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Pole Placement | State Space, Part 2

From the series: State Space

Brian Douglas

This video provides an intuitive understanding of pole placement, also known as full state feedback. This is a control technique that feeds back every state to guarantee closed-loop stability and is the stepping stone to other methods like LQR and H infinity.

We’ll cover the structure of the control law and why moving poles or eigenvalues around changes the dynamics of a system. We’ll also go over an example in MATLAB and touch on a few interesting comparisons to other control techniques.

In this video, we’re going to talk about a way to develop a feedback controller for a model that’s represented using state-space equations. And we’re going to do that with a method called pole placement, or full state feedback. Now, my experience is that pole placement itself isn’t used extensively in industry; you might find that you’re using other methods like LQR or H infinity more often. However, pole placement is worth spending some time on because it’ll give you a better understanding of the general approach to feedback control using state-space equations and it’s a stepping stone to getting to those other methods. So I hope you stick around. I’m Brian, and welcome to a MATLAB Tech Talk.

To start off, we have a plant with inputs u and outputs y. And the goal is to develop a feedback control system that drives the output to some desired value. A way you might be familiar with doing this is to compare the output to a reference signal to get the control error. Then you develop a controller that uses that error term to generate the input signals into the plant with the goal of driving the error to zero. This is the structure of the feedback system that you’d see if you were developing, say, a PID controller.

But for pole placement, we’re going to approach this problem in a different way. Rather than feed back the output y, we’re going to feed back the value of every state variable in our state vector, x. We’re going to claim that we know the value of every state even though it’s not necessarily part of the output y. We’ll get to that in a bit, but for now, assume we have access to all of these values. We then take the state vector and multiply it by a matrix that is made up of a bunch of different gain values. The result is subtracted from a scaled reference signal, and this result is fed directly into our plant as the input. 

Now you’ll notice that there isn’t a block here labeled “controller” like we have in the top block diagram. In this feedback structure, this whole section is the controller. And pole placement is a method by which we can calculate the proper gain matrix to guarantee system stability, and the scaling term on the reference is used to ensure that steady state error performance is acceptable. I’ll cover both of these in this video.

In the last video, we introduced the state equation x dot = Ax + Bu. And we showed that the dynamics of a linear system are captured in this first part, Ax. The second part is how the system responds to inputs, but how the energy in the system is stored and moves is captured by the Ax term. So you might expect that there is something special about the A matrix when it comes to controller design. And there is: Any feedback controller has to modify the A matrix in order to change the dynamics of the system. This is especially true when it comes to stability.

The eigenvalues of the A matrix are the poles of the system, and the location of the poles dictates stability of a linear system. And that’s the key to pole placement: Generate the required closed-loop stability by moving the poles or the eigenvalues of the closed-loop A matrix.

I want to expand a bit more on the relationship between poles, eigenvalues, and stability before we go any further because I think it’ll help you understand exactly how pole placement works.

For this example, let’s just start with an arbitrary system and focus on the dynamics, the A matrix. We can rewrite this in non-matrix form so it’s a little bit easier to see how the state derivatives relate to the states. In general, each state can change as a function of the other states. And that’s the case here; x dot 1 changes based on x2 and x dot 2 changes based on both x1 and x2. This is perfectly acceptable, but it makes it hard to visualize how eigenvalues are contributing to the overall dynamics. So what we can do is transform the A matrix into one that uses a different set of state variables to describe the system.

This transformation is accomplished using a transform matrix whose columns are the eigenvectors of the A matrix. What we end up with after the transformation is a modified A matrix consisting of the complex eigenvalues along the diagonal and zeroes everywhere else. These two models represent the same system. They have the same eigenvalues, the same dynamics; it’s just the second one is described using a set of state variables that change independently of each other. 

With the A matrix written in diagonal form, it’s easy to see that we’re left with a set of first-order differential equations where the derivative of each state is only affected by that state and nothing else. And here’s the cool part: The solution to a differential equation like this is in the form Z = a constant times e ^ lambda t. Where lambda is the eigenvalue for that given state variable. 

Okay, let’s dive into this equation a little bit more. Zn shows how the state changes over time given some initial condition, C. Or another way of thinking about this is that if you initialize the state with some energy, this equation shows what happens to that energy over time. And by changing lambda, you can affect how the energy is dissipated or, in the case of an unstable system, how the energy grows.

Let’s go through a few different values of lambda so you can visually see how energy changes based on the location of the eigenvalue within the complex plane.

If lambda is a negative real number, then this mode is stable since the solution is e raised to a negative number, and any initial energy will dissipate over time. If it’s positive, then it’s unstable because the energy will grow over time. And if there is a pair of imaginary eigenvalues, then the energy in the mode will oscillate, since e ^ imaginary number produces sines and cosines. And any combination of real and imaginary numbers in the eigenvalue will produce a combination of oscillations and exponential energy dissipation.

I know this was all very fast, but hopefully it made enough sense that now we can state the problem we’re trying to solve. If our plant has eigenvalues that are at undesirable locations in the complex plane, then we can use pole placement to move them somewhere else. Certainly if they’re in the right half plane it’s undesirable since they’d be unstable, but undesirable could also mean there are oscillations that you want to get rid of, or maybe just speed up or slow down the dissipation of energy in a particular mode.

With that behind us, we can now get into how pole placement moves the eigenvalues. Remember the structure of the controller that we drew at the beginning? This results in an input u = r*Kr - k*x. Where r Kr is the scaled reference, which again we’ll get to in a bit. And kx is the state vector that we’re feeding back multiplied by the gain matrix.

Here’s where the magic happens. If we plug this control input into our state equation, we are closing the loop and we get the following state equation: Notice that A and -Bk both act on the state vector so we can combine them to get modified A matrix.

This is the closed-loop A matrix and we have the ability to move the eigenvalues by choosing an appropriate K. And this is easy to do by hand for simple systems. Let’s try an example with a second-order system with a single input. We can find the eigenvalues by setting the determinant of A - lambda I to zero and then solve for lambda. And they are at -2 and +1. One of the modes will blow up to infinity because of the presence of the positive real eigenvalue and so the system is unstable. Let's use pole placement to design a feedback controller that will stabilize this system by moving the unstable pole to the left half plane.

Our closed-loop A matrix is A - BK and the gain matrix, k, is 1x2 since there is one output and two states. This results in - K1, 1 - k2, 2 and -1. We can solve for the eigenvalues of Acl like we did before and we get this characteristic equation that is a function of our two gain values.

Let’s say we want our closed-loop poles at -1 and -2. In this way, the characteristic equation needs to be L^2 + 3L + 2 = 0. So at this point, it’s straightforward to find the appropriate K1 and K2 that make these two equations equal. We just set the coefficients equal to each other and solve. And we get K1 = 2, and K2 = 1 and that’s it. If we place these two gains in the state feedback path of this system, it will be stabilized with eigenvalues at -1 and -2. 

Walking through an example by hand, I think, gives you a good understanding of pole placement; however, the math involved starts to become overwhelming for systems that have more than two states. The idea is the same; just solving the determinant becomes impractical. But we can do this exact same thing in MATLAB with pretty much a single command. 

I’ll show you quickly how to use the place command in MATLAB by recreating the same system we did by hand. I’ll define the four matrices, and then create the open-loop state-space object. I can check the eigenvalues of the open-loop A matrix just to show you that there is, in fact, that positive eigenvalue that causes this system to be unstable.

That’s no good, so let’s move the eigenvalues of the system to -2 and -1. Now solving for the gain matrix using pole placement can be done with the place command. And we get gain values 2 and 1 like we expected.

Now the new closed-loop A matrix is A - BK, and just to double check, this is what Acl looks like and it does have eigenvalues at -1 and -2. Okay, I’ll create the closed-loop system object and now we can compare the step responses for both. 

The step response of the open-loop system is predictably unstable. The step response of the closed-loop system looks much better. However, it’s not perfect. Rather than rising to 1 like we’d expect, the steady state output is only 0.5. And this is finally where the scaling on the reference comes in. So far, we’ve only been concerned with stability and have paid little attention to steady state performance. But even addressing this is pretty straightforward. If the response to the input is only half of what you expect, why don’t we just double the input? And that’s what we do. Well, not double it, but we scale the input by the inverse of the steady state value. 

In MATLAB, we can do this by inverting the DC gain of the system. You can see that the DC gain is 0.5, and so the inverse is 2. Now we can rebuild our closed-loop system by scaling the input by Kr. and checking the step response. No surprise; its steady state value is 1. 

And that’s pretty much what there is to basic pole placement. We feed back every state variable and multiply them by a gain matrix in such a way that moves the closed-loop eigenvalues, and then we scale the reference signal so that the steady state output is what we want.

Of course, there’s more to pole placement than what I could cover in this 12-minute video, and I don’t want to drag this on too long, but I also don’t want to leave this video without addressing a few more interesting things for you to consider. So in the interest of time, let’s blast through these final thoughts lightning-round style. Are you ready? Let’s go!

Pole placement is like fancy root locus. With root locus you have one gain that you can adjust that can only move to the poles along the locus lines. But with pole placement, we have a gain matrix that gives us the ability to move the poles anywhere in the complex plane, not just along single-dimensional lines.

A two-state pole placement controller is very similar to a PD controller. With PD, you feed back the output and generate the derivative within the controller. With pole placement, you are feeding back the derivative as a state, but the results are essentially the same: 2 gains, one for a state and one for its derivative.

Okay, we can move eigenvalues around, but where should we place them? The answer to that is a much longer video, but here are some things to think about. If you have a high-order system, consider keeping two poles much closer to the imaginary axis than the others so that the system will behave like a common second-order system. These are called the dominant poles since they are slower and tend to dominate the response of the system.

Keep in mind that if you try to move a bunch of eigenvalues really far left in order to get a super-fast response, you may find that you don’t have the speed or authority in your actuators to generate the necessary response. This is because it takes more gain, or more actuator effort, to move the eigenvalues further from their open-loop starting points.

Full state feedback is a bit of a misnomer. You are feedback every state in your mathematical model, but you don’t, and can’t, feed back every state in a real system. For just one example, at some level, all mechanical hardware is flexible, which means additional states, but you may choose to ignore those states in your model and develop your feedback controller assuming a rigid system. The important part is that you feed back all critical states to your design so that your controller will still work on the real hardware.

You have to have some kind of access to all of the critical states in order to feed them back. The output, y, might include every state, in which case you’re all set. However, if this isn’t the case, you will either need to add more sensors to your system to measure the missing states or use the existing outputs to estimate or observe the states you aren’t measuring directly. In order to observe your system, it needs to be observable, and similarly, in order to control your system it needs to controllable. We’ll talk about both of those concepts in the next video.

So that’s it for now. I hope these final few thoughts helped you understand a little more about what it means to do pole placement and how it’s part of an overall control architecture.

If you want some additional information, there are a few links in the description that are worth checking out that explain more about using pole placement with MATLAB.

If you don’t want to miss the next Tech Talk video, don’t forget to subscribe to this channel. Also, if you want to check out my channel, control system lectures, I cover more control theory topics there as well. Thanks for watching. I’ll see you next time.

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This video helps you answer two really important questions that come up in control systems engineering: Is your system controllable? And is it observable? In this video, we’re going to approach the answers from a conceptual and intuitive direction.

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Create and analyze state-space models using MATLAB and Control System Toolbox. State-space models are commonly used for representing linear time-invariant (LTI) systems.

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pole placement wiki

Introduction: State-Space Methods for Controller Design

In this section, we will show how to design controllers and observers using state-space (or time-domain) methods.

Key MATLAB commands used in this tutorial are: eig , ss , lsim , place , acker

Related Tutorial Links

  • LQR Animation 1
  • LQR Animation 2

Related External Links

  • MATLAB State FB Video
  • State Space Intro Video

Controllability and Observability

Control design using pole placement, introducing the reference input, observer design.

There are several different ways to describe a system of linear differential equations. The state-space representation was introduced in the Introduction: System Modeling section. For a SISO LTI system, the state-space form is given below:

$$
\frac{d\mathbf{x}}{dt} = A\mathbf{x} + Bu
$$

To introduce the state-space control design method, we will use the magnetically suspended ball as an example. The current through the coils induces a magnetic force which can balance the force of gravity and cause the ball (which is made of a magnetic material) to be suspended in mid-air. The modeling of this system has been established in many control text books (including Automatic Control Systems by B. C. Kuo, the seventh edition).

pole placement wiki

The equations for the system are given by:

$$
m\frac{d^2h}{dt^2} = mg - \frac{Ki^2}{h}
$$

From inspection, it can be seen that one of the poles is in the right-half plane (i.e. has positive real part), which means that the open-loop system is unstable.

To observe what happens to this unstable system when there is a non-zero initial condition, add the following lines to your m-file and run it again:

pole placement wiki

It looks like the distance between the ball and the electromagnet will go to infinity, but probably the ball hits the table or the floor first (and also probably goes out of the range where our linearization is valid).

$u(t)$

Let's build a controller for this system using a pole placement approach. The schematic of a full-state feedback system is shown below. By full-state, we mean that all state variables are known to the controller at all times. For this system, we would need a sensor measuring the ball's position, another measuring the ball's velocity, and a third measuring the current in the electromagnet.

pole placement wiki

The state-space equations for the closed-loop feedback system are, therefore,

$$
\dot{\mathbf{x}} = A\mathbf{x} + B(-K\mathbf{x}) = (A-BK)\mathbf{x}
$$

From inspection, we can see the overshoot is too large (there are also zeros in the transfer function which can increase the overshoot; you do not explicitly see the zeros in the state-space formulation). Try placing the poles further to the left to see if the transient response improves (this should also make the response faster).

pole placement wiki

This time the overshoot is smaller. Consult your textbook for further suggestions on choosing the desired closed-loop poles.

Note: If you want to place two or more poles at the same position, place will not work. You can use a function called acker which achieves the same goal (but can be less numerically well-conditioned):

K = acker(A,B,[p1 p2 p3])

Now, we will take the control system as defined above and apply a step input (we choose a small value for the step, so we remain in the region where our linearization is valid). Replace t , u , and lsim in your m-file with the following:

pole placement wiki

The system does not track the step well at all; not only is the magnitude not one, but it is negative instead of positive!

$K\mathbf{x}$

and now a step can be tracked reasonably well. Note, our calculation of the scaling factor requires good knowledge of the system. If our model is in error, then we will scale the input an incorrect amount. An alternative, similar to what was introduced with PID control, is to add a state variable for the integral of the output error. This has the effect of adding an integral term to our controller which is known to reduce steady-state error.

$y = C\mathbf{x}$

From the above, we can see that the observer estimates converge to the actual state variables quickly and track the state variables well in steady-state.

Published with MATLAB® 9.2

pole placement wiki

Robust Stability and Pole Placement

An Application of Parametric Interval Analysis

  • Published: 07 September 2021
  • Volume 32 , pages 1498–1509, ( 2021 )

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  • Heloise Assis Fazzolari   ORCID: orcid.org/0000-0003-4113-2598 1   nAff2 &
  • Paulo Augusto Valente Ferreira 1  

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In this paper, we propose an integration of classic and parametric interval analysis methods for addressing robust stability and robust pole placement problems associated with linear dynamic systems with interval parameters. In order to reduce the conservatism of classic interval analysis and synthesis methods due to the parameter dependency phenomenon, we adopt a less conservative approach that explicitly considers multi-incident interval parameters. The paper includes numerical experiments which illustrates the characteristics and properties of the proposed methods, as well as their application to the control of an interval gyroscope system.

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Acknowledgements

This work was sponsored by the National Council for Scientific and Technological Development (CNPq), Brazil, Grants 159829/2013-5 and 307926/2009-5.

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Heloise Assis Fazzolari

Present address: Engineering, Modeling and Applied Social Sciences Center - CECS, Federal University of ABC - UFABC, Av. dos Estados, 5001, Santo André- SP, Brazil

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School of Electrical and Computer Engineering, University of Campinas, Campinas, 13083-852, Brazil

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Fazzolari, H.A., Ferreira, P.A.V. Robust Stability and Pole Placement. J Control Autom Electr Syst 32 , 1498–1509 (2021). https://doi.org/10.1007/s40313-021-00798-7

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DOI : https://doi.org/10.1007/s40313-021-00798-7

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LMIs in Control/pages/Switched Systems Pole Placement

Pole Placement for Switched Systems

This LMI lets you provide specifications of the switched system closed loop poles. Note that arbitrarily switching between stable systems can lead to instability whilst switching can be done between individually unstable systems to achieve stability.

  • 1 The System
  • 3 The Optimization Problem
  • 4 The LMI: An LMI for Pole Placement
  • 5 Conclusion:
  • 6 Implementation
  • 7 External Links
  • 8 Related LMIs
  • 9 Return to Main Page:

The System [ edit | edit source ]

Suppose we were given the switched system such that

{\displaystyle {\begin{aligned}{\dot {x}}(t)&=A_{i}x(t)+B_{i}u(t)\\y(t)&=C_{i}x(t)+D_{i}u(t)\end{aligned}}}

The Data [ edit | edit source ]

In order to properly define the acceptable region of the poles in the complex plane, we need the following pieces of data:

{\displaystyle A_{i}}

Having these pieces of information will now help us in formulating the optimization problem.

The Optimization Problem [ edit | edit source ]

{\displaystyle z}

The LMI: An LMI for Pole Placement [ edit | edit source ]

{\displaystyle P>0}

Conclusion: [ edit | edit source ]

The resulting controller can be recovered by

{\displaystyle K=ZP^{-1}}

Implementation [ edit | edit source ]

The implementation of this LMI requires Yalmip and Sedumi /MOSEK [1]

External Links [ edit | edit source ]

A list of references documenting and validating the LMI.

  • LMI Methods in Optimal and Robust Control - A course on LMIs in Control by Matthew Peet.
  • LMI Properties and Applications in Systems, Stability, and Control Theory - A List of LMIs by Ryan Caverly and James Forbes.
  • LMIs in Systems and Control Theory - A downloadable book on LMIs by Stephen Boyd.

Related LMIs [ edit | edit source ]

Return to main page: [ edit | edit source ].

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Intro to Control Theory Part 6: Pole Placement

In Part 4 , I covered how to make a state-space model of a system to make running simulations easy. In this post, I'll talk about how to use that model to make a controller for our system.

For this post, I'm going to use an example system that I haven't talked about before - A mass on a spring:

A simple mass on a spring. Image Credit: University of Southern Queensland

If we call \(p\) the position of the cart (we use \(p\) instead of \(x\), since \(x\) is the entire state once we're using a state space representation), then we find that the following equation describes how the cart will move:

\[ \dot{p} = -\frac{k}{m}p \]

Where \(p\) is position, \(k\) is the spring constant of the spring (how strong it is), and \(m\) is the mass of the cart.

You can derive this from Hooke's Law if you're interested, but the intuitive explanation is that the spring pulls back against the cart proportionally to how far it is away from the equilibrium state of the spring, but gets slowed down the heavier the cart is.

This describes a ideal spring, but one thing that you'll notice if you run a simulation of this is that it will keep on oscillating forever! We haven't taken into account friction. Taking into account friction gets us the following equation:

\[ \dot{p} = -\frac{k}{m}p - \frac{c}{m}\dot{p} \]

Where \(c\) is the "damping coefficient" - essentially the amount of friction acting on the cart.

Now that we have this equation, let's convert it into state space form!

This system has two states - position, and velocity:

\[ x = \begin{bmatrix} p \\ \dot{p} \end{bmatrix} \]

Since \(x\) is a vector of length 2, \(A\) will be a 2x2 matrix. Remember, a state space representation always takes this form:

\[ \dot{x} = Ax + Bu \]

We'll find \(A\) first:

\[ \begin{bmatrix} \dot{p} \\ \ddot{p} \end{bmatrix} = \begin{bmatrix} ? & ? \\ ? & ? \end{bmatrix} \begin{bmatrix} p \\ \dot{p} \end{bmatrix} \]

The way that I like to think about this is that each number in the matrix is asking a question - how does X affect Y? So, for example, the upper left number in the A matrix is asking "How does position affect velocity?". Position has no effect on velocity, so the upper left number is zero. Next, we can look at the upper right number. This is asking "How does velocity affect velocity?" Well, velocity is velocity, so we put a 1 there (since you need to multiply velocity by 1 to get velocity). If we keep doing this process, we get the following equation:

\[ \begin{bmatrix} \dot{p} \\ \ddot{p} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -\frac{k}{m} & -\frac{c}{m} \end{bmatrix} \begin{bmatrix} p \\ \dot{p} \end{bmatrix} \]

For the sake of this post, I'll pick some arbitrary values for \(m\), \(k\), and \(c\): \(m = 1 \text{kg}\), \(k = 0.4 \text{N/m}\), \(c = 0.3 \text{N/m/s}\). Running a simulation of this system, starting at a position of 1 meter, we get the following response:

Open loop response of mass on spring system

Notice that this plot shows two things happening - the position is oscillating, but also decreasing. There's actually a way to quantify how much the system will oscillate and how quickly it will converge to zero (if it does at all!). In order to see how a system will act, we look at the "poles" of the system. In order to understand what the poles of a system mean, we need to take a quick detour into linear algebra.

Our matrix \(A\) is actually a linear transformation . That means that if we multiply a vector by \(A\), we will get out a new, transformed vector. Multiplication and addition are preserved, such that \( A(x \times 5) = (Ax) \times 5 \) and \( A(x_1 + x_2) = Ax_1 + Ax_2 \). When you look at \(A\) as a linear transformation, you'll see that some vectors don't change direction when you apply the transform to them:

The vectors that don't change direction when transformed are called "eigenvectors". For this transform, the eigenvectors are the blue and pink arrows. Each eigenvector has an "eigenvalue", which is how much it stretches the vector by. In this example, the eigenvalue of the blue vectors is 3 and the eigenvalue of the pink vectors is 1.

So how does this all relate to state space systems? Well, the eigenvalues of the system (also called the poles of a system) have a direct effect on the response of the system. Let's look at our eigenvalues for our system above. Plugging the matrix into octave/matlab gives us:

So we can see that we have two eigenvalues, both of which are complex numbers. What does this mean? Well, the real component of the number tells you how fast the system will converge to zero. The more negative it is, the faster it will converge to zero. If it is above zero, the system is unstable, and will trend towards infinity (or negative infinity). If it is exactly zero, the system is "marginally stable" - it won't get larger or smaller. The imaginary part of the number tells you how much the system will oscillate. For every positive imaginary part, there is a negative one of the same magnitude with the same real part, so it's just the magnitude of the imaginary part that determines how much the system will oscillate - the higher the magnitude of the imaginary part, the more the system will oscillate.

Why is this the case? Well, as it turns out, the derivative of a specific state is the current value of that state times the eigenvalue associated with that state. So, a negative eigenvalue will result in a derivative that drives the state to zero, whereas a positive eigenvalue will cause the state to increase in magnitude forever. A eigenvalue of zero will cause the derivative to be zero, which obviously results in no change to the state.

That explains real eigenvalues, but what about imaginary eigenvalues? Let's imagine a system that has two poles, at \(0+0.1i\) and \(0-0.1i\). Since this system has a real component of zero, it will be marginally stable, but since it has an imaginary component, it will oscillate. Here's a way of visualizing this system:

The blue vector is the position of the system. The red vectors are the different components of that position (the sum of the red vectors will equal the blue vector). The green vectors are the time derivatives of the red vectors. As you can see, the eigenvalue being imaginary causes each component of the position to be imaginary, but since there is always a pair of imaginary poles of the same magnitude but different signs, the actual position will always be real.

So, how is this useful? Well, it lets us look at a system and see what it's response will look like. But we don't just want to be able to see how the system will respond, we want to be able to change how the system will respond. Let's return to our mass on a spring:

Now let's say that we can apply an arbitrary force \(u\) to the system. For this, we use our \(B\) matrix:

\[ \begin{bmatrix} \dot{p} \\ \ddot{p} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -\frac{k}{m} & -\frac{c}{m} \end{bmatrix} \begin{bmatrix} p \\ \dot{p} \end{bmatrix} + \begin{bmatrix} 0 \\ \frac{1}{m} \end{bmatrix} u \]

Now, let's design a controller that will stop there from being any oscillation, and drive the system to zero much more quickly. Remember, all "designing a controller" means in this case is finding a matrix \(K\), where setting \( u = Kx \) will cause the system to respond in the way that you want it to. How do we do this? Well, it turns out that it's actually fairly easy to place the poles of a system wherever you want. Since we want to have no oscillation, we'll make the imaginary part of the poles zero, and since we want a fast response time, we'll make the real part of the poles -2.5 (this is pretty arbitrary). We can use matlab/octave to find what our K matrix must be to have the poles of the closed loop system be at -2.5:

Which gives us the output:

So our K matrix is:

\[ K = \begin{bmatrix} 5.85 \\ 4.7 \end{bmatrix} \]

And running a simulation of the system with that K matrix gives us:

Closed loop response of mass on spring system

Much better! It converges in under five seconds with no oscillation, compared with >30 seconds with lots of oscillations for the open-loop response. But wait, if we can place the poles anywhere we want, and the more negative they are the faster the response, why not just place them as far negative as possible? Why not place them at -100 or -1000 or -100000? For that matter, why do we ever want our system to oscillate, if we can just make the imaginary part of the poles zero? Well, the answer is that you can make the system converge as fast as you want, so long as you have enough energy that you can actually apply to the system. In real life, the motors and actuators that are limited in the amount of force that they can apply. We ignore this in the state-space model, since it makes the system non-linear, but it's something that you need to keep in mind when designing a controller. This is also the reason that you might want some oscillation - oscillation will make you reach your target faster than you would otherwise. Sometimes, getting to the target fast is more important than not oscillating much.

So, that's how you design a state space controller with pole placement! There are also a ton of other ways to design controllers (most notably LQR) which I'll go into later, but understanding how poles determine the response of a system is important for any kind of controller.

If you're in NYC and want to meet up over lunch/coffee to chat about the future of technology, get in touch with me .

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Engineering LibreTexts

10.2: Controllers for Discrete State Variable Models

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  • Page ID 24434

  • Kamran Iqbal
  • University of Arkansas at Little Rock

Emulating an Analog Controller

The pole placement controller designed for a continuous-time state variable model can be used with derived sampled-data system model. Successful controller emulation requires a high enough sampling rate that is at least ten times the frequency of the dominant closed-loop poles of the system.

In the following we illustrate the emulation of pole placement controller designed for the DC motor model (Example 8.3.4) for controlling the discrete-time model of the DC motor. The DC motor model is discretized at two different sampling rates for comparison, assuming ZOH at the plant input.

Example \(\PageIndex{1}\)

The state and output equations for a DC motor model are given as:

\[\frac{\rm d}{\rm dt} \left[\begin{array}{c} {i_a } \\ {\omega } \end{array}\right]=\left[\begin{array}{cc} {-100} & {-5} \\ {5} & {-10} \end{array}\right]\left[\begin{array}{c} {i_a } \\ {\omega } \end{array}\right]+\left[\begin{array}{c} {100} \\ {0} \end{array}\right]V_a , \;\;\omega =\left[\begin{array}{cc} {0} & {1} \end{array}\right]\left[\begin{array}{c} {i_a } \\ {\omega } \end{array}\right]. \nonumber \]

The motor model is discretized at two different sampling rates in MATLAB. The results are:

\[T=0.01s: A_{\rm d} =\left[\begin{array}{cc} {0.367} & {-0.030} \\ {0.030} & {0.904} \end{array}\right],\; \; B_{\rm d} =\left[\begin{array}{c} {0.632} \\ {0.018} \end{array}\right],\; \; C_{\rm d} =\left[\begin{array}{cc} {0} & {1} \end{array}\right]. \nonumber \]

\[T=0.02s: A_{\rm d} =\left[\begin{array}{cc} {0.134} & {-0.038} \\ {0.038} & {0.816} \end{array}\right],\; \; B_{\rm d} =\left[\begin{array}{c} {0.863} \\ {0.053} \end{array}\right],\; \; C_{\rm d} =\left[\begin{array}{cc} {0} & {1} \end{array}\right]. \nonumber \]

For a desired characteristic polynomial: \(\Delta _{\rm des} (s)=s^{2} +150\,s+5000\), a state feedback controller for the continous-time state variable model was obtained as (Example 9.1.1): \(k^{T} =\left[\begin{array}{cc} {0.4} & {7.15} \end{array}\right]\).

We can use the same controller to control the corresponding sample-data system models.

The unit-step response of the closed-loop system is simulated in Figure 10.2.1, where both state variables, \(i_a\left(t\right)\) and \(\omega \left(t\right)\), are plotted.

clipboard_e29ed3918c27425662d5120784e9bbfac.png

We observe from the figure that the armature current has a higher overshoot at the lower sampling rate, though both models display similar settling time of about 100 msec.

Pole Placement Design of Digital Controller

Given a discrete state variable model \(\left\{A_{\rm d},\ B_{\rm d}\right\}\), and a desired pulse characteristic polynomial \(\Delta _{\rm des} (z)\), a state feedback controller for the system can be designed using pole placement similar to that of the continuous-time system (Sec. 9.1.1).

Let the discrete-time model of a SISO system be given as:

\[{\bf x}_{k+1} ={\bf A}_{\rm d} {\bf x}_{k} +{\bf b}_{\rm d} u_{k} , \;\; y_{k} ={\bf c}^T {\bf x}_{k} \nonumber \]

A state feedback controller for the discrete state variable model is defined as:

\[u_k=-{\bf k}^T{\bf x}_k+r_k \nonumber \]

where \({\bf k}^{T}\) represents a row vector of constant feedback gains and \(r_k\) is a reference input sequence. The controller gains can be obtained by equating the coefficients of the characteristic polynomial with those of a desired polynomial:

\[\Delta (z)=\left|z{\bf I-A}_{\rm d} \right|=\Delta _{\rm des} (z) \nonumber \]

The \(\Delta _{\rm des} (z)\) above is a Hurwitz polynomial (in \(z\)), with roots inside the unit circle that meet given performance (damping ratio and/or settling time) requirements. Assuming that desired \(s\)-plane root locations are known, the corresponding \(z\)-plane root locations can be obtained from the equivalence: \(z=e^{Ts}\).

Closed-loop System

The closed-loop system model is given as:

\[{\bf x}_{k+1} ={\bf A}_{\rm cl} {\bf x}_{k} +{\bf b}_{\rm d} r_{k} , \;\; y_{k} ={\bf c}^T {\bf x}_{k} \nonumber \]

where \({\bf A}_{\rm cl} =({\bf A}_{\rm d}-{\bf b}_{\rm d}{\bf k}^T)\).

Assuming closed-loop stability, for a constant input \(r_k=r_{\rm ss}\), the steady-state response, \({\bf x}_{\rm ss}\), of the system obeys: 

\[{\bf x}_{ss} ={\bf A}_{\rm cl} {\bf x}_{ss} +{\bf b}_{\rm d} r_{ss} ,\;\; y_{\rm ss} ={\bf c}^T {\bf x}_{ss} \nonumber \]

Hence, \(y_{\rm ss}={\bf c}^T\,({\bf A}_{\rm cl}-{\bf I})^{-1}\,{\bf b}_{\rm d}\,r_{\rm ss}\).

Example \(\PageIndex{2}\)

The discrete state variable model of a DC motor (\(T=0.02\)s) is given as: \[\left[\begin{array}{c} {i_{k+1} } \\ {\omega _{k+1} } \end{array}\right]=\left[\begin{array}{cc} {0.134} & {-0.038} \\ {0.038} & {0.816} \end{array}\right]\left[\begin{array}{c} {i_{k} } \\ {\omega _{k} } \end{array}\right]+\left[\begin{array}{c} {0.863} \\ {0.053} \end{array}\right]V_{k} , \;\; y_{k} =\left[\begin{array}{cc} {0} & {1} \end{array}\right]\left[\begin{array}{c} {i_{k} } \\ {\omega _{k} } \end{array}\right] \nonumber \]

The desired \(s\)-plane root locations for the model are given as: \(s=-50,\; -100.\)

The corresponding \(z\)-plane roots (\(T=0.02s\)) are obtained as: \(z=e^{-1} ,\; e^{-2}\).

The desired characteristic polynomial is given as: \(\Delta _{\rm des} (z)=z^{2} -0.95z+0.05.\)

The feedback gains \(k^T =[k_{1} ,\; k_{2} ]\), computed using the MATLAB ‘place’ command, are given as: \(k_{1} =0.247,\; k_{2} =4.435.\)

The closed-loop system matrix is given as: \(A_\rm d)= \left[\begin{array}{cc} {-0.080} & {-3.867} \\ {0.025} & {0.583} \end{array}\right]\).

An update rule for implementation of the controller on computer is obtained as: \(u_{k} =-0.247\, i_{k} -4.435\, \omega _{k} .\)

The closed-loop response has steady-state value of \(\omega _{\rm ss}=0.143 \;\rm rad/s\).

The step response of the closed-loop system is plotted in Figure 10.2.2, where the discrete system response was scaled to match the analog system response. The step response of the continuous-time system and that for the emulated controller gains are plotted alongside.

clipboard_e9ccc6b43c4082491c53a681502ca8cb5.png

Deadbeat Controller Design

A discrete-time system is called deadbeat if all closed-loop poles are placed at the origin \((z=0)\).

A deadbeat system has the remarkable property that its response reaches steady-state in \(n\)-steps, where \(n\) represents the model dimension.

The desired closed-loop pulse characteristic polynomial is selected as \(\Delta _{\rm des} (z)=z^{n}\).

To design a deadbeat controller, let the closed-loop pulse transfer function be defined as: \[T(z)=\frac{K(z)G(z)}{1+K(z)G(z)} \nonumber \]

The above equation is solved for \(K(z)\) to obtain: \[K(z)=\frac{1}{G(z)} \frac{T(z)}{1-T(z)} \nonumber \]

Let the desired \(T(z)=z^{-n}\); then, the deadbeat controller is given as: \[K(z)=\frac{1}{G(z)(z^{n} -1)} \nonumber \]

Example \(\PageIndex{3}\)

Let \(G(s)=\frac{1}{s+1} ;\) then \(G(z)=\frac{1-e^{-T} }{z-e^{-T} }\).

A deadbeat controller for the model is obtained as: \(K(z)=\frac{z-e^{-T} }{(1-e^{-T} )(z-1)}\).

Example \(\PageIndex{4}\)

The discrete state variable model of a DC motor for \(T=0.02\; \rm s\) is given as: \[\left[\begin{array}{c} {i_{k+1} } \\ {\omega _{k+1} } \end{array}\right]=\left[\begin{array}{cc} {0.134} & {-0.038} \\ {0.038} & {0.816} \end{array}\right]\left[\begin{array}{c} {i_{k} } \\ {\omega _{k} } \end{array}\right]+\left[\begin{array}{c} {0.863} \\ {0.053} \end{array}\right]V_{k} , \;\;y_{k} =\left[\begin{array}{cc} {0} & {1} \end{array}\right]\left[\begin{array}{c} {i_{k} } \\ {\omega _{k} } \end{array}\right] \nonumber \]

The state feedback controller is given as: \(u_{k} =-\left[k_{1} ,\, \, k_{2} \right]x_{k}\).

The closed-loop characteristic polynomial is obtained as: \[\Delta (z)=z^{2} +(0.863k_{1} +0.053k_{2} -0.95)z-0.707k_{1} +0.026k_{2} +0.111 \nonumber \]

For pole placement design, let \(\Delta _{\rm des} (z)=z^{2}\). By equating the polynomial coefficients, the deadbeat controller gains are obtained as: \(k_{1} =0.501,\; k_{2} =9.702\).

The update rule for controller implementation is given as: \[u_{k} =0.501\, \, i_{k} +9.702\, \, \omega _{k} \nonumber \]

The step response of the deadbeat controller (Figure 10.2.3) settles in two time periods. The response was scaled to match that of the continuous-time system.

An approximate deadbeat design can be performed by choosing distinct closed-loop eigenvalues close to the origin, e.g., \(z=\pm {10}^{-5}\), and using the 'place' command from the MATLAB Control Systems Toolbox.

The feedback gains for the approximate design are obtained as: \(k_{1} =0.509,\; k_{2} =9.702\). The resulting closed-loop system response is still deadbeat.

clipboard_ebdfc98111c6648a5cf1713d4bf80c26c.png

Feedforward Tracking System Design

A tracking system was previously designed by using feedforward cancelation of the error signal (Section 9.2.1). A similar design can be performed in the case of discrete systems.

Towards this end, let the discrete state variable model be given as: \[{\bf x}_{k+1} ={\bf A}_{\rm d} {\bf x}_{k} +{\bf b}_{\rm d} u_{k} , \;\;y_{k} ={\bf c}^T {\bf x}_{k} \nonumber \]

A tracking controller for the model is defined as: \[u_k=-{\bf k}^T{\bf x}_k+k_rr_k \nonumber \] where \({\bf k}^{T}\) represents a row vector of feedback gains, \(k_r\) is a feedforward gain, and \(r_k\) is a reference input sequence.

Assuming that a pole placement controller for the discrete system has been designed, the closed-loop system is given as: \[{\bf x}_{k+1}=\left({\bf A}_{\rm d}-{\bf b}_{\rm d}{\bf k}^T\right){\bf x}_k+{\bf b}_{\rm d}k_rr_k \nonumber \]

The closed-loop pulse transfer function is obtained as: \[T\left(z\right)={\bf c}^T_{\rm d}{\left(z{\bf I-A}_{\rm d}+{\bf b}_{\rm d}{\bf k}^T\right)}^{-1}{\bf b}_{\rm d}k_r \nonumber \] where \({\bf I}\) denotes an identity matrix. The condition for asymptotic tracking is given as: \[T\left(1\right)={\bf c}^T_{\rm d}{\left({\bf I-A}_{\rm d}+{\bf b}_{\rm d}{\bf k}^T\right)}^{-1}{\bf b}_{\rm d}k_r=1 \nonumber \]

The feedforward gain for error cancelation is obtained as: \(k_r=\frac{1}{T\left(1\right)}\).

Example \(\PageIndex{5}\)

The discrete state variable model of a DC motor (\(T=0.02\)s) is given as: \[\left[\begin{array}{c} {i_{k+1} } \\ {\omega _{k+1} } \end{array}\right]=\left[\begin{array}{cc} {0.134} & {-0.038} \\ {0.038} & {0.816} \end{array}\right]\left[\begin{array}{c} {i_{k} } \\ {\omega _{k} } \end{array}\right]+\left[\begin{array}{c} {0.863} \\ {0.053} \end{array}\right]V_{k} , \;\;y_{k} =\left[\begin{array}{cc} {0} & {1} \end{array}\right]\left[\begin{array}{c} {i_{k} } \\ {\omega _{k} } \end{array}\right] \nonumber \]

A state feedback controller for the motor model was previously designed as: \(k^T =[k_{1} ,\; k_{2} ]\), where \(k_{1} =0.247,\; k_{2} =4.435.\)

The closed-loop system is defined as: \[T\left(z\right)=\frac{0.367z+0.179}{z^2-0.503z+0.05}k_r \nonumber \]

From the asymptotic condition, the feedforward gain is solved as: \(k_r=6.98\).

The step response of the closed-loop system is shown in Figure 10.2.4.

clipboard_edca113b5d7dfbf45077bd40bafc3153a.png

Example \(\PageIndex{6}\)

The discrete state variable model of a DC motor (\(T=0.02\)s) is given as:

\[\left[\begin{array}{c} {i_{k+1} } \\ {\omega _{k+1} } \end{array}\right]=\left[\begin{array}{cc} {0.134} & {-0.038} \\ {0.038} & {0.816} \end{array}\right]\left[\begin{array}{c} {i_{k} } \\ {\omega _{k} } \end{array}\right]+\left[\begin{array}{c} {0.863} \\ {0.053} \end{array}\right]V_{k} , \;\;y_{k} =\left[\begin{array}{cc} {0} & {1} \end{array}\right]\left[\begin{array}{c} {i_{k} } \\ {\omega _{k} } \end{array}\right] \nonumber \]

A dead-beat controller for the motor model was designed as: \(k^T =[k_{1} ,\; k_{2} ]\), where \(k_{1} =0.501,\; k_{2} =9.702\).

The closed-loop system is defined as: \[T\left(z\right)=\frac{0.672z+0.328}{z^2}k_r \nonumber \]

From the asymptotic condition, the feedforward gain is solved as: \(k_r=12.77\).

The step response of the closed-loop system is shown in Figure 10.2.5.

clipboard_eaf6c75fb62a0c3b9c3294150ad9f04e5.png

Tracking PI Controller Design

A tracking PI controller for the discrete state variable model is designed similar to the design of continuous-time system (Figure 9.3.1). The tracking PI controller places an integrator in the feedback loop, thus ensuring that the tracking error goes to zero in the steady-state.

In the case of continuous-time system, the tracking PI controller was defined as: \(u=-{\bf k}^{T} {\bf x}+k_{i} \int (r-y)\rm dt\).

Using the forward difference approximation to the integrator, given as: \(v_k=v_{k-1}+Te_k\), an augmented discrete-time system model including the integrator state variable is formed as:

\[\left[\begin{array}{c} {{\bf x}(k+1)} \\ {v(k+1)} \end{array}\right]=\left[\begin{array}{cc} {{\bf A}_{\rm d} } & {\bf 0} \\ {-{\bf c}^T T} & {1} \end{array}\right] \left[\begin{array}{c} {{\bf x}(k)} \\ {v(k)} \end{array}\right]+\left[\begin{array}{c} {{\bf b}_{\rm d} } \\ {0} \end{array}\right]u+\left[\begin{array}{c} {\bf 0} \\ {T} \end{array}\right]r \nonumber \]

The state feedback controller for the augmented system is defined as:

\[u(k)=\left[\begin{array}{cc} {-{\bf k}^T } & {k_ i } \end{array}\right]\, \left[\begin{array}{c} {{\bf x}(k)} \\ {v(k)} \end{array}\right] \nonumber \]

where \(k_ i\) represents the integral gain. With the addition of the above controller, the closed-loop system is described as:

\[\left[\begin{array}{c} {{\bf x}(k+1)} \\ {v(k+1)} \end{array}\right]=\left[\begin{array}{cc} {{\bf A}_{\rm d} -{\bf b}_{\rm d} k^{T} } & {{\bf b}_{\rm d} k_{i} } \\ {-{\bf c}^T T} & {1} \end{array}\right] \left[\begin{array}{c} {{\bf x}(k)} \\ {v(k)} \end{array}\right]+\left[\begin{array}{c} {\bf 0} \\ {T} \end{array}\right]r(k) \nonumber \]

The closed-loop characteristic polynomial of the augmented system is formed as:

\[{\mathit{\Delta}}_a\left(z\right)=\left| \begin{array}{cc} z{\bf I-A}_{\rm d}+{\bf b}_{\rm d}k^T & -{\bf b}_{\rm d}k_i \\ -{\bf c}^T_{\rm d}T & z-1 \end{array} \right| \nonumber \]

where \({\bf I}\) denotes an identity matrix of order \(n\).

Next, we choose a desired characteristic polynomial of \((n+1)\) order, and perform pole placement design for the augmented system. The location of the integrator pole in the \(z\)-plane may be selected keeping in view the desired peformance criteria for the closed-loop system.

\[\left[ \begin{array}{c} i_{k+1} \\ {\omega }_{k+1} \end{array} \right]=\left[ \begin{array}{cc} 0.134 & -0.038 \\ 0.038 & 0.816 \end{array} \right]\left[ \begin{array}{c} i_k \\ {\omega }_k \end{array} \right]+\left[ \begin{array}{c} 0.863 \\ 0.053 \end{array} \right]V_k,\ \ {\omega }_k=\left[ \begin{array}{cc} 0 & 1 \end{array} \right]\left[ \begin{array}{c} i_k \\ {\omega }_k \end{array} \right] \nonumber \]

The control law for the tracking PI controller is defined as:

\[u_k=-k_1i_k-k_2{\omega }_k+k_iv_k \nonumber \]

where \(v_{k} =v_{k-1} +T(r_{k} -\omega _{k} )\) describes the output of the integrator. The augmented system model for the pole placement design using integral control is given as:

\[\left[ \begin{array}{c} i_{k+1} \\ {\omega }_{k+1} \\ v_{k+1} \end{array} \right]=\left[ \begin{array}{ccc} 0.134 & -0.038 & 0 \\ 0.038 & 0.816 & 0 \\ 0 & -0.02 & 1 \end{array} \right]\left[ \begin{array}{c} i_k \\ {\omega }_k \\ v_k \end{array} \right]+\left[ \begin{array}{c} 0.863 \\ 0.053 \\ 0 \end{array} \right]V_k+\left[ \begin{array}{c} 0 \\ 0 \\ 0.02 \end{array} \right]r_k \nonumber \]

The desired \(z\)-plane pole locations for a desired \(\zeta=0.7\) are selected as: \(z=e^{-1} ,\; e^{-1\pm j1}\).

The controller gains, obtained using the MATLAB ‘place’ command, are given as: \(k_{1} =0.43,k_{2} =15.44,\; k_{i} =-297.79.\)

An update rule for controller implementation on computer is given as:

\[u_k=-0.43i_k-15.44{\omega }_k+297.8v_k \nonumber \]

\[v_k=v_{k-1}+0.02\left(r_k-{\omega }_k\right) \nonumber \]

The step response of the closed-loop system is plotted in Figure 10.2.6. The step response of the continuous-time system (Example 9.1.1) is plotted alongside. The output in both cases attains steady-state value of unity in about 0.12sec.

clipboard_ea8f695d9639c458f7619cc7b76b4d269.png

A pole-placement design approach for systems with multiple operating conditions

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Poll Ranks Biden as 14th-Best President, With Trump Last

President Biden may owe his place in the top third to his predecessor: Mr. Biden’s signature accomplishment, according to the historians, was evicting Donald J. Trump from the Oval Office.

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President Biden standing at the top of the steps leading to Air Force One.

By Peter Baker

Peter Baker has covered the past five presidents, ranked seventh, 12th, 14th, 32nd and 45th in the survey.

President Biden has not had a lot of fun perusing polls lately. He has a lower approval rating than every president going back to Dwight D. Eisenhower at this stage of their tenures, and he trails former President Donald J. Trump in a fall rematch. But Mr. Biden can take solace from one survey in which he is way out in front of Mr. Trump.

A new poll of historians coming out on Presidents’ Day weekend ranks Mr. Biden as the 14th-best president in American history, just ahead of Woodrow Wilson, Ronald Reagan and Ulysses S. Grant. While that may not get Mr. Biden a spot on Mount Rushmore, it certainly puts him well ahead of Mr. Trump, who places dead last as the worst president ever.

Indeed, Mr. Biden may owe his place in the top third in part to Mr. Trump. Although he has claims to a historical legacy by managing the end of the Covid pandemic; rebuilding the nation’s roads, bridges and other infrastructure; and leading an international coalition against Russian aggression, Mr. Biden’s signature accomplishment, according to the historians, was evicting Mr. Trump from the Oval Office.

“Biden’s most important achievements may be that he rescued the presidency from Trump, resumed a more traditional style of presidential leadership and is gearing up to keep the office out of his predecessor’s hands this fall,” wrote Justin Vaughn and Brandon Rottinghaus, the college professors who conducted the survey and announced the results in The Los Angeles Times .

Mr. Trump might not care much what a bunch of academics think, but for what it’s worth he fares badly even among the self-identified Republican historians. Finishing 45th overall, Mr. Trump trails even the mid-19th-century failures who blundered the country into a civil war or botched its aftermath like James Buchanan, Franklin Pierce and Andrew Johnson.

Judging modern-day presidents, of course, is a hazardous exercise, one shaped by the politics of the moment and not necessarily reflective of how history will look a century from now. Even long-ago presidents can move up or down such polls depending on the changing cultural mores of the times the surveys are conducted.

For instance, Barack Obama, finishing at No. 7 this year, is up nine places since 2015, as is Grant, now ranked 17th. On the other hand, Andrew Jackson has fallen 12 places to 21st while Wilson (15th) and Reagan (16th) have each fallen five places.

At least some of that may owe to the increasing contemporary focus on racial justice. Mr. Obama, of course, was the nation’s first Black president, and Grant’s war against the Ku Klux Klan has come to balance out the corruption of his administration. But more attention today has focused on Jackson’s brutal campaigns against Native Americans and his “Trail of Tears” forced removal of Indigenous communities, and Wilson’s racist views and resegregation of parts of the federal government.

As usual, Abraham Lincoln, Franklin D. Roosevelt, George Washington, Theodore Roosevelt and Thomas Jefferson top the list, and historians generally share similar views of many presidents regardless of their own personal ideology or partisan affiliation. But some modern presidents generate more splits among the historians along party lines.

Among Republican scholars, for instance, Reagan finishes fifth, George H.W. Bush 11th, Mr. Obama 15th and Mr. Biden 30th, while among Democratic historians, Reagan is 18th, Mr. Bush 19th, Mr. Obama sixth and Mr. Biden 13th. Other than Grant and Mr. Biden, the biggest disparity is over George W. Bush, who is ranked 19th among Republicans and 33rd among Democrats.

Intriguingly, one modern president who generates little partisan difference is Bill Clinton. In fact, Republicans rank him slightly higher, at 10th, than Democrats do, at 12th, perhaps reflecting some #MeToo era rethinking and liberal unease over his centrist politics.

The survey, conducted by Mr. Vaughn, an associate professor of political science at Coastal Carolina University, and Mr. Rottinghaus, a professor of political science at the University of Houston, was based on 154 responses from scholars across the country.

Peter Baker is the chief White House correspondent for The Times. He has covered the last five presidents and sometimes writes analytical pieces that place presidents and their administrations in a larger context and historical framework. More about Peter Baker

Our Coverage of the 2024 Presidential Election

News and Analysis

At the influential Conservative Political Action Conference, which is currently underway in Washington, the question is not which Republican will face off against President Biden in November, but rather who will join former President Donald Trump atop the ticket as his running mate .

Speaking at a Christian media convention in Nashville, Trump claimed that a “radical left, corrupt political class” was persecuting Christians and framed the election as a battle against a “wicked” system .

Anger within the Democratic Party over Biden’s support for Israel in the war in Gaza has been building for months. Michigan’s upcoming primary will put that discontent on the ballot for the first time .

Nikki Haley’s struggles to gain traction ahead of South Carolina’s Republican primary stem in part from a demographic fact : Nearly 10% of the state’s voters were not there when she left the governor’s mansion in 2017, and many of the newcomers have an affection for Trump. How did the state become Trump country ?

Fact-Checking Biden: During campaign and public events in recent weeks, Biden has made some misleading statements  about taxes, industry, jobs and more.

A Right-Wing Nerve Center:  The Conservative Partnership Institute has become a breeding ground for the next generation of Trump loyalists and an incubator for policies he might pursue. Its fast growth is raising questions .

 On Wall Street:  Investors are already thinking about how financial markets might respond to the outcome of a Biden-Trump rematch , and how they should trade to prepare for it.

  • International

Odysseus becomes first US spacecraft to land on moon in over 50 years

By Elise Hammond and Jackie Wattles , CNN

This is the NASA instrument that saved Odysseus' mission

From CNN's Jackie Wattles

NASA's Navigation Doppler Lidar.

Odysseus has now officially made history with its successful lunar touchdown — and none of it could have happened without some fast work from engineers on the ground and a breath-catching save from a NASA payload.

Before descent, Intuitive Machines, which developed the Odysseus lunar lander, revealed crucial pieces of the vehicle's navigation equipment were not working.

Fortunately, NASA — which considers itself one of many customers on this mission — had an experimental instrument already on board Odysseus that could be swapped in to make up for the malfunctioning equipment.

Engineers were able to bypass Odysseus' broken pieces and land using two lasers that are part of NASA's Navigation Doppler Lidar, or NDL, payload.

Here's how the NDL is described in IM-1's press kit:

The NDL is a LIDAR-based (Light Detection and Ranging) sensor composed of an optical head with three small telescopes and a box with electronics and photonics. NDL uses lasers to provide extremely precise velocity and range (distance to the ground) sensing during the descent and landing of the lander. This instrument operates on the same principles of radar, similar to a police radar detector, but uses pulses of light from a laser instead of radio waves and with very high accuracy. This will enhance the capabilities of space vehicles to execute precision navigation and controlled soft landings.

Odysseus is "upright and starting to send data"

After some intense waiting, Intuitive Machines, the company behind the Odysseus lunar landing mission, has confirmed the spacecraft is "upright and starting to send data."

That's a major milestone.

An upright landing potentially puts Odysseus in a better position than even Japan's SLIM "Moon Sniper" mission. SLIM was deemed a success as it made a soft touchdown , but later was revealed to have landed in a position that left its solar panels pointed in the wrong direction, causing that spacecraft to quickly lose power.

"Right now, we are working to downlink the first images from the lunar surface," Intuitive Machines said in a post on social media platform X.

Art in space: Sculpture hitches a ride to the moon on Odysseus lunar lander

From CNN's Jacqui Palumbo

Jeff Koons' "Moon Phases" is seen on the Odysseus lunar lander as it flies over the near side of the moon on Wednesday.

Exchanging the gallery space for a transparent box in  space, the American artist Jeff Koons now has one of his works of art on the moon .

On Thursday, a sculpture called “Moon Phases" hitched a ride on the Odysseus lunar lander as it touched down on the moon. It marked the United States' first landing on the lunar surface in more than 50 years.

The artwork depicts 125 mini-sculptures of the moon contained in a box, measuring about one inch in diameter. “Moon Phases" shows 62 phases of the moon as seen from Earth, 62 phases visible from other viewpoints in space, and one lunar eclipse.

Jeff Koons holds "Moon Phases" before it was attached to the lunar lander.

Each sculpture is inscribed with the name of a groundbreaking figure in human history, including Aristotle, David Bowie, Leonardo da Vinci, Gandhi, Billie Holiday, Gabriel García Márquez, Andy Warhol and Virginia Woolf. Koons “has drawn inspiration from the Moon as a symbol of curiosity and determination,” according to a statement from his gallery, Pace.

But the art market wouldn’t be able to do much with far-flung sculptures "exhibited" in outer space, so there’s a commercial component to Koons’ project as well. Pace Verso, the NFT wing of Pace, is also offering NFTs of each sculpture, while Koons has produced larger, coinciding physical sculptures of his “Moon Phases” to remain on Earth.

NASA reacts to lunar landing: "Great and daring quest"

NASA posted a reaction to the moon mission on social media, saying "Your order was delivered… to the Moon!"

"(Intutive Machines') uncrewed lunar lander landed at 6:23pm ET (2323 UTC), bringing NASA science to the Moon's surface. These instruments will prepare us for future human exploration of the Moon under #Artemis ," the space agency posted on X, the website formerly known as Twitter.

NASA Administrator Bill Nelson added during the webcast: "Today for the first time in more than a half-century, the US has returned to the moon." "Today is a day that shows the power and promise of NASA's commercial partnerships," he added. "Congratulations to everyone involved in this great and daring quest."

Applause and celebrations could be heard on the Intuitive Machines webcast of the event before the live coverage concluded.

CNN is standing by for additional updates on the spacecraft's status.

Odysseus becomes first US lander to touch down on the moon in over 50 years

Intuitive Machines mechanics, friends and family cheer after confirmation the lunar lander made a touchdown on the moon, in this still from the webcast.

The US-made  Odysseus lunar lander  has made a touchdown on the moon, surpassing its final key milestones — and the odds — to become the first commercial spacecraft to accomplish such a feat, but the condition of the lander remains in question.

Intuitive Machines, however, says the mission has been successful.

"I know this was a nail-biter, but we are on the surface, and we are transmitting," Intuitive Machines CEO Steve Altemus just announced on the webcast. "Welcome to the moon."

Odysseus is the first vehicle launched from the United States to land on the moon’s surface since the  Apollo 17 mission  in 1972.

Mission controllers from Intuitive Machines, the Houston-based company that developed the robotic explorer, confirmed the lander reached the lunar surface Thursday evening.

The uncrewed spacecraft traveled hundreds of thousands of miles from its Florida launch site at NASA’s Kennedy Space Center to the moon before making its final, perilous swoop to the lunar surface.

Odysseus: "Welcome to the moon"

The Odysseus lunar lander, nicknamed  “Odie” or IM-1 , is on the moon's surface and transmitting, Intuitive Machines CEO Steve Altemus just announced on the webcast.

"I know this was a nail-biter, but we are on the surface, and we are transmitting," Altemus said. "Welcome to the moon."

The exact state of the lander is not yet clear. But the company has confirmed it has made contact.

The Odysseus lander is "not dead yet"

Mission control is seen in this still from the livestreamed webcast.

Lunar landing missions typically offer moments of uncertainty. And though we're waiting for confirmation of communications, there have been some promising updates:

"We have an onboard fault detection system for our communications that after 15 minutes with lack of communication will power cycle the radios and then after that for another 15 minutes it will then switch antenna pairs, so we have some time here to evaluate," an Intuitive Machines flight controller said on the stream.

"We're not dead yet," they emphasized.

The company has also confirmed a "faint signal" — potentially representing signs of life from the spacecraft.

Intuitive Machines is troubleshooting communications after the expected landing time passes

The 6:24 p.m. ET landing time has come and gone.

Intuitive Machines knew it couldn't make contact with the lander right at the moment of touchdown, but expected to potentially have an answer shortly after.

It's now a waiting game to see whether Intuitive Machines can establish communications.

A good performance from Odysseus' engine

The webcast just announced that the engine is "nominal" — aerospace parlance for working as expected.

The spacecraft is functioning all on its own.

The expected landing time is 6:24 p.m. ET, though there could be wiggle room.

We could learn right at that time if Odysseus made a safe touchdown, or it could take a few minutes, according to the webcast.

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IMAGES

  1. Class 22 Pole Placement: Control Canonical Form and Pole Placement

    pole placement wiki

  2. Hands-free pole placement| Concrete Construction Magazine

    pole placement wiki

  3. Most Important Positions in the Pole Vault

    pole placement wiki

  4. Pole placement design

    pole placement wiki

  5. Unit4_3-Pole Placement Using Direct Substitution Method

    pole placement wiki

  6. PPT

    pole placement wiki

VIDEO

  1. The Concept of Pole Placement in Classical and Modern Control, 30/3/2016

  2. Pole Placement Method Theory Part 2

  3. Build Strength And Progress With Pole At Home. #poleconditioning #beginnerpole #poledance

  4. Pole Placement in Matlab using the "place" Command, 11/4/2016

COMMENTS

  1. Full state feedback

    Full state feedback (FSF), or pole placement, is a method employed in feedback control system theory to place the closed-loop poles of a plant in pre-determined locations in the s-plane. [1]

  2. PDF 8.2 State Feedback and Pole Placement

    Pole Placement Design Technique 8.2 State Feedback and Pole Placement Consider a linear dynamic system in the state space form In some cases one is able to achieve the goal (e.g. stabilizing the system or improving its transient response) by using the full state feedback, which represents a linear combination of the state variables, that is

  3. Pole Placement

    Trial Software Product Updates Pole Placement Closed-loop pole locations have a direct impact on time response characteristics such as rise time, settling time, and transient oscillations. Root locus uses compensator gains to move closed-loop poles to achieve design specifications for SISO systems.

  4. Control theory

    The objective is to develop a model or algorithm governing the application of system inputs to drive the system to a desired state, while minimizing any delay, overshoot, or steady-state error and ensuring a level of control stability; often with the aim to achieve a degree of optimality .

  5. PDF 16.30 Topic 11: Full-state feedback control

    Pole Placement Examples • Simple example: 8 14 20 G(s) = · · (s + 8)(s + 14)(s + 20) • Target pole locations −12 ± 12i, −20 Fig. 2: Response to step input with and without the N correction. Gives the desired steady-state behavior, with little difficulty! Fig. 3: Closed-loop frequency response. Clearly shows unity DC gain October 17, 2010

  6. Pole placement design

    Pole placement is a method of calculating the optimum gain matrix used to assign closed-loop poles to specified locations, thereby ensuring system stability. Closed-loop pole locations have a direct impact on time response characteristics such as rise time, settling time, and transient oscillations. For more information, see Pole Placement.

  7. PDF Pole Placement Design

    Introduction Simple Examples Polynomial Design State Space Design Robustness and Design Rules Model Reduction Oscillatory Systems Summary Theme: Be aware where you place them! Control of First Order Systems State: variables required to characterize storage of mass, momentum and energy Many systems are approximately of first order

  8. PDF 16.30 Topic 12: Pole placement approach

    Pole Placement Approach So far we have looked at how to pick K to get the dynamics to have some nice properties (i.e. stabilize A) λi(A) λi(A − BK) Question: where should we put the closed-loop poles? • Approach #1: use time-domain specifications to locate dominant poles - roots of: s 2 + 2ζωns + 2 ω = 0 n

  9. What is Pole Placement (Full State Feedback)

    Check out the other videos in the series: https://youtube.com/playlist?list=PLn8PRpmsu08podBgFw66-IavqU2SqPg_wPart 1 - The state space equations: https://you...

  10. Pole Placement

    Pole Placement | State Space, Part 2. This video provides an intuitive understanding of pole placement, also known as full state feedback. This is a control technique that feeds back every state to guarantee closed-loop stability and is the stepping stone to other methods like LQR and H infinity. We'll cover the structure of the control law ...

  11. Introduction: State-Space Methods for Controller Design

    Control Design Using Pole Placement. Let's build a controller for this system using a pole placement approach. The schematic of a full-state feedback system is shown below. By full-state, we mean that all state variables are known to the controller at all times. For this system, we would need a sensor measuring the ball's position, another ...

  12. PDF Module 2-1: Pole Placement

    Control:continuously operating dynamical systems. Frequency response methods made it possible to design linear closed-loop Norbert Wiener (Cybernetics) State space methods Minorsky worked on automatic controllers (PID) for steering ships. linear model-based (MIMO) optimal/stochastic/ adaptive control Rudolf Kalman (Apollo) modern control theory ...

  13. Robust Stability and Pole Placement

    Abstract. In this paper, we propose an integration of classic and parametric interval analysis methods for addressing robust stability and robust pole placement problems associated with linear dynamic systems with interval parameters. In order to reduce the conservatism of classic interval analysis and synthesis methods due to the parameter ...

  14. PDF 1 Robust pole placement with Moore's algorithm

    Robust pole placement with Moore's algorithm Robert Schmid, Amit Pandey and Thang Nguyen Abstract We consider the classic problem of pole placement by state feedback. We adapt the Moore eigen-structure assignment algorithm to obtain a novel parametric form for the pole-placing gain matrix, and

  15. Ackermann's formula

    In control theory, Ackermann's formula is a control system design method for solving the pole allocation problem for invariant-time systems by Jürgen Ackermann. [1]

  16. LMIs in Control/pages/Switched Systems Pole Placement

    Pole Placement for Switched Systems. This LMI lets you provide specifications of the switched system closed loop poles. Note that arbitrarily switching between stable systems can lead to instability whilst switching can be done between individually unstable systems to achieve stability.

  17. Intro to Control Theory Part 6: Pole Placement

    Wesley Aptekar-Cassels | Intro to Control Theory Part 6: Pole Placement Intro to Control Theory Part 6: Pole Placement May 7, 2017 In Part 4, I covered how to make a state-space model of a system to make running simulations easy. In this post, I'll talk about how to use that model to make a controller for our system.

  18. PDF 4 Pole placement using polynomial methods

    fpigm i=0 and fligm i=0, are to be chosen. Once we choose m, the characteristic poly-nomial is given by: ¢(s) = A(s)L(s) + B(s)P(s); which is an (n + m)th order polynomial. We can select n + m desired closed-loop pole locations, leading to a desired characteristic polynomial: ¢d(s) = dn+msn+m + dn+m¡1sn+m¡1 + ¢ ¢ ¢ + d0;

  19. Pole Placement Method

    An assisted pole placement method has been proposed to help the designers in their choice of the design parameters. Once the desired rise time, settling time and percentage of overshoot of the step response of the closed-loop system with respect to a reference change have been specified, the computation of the controller is performed automatically.

  20. 10.2: Controllers for Discrete State Variable Models

    The pole placement controller designed for a continuous-time state variable model can be used with derived sampled-data system model. Successful controller emulation requires a high enough sampling rate that is at least ten times the frequency of the dominant closed-loop poles of the system.

  21. A pole-placement design approach for systems with multiple operating

    The author proposes design procedures based on state-space pole-placement techniques for systems with multiple operating conditions. This is the so-called simultaneous pole-placement problem. First, the full state feedback problem is studied, in which a nonlinear local pole-placement solution is proposed. The design condition is formulated in terms of the rank condition of a multimode ...

  22. Linear-quadratic regulator

    General description The settings of a (regulating) controller governing either a machine or process (like an airplane or chemical reactor) are found by using a mathematical algorithm that minimizes a cost function with weighting factors supplied by a human (engineer).

  23. IET Digital Library: Controller synthesis by pole placement

    The pole placement synthesis technique allows placing all closed-loop poles at desired locations, so that the system closed-loop specifications can be met. Thus, the main advantage of pole placement over other classical synthesis techniques is that we can force both the dominant and the non-dominant poles to lie at arbitrary locations.

  24. Poll Ranks Biden as 14th-Best President, With Trump Last

    President Biden may owe his place in the top third to his predecessor: Mr. Biden's signature accomplishment, according to the historians, was evicting Donald J. Trump from the Oval Office.

  25. Odysseus becomes first US spacecraft to land on moon in over 50 years

    The Odysseus lander's mission is designed to assess the lunar environment of the moon's south pole ahead of NASA's current plan to return a crewed mission there in late 2026. Our live coverage ...