Nyquist 2 - sketching from gain and phase information
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- เผยแพร่เมื่อ 8 ก.ค. 2024
- Introduces the idea that an effective means of sketching of a Nyquist diagram is to transcribe frequency response gain and phase information. A few useful insights are presented to allow viewers to form sketches quickly from key trends in the gain and phase.
Lectures aimed at engineering undergraduates. Presentation focuses on understanding key prinicples, processes and problem solving rather than mathematical rigour.
The best video on this topic by far and I am not joking. I swear you can't imagine how much stress and time I spent on trying to understand but only your videos have finally managed to help me. Bless you
If I had found you sooner I'd be killing it in control systems. It's a shame that, at university, these topics are taught often by researches who begrudgingly teach. By the end of the quarter, only about a third of the students in my control system class were even attending lectures anymore.
Thank you for putting videos like this on TH-cam.
Nice video! It's the best video on the internet which explains how to plot nyquist from bode diagram.
True!
Thank you, sir! Very helpful insights on a poorly taught topic.
thank you sir, as a French student it was clearer than the French videos!
Very good video. I think it illustrates very clearly how to obtain the Nyquist diagram from Bode and clarifies several things along the way. It is also very easy to follow, even for someone who is not a native English speakers like me.
very very useful and informative video. all my doubts and queries are cleared watching this vid! hatsoff to John Rossiter for this awesome explanation :)
Thank you for this explanation, exspecially for the guidline for quick sketches :)
seven years ago and this video is very helpful thanks you and I'll subscribe
thank you man you helped me so much
thank you very much ... the turorial was useful and understandable...
Not sure if you forgot to mention or if I missed it but the plot he derives you need to reflect that about the real axis for the full Nyquist plot. Super helpful though here ive been trying to figure it out through Nyquest criteria and that is painful to say the least when all you want to determine is regions of stability...
Saviour thank you!!
How can you calculate or estimate 6/4*sqrt(3) = 0.9 so quickly?
Perfect video
Thank you!
One of the best videos I-ve ever seen! beautiful voice, very clear explination, very nice how you go from easy things to hard things and explain SO WELL EVERYTHING!! THANX
really, really helpful. I whis my teacher could exaplain like you do sir. Thank you!
how can draw bode plot easily ,to find out inwhich gain verus corner frequency
Excelent! thanks!
nice and helpful thanks
Thanks a lot
i like this
Hello,
At 10:43 how do you get so quickly at sqrt(3) as the middle?
Regards,
Olaf
Thx for sharing these videos
+Olaf van Buul Because there are only 2 poles it is easy to show that the phase of (s+a)(s+b) = 90 degrees when w^2=ab.Anthony
This has not been incredibly obvious for me from the start, but the gain and phase "arrows" are meant to be perpendicular to each other, right?
Hello,sorry what is the method to get 2 at the y axis at the moment 10:00? Thank you very much!
This is not in dB but just the steady-state gain ( that is 6/3)Anthony
I understood.Thank you!And have a nice day!
sorry if i don't see it, but how did you get sqr(4*12)?
+Diego Ballesteros sqr(4*12)=sqr(4*4*3)=4*sqr(3)
+John Rossiter Where does the 4 come from?
+lorenz ed I had same problem also. check previous video for details. G(jw)=6/sqrt((w^2+1)*(w^2+3)). Then just put w=sqrt(3). enjoy.
+lorenz ed I had same problem also. check previous video for details. |G(jw)|=6/sqrt((w^2+1^2)*(w^2+3^2)). Then just put w=sqrt(3). enjoy. discard my older comment.
where did you get the square root from in the calculations you do at 11:24?
A well known result (obvious from right angle triangle) is that atan(w/a)+atan(w/b)=90 when w^2=abAnthony
it should be 6/square root(4*6) at 11:20, right ?
Sorry some obvious typos are in some videos.
belle perf
I still do not know how to use the frequency while graphing?
The frequency is omega, we define s=j*omega. Then s is defined, in this paticular example as s=sqrt(3)*j. When you calculate the gain and phase you exclude the j.
Great explanation, but please correct 6/square root(4*7) this instead of 4*12
6/square root(4*6)