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5

Suppose we have a third order polynomial in the form : $$ s^3+a_2s^2+a_1s+a_0 = 0$$ There is nice caveat for third order systems which is derived from the Routh-Hurwitz stability criterion. In order for this polynomial to be stable the following three conditions have to be met (trying to derive the Routh-Hurwitz table will be a total mess for this particular ...


4

Your closedloop crossover frequency (when the magnitude of $P(s)\,C(s)$ is equal to one) lies at roughly 1 rad/s. This means that the feedback controller already causes the system to track reference signals that have a frequency content sufficiently below that. So adding feedforward that only approximates the inverse of the plant ($C_f(s)\,P(s)\approx1$) ...


4

See doc nyquist: The nyquist function has support for M-circles, which are the contours of the constant closed-loop magnitude. M-circles are defined as the locus of complex numbers where $$T(j\omega) = \left|\dfrac{G(j\omega)}{(1+G(j\omega))}\right|$$ is a constant value. In this equation, $\omega$ is the frequency in radians/TimeUnit, ...


4

To my knowledge the simplest way to accomplish the transfer function... $$ G(s) = \frac{E_o(s)}{E_i(s)} = \frac{0.0364s}{0.0002s + 1} = \frac{KTs}{Ts +1} $$ ...is the following high-pass filter circuit: The transfer function of this circuit is: $$ \frac{E_o(s)}{E_i(s)} = \frac{R_2+R_3}{R_3}\frac{R_1 C s}{R_1 C s + 1} $$ So let $R_1C = T = 0.0002$ and $\...


4

With your code, and with $c = 0.3$, I get the following results: The amplitude and phase look OK. But the displacement does not show any damping. To see why read on below. I'm not sure about your notation. So the equations below may differ from those in your textbook. ODE The ODE you are trying to solve is $$ \ddot{u} + 2\xi\omega_{\text{res}}\dot{u}...


4

Short answer: you can't directly import CATIA assemblies in SimMechanics. Long answer: SimMechanics supports import of CAD models (assemblies or parts) from Pro/Engineer, SolidWorks or Autodesk Inventor via SimMechanics Link (see Importing CAD Models on the MathWorks website). The SimMechanics Import XML Schema should allow import from any CAD system, but ...


3

The trick to this one is at the end there. You took the reciprocal of your y data but what you need is the reciprocal of the entire equation. Original: $y = a\frac{x}{b+x}$ Reciprocal: $\frac{1}{y}=\frac{1}{a}\frac{b+x}{x}$ So, when you simplify that a little, you'll get: $y^{-1}=\frac{b}{a}x^{-1}+\frac{1}{a}$ Hopefully, you can see that this is a linear ...


3

For a MIMO system $y(s) = G(s)d(s)$, with $m$ inputs and $l$ outputs. Consider a fixed frequency $\omega$ where $G(j\omega)$ is a constant $l \times m$ complex matrix. For the sake of simplicity the matrix $G(j\omega)$ is written as $G$. In short, the singular value decomposition (SVD) states that any matrix $G$ may be decomposed into an input rotation $V$, ...


3

The general form of a transfer function for a first order system is the following: $$ T(s) = \frac{K}{\tau s+1} $$ where: $\ K \rightarrow $ DC Gain of the system $\ \tau \rightarrow $ Time constant of the system The above form can also be written in another way as described below: $$ T(s) = \frac{K}{\tau s+1} = \frac{b_0}{s+a_0} $$ By matching the ...


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 ...


2

The plant and controller: $$\text{sys}=\frac{4700 s^2+4393 s+3.245\times 10^8}{s^4+7.574 s^3+120200. s^2}$$ $$pid=0.287\, +0.008 s+\frac{0.5}{s}$$ The closed-loop system obtained as $\frac{pid*sys}{1+pid*sys}$: $$csys=\frac{37.6 s^4+1384.04 s^3+2.59961\times 10^6 s^2+9.31337\times 10^7 s+1.6225\times 10^8}{1. s^5+45.174 s^4+121584. s^3+2.59961\times 10^...


2

For the pole at $5 i$, the contour that has to be considered is $5 i+\epsilon e^{i \theta }$. Here $\epsilon \to 0$ and $\theta \in[-\frac{\pi }{2},\frac{\pi }{2}]$. The denominator of the transfer function becomes $\left(5 i+\epsilon e^{i \theta } \right)^2+25$. Expanding and neglecting higher order terms, we get $-10 \epsilon \sin (\theta )+10 i \epsilon ...


2

Your implementation is: for i = 1: size(v_phi,1) y(i) = r(i)*((v_phi(i)/r(i) - v_phi(i+1)/r(i+1)) / (r(i+1) - r(i))); y1(i) = (v_phi(i)/r(i) - v_phi(i+1)/r(i+1)) * (v_phi(i)/r(i)); end That means that the discretization you are using is $$ y_i = r_i \left(\frac{v_{\phi,i}}{r_i} - \frac{v_{\phi, i+1}}{r_{i+1}}\right) \left(\frac{1}{r_{i+1} - r_i}\...


2

You haven't specified which method you are using, so i am going to assume Finite Volume. In that case you have a staggered grid like so: The vertical lines are the faces of you cells and the circles are the centers of you cells. Your left boundary is located at the face at $x=0$ and your right boundary is located at the face at $x=L$. The domain of size $L$ ...


2

So assuming that we are dealing with LTI systems of the form $$ \begin{align} \dot{x} &= A\,x + B\,u \\ y &= C\,x + D\,u \end{align} \tag{1} $$ with $x\in\mathbb{R}^{n_x}$, $u\in\mathbb{R}^{n_u}$, $y\in\mathbb{R}^{n_y}$ and which is stabilizable and detectable, however I will assume that the full state is known so no need for constructing an actual ...


2

"Windows 2010" is not a Windows version so i'm guessing that you are referring to Windows 10. In general all software that came out since Windows Vista (2007) also runs fine on newer Windows versions like Windows 7, 8 or 10. Since Matlab 2008 requires only a Pentium 4 with 512MB memory it will probably run fine on any current system


2

So, look at the time history that you added in edit 3. Notice how it is never ever ever below 1. That's very telling. Let us assume that you have a MEMS accelerometer is able to read down to 0 Hertz. So in the vertical direction, we expect to see 1g due to gravity PLUS the vibration. In other words, we expect to see $y(t) = 1 + A cos(\omega t)$, where A ...


2

I think I understand where you are stuck - it appears (I don't "speak" Matlab) you're trying to solve u directly, without adressing that u goes into Re. Here's what I'd do: Guess $u$ for the smallest diameter, calculate $Re$ from this $u$ to arrive at a new $u'$. Repeat/Reiterate by calculating $Re$ and a new $u'$ until the difference between $u$ and $u'$ ...


2

There are two aspects to this problem that are each significantly more complicated than what you seem to have tried so far: The angle of 45 degrees by which the cross section is changing dimension is just too much for the usual approximations to make sense. At such a large angle, plane section will not remain plane and the standard beam equations no longer ...


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

It means that there is a positive offset of one. So your signal is $$ x(t) = 1 + 2 \cdot \sin(\omega t)$$ where: 1 is the offset 2 is the amplitude of your signal (not 4).


2

Taking the norm of a system is something different that taking the norm of a matrix. Luckily, matlab has you covered: https://nl.mathworks.com/help/control/ref/lti.norm.html This page even shows an example in discrete time. However, what I think is missing in your example is that you should provide a sampling time. As the norm is calculated for every value ...


2

The natural frequency $\omega_n$ and damping ratio $\zeta$ of a system are computed by using the two dominant poles of the system. Dominant poles are the ones that are closer to the imaginary axis (i.e. their real part is closer to zero) and define the dynamic behaviour of the system. Since your system is of order $4$ the characteristic polynomial (...


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 ...


1

In general, the problem of finding a controller that minimizes some objective function falls under an extremely deep field of control theory called “optimal control”. Optimal control theory deals with minimizing functions of the state and controls (in your case it’s the squared error, which would fall under state) subject to some dynamics (in your case the ...


1

You don't need a computer program to solve this. You can almost do it in your head. The last equation is satisfied by any values of $x, y, z$ so you can forget about it. If you multiply out the first row of the matrix product, you get $$1 + 2.4 + x + 0 -5.6 + 0 + 0 = 0$$ which is pretty easy to solve for $x$. The second row is just as easy to solve for $y$...


1

The answer really boils down to how you define your system. If you say "this is my plant" and give that equation, then as long as $u$ does not depend on $y$ then the system is operating in open loop. Yes, there is some inherent feedback internal to your plant, but that's the case with just about anything that might be described with a differential equation....


1

There are various system performance measures which a designer can use in order to compare the closed loop performance of a system when he has, for example, designed two different controllers $G_1(s)$ and $G_2(s)$. Obviously, the controller which in the closed loop minimizes the chosen performance measure is better. Such performance measures are the ...


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