What are Transfer Functions? | Control Systems in Practice
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- เผยแพร่เมื่อ 29 มิ.ย. 2024
- This video introduces transfer functions - a compact way of representing the relationship between the input into a system and its output. It covers why transfer functions are so popular and what they are used for.
Learn more:
- MATLAB Tech Talk: 4 Ways to Implement a Transfer Function in Code: bit.ly/3CDCwdV
- MATLAB Tech Talk: Nichols Chart, Nyquist Plot, and Bode Plot: bit.ly/3RhtZSb
- Gain a better understanding of Root Locus using MATLAB: • Gain a better understa...
- Laplace Transform - Chapter 32 (dspguide.com): www.dspguide.com/CH32.PDF
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10 minute video teaches me more than 4 hours of lectures. Bravo.
EXACTLY!
🤗
I have to say, you are the King of explaination of control system.
Control System Concept of full one semester course in 10 minutes
One problem with professors in Universities is many of them have little to no professional experience so they just run through the course with out explaining in details like this
Correct!! The REAL World is super extra fascinating and a book or Theory can't explain it all. Theory is a wonderful and useful tool..but Real world surprises us all the time.
All mine are generally very good at my Uni
Yeah, that's the problem I had with it. Most professors seem to teach their subjects in the most technical way possible, and because of that I always struggled to really learn anything until I had to apply it to a more advanced course.
How I wish I had watched this video when I was in the university, which might have saved me two years of gaining the same level of understanding! Thank you so much, Brian!!!
For anyone curious about the definition of gain (like I was) here's a more detailed explanation:
The final steady state value of y(t) is given by the formula:
y_ss = (lim s-->0) s*Y(s)
Let's say the input is a constant step: r(t) = A. Therefore R(s) = A/s. Also Y(s) = G(s) * R(s). So
y_ss = (lim s-->0) s*G(s) * R(s)
y_ss = (lim s-->0) s*G(s) * A/s
y_ss = (lim s-->0) A * G(s)
y_ss = A*G(0)
Gain is defined in the time domain as the final output value / input value = y_ss / r(t)
gain = A*G(0)/A
gain = G(0)
Wow. I like it more when gain is simply output amplitude over input amplitude.
yes. this is the king of explanation. i have been lost on this since my UG days 20 over years ago...it is so well explained. wish my lecturers could do the same.
I really like and find your video easy to follow and very helpful. Thank you
Your explanations are magic
What are transfer functions? More like "Wonderful knowledge and instruction!" Thank you so much for sharing.
You're getting me through my Robot control systems module at university Brian, not all heroes wear capes!
It helps a lot, Thank you Brian! You are the best.
For the Laplace Transform demos, I can recommend Khan Academy playlist. In a nutshell, Laplace Transform scans (like a sensor) for exponentials (natural/proportional to itself evolutions of something like bacteria growth, radioactive decay, compound interest stuff that can be modeled using exp(x)) and sinusoids (cos, sin and a mix of them, Fourier Transform does that too), poles are the exponentials, sinusoids or mix of both that your function contains, a resonance that creates a peak (the pole). e.g.: a wind can cause a bridge to sway at its natural frequency, which can break it (this has already happened). If we can model the bridge mathematically and use the Laplace transform, we can find out the bridge's natural frequency and prevent a future tragedy.
Looking forward to see more videos Brian :)
please more videos like this.
Great!
Brian u are awesom ❤
Well done
amazing
Loved this explanation
I appreciate it!
Not sure what is been done in the 3D plot at the 5:52 time mark, is it the magnitude or just for s=-0.1+0.2j, the values are 0.25+0.75j and so on across the Re-Im plane?
Your x and y axes (assuming that z is taken to be height) are the different combinations of sigma and omega, representing different s combinations. He then plots the real and imaginary portions of G(s) given that specific sigma and omega coordinate in the z direction with the red and blue markers. This ends up showing the notions of zeroes and poles, sending G(s) to infinity or 0. Hope this helps.
Greatly appreciate the lecture. Question: let's say that you have a circuit in a 'black box' and you know it is likely to be represented by a transfer function, but you have no clue as to the circuit, components or values. Can you do a few tests and from the output, figure out what the black box acts like? Like searching for an equivalent circuit.
Thanks in advance.
yes, look up system identification
*Where is the link to the Laplace movie mentioned in that movie?*
Hi. I am new to control systems. I have one question. Can transfer functions be used for MIMO systems?
They only work for SISO systems, if memory serves.
Oh oh oh! So basically the 2 numbers output by the transfer function are each mapped to a separate vertical axis to make 2 3D models with the same xy base. (if z is the vertical axis) !!! Thankfully figured that out now. Could have explained that better in the video. No offense.
Where are the said links about Laplace transformation?
Hmmm, it got left off. Thanks for letting me know! I'll get it added. In the meantime, this is my favorite reference for the Laplace Transform: www.dspguide.com/CH32.PDF
Is learning about control systems through one's EE curriculum?
That jump at about 5:49 was too large. So basically s is xy and the transfer function is z? Why is it outputting 2D then and how are those being plotted as 2 different z's?
So by "gain", UNITY GAIN
it travels 7.5m... I don't know rick...
Polynomial multiplication IS convolution. Maybe you meant some other pointwise multiplication?
I need more useful video to fight against dictators politic in VietNam
thats a pretty bad explanation; its just a bunch of statements
If Y(S) is the Laplace transformed signal of input u(t) and u(S) is the Laplace trasformed differencial equation of the model of the system d/dt any^n+an-1y^n-1...a0y^0 so why we tell that transfer function is the laplace trasformed signal of syetem output diveded by the laplace trasformed signal of system input and not telling instead that is the laplace transformed signal of system input divided by the Laplace transformed homogenus equation of the system?
I really dont understand that.
anyone explain how that deferential equation is written 01:30