Very helpful thanks. I think the transfer function at the end can be simplified further. If you expand out the brackets there are two m^2*l^2 terms which cancel out to give -(J(M+m)+M*m*l^2) for the s^2 term.
in my logic, the horizontal displacement of the pendulum should be: x - l*sin(theta), because when we assume that theta is small, sin(theta) ~ theta, it means that the rotation of the pendulum will opposite to the movement of the cart.
The direction of movement of pendulum depends upon the direction of displacement of the cart (opposite, as you mentioned) and also up on the rotational velocity of the pendulum. The Figure on the slide depicts the situation at a particular time instant, and at this particular time instant, the displacement of the CoG of pendulum is x+l*sin(theta).
Very useful. Thank you♥ What if the pendulum is at the left side? What are the changes in the equation? Like at the right, x + lcos(theta) → x - lcos(theta) at the left.
Yes, you are right. However, there is another way to look at it. There are two things associated with angle theta, that is, it's magnitude and it's direction. If the pendulum is inclined towards left, we can still consider it with right inclination but with a negative magnitude. And in that case, the expressions written will be applicable for both, the right and the left, inclinations of the pendulum. This situation is similar to the strategy utilized for the analysis of electrical circuits. We assume a direction of current and if actual direction of current is opposite to the assumed direction, it's magnitude comes out to be negative.
Simple! You will derive a relation between the voltage applied to the motor and the force (force is related with torque ) generated by the motor and then cascade it with the model of the pendulum on a cart system.
Example • A 460-V, 25-hp, 60-Hz, 4-pole, Y-connected induction motor has the following impedances in ohms per phase referred to the stator circuit: •R1=06.41 • R2 x1=1.106ohm xm=26.3 = 0.332 ohm X2= 0.464 ohm Find the maximum torque and the corresponding speed and slip • Find the starting torque When the rotor resistance is doubled, determine the speed at which the maximum torque occurs and the new starting torque • Plot the torque-speed characteristics of the original and doubled rotor resistances
I have a question: when modeling angular movement, can we set the center of rotation at the joint between pendulum and the cart instead of center of mass of the pendulum? If so is there advantage or disadvantage for each method?
At 15:42, for the simplification of 3rd dynmaic eqn (m (x+lsin(theta))''), can you justify replacing sin(theta) with theta? Shouldn't you take 2nd order derivatives of sin(theta) first then proceed the approximation?
The justification of this replacement is at 13:45. We know from our knowledge on trigonometric relations that, if theta is small, then sin(theta) is nearly equal to theta (this can also be verified from the Taylor series expansion of sine(theta)). Our objective in replacing sin(theta) with theta is to obtain a simplified model (linearized model). If we first take the derivative and then make the approximation, this will complicate the things. We know that d/dt (sin (theta) ) = cos (theta) * d/dt (theta) and then second derivative is even more complex. Thus to avoid complexity , we have first made approximation.
@@MAFarooqi Thanks for the response, however I am not quite convinced for the approximation. My argument is that \theta \approx \sin\theta does NOT imply \theta'' \approx (\sin\theta)'' near \theta = 0. I have an example, if theta = a*sin(1/a^3), then theta -> 0 when a->0. However -theta'' * sin(theta) explodes when a->0. Therefore \theta'' \approx (\sin\theta)'' cannot hold.
sin(theta) ~ theta --------- (i) Lets check it for small theta (please set your calculator in radian mode) sin (0) = 0 (Equation (i) is satisfied) sin(0.01) = 0.009999 (Equation (i) is satisfied) sin(0.1) = 0.0998 (Equation (i) is satisfied) sin(0.5) = 0.479 (since theta is becoming larger, Equation(i) is becoming less applicable) Now coming back to your observation: We cannot say that theta ~ a*sin(1/a^3), whatever is the argument of sine, the same should be on the other side of the inequality. For example, you may check (1/a^3) ~ a*sin(1/a^3), and it will be satisfied.
![Inverted Pendulum on a Cart [Control Bootcamp]](/img/1.gif)
Very helpful thanks. I think the transfer function at the end can be simplified further. If you expand out the brackets there are two m^2*l^2 terms which cancel out to give -(J(M+m)+M*m*l^2) for the s^2 term.
Man, u really really really saved me a lot! Very great explanation!!! Thank youS2
That the explanation i required...Its perfect.
Thank u sir 👌
Thanks for such a good explanation.
thanks for this lecture
mash Allah thank you sir
Many thanks for this excellent explanation!
You are welcome!
very nice
Is there no state-space for this?
Yes, you can obtain state space model for it. State space will be discussed in a later lecture.
Me too. I need it to be passed this october13.
We need to make a state model. I wonder how to apply the transfer function to state-space.
perfect explanation!
Thanks.
Thank you for such excellent explanation
You are welcome!
thanks sir
in my logic, the horizontal displacement of the pendulum should be: x - l*sin(theta), because when we assume that theta is small, sin(theta) ~ theta, it means that the rotation of the pendulum will opposite to the movement of the cart.
The direction of movement of pendulum depends upon the direction of displacement of the cart (opposite, as you mentioned) and also up on the rotational velocity of the pendulum. The Figure on the slide depicts the situation at a particular time instant, and at this particular time instant, the displacement of the CoG of pendulum is x+l*sin(theta).
Thanks a lot for the great class. Could you teach how to model the inverted double pendulum too?
Yes, soon
Hello bro do you get material or documents or videos about (inverted double pendulum )?? If yes, please share it
Sir Could you teach how to model the inverted double pendulum too? please
Very useful. Thank you♥
What if the pendulum is at the left side?
What are the changes in the equation?
Like at the right, x + lcos(theta) → x - lcos(theta) at the left.
Will it be? if the pendulum is placed at the left side.
Hcos(theta) x l - Vsin(theta) x l
Yes, you are right. However, there is another way to look at it. There are two things associated with angle theta, that is, it's magnitude and it's direction. If the pendulum is inclined towards left, we can still consider it with right inclination but with a negative magnitude. And in that case, the expressions written will be applicable for both, the right and the left, inclinations of the pendulum.
This situation is similar to the strategy utilized for the analysis of electrical circuits. We assume a direction of current and if actual direction of current is opposite to the assumed direction, it's magnitude comes out to be negative.
@@MAFarooqi Thank you so much. You are great♥
Legends watch this 3hr before the exam
love your video! How would I proceed if the force applied is coming from a motor and I want a equation with the voltage V_a in term of x?
Simple! You will derive a relation between the voltage applied to the motor and the force (force is related with torque ) generated by the motor and then cascade it with the model of the pendulum on a cart system.
Thank you sir... Very informative...
Thanks
Example • A 460-V, 25-hp, 60-Hz, 4-pole, Y-connected induction motor has the following impedances in ohms per phase referred to the stator circuit: •R1=06.41 • R2 x1=1.106ohm xm=26.3 = 0.332 ohm X2= 0.464 ohm Find the maximum torque and the corresponding speed and slip • Find the starting torque When the rotor resistance is doubled, determine the speed at which the maximum torque occurs and the new starting torque • Plot the torque-speed characteristics of the original and doubled rotor resistances
I have a question: when modeling angular movement, can we set the center of rotation at the joint between pendulum and the cart instead of center of mass of the pendulum? If so is there advantage or disadvantage for each method?
Yes, you can also take the joint as the center of rotation. In both the cases, you will finally get the same mathematical model.
At 15:42, for the simplification of 3rd dynmaic eqn (m (x+lsin(theta))''), can you justify replacing sin(theta) with theta? Shouldn't you take 2nd order derivatives of sin(theta) first then proceed the approximation?
The justification of this replacement is at 13:45. We know from our knowledge on trigonometric relations that, if theta is small, then sin(theta) is nearly equal to theta (this can also be verified from the Taylor series expansion of sine(theta)).
Our objective in replacing sin(theta) with theta is to obtain a simplified model (linearized model). If we first take the derivative and then make the approximation, this will complicate the things. We know that d/dt (sin (theta) ) = cos (theta) * d/dt (theta) and then second derivative is even more complex. Thus to avoid complexity , we have first made approximation.
@@MAFarooqi Thanks for the response, however I am not quite convinced for the approximation. My argument is that \theta \approx \sin\theta does NOT imply \theta'' \approx (\sin\theta)'' near \theta = 0.
I have an example, if theta = a*sin(1/a^3), then theta -> 0 when a->0. However -theta'' * sin(theta) explodes when a->0. Therefore \theta'' \approx (\sin\theta)'' cannot hold.
sin(theta) ~ theta --------- (i)
Lets check it for small theta (please set your calculator in radian mode)
sin (0) = 0 (Equation (i) is satisfied)
sin(0.01) = 0.009999 (Equation (i) is satisfied)
sin(0.1) = 0.0998 (Equation (i) is satisfied)
sin(0.5) = 0.479 (since theta is becoming larger, Equation(i) is becoming less applicable)
Now coming back to your observation: We cannot say that theta ~ a*sin(1/a^3), whatever is the argument of sine, the same should be on the other side of the inequality. For example, you may check (1/a^3) ~ a*sin(1/a^3), and it will be satisfied.
so this is the newtonian way to describe/analyse the system?
Yes.
Salam I am a aerospace master student
Thank you sir...
Most welcome
👍🏼