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Half of your statement is correct: If we are considering the following system: G : \begin{aligned} \dot{x} &= Ax+Bu \\ y &= Cx \end{aligned} then $A$, $B$, and $C$ all influence the zeros of $G$, but $C$ certainly does not affect the poles. The poles $G$ are the eigenvalues of $A$. This makes sense from a purely common-sense consideration: the ...

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Let's first compute the model. The control design is a separate effort. The torque applied to the drum is $n T_M$, where n is the gear ratio and $T_M$ is the output produced by the motor. $T_M= K_T i(t)$, where $K_T$ is a proportionality constant and $i(t)$ is the motor current. Now we can write the equations for the mechanical system: $$m y''(t)+m g-k (... 5 If you want to evaluate a continues time transfer function at a specific frequency \omega in rad/s you substitute s with j\,\omega. For a discrete time transfer function you substitute z with e^{T_e\,j\,\omega}. In order to see why you have to substitute z with e^{T_e\,j\,\omega} you can consider the transfer function z^{-1}, which is a ... 5 I thought I would expand a little on the answer offered by Karlo. Long story short, I would not try to calculate the analytical time response of a system to a square wave. That would be a serious pure-maths exercise, and not necessary for most engineering applications. Instead, I would suggest using the step response analytical vs. simulated to validate ... 5 There is a mistake in your expressions. The coefficient of s is 2+k_p. The conditions are:$$\frac{1}{3} \left(3 \left(k_p+2\right)-k_i\right)>0k_i>0k_p>0$$This simplifies to:$$k_p>00<k_i<3 k_p+6$$Update: If the transfer function is$$ \frac{1}{s^2+3 s + c} $$where c is some positive constant, then the conditions ... 5 Zeros are very important when you are looking at how to control a system. A good way to look at this is to examine a root locus plot. There are several rules on how to sketch a root locus, but the basic idea is to first plot the poles and zeros of the open loop system on the real-imaginary plane. Take this system as an example (code for matlab): G = tf([1 ... 4 Stretch in the spring delta  Y = A.sin(\omega.t) = A.sin\sqrt(k/m) . t  So the delta Y is not constant but if you are interested in delt Y_max delta Y max = m/k, by Hooks law. Because your system doesn't accelerate except at the beginning and end assuming the pulley starts and stops suddenly that's you maximum. Any gradual start/stop acceleration ... 4 As I remember (learned it 15 yrs ago) You need to have an experimental data, which will help you to find "linear" sections and the their limits (with taking into account the things like hysteresis), so you will be able to find transfer function for each band. P.S. Why do you think that the servo motor system is nonlinear? Please describe the the mechanism ... 4 Step 1: Draw the root locus of the system. Here you can see the two poles of your plant G(s) (marked with an x), at p_1=-9 and p_2=-1, the pole of your controller C(s) at p_c = 0 and the zero (marked with an o) at z_c = -c (for now just at a random location). The purple squares indicate the poles of the closed loop system for a certain gain K_p... 4 Your systems shows extremely close pole-zero cancellation. So much even that it nearly removes 4 poles and zeros. Lets look at why, starting with the Bode plot: The magnitude plot is constantly decreasing with a slope of -40dB/decade. Following basic rules this already implies that at the given frequencies, the system can be approximated using a double ... 4 The forward path of your control diagram consists of the following transfer functions connected in series:$$ \begin{align*} G(s) &= \frac{s+8}{s^2-2s-3}\rightarrow \text{Plant}\\ K(s) &= K \rightarrow \text{Controller} \end{align*} $$The Loop Transfer Function then is (by considering all other components' dynamics equal to 1):$$ L(s) = K\cdot \...

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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 ...

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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 $V$, ...

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A lead filter implies that the zero has a lower frequency than the pole. While a PD controller with a low-pass-filter does not necessarily imply that order. Also a lead filter (usually) does not have the zero and pole to far apart from each other, meaning that the bode diagram does not get very close to the asymptotes of +1 slope for magnitude and 90° for ...

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Why would a linear, time-invariant system require initial conditions to be zero? This is completely incorrect. A linear, time-invariant system is any system that is linear (no state terms multiplying one another or themselves) and time-invariant, meaning that the coefficients don't change with respect to time. A simple system would be an RLC circuit. ...

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Checking the units is an excellent way to double check your work; kudos for doing so. However, the next step in checking to see if your results make sense is to check limits. In your case, you can use physical intuition to identify how the system would act at very low frequencies, and how it would act without any damping. At very low frequencies ($s\... 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 In mathematics,which are very closely related and broadly applied in control engineering, the derivatives with respect to time are often notated using the so-called: dot notation. You can check this link for more information: Derivative. This means that the particular equation (the one at the question) can be rewritten in the following form: $$\ddot{Θ}_ο+4\... 3 Based on the information you've given, I believe your professor is suggesting that a friction term can be represented as shown in the following block diagram. The transfer function G(s) relates force (F(s), the input in the diagram) to velocity (sX(s), the output in the diagram) for a mass-spring system. The damping (\rho) is represented in the ... 3 You are going in the right direction! Lets take these two equations:$$(1) \quad in = \alpha+a_1\alpha z^{-1}+a_2\alpha z^{-2}(2) \quad out = b_0\alpha+b_1\alpha z^{-1}+b_2\alpha z^{-2}$$now rewrite (1) such that it becomes a function of \alpha:$$in = \alpha\left(1+a_1z^{-1}+a_2z^{-2}\right)\alpha = \frac{in}{1+a_1z^{-1}+a_2z^{-2}}$$Substitute ... 2 Imagine hitting a pendulum by a hammer with same force and same direction. The pendulum's response will be always same, yesterday, today, tommorow, and 1 year after. It means, between your impact force by hammer and pendulum's respone, there will be an unchanged relation independent on time. Here, you can model this system as LTI system. Hammer impact and ... 2 I would refer you to "Poles and Zeros of Linear Multivariable Systems: A Survey of the Algebraic, Geometric and Complex Variable Theory", A. G. J. Macfarlane, N. Karcanias for a very detailed analysis of zeros. In picturesque terms, poles can be thought of as system resonances coupled to input and output, and zeros as associated with anti-resonances at ... 2 You can think of poles as low pass filters and zeros as high pass filters. They are in some way two sides of the same coin. You can't decide stability solely by looking at poles, any more than solely looking at zeros. For one thing, a zero can be used to offset the affects of a pole, or the other way around. Each has the opposite affect on the response ... 2 Observe that G(s) is formed by$$\frac{1}{s^2} \to \frac{1}{(s+1)^2}$$with the corresponding Z-transform/Laplace pairs$$\frac{1}{s^2} \to \frac{Tz}{(z-1)^2}$$and$$\frac{1}{s+a} \to \frac{z}{z-e^{-aT}}$$With this information you should be able to make the conversion adhering to all the rules of linearity, e.t.c, thats if you wanna find the ... 2 In order to solve it, you have to divide G_s(s) by s first and then you have it in the form on which you can perform the Z-transform. 2 The plant and controller:$$\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 The transfer function you provide is merly a model of the egg incubation system it is not a controller. In addition a lot of assumptions are made, for instance, heat does spread evenly through the egg incubator. Heat is lost evenly throughout the egg incubator, et cetera. The model is derived from basic heat equations, this is described in the article. Tau ... 2 I realize this is an old thread, and I am not sure how deep of a dive you finally took on this, but one thing I don't see accounted for in your equations is drum/cable friction. This will be small, and like the accumulated mass of the wound steel wire rope you did not include, it may not be on your list. The cable could be pre-stretched and pre-loaded, ... 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 You want to look for the Laplace transform of a square wave. Note: The transfer function$H(s)$is the ratio of the Laplace transforms of output$Y(s)$and input$U(s)$:$Y(s)=H(s)\cdot U(s)\$ If you take the Laplace transform of a function (input our output), it is not called a transfer function.

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