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Yes, it is!

❤🎉

Thanks for your (wait for it) ... feedback!

Good! I'm glad it's helpful :) Thanks for your feedback; finding out what aspect of the video worked for you is good. Regarding your comment, I agree, it's important to keep the big picture in mind --- that PID and state space controllers are designed in order to modify the characteristic equation --- but it's also easy to "miss the forest for the trees" and get too focused on the details of how to use the tools, such as root locus techniques etc.

You're welcome! I wish TH-cam didn't play ads in the video, but I guess they gotta keep the lights on :)

If you still have questions about this video, feel free to write back. Your question is not clear to me. Are you referring to the figure that comes up at 14:19?

Thank you for the feedback:) I hope that you enjoy continuing to learn about feedback controls!

Thanks for taking the time to comment :)

My guess is that you're referring to the equation at 15:42. That is the transfer function, so it's not that \{Theta}_1 equals 1, it's that \{Theta}_1 / \{Tau} = 1 / (J_e s^2 + c_e s + k_e). To get this transfer function, apply Newton's second law to the rotational system shown in the middle of the page. (That is, the second figure down, the one with J_e, c_e, and k_e.) Take the Laplace transform of the resulting differential equation. Factor out the \{Theta}_1 and rearrange the equation to get the transfer function shown at 16:01. If you still have questions, please feel free to comment.

That is a great question! The dimensions of 8x3 for the constraint matrix are correct. An explanation for why it's not invertible is that with four standard wheels you have an "overconstrained" system. There are only special cases where the wheel speeds are non-zero and also satisfy (all eight of) the constraint equations. For example, when all the wheel axes are parallel and all the wheel speeds are equal, the equations are satisfied. This video on "instant centers" might help with visualizing this. th-cam.com/video/FAJljU9HG9I/w-d-xo.html To satisfy the no-sliding constraint on all four wheels (none of the wheels slides sideways), the four wheel axes have to intersect at a common point. For a car, the axes do not all meet at a single point, and so we know that there is some sideways sliding going on during driving. In a case where there is no exact solution (like in a car) you can use the "pseudoinverse" of the constraint matrix to find the wheel velocities that come closest (in a least-squares sense) to satisfying the constraints.

Thanks for your feedback. Since Google has started showing advertisements even on non-monetized videos, I've been considering whether or not to remove my content. The fact that you and others are getting something out of the videos is an argument in favor of leaving them up. I wish you all the best with your modeling studies!

You're welcome. It's a neat topic, so I enjoyed creating the video. It should have a discussion of drawbacks to the pole-placement method :(

No, I've never made any material for those topics. All the best!

Alright ,Thank you 😁☺

To quote Jemaine from Flight of the Conchords: "Why exactly? Be more constructive with your feedback. Please." This seems like a decent presentation to me. Perhaps it's a bit slow, but choosing higher-speed playback fixes that.

Thanks for posting this question. The only way I know how to answer is pretty long and wordy, so stick with me and parse out all I'm about to say. Imagine a vector from the origin to the wheel; call this r_p. Imagine a second vector showing the angular velocity of the wheel when the wheel is rolling in the direction. This vector will always be perpendicular to the wheel's linear velocity (the yellow vector in the figure). Its orientation is based on the right-hand-rule. If you let your right hand curl in the direction of the wheel then the thumb of that hand will point in the direction of the angular velocity. In the figure, this would mean your right thumb would lie along the dashed line labeled \beta. Now finally we can define the angle \beta: it is the angle from the vector r_p to the vector showing the angular velocity of the wheel when the wheel is rolling in the direction denoted as positive. (That last part is important because if the wheel was rolling in its negative direction then the angular velocity would point in the opposite direction and the angle \beta would change by pi radians.) I hope that helps. It's really wordy, but the point is to imagine the two vectors and then let \beta be the angle from r_p to the other one.

@@drmattmarshall6545 thanks for taking the time to answer during New Year's Eve. I believe I understood the definition of beta, but my issue was visualizing a zero beta. I think I got it ... feel free to correct me. If alpha is 0 to start with, I can drag the wheel down, to the x axis, where its center will intersect with the x axis. At this point, the wheel's orientation is still south-east facing, the perpendicular ( normal to wheel) is still going in the north-east direction. Now, if I make beta angle 0, which is now the angle between Xr and the normal to the wheel, the wheel will be forced to move vertical, parallel to Yr. Right? ( thanks)

​@@missmahik, you're welcome to the reply. Your summary is exactly right! In the scenario you describe, positive wheel rotation would correspond to translation in the negative-y direction.

All the images came from mobilerobots.org, if that's what you're asking.

What aspect of the video causes you to say that?

It's possible to call sei() in an interrupt service routine (ISR) and thereby enable nested interrupts. Otherwise, if the button gets pressed during the LED ISR then the button interrupt will just get handled as soon as the LED ISR finishes.

@@drmattmarshall6545 I tried also your sugestion but still, the ISR for blinking a led is executed when the button is pressed (when PD2 is tied to ground). Otherwise, the led is in sort of a bouncing effect, similiar to not having pull up resitors( is either on, or off, or it flickers..its weird). github.com/tandrei96/Interrupts-Atmega328p_personal-projects/blob/main/double_interrupts/double_interrupts.c .I appreciate the help !

@@SteaM992 Your code looks nice, IMO. There's nothing I see that might cause weird behavior. I should think that nested interrupts would be unnecessary since your ISRs are so short that even if the button-press routine has to wait for the LED-blink routine the performance should be fine. My only suggestion right now is to declare cnt as a volatile variable but I'm pretty sure that's not causing any problems for you.

@@drmattmarshall6545 yup, is ok even though is not initialized as volatile... would be cool a future video that tackles 2 or 3 different interrupts, maybe you get the same error as me haha:) Cheers!

Good! The information comes from The AVR Microcontroller and Embedded Systems, by Mazidi, Naimi, and Naimi. It's a pretty readable book.

Application of the principles and equations is very important for learning, I agree. Thanks for the feedback.

@@drmattmarshall6545 Your videos are incredibly helpful to many students including myself. Thank you so much Dr. Marshall for creating these. I have an exam coming up and feel very prepared after watching every video in this playlist. God Bless.

@@kiwi.10th, that's great to hear! Thanks for taking the time to offer feedback.

Glad it helped :)

The second block is free to rotate about the contact point in any direction. Maybe me dragging it around does not show that very well, but you can imagine a coordinate system at the contact point and picture the part spinning about each axis, one at a time, and see that the point remains in contact with the plane (which is the only constraint imposed). I hope that helps.

Thanks for the feedback.

Oh, most definitely. The only difference is that the characteristic equations, that is, steps (3) and (4) at about 7:30 will be fourth-order.

And another difference is that the desired characteristic equation will be based on a second-order approximation. That is, the real parts of two of the poles will be at least about five times farther from the imaginary axis than the real part of the dominant, second-order pair of poles.

Well, great! Thanks :)

also when diving 20/2 to find theta p I feel like there should be pi involved somewhere?

Yes! Good catch.

Good, I'm glad it helped!

Good!