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In nonlinear control theory, you will recognize most concepts such as controllability and observability where the linear case is often a special case of the nonlinear case. I would highly recommend digging into linear control theory first if you have not done so. Depending on the course you take, concepts such as Lyapunov stability are discussed here, which ...


4

Your systems shows extremely close pole-zero cancellation. So much even that it nearly removes 4 poles and zeros. Lets look at why, starting with the Bode plot: The magnitude plot is constantly decreasing with a slope of -40dB/decade. Following basic rules this already implies that at the given frequencies, the system can be approximated using a double ...


3

IMHO the problem you observe is mainly related to limitations of your system. Notice, that up until 55 degrees C the heating element can follow the gradient of temperature wrt to time. Just after 55 degrees C, the system is unable to heat fast enough the wooden box. (probably the insulation is not sufficient). So beyond that point, the error accumulated and ...


3

When identifying a system, you effectively compute a system that happens to have the same response to your inputs as the physical system. However, this does not mean that the states do physically represent the same. You find a transformed system that can be represented as such: $$T\dot{x} = (TAT^{-1})Tx + TBu$$ $$y = (CT^{-1})Tx + Du$$ where $T$ is a ...


3

Based on the information you've given, I believe your professor is suggesting that a friction term can be represented as shown in the following block diagram. The transfer function $G(s)$ relates force ($F(s)$, the input in the diagram) to velocity ($sX(s)$, the output in the diagram) for a mass-spring system. The damping ($\rho$) is represented in the ...


3

Let's consider the transfer function of a dc motor: $$ \frac{\dot{\Theta}(s)}{U(s)} = \frac{K}{(Js+b)(Ls+R)+K^2} $$ where the variables represent the following terms: $\dot{\Theta} \rightarrow$ Motor's Rotational Speed (output) $U \rightarrow$ Armature Voltage (input to motor via some motor drive) $J\rightarrow$ Moment of Inertia $b \rightarrow$ Viscous ...


2

First of all you could take a look at my answer here Is there any other controller than PID Controller? . It kind of deals with the fact of the various controllers out there. The problem of the best controller to use is something really debatable. The best controller always depends on what you have to control and in most cases the engineer has to design some,...


2

When dealing with block diagrams be aware of the fact that you can not only derive the general transfer function (input-outuput) but also the disturbance transfer function, the error transfer function, the noise transfer function and some others. The one which you should derive depends on what you would like to study for the particular given system. Each one ...


2

Due to the fact that you are mentioning the $A,B,C,D$ matrices I assume that your model is a linear one of the form: $$ \dot{x} = Ax + Bu$$ $$ y = Cx + Du $$ where $x$ is the state vector and $y$ the output of the system. One thing you can do, at least I would go like this, is transform your system to the $\text{s-domain}$, that is obtain the input to output ...


2

TL;DR: Use only the things/concepts you need for an application. First I'll just say, that the D term is not recommended for the particular system that you are describing. A PI control would be much more appropriate. I'll try to reiterate some points already made here from my point of view. In order for a control system to work properly the time between ...


2

Basic control theory is very useful in a wide range of situations. Really advanced control theory is specialized and useful to theory and complex design of certain actual control systems. I had math in grad school, and EE undergrad, worked as a Systems Engineer. I had control theory in both undergrad and grad school. I do not see any use for control theory ...


2

Let's start by obtaining the state space form of the closed-loop system (closed loop means that you plug in the equations the expression of the controller). The controller of this specific system has the following form: $$ u = -Kx+r $$ This is a full state feedback controller with feedforward gain of $1$ (feedforward is the gain by which the input signal is ...


2

It can be noted that the high-pass filter can also be written as $$ \frac{s}{s + \omega} = 1 - \frac{\omega}{s + \omega}, $$ such that $$ \frac{s}{s + \omega} Y(s) = Y(s) - \underbrace{\frac{\omega}{s + \omega} Y(s)}_{\mathcal{L}\{\eta(t)\}(s)}, $$ with $Y(s) = \mathcal{L}\{y(t)\}(s)$ denoting the Laplace transform of $y(t)$. In order to show that the second ...


2

Input signal: ±U V. Scaled output signal: ±1 V. Required gain: $ \frac 1 U $. Figure 1. Possible solutions. If U > 1 then a simple potential divider may suffice. $ V_{OUT} = \frac {R_2}{R_1+R_2} $. Bear in mind that whatever follows this circuit may load it somewhat so keep the parallel combination value of R1 || R2 < 10% of the value of the ...


2

I am having trouble imaging a PID controller without a feedback loop. The reason is that P, I and D controllers each calculate their response based on the error between the target and the current state. If there is no feedback to provide the current state, then I can't imagine how they would work in principle. UPDATE: After reading Abel's post, I realized (...


2

Testing physical controller hardware in "open loop" isn't uncommon, but it would just be part of a test, and the feedback line can be split at a point and treated as both input and output of a test setup. For example you can use the feedback sensor to measure what the system is doing and inject a fake sensor signal along with a desired input to ...


2

As you already stated at your question, PID-Controllers incorporate the I-Term in order to able to reject any external disturbance (unmodeled dynamics, linearizations, etc) and as a result achieve the desired output performance by driving the steady state error to $0$ ($e_{ss} \rightarrow 0$). Full state feedback control is like having only the P and D term ...


2

The arduino does impose some limits. But a classical frequency identification is build from a few steps: As the other comments already suggest, the design of a suitable input. As mentioned, a sinusoidal input (I actually recommend multisine) is suitable to accurately identify the response of a few frequencies. White noise or band-limited white noise can ...


2

How do I proceed with finding the transfer function? The concept of (Laplace transform) transfer functions are applicable for linear systems. So, you need to linearise your equations of motion. Linearisation uses the concept of Taylor series. In the example given in your question, that reduces to removing the term $\dot{\theta}^2$. This comes from the ...


1

The equation of motion after the impulse is $$m\ddot x + c\dot x + kx = 0$$ with initial conditions $$x = 0, \dot x = \dot x_0 \text{ when } t = 0.$$ You have found the frequency and damping ratio already, so you can rewrite the EOM as $$m(\ddot x + \beta \omega\dot x + \omega^2 x) = 0$$ where you know $\beta$ and $\omega$. Scale your measured results to ...


1

uselessly general procedure: write out equations for the kinematics, then the forces, then linearize the resulting differential equation, then transform the linearized version to frequency domain, which is your transfer function. This result will vary (i.e. is parametrized by) the range of positions. A "good" mechanism would have a transfer ...


1

Shouldn't the error equation have $-H_iq_i$ instead of $H_iq_i$? $$e_i=−d_{r,i}+d_i$$ $$d_i=q_{i−1}−q_i$$ $$−d_{r,i}=−H_iq_i$$ Therefore (modified equation): $$e_i=-H_iq_i+q_{i−1}−q_i$$ $$q_i=e_iPDG_i$$ $$q_i=PDG_i(-H_iq_i+q_{i−1}−q_i)$$ $$q_i=-PDG_iH_iq_i+PDG_iq_{i−1}−PDG_iq_i$$ $$\frac{q_i}{q_{i−1}}=\frac{PDG_i}{1+H_iPDG_i+PDG_i}$$


1

From the Nyquist plot we see that the maximum angle, $\angle H(s) = -360^{\circ}$. As the transfer function $$ H(s) = \frac{k}{(s+a)^2(s+1)^n} $$ has no zeros and all the poles are in the open left half-plane (LHP) as $a > 0$, the phase of $\angle H(s) = -360^{\circ}$ can only be achieved if the transfer function has 4 poles in the LHP, i.e. $n = 2$. The ...


1

Taking the time derivative of $y$ yields: $$ \dot{y}=\frac{\partial f}{\partial x}(f(x)+u) $$ We need $y(t)=y(0) e^{-\beta t}$ but this is possible if and only if $\dot{y}=-\beta y $ , if the hessian is invertible then this is possible if and only if $u=-f(x)-\left(\frac{\partial f}{\partial x}\right)^{-1}\beta y$. To get an intuition on why the hessian must ...


1

You can show it with Laplace transform. Let's say Y = ℒ[y], H = ℒ[η], and ω_p is the filter pole. (Let's not simply call it ω -- when omega is used as a parameter rather than a variable, a subscript should be added for clarity) dη / dt = ω_p (y - η) ... transform ... s * H = ω_p ( Y - H ) s * H / ω_p = Y - H Y = H ( 1 + s/ω_p ) H/Y = 1 / ( 1 + s/ω_p ) which ...


1

MPC finds the optimal input $u^*$ which is the input that minimizes the cost function $J$ or $c$. This means that regardless of what this actual value is, its proven to be its minimum. As such, multiplying the cost function with any constant value does not change this minimum, it just scales the value. Therefore, $\frac{t_{hor}}{N}$ does not affect the ...


1

The derivative is the rate of change of the error, $ \frac {de}{dt} $. The current position (at t = Ns) is 80 m, and the previous position (at t = (N-1) s) was 60 m. ... this gives a derivative error of (20 - 40)/1s = -20 m/s. Correct error is negative. As you pass through the target position (with the figures you've given us) the error will be zero but ...


1

I would think that this refers to using mathematical and physical principles and equations to predict the behaviour of a control system. The opposite would be to empirically design a control system, by implementing it and measuring it.


1

Suppose we have a system $G(s) = \frac{1}{s(s+1)}$ and controller $K$ (this is purely a gain) and we close the loop: $$T(s) = \frac{KG(s)}{1+KG(s)} = \frac{K\frac{1}{s(s+1)}}{1+K\frac{1}{s(s+1)}}$$ $$ = \frac{K}{s^2+s+K} $$ As you might notice, the poles of this closed-loop equation depend on the value of $K$: $$s^2 + s + K = 0 \rightarrow s = -0.5\pm\sqrt{0....


1

your question is ill-conditioned: If $A$ must have a higher degree than $B$, 2 things can happen: $A$ is a constant, meaning $B$ must be 0: which means you cannot solve the equation as there is no $s$ term in $A(s)(s^2-1)$. Or $A$ is of order one (atleast one $s$) and $B$ is a constant: Which means the function can not be solved as there will be a $s^3$ ...


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