Bode Plots by Hand: Poles and Zeros at the Origin
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- เผยแพร่เมื่อ 2 มิ.ย. 2024
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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 pole or zero at the origin. This is the second of several videos where I will 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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Brian have you done any videos about what your work life is/was when you were a control engineer? I would be really interested to find out what your day to day tasks were and what sort of projects you have been involved in to give some insight into the career side of Control.
Thanks for the tutorials, they are a God send.
You have deep understanding of this subject sir, I just took this class earlier and now have to design simulink exercices for this same class, as a project. I watch all of your videos and they fill me with comprehension.
your videos are extremly wonderful and they will definitly help me out in my enginneering...thamk you for making control engineering look so easy for all of us..
Thanks a lot I've exams and this is gonna help a lot God bless you 😊
These videos are just amazing! Even though my professor is world-class, given the time constraint, he often has to shorten his explanations, so this is where I come. Thank you Brian! Hope these videos remain on youtube for decades to come.
Brian,
Thank you for doing this. Your videos have been a HUGE help (right before finals week).
(I'll comment more later) This lecture series on Bode plots is - SO - USEFUL - Mr. Douglas - thank you so much! You're really a great lecturer!
One year studying it at college with exercises and more exercises. To finally be able to understand the whole thing in about one afternoon watching everything. Brian, You are my fucking hero!!!
Terrific videos. You're helping million of students out there , bless up !
I like your style of presentation of control theory. Thumbs up. keep up the good work.
I've been looking all night for a video that explains this concept well. Finally found it. Thanks man, really well done.
in Tamil they say "ezhutharivithavan iraivanavan" which means "who he teaches is equivalent to God"; These lectures are of very high quality; thanks
I love your video. Explains everything from top to bottom :)
YOU ARE AWESOME. I LEARNED A LOT AND HELPED ME TO UNDERSTAND EVERY TERM I HAVE SEEN IN CONTROL SYSTEM SUBJECT SO FAR. THANK YOU VERY MUCH. GREAT WORK!!!
Love you dude, missed this lecture and you probably explained it better than my professor
I just look for bode plot, thanks for you teaching.
nice teaching !
God bless you man. What a playlist this is!
Best video of control system in TH-cam
Thanks for putting these up! Subbed
Woah! Everything now makes sense! Thanks! :)
Wow!!! Brilliantly explained. Thanks a lot.
man, your videos are the best!
Спасибо большое за урок! Подача материала и объяснение на интуитивно понятны и просты!!
This is a very clear explanation. Well done.
Best Sir Who I ever Met in My life🙏
Theres a reason why my professor forms his lectures around your videos. Youre an amazing teacher.
Thanks man this is very helpful you are helping out a lot students out here
Very Clear. Thank you so much! Expect your next video.
This is the clearest Bode plot explanation I have found anywhere, thank you. I hope the next video is up before my exam on 10/25!
How was your exam ? :)
one of the best teachers in youtube, keep doing these God works
This is a crystal clear explanation of Bode plot. I had a question about the line that you draw by adding a constant term and linear line 7.04-7.10. When you add both lines don't you get a parallel line to the linear line as your output.
Great video! Good explanation! Cheers!
I don't think I would be able to pass control systems tomorrow if I didn't watch this, thank you so much Brian.(a third year Electric and Electronic engineer from Stellenbosch, South Africa)
in uct doing mine today and fingers crossed. Hope yours went well :)
thank you. this is so well explained
MAAAAAAANNNNNNNNNNNNN!!! Where have u been... oh, rather whether where have I been!!! Awesome !! Taking up studies after 4 long years, memory is sketchy.. thanks a lot!!!
Thank you for this awesome quality video.
side note explaining why jw is -ve !! now it finally makes sense!! thank you
Thanks Brian! It does make sense now.
Excellent explanation！Cheers!👍
Thank you! This is my teacher's first semester, and your explanation beat hers by a long shot...
Very good explanation
Thanks!
only a note: when you write H(s) -> infinity : because of H(s) is a complex number and the complex field is not ordered, it would be more correct to write |H(s)| --> infinity. Thank you for your awesome work!
I don't think you are aware of how much I love you
3:48 the response function should be multiplied by 1/w, so the amplitude of the graph should also be multiplied by 1/w .while we see that both input function graph and output are having the same amplitude(Correction)
awesome lectures! helped me allot!!
Damn thanks man! this is very helpful!
What a fricking hero you are ❤🎉
Love the videos. Any chance of posting a PDF of notes in the description for each of the BODE plot videos?
Thanks, these videos are great.
you helped me a lot with you videos thx a lot!
You are a Legend. Thank you!
I love you man!
God bless you!
you are great, brian
i am in love
You sir are brilliant.
Interesting. Thank you very much.
Thank you for the explanation I see it now ! Green and blue line looked the same in my screen that's why.
The argument of a complex number is the phase (or angle from the positive real line). For example, the complex number 1 + 0i has an argument of 0 degrees since it lies on the positive real line. But a number of 0 + 1i has an argument of +90 degrees since it's on the positive imaginary axis. It's the inverse tangent of the imaginary part divided by the real part.
Thank you very much Brian Douglas. I have a doubt. You said we can estimate a transfer function from bode plot, and I know how do it and I have done similar problems during my control system course. But why we need it in practical scenario such as in a plant or process?
Hello Brian, I love your lectures and it helps me a lot, thank you...I wanted to know what kind of software you are using for those, I like it and can not figure out which it is...
thank you
Thank you for the amazing clarifications but I am confused on How to add the 3 lines together? what am I missing here?
you are the best bro
Hello Divyagash, the transfer function was 3/s^2, which can be rewritten as 3 * 1/s * 1/s. There is a double pole at the origin. I actually drew three lines, a pink line for the constant, and a green line and a blue line for the two poles (although they are hard to distinguish). When you add all three lines together you get the red line which has a slope that is twice as steep as a single pole (or -40 dB/decade). Hope that Helps.
Shouldn't this 1/s graph start from zero not from a offset
This is a great explanation. Also, you sound a LOT like Adam Scott.
thank you, your videos are excellent and very helpful, however, in my opinion it would be great if you could keep the pace a bit slower as not everybody can catch up with this tempo.
Thanks for the great comment. Look for the next video on 10/24 and good luck on your exam!
Hi Brian, can I ask you something? Recently I bought a penpad to create a video like you but the quality is not that good. If you don't mind can you recommend the one you are using right now? Thanks you.
Very helpful video! Thank you very much for making it.
I have a question about the additive nature of multiple transfer functions and zeros (recipricol poles) when simplifiying the plotting process of the bode plots. Does this multiplcation rule of the TF work with complex roots? You mentioned in this video this was possible due to the logarithmic nature of the bode plot analysis. While I can see this working for the just real roots to the TF, I fail to see how this would work out for complex roots to the TF. To calculate the amplitude, you need to add the square of the real and imagineary components, which cannot be simplified by logarithm rules. i.e. 0.5 log (Real^2+Img^2) is already in its most simplified form.
if I ever graduate from electrical engineering it will be thanks to you
Excellent.
Hey Brian, Thanks for the amazing lectures..just a question.. at 2:23, why is the arg(1/w , 0) = -90 degrees?
kindly add link of next video in description as You tube does not suggest it automatically
Its difficult to take lectures this way
Does anyway know what is the software used in these tutorials? I wanna try it very much.
Can you explain the "just add the lines together" part when doing bode plot on 3/s^2? I basically got lost from there on out. Like the zero S wasnt thoroughly explained enough....that is for me.
thank you so much!
Might be a silly question but why don't we have a constant "+c" when integrating sin(wt) at 3:00? Also is this integration equivalent to convolving sin(wt) with the heaviside function in the time domain (i.e. is integration and convolving with heaviside the same thing?). Thanks for this video btw!
So helpful thank you, have my Controls final tomorrow, and may not fail anymore!!
Drew Stiffler did you pass?
I have em tomorrow xD
I did pass. Controls is rough, hope you do well!!
Drew Stiffler at which level do u study bro
im undergrad
I graduated with a BS in Aerospace Engineering in May 2016.
Drew Stiffler u think I could contact u for some info if u don't mind?
amazing ♥️
I'm a little confused on why you can add the phase shift responses together for a product of simpler transfer functions (the 3/s^2 example). There's no log operator when you calculate the argument, so how can you just add them together?
I believe at 2:17 the correct order of the phase eqtn should be "phase = arg( 0 , -1/omega)" because 0 lies in the real (x) axis and -1/omega in the imaginary (y) axis.
As you told us in the video that an integrator would bring a phase shift of 90 degrees and proved the same by giving us an example of sin wt and its corresponding integral cos wt..............can you show the samething for a ramp signal (kt) and its corresponding integral
how do i start by drawing a bode plot
when i have 1 pole at the origin 1/s = 0
and a 20 log k , k=100, 20 log k = 40
do i start the bode plot at 40 or 0??
Thanks very much
you re great!
Thank u very much
Isnt it all the other way? I mean the lines should be when w= 10 the decade is 10^(-1) and the one used is 10^(1), so the graph of zeros is the one of poles... or not?
Would the gain not be 1 over omega squared?
I think it would be good to show how to add the three lines together.
I would also appreciate this help... or could someone point me in the right direction?
Just sum them and you are there. You have some sites on the internet where you could play with grafs...
fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIxIiwiY29sb3IiOiIjRkYwMEZGIn0seyJ0eXBlIjowLCJlcSI6IigyLXgvMikiLCJjb2xvciI6IiMyMUMwRkYifSx7InR5cGUiOjAsImVxIjoiMSsoMi14LzIpKygyLXgvMikiLCJjb2xvciI6IiNGRjAwMDAifSx7InR5cGUiOjEwMDAsIndpbmRvdyI6WyItMTAuMTk1ODgzNTUxMzU5OTgyIiwiMTAuMTE2NjE2NDQ4NjM5OTgzIiwiLTUuODQzOTM0MzU2NDc5OTkxIiwiNi42NTYwNjU2NDM1MTk5OTMiXX1d
Hello Grisel, you have the equation slightly wrong. It is s = 1/(1/s). And if you do the algebra you'll see that the right side simplifies to s. All that I'm saying is that a pole at the origin is 1/s. And a zero at the origin is s, which is just 1 over the pole. And since we're plotting this on a log-log scale division becomes subtraction. So that a pole and a zero at the origin are just negatives of each other when you plot them on a Bode plot.
How do you add the 3 lines together to get the resulting line on 7:56?
Here you can see how grafs look like without using some softwer like matlab, etc...
fooplot.com/#W3sidHlwZSI6MCwiZXEiOiJ4XjIiLCJjb2xvciI6IiMwMDAwMDAifSx7InR5cGUiOjEwMDB9XQ--
Add their algebraic distance from jw axis
@@rohitchaturvedi2271 you mean Re axis?
How come phase also gets multiplied by -1? I get why we have the subtraction due to the property of log in the equation for gain but I don't get it for phase
Here I was wondering at 7:20, how the f will Douglas manage to cover the bode plot of Zeros within the remaining ONE MINUTE! wow man, well done, you handled it beautifully.
Could you explain why the response of s = response of 1 - response of 1/s please
@@rohlay00 s is basically differentiation while 1/s is integration.
If you take logarithm of the response of s, then this is the same as taking the log of the response of 1/(1/s) which is the response of 1 - response of 1/s
This is using properties of Log,
Log a/b = log a - log b
As he explained in 7:50
So imagine a transfer function of a car, input is gas pedal and output is velocity. I can interpret Nyquist plot here intuitively. But how would bode plot work here? I mean. should i imagine the gas pedal being depressed with variable frequency and then look a corresponding output :/ I am a bit confused, can you please help he clarify it.
Regarding zeroes, why is s= s/(1/s)? Thanks.
where can i get the book?
so arg is the same as polar notation?
brian nice work.... you made my life easier.
I got a query, what if G=s^2
so G(jw)= -w^2
this tells a negative phase of -180 degrees. but bode plot in matlab gives +180 degrees. bode plot for single zero came out correctly.
Syed Tirmizi Hi buddy, -180 or +180 both the same, -ve comes into picture when you measure the angle in clockwise direction and +ve in anticlockwise direction. Example +270 is equal to -90 degrees.
Have you completed ur book ?
Phase is liner...! Can you just add it at to get the total response as in 7.25?
4:44 1/w is NOT a linear equation, but it looks linear when plotted logarithmically
For those curious it's because the degree of w is -1, and linear equations require a degree of 1 or 0. (I could be wrong on the 0 degree part)