can you do an example like this for the nyquist please.

[@MrNuigit]

Thank you Brian/God.

[@SaideepRokzz]

I have a doubt: When W(omega)/W0(natural frequency) is tending to zero the complex equation is negative((1-j(small number)) In that case arctan2(value) of this complex number is 180 degrees as you told in previous lectures but why it is zero degrees in this case? Thanks for spending your valuable time

[@MasoomaAli-eh3qo]

because imaginary part is very small almost zero so arctan is also 0 degree

[@FORESTER2011]

This correction is an improvement from last videos, which are more clear :)

[@jassimjaved3309]

Thank you for making this!

[@islamnazmy302]

I think a more mathematically rigorous way to get the limit at 4:10 is with L'Hopital's Rule

[@asmarali8579]

your methodology is awesome :)

[@danningzhao7361]

You saved my life😊

[@bsaxon715]

please add this to your bode plots playlist

[@SohilShah_Melodyman]

Very well explained....thanks a ton!

[@steven28091]

This is an excellent video, subbed. However, can some one clarify where the equations for the imaginary and real components came from?

[@abcddd580]

This was extremely helpful. thanks.

[@nurdeens]

hey did you do a nyquist by Hand: Complex Poles or Zeros video? Or do you know a good website or video that explains with a few examples?

[@doctorbobstone]

It would be helpful if you included this video in the "All Control Systems Videos" playlist.

[@tvpedroso]

Great videos, it has helped many people. Just one question. Using -180º to the phase angle is equivalent to using 180º? At 7:00 you used the negative one, but is it different if I use positive?

[@MsHShuaib]

Yes, negative is for 1/s^2 and positive is for s^2. I THINK. Please do NOT take my word for it. watch his complete playlist where he mentions the difference.

[@matthewjames7513]

Can someone explain why we let s = jw? He mentioned in a previous video it's because we're interested in the steady state response, but I don't get why :'(

[@JamesS221193]

s=sigma +jw, at steady state sigma=0

[@aparnasadhukhan5567]

Jw are the phisically realizable frequencies

[@kindjupiter]

thank you for very useful videos. I have a question. By OMEGA not (W0) u mean Cutoff frequency , Am I right?

[@psonney]

can you please explain why you took 5^zeta approximation in the last section?

[@jimbalta]

Do you have any useful links on how to estimate the transfer function from a bode plot? For example, how many poles/zeros your system likely has and where they might be located?

[@jimbalta]

+James Finch *sighs* never mind. The number of poles + number of zeros = the number of break frequencies that can be seen. Any time you have a break frequency that takes you up in magnitude, that indicates a zero, and any time you have a break frequency that takes you down, you have a pole. Furthermore, your high frequency slope is related to how many more poles than zeros you have. -20dB/decade = 1 more pole, -40dB/decade = 2, etc.

[@hazirahmoon8795]

sir can i know how to get transfer function from my bode plots if my high frequency asymptote at -270°?

[@imadalrouh41]

thanks for these amazing video, But could you please tell me which software you used to make these video. it will be very helpful for me because i am professor of power electronics. Thanks again

[@MEHJABIN]

Could You please tell me how to Find Asymptote on the bode plot??

[@abdullahalmubin]

Dude! why don't you put a popup info message about the mistake? I understood the last one well, and now you are telling me there was a mistake, and what I've learned was wrong! But anyway! you explained here very well; thanks!

[@GeneralSeptem]

Is 5^\zeta a better approximation than 10^\zeta for the phase breakpoints? It seems if you had \zeta = 1 then you'd want the straight line to start at \omega_0/10 and end at \omega_0*10.

[@jerred2275]

Very good video, what would be the form of the transfer function for complex zeros? Would it be exactly the same as that for the poles, but flipped? So would you find w0^2 at the end also?

[@carultch]

Yes, it will be flipped, and w0^2 will still equal the final term of the quadratic expression.

[@arunpandian294]

5:59 How can you say that the gain falls of at -40 dB *per Decade* Isn't the frequency response plotted for powers of 10, i.e., 10^-2, 10^-1, 10^0, 10^1, 10^2, 10^3...... and *not* 10^1*w0, 10^2*w0 ?? (Here w0 -> Omega nought) As far as I know, the Frequency response is simply plotted for powers of 10 (i.e. 0.01, 0.1, 1, 10, 100, 1000,...) and not for multiples of 10 for w0. I mean *that's why the name DECADE* right??

[@arunpandian294]

*multiples of w0 by powers of 10* is not Decades right??

[@Sarang4401]

Its really helpful.. Thanks a ton :)

[@96izsi]

what program do you use?

[@saransh85]

At 8:50 what do u mean by peak

[@chvishnuvardhanreddy3168]

can u please explain how the 20db/decade = 6db/octave ?

[@khuldoonusman8582]

CH Vishnu Well, I think, that is simply converting from a base of 10 to a base of 8.

[@Krsnda]

I've seen people using w0/(5^zeta) and w0*5^zeta and w0/(10^zeta) and w0*10^zeta. Which one is correct ?

[@ianchen8718]

it's do really help me on my final exam lol !!

[@mustafaadel8194]

at 5:35 I found a contradiction the equation that you deduced depending on w >> w0 is used at 5:35 to evalute the response when w = w0 which doesn't make any sense ! and BTW thanks a lot for your videos they're really useful

[@koukaalgerienboy]

you're the best thank you

[@Kletustheslackjawedyokel]

Hi. I don't understand how to deal with the DC gain for these. If we had for instance 10/(s^2 + s + 4) the break frequency would be 2 and the slope at 2 would be -40dB/decade, but how do you calculate the DC gain from this?

[@BrianBDouglas]

Ahh, DC gain is very simple to calculate from a transfer function. DC has a frequency of w = 0 Hz. And since for steady state responses S=jw, all you have to do is set s = 0 and see what you get. In this case it's 10/4 or 2.5. Another way to view it is like this: The classic 2nd order transfer function is w^2/(s^2 + 2zw*s+w^2). If you set S to 0 you can see it has a DC gain of 1 (0dB). If you match your transfer function to the standard TF you'll get (2.5 * w^2) / (s^2 + 2zw*s+w^2), where w=2, and z = 1/4, and the DC gain is 2.5. These are two transfer functions, a 2nd order one and a constant one. So to draw the Bode plot start with the standard TF and then move the whole curve up 2.5 ( ~8dB). Hope that helps you.

[@Kletustheslackjawedyokel]

Brian Douglas helps very much. thank you. very clear and helpful videos.

[@Kletustheslackjawedyokel]

James Gilson Hi Brian. I looked over your Nyquist videos and I am still confused what to do for a certain type of transmittance: specifically one where we have a pole at the origin, a pole on the real axis, and a pair of complex poles on the Im axis. The one at the origin and on the real axis I get when they are alone, but the Im axis poles I am confused on. Not sure how to bypass them?

[@lesliegilson4891]

That's incredible James! Lm

[@JohnMillner]

you are amazing, thank you!

[@tajammalnawaz5552]

atleast you need to give when one example that utilize all these stuff.

[@TaahirBhaiyat]

why divide and multiply 5^zeta to get the limits for slope? why 5?

[@watanake]

OMG Thank you so much!!!!

[@ronpearson1912]

How to calculate an exact zeta function?

[@ningfengzong5513]

you are the man !!!!!!!!

[@bje1189]

Truly Excellent videos . Brian you have a gift for communication . Is there a good text book you can recommend that presents these topics for electronic circuits

[@juliansydenham9371]

Good effort up-loader, but for some reason I can’t keep up with your output.

[@amanjain2549]

why would you have to thank i thank you for putting this lecture

[@Muhammad_Saleh754]

how can i draw the line in gain and magnitude when the zeta equals zero ?

[@BrianBDouglas]

As zeta goes to zero then then 1/(2*zeta) goes to infinity, and 5^zeta goes toward 1. Therefore, gain would spike up to infinity at the break frequency and magnitude would jump from 0 deg to -180 instantly (vertical line). Otherwise, the approximation much greater than and much less than the break frequency would stay more or the less the same.

[@Muhammad_Saleh754]

thanks man

[@Shuey272]

wait you guys mean to say '... at the break frequency and phase would jump from 0 deg...' right?

[@egyMag1990]

what do you mean by the gain is more negative at w.

[@carultch]

The gain isn't really negative, because it is on a logarithmic scale. The decibel value which is a logarithmic representation of the gain is negative, which simply means it is less than unity.

[@egyMag1990]

@@carultch this was 7 years ago😀😀😀😀. Thank you

[@nimeshsingh9271]

Quality !

[@mr.hibster9603]

Could someone tell me what w0 stands for?

[@carultch]

It's an omega, and it means natural frequency.

[@jessstuart7495]

5:44 You don't need to calculate the Real and Imaginary parts of the entire transfer function! Just figure out the Re and Im parts of the Numerator and Denominator. Your approach is kind-of nuts, and way too much work! You can write these out almost by inspection... |T(jω)| = |Numerator|/|Denominator| = (ω0)^2 / | -ω^2 + j*2*ζ*ω0*ω + (ω0)^2| = (ω0)^2 / sqrt( ((ω0)^2-ω^2)^2 + (2*ζ*ω0*ω)^2 ) Magnitude at (ω = ω0): |T(jω0)| = (ω0)^2 / sqrt(0+ (2*ζ*ω0*ω0)^2 ) = (ω0)^2 / 2*ζ*ω0*ω0)^2 = 1/(2*ζ) arg(T(jω)) = arg(Numerator) - arg(Denominator) = 0 - arctan(2*ζ*ω0*ω/((ω0)^2-ω^2) = atan2(-2*ζ*ω0*ω, (ω0)^2-ω^2) Phase at small ω: atan2(-2*ζ*ω0*ω, (ω0)^2)) = -arctan(-2*ζ*ω, ω0) goes to -0 deg. Phase at large ω: atan2(-2*ζ*ω0*ω, -ω^2) = atan2(-2*ζ*ω0, -ω) goes to -180 deg. Phase angle at ω0 + e: atan2(-2*ζ*ω0*(ω0+e),(ω0)^2-(ω0+e)^2) = atan2(-2*ζ*ω0*(ω0+e),-2*ω0*e) = atan2(-ζ*(ω0+e), -e) = atan2(-ζ*ω0, -e) = atan2(any negative numer, -0) goes to -90 deg.

[@محمودامجد-ب8ش]

WHO WE FIND FIND PHASE -180 DEGREE ??!

[@atayavuzz]

who let the dogs out? who who who who

[@lonelymechanic3688]

it's +40log(w/wn) not -40log(w/wn)