10

Assuming all the wheels are evenly spaced on the same circle, then more wheels is always more stable than less wheels. However, there is diminishing return as the number of wheels gets large. The metric of stability is how far from the center of the circle the center of mass can be before the chair tips over. The chair is stable whenever the center of ...


5

Without any more info, I think your problem arises from the values of $a_0$ and $a_1$. The answer is a little involved, so a bit of systems background is necessary. The short answer Your system should be stable, but I don't think you are simulating your system long enough to see it stabilize. Calculate your settling time and simulate it for at least that ...


4

I think the "health and safety" regulation about 5 wheels is a compromise between stability and cost. If your weight is on the edge of the chair seat and the chair has only 3 wheels, it is much less stable if you are in line with one of the wheels than if you rotate through 60 degrees to be mid-way between two wheels. That might happen (1) because the ...


4

Consider what happens in front wheel steering, such as an automobile: Turn the steering wheel slightly clockwise. The front wheels each turn slightly clockwise. Their rolling action causes the wheels to pull the front end of the vehicle slightly to the right. The back wheels follow the track of the front wheels, but in a slightly tighter turn. Holding the ...


3

If the algebraic multiplicity of an eigenvalue is greater than the geometric multiplicity, then the system matrix is not diagonalizable and there are vectors which are not linear combinations of the eigenvectors of the matrix. If you interpret the Lyapunov matrix as a "generalized potential energy function" (but not necessarily the real physical potential ...


3

A set is invariant with respect to its dynamics if $x(t_0)\in M \implies x(t)\in M$, this is not the case for the set $E$. The set $E = \left\lbrace x\in \Omega \mid \dot{V}(x) = 0 \right\rbrace$ does not need to be an invariant set, since it does not consider the solution of $x(t)$. I'll show this statement using the well known pendulum example. The ...


3

I would like to extent the already given answer by MrYouMath. So question 1 is pretty straight forward and you already got it right. If there's no right half plane (RHP) pole then it doesn't matter what gain you chose. Even for $A = B$, $G(s) = K$ yields a finite response. For Question 2 have a look at the Routh Hurwitz Array \begin{array} {|r|r|} \hline ...


3

Step 1 The first thing to do is to determine the desired poles. The natural frequency can be computed using $\omega =\frac{4}{\zeta T_s}$. This is a rule-of-thumb calculation for underdamped systems. The system here is slightly overdamped and is nonlinear as well. If the desired settling time is not obtained in the end, we have to come back and increase ...


2

Your system is open loop stable as the poles are at $s=-1$, $s=-3$ and $s=0$. Note, that if the order of the pole at $s=0$ is greater then 1, then the open loop system is also unstable. But closing the loop changes the poles of the system. If $F(s)$ is your transfer function of the open loop system, then the transfer function of the closed loop system is: $...


2

First, create the free body diagram for this system. If you cut through the spring $k_1$ and the damper $b_1$ you will get two forces $F_{k_1}=k_1(x_i-x_0)$ and $F_{b_1}=b_1(\dot{x}_i-\dot{x}_0)$ opposing the direction of $x_i$. Writing down Newton's second law of motion for the mass $m_i$ will result in: $$m_i\ddot{x}_i=-b_1(\dot{x}_i-\dot{x}_0)-k_1(x_i-...


2

A short answer for each of your questions in order: How do I describe passivity without the storage function? Passivity analysis is based around the idea of energy flowing into and out of the system. When your system is passive it means it is not generating energy, only storing it or 'passing it through'. Why is passivity useful? When dealing with nonlinear ...


2

I'm also a software guy and not completely master of the subject, but I tried to model your system. Each step is shown, so you can catch mistakes I did. Here is the schematic diagram of DC motor: Kirchoff's voltage law for electrical circuit: \begin{equation}\label{eq:kirchoff} \tag{1} V_s = V_l + V_r + V_e \\ \end{equation} Newton's $2^{nd}$ law of ...


2

I would take the wooden slats from the old bed and cut them so they can stand on end and be trapped between the steel tubing and the floor, thereby supporting the load that the steel tubing by itself cannot. If you wish, you can cut a round notch in one end of the slat so as to capture the tubing; this will help prevent the slats from popping out of position....


2

To solve this problem I would: Sketch the block diagram of the control network. Derive the required closed loop transfer function in terms of $G(s)$ and $C(s)$. Substitute the expressions for $G(s)$ and $C(s)$ to obtain the closed loop transfer function in terms of $s$. Calculate the poles of the closed loop transfer function using your preferred method. ...


2

This is because even if a system is Lyapunov stable it doesn't mean that $\|x(0) - x^*\| \leq \|x(t) - x^*\|\ \forall\,t \geq 0$. For example consider a simple mass spring system with unit mass and spring constant $k>0$ $$ \ddot{x} = -k\,x. $$ For this system one can use the following Lyapunov function to also show Lyapunov stability $$ V = \dot{x}^2 +...


2

Forget about the fact that it is supposed to be a PID controller, and think about how your circuit works for a constant (DC) input. Because of the capacitor C2, there is no feedback at all from the output of the op amp to its input for DC, so there is nothing controlling the output voltage and setting the DC gain of the circuit. Presumably, the output slowly ...


2

The sets $M$ and $E$ can be different. The set $E$ only considers $\dot{V}=0$ while $M$ also considers $f(x)$. Namely, invariant set means that for all $x(0) \in M$ the solution $x(t)\in M$ for all $t>0$, which can also be written as $x(0)+\delta\,f(x(0)) \in M$ for infinitesimally small $\delta$. For example consider the system \begin{align} \dot{x}_1 &...


1

You could get 3 compression springs for your 1/2/3 levels of tip. Very high spring rate for #1, medium for #2 and low for #3, depending on what you need. The limit could be once you fully compress the spring (all the coils touch), that much force is how much should be needed to tip over whatever you're talking about.


1

I suggest that you take a closer look at typical noise. I have highlight the area of the specification for you reference. The best method is to confirm is measure the output to confirm the output.


1

The gain and phase margins of a system are characteristics that can be obtained by studying one of the following well-known plots: Nyquist Plot, Bode Plot, Nichols Chart, Disk Margin which gives details in a different way (may be others that I'm not aware of). I will address the issue of your question by using the bode plot tool. When looking into a bode ...


1

You are probably using an FMD type printer. No matter how good your printer, or the density of the print, you will have porosity. You need the temperature to denature the proteins of the virus. If someone is wearing a mask all day breathing hot air into it, especially with PLA, there will be absorption into the plastics. I think you cannot assume that a ...


1

You could just multiply $H$ with minus one, change the positive feedback to negative feedback and proceed with the standard analysis for the Nyquist stability criterion. Here $H$ is the transfer function in the feedback path as shown in this image.


1

The object of the game is find an expression for the forces, X,Y,Z; and the moments L,M,N on the airframe expressed as a function of the state vector {x,y,z, phi,theta,psi, u,v,w, p,q,r} and the control vector {aileron,elevator,rudder,flap,thrust}. This is done by first picking a steady trim state and linearizing the dynamic system about this trim state. ...


1

Nope. A system can have better local stability than global. As an extreme example, a system can be stable within some boundary in the state space, and unstable beyond it.


1

Your (1) and (3) are along the right lines, but I'm not sure what (2) really means and some limit cycles don't look much like "regular orbits" anyway. There is always a danger in trying to produce a very general explanation of something - hence the (probably apocryphal) quote "This theory is so perfectly general that no particular application of it is ...


1

A root locus plot shows how the poles of the transfer function change, when some parameter of the system changes. The parameter is often the loop gain of the system, but it could be anything. A Nyquist plot shows how the response of the system changes with frequency, for one fixed value of the parameter and one fixed set of pole positions. So, you can't ...


1

Since you're worried about an open-loop system, in theory all you need to do is perform standard stability analysis on the open-loop system alone. The source of the input signal to the open loop system is irrelevant if you can prove that it is stable for a known/expected set of input signals. For example, for a linear system you would need to demonstrate its ...


1

This approach is just assuming a roll-out policy, so using a control policy after and including time step $k+p$ which is known to stabilize the system. For this roll-out policy the input and output constraints are not taken into consideration. So in order to guarantee stability $p$ has to be chosen to be sufficiently large such that these constraints will ...


1

Can you cross brace between the left and right side of the mechanism. If I understand correctly there are two parallel four bar mechanism that in tandem. I am afraid that if you place more constraints in the joints, if they are not aligned correctly the mechanism will not work correctly as it will try to bend some of the linkages. Proposal: Weld the red ...


1

How do you have an LQR with open loop control? I know that there would be these requirements if I were to implement a closed loop control A linear quadratic regulator is basically a multi input version of a state feedback controller; your $Q$ and $R$ matrices determine the "cost" or priority each input should take. With the $Q$ and $R$ matrices, you ...


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