I want the system to stabilize the pendulum while keeping the cart toward the center of the track.
So it needs to balance the pendulum in its inverted state and if you tap the cart the pendulum needs to stay stabilized. That's it.
That is what I thought, but my professor has argued with me about how I need a reference input. So I simply set the reference input to zero - i.e. just leave the reference input out?
So how would you suggest I account for a step input? Does the state feedback already account for that, or...
No, that is what I thought, but I wasn't sure. So for my application I am implementing the control system described on that page - but I want all of the states to go to zero, even after a step input (bumping the cart).
So do I just play around with that Nbar precompensator value until I get...
But before they add that it only deviates 3mm from the initial point, which is much less than 200mm. Am I interpreting the graphs correctly: the cart moves to the left -3mm in the second graph, and to the right 200mm in the third graph?
Yes, but I can't find a good explanation of how the reference input works. I was trying to ask a very general question. Let me ask a very specific one instead.
My question is in regards to the section of "Adding precompensation" on this page...
In the design of a State Space controller using state feedback the input to the plant is given as (where y=x & D=0): u=-Kx
If there is a "reference input" then it would be:
u=-Kx+r
So the state feedback without the reference input simply drives all of the state variables to zero. When there is...
Okay, so the mapping from the s-plane to z-plane is as follows:
z=e^{sT}
And from s-plane to z-plane:
s=\frac{1}{T}ln(z)
Where T is the sampling period.
So if the observer poles are supposed to be "10 time faster" than the controller poles can I do the following, given a digital control...
When designing a State Observer for a control system the observer poles (eig(A-LC)) should typically be about 10 times faster than the controller poles (eig(A-BK)).
But when designing a digital control system what does it mean for the poles to be faster? For the analog case it simply means...
Okay, so referring to my previous post, I am getting to this point:
(-\frac{L}{K}\frac{d}{dt} \dot{i} - \frac{R}{K}\frac{d}{dt}i +\frac{1}{K}\frac{d}{dt}V) = -\frac{b}{J} (-\frac{L}{K} \dot{i} - \frac{R}{K}i +\frac{1}{K}V)+\frac{K}{J}i
I can simplify this further to get the last row of the...
Okay, I will walk through what I have done to try to figure out i in terms of V :
\frac{d}{dt}i=-\frac{R}{L}i + \frac{1}{L}V -\frac{K}{L}\dot{\theta}\\
\frac{d}{dt} \dot{\theta} = -\frac{b}{J} \dot{\theta}+\frac{K}{J}i
Solve the first equation for \dot{\theta}:
\dot{\theta}=-\frac{L}{K}...
Well, from your equations that separate the electrical and mechanical description of the motor:
\frac{\mathrm{d}}{\mathrm{dt}}i = -\frac{R}{L}i +
\begin{bmatrix}
\frac{1}{L} & -\frac{K}{L}
\end{bmatrix}
\begin{bmatrix}
V\\
\dot{\theta}
\end{bmatrix}
\frac{\mathrm{d}}{\mathrm{dt}}...