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The typical steps followed to linearise the system \$\dot{x} = f(x, u)\$ is to split the state variable into two parts; a steady part (operating point) and a small-signal part. This can be done with the help of Taylor series. Only the first derivative contributes to the linearisation. $$ x \triangleq x_0 + \delta x$$ $$ \dot{x} = f(x,u) = f(x_0, u_0) + \frac{...


4

Generally in DC-DC circuit models, the converter load is modeled as a current source (i2) with a resistance in parallel. This is to generalize the model as much as possible and to take into account not only the passive loads, but also the active loads such as batteries. You can see i2 as current source in load side in the aforementioned paper (Fig. 5).


3

If the system has memory, you need to know what state the memory is in. For example: a system with a low pass filter of a resistor and a capacitor will respond differently if the capacitor is fully charged or if the capacitor starts with zero voltage. The system will respond differently depending on the initial conditions or state It isn't simply enough to ...


3

This is more of an extended comment than an answer. The system may be inherently discrete-time. It may not make sense to find the continuous time plant model, as it may not exist. I am not familiar with atomic clock plant modeling, but the following points in the references indicate that the system is inherently discrete-time and that the input to the ...


2

My question is that even if we have control over eigen values of this new vector how does that help us control the system i.e. to bring our system to our desired state? If you consider your system as just the plant, then all you can achieve by state feedback control is stabilization (if the system is stabilizable) with pole-placement (if the system is ...


2

It is basic block diagram algebra. First, you write out the algebraic equations and solve for the unknowns \$x1\$ and \$x2\$. $$\text{x1}=\frac{0.05 (\text{Ua}-0.1 \text{x2})}{0.01 s+1}$$ $$\text{x2}=\frac{-\text{Mx}+\text{x1}-2 Y}{0.5 s+1}$$ Next you write the output equation as \$Y= \frac{1}{s} x2\$ and solve for \$Y\$ in terms of the two inputs \$Ua\$ ...


2

A possible solution: Choosing state variables as current in the inductor and the voltage in the capacitor: $$ x_1 = i_L = i_1 $$ $$x_2 = v_C $$ Applying KVL and KCL to the left mesh and top node, respectively: $$ v_1 - R_1x_1 -L\dot{x_1}-x_2=0 $$ $$ -x_1 + C\dot{x_2} + \frac{x_2-v_2}{R_2} = 0 $$ Rearranging: $$ \dot{x_1}= -\frac{R_1}{L}x_1 -\frac{1}{...


2

Sometimes we have systems where a transfer function representation, called an external description, is insufficient to completely characterize it. As an example consider the circuit below, taken for the most part from Lathi's Principles of Linear Systems and Signals. simulate this circuit – Schematic created using CircuitLab Where out input is x(t) ...


2

The similarity transformation is only a change of basis (co-ordinates) and doesn't change the input to output relations, viz, the transfer function (TF). The similarity transformation preserves the eigen values; i.e. poles of the system; i.e. denominator of the TF. However, to show that the TF doesn't change, we need to show that the zeroes do not change ...


2

How come knowing the initial conditions with the input 'determine completely the future behaviour of the system'? From such knowledge, I can only approximate the 'next' state at 0+dt... If you have \$u(t)\$ for future \$t\$, you can continue calculating the output at \$(0+dt+dt)\$, \$(0+dt+dt+dt)\$ etc. If I am not mistaken, being able to propagate the ...


2

At a given time this circuit has several solutions. Of course, once all voltages and currents are stabilised, there is only one solution ( current through L1 and R1 = 0.01A, C2 voltage = 1 volt). But otherwise, when you apply power to this circuit, the values of the currents and voltages at a given time depends on how the power was applied : When ? Suddenly ?...


2

You know that \$V_{R_1} = V_{C_2}\$ So replacing \$V_{R_1}\$ by \$V_{C_2}\$ in your equation \$V_{C_1} + V_{R_1} - V_1 = 0\$ leads to \$V_{C_1} + V_{C_2} = V_1\$ Differentiating results in \$d(V_{C_1}) /dt + d(V_{C_2})/dt = 0\$ or \$d(V_{C_1}) /dt = -d(V_{C_2})/dt\$ Inserting that in your last equation leaves you with only one derivative.


2

I think the problem might be that your cap currents don't follow your +/- convention. Here I've drawn all of the +/- indicators and current arrows: simulate this circuit – Schematic created using CircuitLab The resulting KCL and KVL equations are: $$ \begin{align} C_1\frac{dv_{C_1}}{dt} + \frac{v_{C_1}}{R_1} + i_L &= 0 \\ C_2\frac{dv_{C_2}}{dt} + \...


2

At this moment, on the second bullet, is there a reason not to consider if time-varying (shouldn't it be \$i_f(t)\$ or was it just a typo)? I would consider it a typo. I've tried to put together the differential equations in the following way (considering if, \$i_a\$ and \$\omega_m\$ as state variables respectively): If you are going to allow \$i_f\$ and \...


1

The inductor equation: $$L i_3'=v_1-v_C$$ The capacitor equation: $$c v_c'=i_3\ \ \ (1)$$ Kirchoff's current law: $$\frac{u-v_1}{R_1}=i_3+\frac{v_1}{R_2}$$ which can be solved for \$v_1\$ to get $$v_1=\frac{R_2 \left(u-i_3 R_1\right)}{R_1+R_2}$$ Substitute this in the inductor equation $$L i_3'=\frac{R_2 \left(u-i_3 R_1\right)}{R_1+R_2}-v_c \ (2)$$ The ...


1

There are several ways to convert a transfer function into a state space representation. They lead to apparently different results, but retain the same essential information. Possible representations: _ First companion form (controllable canonical form). _ Jordan canonical form. _ Alternate first companion form (Toeplitz first companion form). _ ...


1

The angle the asymptotes make with the real-axes is \$\frac{(2 k+1)\pi}{2} \$. The denominator is the number of finite poles minus the number of finite zeros (\$3-1\$). This turns out as \$\frac{\pi}{2}\$ and \$\frac{3 \pi}{2}\$. Because of one finite zero, one of the loci ends up at that zero. Only the other two go off to \$\infty\$.


1

There are many nonlinearities that can be impacting your system. The most likely factor is a dead-zone -- a PWM duty cycle too small to make the motor move at all.


1

I knew of a group of students that were building an inverted pendulum, they analyzed the system found the transfer function and programmed it. It refused to work until they put a negative sign on the output to the motor control which put the system in positive feedback. Try putting the system in positive feedback and see if that helps.


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The answer to the question is already provided by other users in the comments, but I think that solved questions should not be marked as unanswered (https://meta.stackoverflow.com/q/251597). I didn't see how to turn this answer into a Community Wiki as suggested on StackOverflowMeta: https://meta.stackoverflow.com/a/251598. I guess I don't have enough ...


1

I believe you are aware of the concept of the Jacobian matrix. This is a case of Linearization of Multistate Models. See, for example, pages 6-7 of Linearization of Nonlinear Models. Yes, that linearization uses the calculation of the partial derivatives of three functions with three variables (two states \$x_1\$, \$x_2\$ at one input \$u\$) around the ...


1

Consider the output \$y(t)\$, as typically represented in state-space models: \$y(t) = Cx(t)+Du(t)\$. If you were to equate one element of \$x(t)\$ to \$u(t)\$, then this means the original representation was incorrect in the first place, and rather a representation \$y(t) = \tilde{C}\tilde{x}(t)+\tilde{D}u(t)\$ was the correct one. Demonstration: if state \$...


1

The transformation back, from a state-space model to a transfer function can be done with \$C(sI-A)^{-1}B\$, irrespective of the number of inputs and outputs. In general, you might not end up with a form \$\begin{pmatrix} G(s) & -1 \\ H(s) & 0 \end{pmatrix}\$ though. The transformation from a transfer function (matrix) to a state-space model is not ...


1

Your notation is confusing. Let \$ \small y(t)\rightarrow Y(s)\$ be the system output signal from the integrator, and let the system input signal be \$\small u(t) \rightarrow U(s)\$. The CLTF is then: $$\small\frac{Y(s)}{U(s)}=\frac{2(s+1)}{3s^2+7s+2}$$ To proceed via controller canonical form (other forms are available - see literature), introduce a dummy ...


1

You will need 4 variables and there are 4 equations $$V_{\text{C1}}-V_{\text{C2}}=L_4 i_1'$$ $$i_1=\frac{V_i-V_{\text{C1}}}{R_2}-C_3 V_{\text{C1}}'$$ $$i_1=C_4 \left(V_0'-V_{\text{C2}}'\right)+\frac{V_{\text{C2}}}{R_5}$$ $$\frac{V_{\text{C2}}-2 V_0}{R_6}=C_4 V_0'$$ And there are 4 unknowns \$\{i_1',V_{\text{C1}}',V_{\text{C2}}',V_0'\}\$ from which you can ...


1

Well, according to 'Faraday's law' in a series RLC-circuit: $$\text{V}_{\space\text{C}}\left(t\right)+0+\text{V}_{\space\text{R}}\left(t\right)-\text{V}_{\space\text{in}}\left(t\right)=-\text{V}_{\space\text{L}}\left(t\right)\tag1$$ Now, we know a few things: $$\text{I}_{\space\text{C}}\left(t\right)=\text{I}_{\space\text{in}}\left(t\right)=\text{V}_{\...


1

1) when you consider a state feedback controller 2) when linear system theory can not be applied ( nonlinear systems) 3)when you don‘t have a sensor for the state you are interested in (estimators) 4) when you have to optimize a state dependent objective function 5) design of tracking applications 6) need to consider initial conditions 7) complex ...


1

I am going to just assume that this is a "mathematical problem" rather than an engineering problem. As @Sparky256 and @Dorian have mentioned, the circuit topology is questionable and not practical. I think that the main cause of non-invertible matrix is the floating capacitors CF1 and CF2 in state 2 and 3 respectively. To avoid this I considered parasitic ...


1

How can this work? I don't see-it anywhere but this should be a DC converter ? There is no DC path to Vo1 and Vo2, the average will always be zero. Are you sure Cf1 and Cf2 are in the right place? Switching to the configuration in Page 2 is not possible, it will lead to infinite current since on the Vin - Cf1 - C1 loop there are no resistive or inductive ...


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