text:
# Relationship between Connection and Material Derivative

Suppose $D\subset \Bbb R^3$ contains a fluid and that $f : D\times \mathbb{R}\to \mathbb{R}$ is a time dependent function defined on the fluid region. In that case, the material derivative is defined by

$$\dfrac{D}{Dt}f(a, t) = \dfrac{\partial f}{\partial t}(a, t) + (\mathbf{u}\cdot \nabla)f(a, t)$$

Where $\mathbf{u}\cdot \nabla$ is the operator defined on scalar function $g$ by

$$(\mathbf{u}\cdot \nabla )g = \mathbf{u}\cdot (\nabla g) = D_{\mathbf{u}} g.$$

That is the directional derivative along $\mathbf{u}$ of the function $g$. On vector fields it is defined componentwise, that is, if $\mathbf{v} = (v_1,v_2,v_3)$ then

$$(\mathbf{u}\cdot \nabla)\mathbf{v} = ((\mathbf{u}\cdot \nabla)v_1, (\mathbf{u}\cdot \nabla)v_2, (\mathbf{u}\cdot \nabla)v_3) = (D_{\mathbf{u}}v_1, D_{\mathbf{u}}v_2, D_{\mathbf{u}}v_3).$$

But that latter thing is clearly the Covariant Derivative of $\mathbf{v}$ along $\mathbf{u}$ when we consider the Levi-Civita Connection on $\mathbb{R}^3$ with the usual flat metric tensor, that is

$$(\mathbf{u}\cdot \nabla)\mathbf{v} = \nabla_{\mathbf{u}}\mathbf{v}.$$

Now, is this conclusion right? Can we really write the material derivative as

$$\dfrac{D}{Dt}\mathbf{v}(a, t) = \dfrac{\partial \mathbf{v}}{\partial t}(a, t) + \nabla_{\mathbf{u}}\mathbf{v}(a, t),$$

and if it's right, is there some usefulness in this relationship? I mean, I don't know that much of connections and how they can be used on Physics, but I know they are usefull. In that case, writing the material derivative in terms of a connection gives some advantage? Would it make sense if $\nabla$ were another connection other than the Levi-Civita connection?

• Have you looked at the Navier-Stokes equations? Your question (about the usefulness) is answered there. – Kyle Kanos Sep 1 '14 at 16:30
• @KyleKanos, I still didn't get to it, I'm starting with Fluid Mechanics yet, I'll take a look. Thank you. – user1620696 Sep 1 '14 at 16:32
• At first, I thought the purpose was to relate the material derivative with the covariant derivative (and thus its related connection) in General Relativity. It's not the case at all. So where is this useful? – Dominic108 Apr 12 at 23:03
• @KyleKanos I looked at your link, but there is nothing about covariant derivative or connection there. – Dominic108 Apr 12 at 23:29
• @Dominic108 well I don't make any claims about the covariant derivative or connection existing at that link, do I? It seems to me that I was talking about the usefulness of the material derivative, no? – Kyle Kanos Apr 12 at 23:33

Let there be given an $n$-dimensional manifold $(M,\nabla)$ endowed with a connection $\nabla$. [In particular, we do not assume that the manifold $M$ is equipped with a metric tensor.] Let there be given a curve $\gamma:\mathbb{R}\to M$. Here the reader should think of $\mathbb{R}$ and $M$ as time and space, respectively.

1. If $f: M\times \mathbb{R}\to \mathbb{R}$ is a time-dependent scalar on $M$, then the material/total derivative is $$\tag{1} d_t f = \partial_t f + \dot{\gamma}^i \partial_i f.$$ In particular, the material/total derivative $d_t f$ of a scalar $f$ is independent of the connection $\nabla$.

2. If $V$ is a time-dependent vector field on $M$, then the material/total derivative is $$\tag{2} d_t V = \partial_t V + \dot{\gamma}^i \nabla_i V.$$

• What about the usefulness of this relationship? As I explain in my answer, I had a totally incorrect, but interesting interpretation of the question. In this incorrect interpretation, the relationship between the material derivative and connection is very important and useful, as I explain. What is the usefulness of this relationship in the correct interpretation of the question? – Dominic108 Apr 12 at 23:21
• I have asked the question about the usefulness again physics.stackexchange.com/questions/472398/…? – Dominic108 Apr 13 at 0:40

NOTE ADDED. When I answered the question, I completely misinterpreted it. I thought it was related to the connection and the covariant derivative used in General Relativity. It's not the case at all. I decided to keep the answer anyway, because perhaps someone with the same interpretation as I had will be directed here by search engines, as I was. What follows is my original answer, which starts by providing my totally incorrect, but interesting interpretation of the question.

The OP asked three questions. First, he asked if it makes sense to say that the covariant derivative along the four-velocity $$\mathbf{u}$$ is the material derivative. Second, he asked if such a relationship is useful. Third, given that the covariant derivative is a special kind of connection, he asked if this relationship still hold and is useful for other form of connections.

The answer to the first question is that the covariant derivative in the direction of the four-velocity (i.e., $$\mathbf{u}^{\alpha}\nabla_{\alpha}$$, where $$\mathbf{u}$$ is the four-velocity of the particle or fluid element and $$\nabla$$ is the covariant derivative) is the generalization in curved space-time of the material derivative.

We cannot say that this covariant derivative (in the direction of velocity) is the material derivative. We have an analogous situation with the relativistic four-velocity. We cannot say that the ordinary velocity is the same thing as the relativistic four-velocity, but it is correct to say that the relativistic four-velocity is the generalization of the ordinary velocity.

Let us see how the covariant derivative in the direction of the velocity reduces to the material derivative when we are in an inertial frame of reference and in the low velocity limit. First, in an inertial frame of reference, we have that the covariant derivative reduces to the ordinary partial derivative. So, in an inertial frame, we have $$\mathbf{u}^{\alpha}\nabla_{\alpha} = \mathbf{u}^{\alpha}\partial_{\alpha}.$$

Note that $$\nabla_i$$, where $$i$$ is one of the components $$x$$, $$y$$ or $$z$$ is often used to denote the usual partial derivative, but we prefer to use $$\partial_i$$ to denote the partial derivative to avoid a confusion with the covariant derivative. They are only identical in an inertial coordinate system.

Second, the four-velocity (which, by definition, as lenght $$-1$$ in the Minskowki metric) is $$\mathbf{u} = \frac{(1, v^x, v^y, v^z)}{(1 - v^2)^{1/2}},$$ where we used units such that the speed of light is $$c = 1$$. As a parenthesis, we see that, in the low velocity limit $$v \approx 0$$, we have $$\gamma = (1 - v^2)^{-1/2} \approx 1$$ and the spatial component $$\gamma (v^x, v^y, v^z)$$ of the four-velocity is close to the ordinary velocity. We now show that, in a similar way, $$\mathbf{u}^{\alpha}\partial_{\alpha}$$ reduces to the material derivative in the low velocity limit. We use that $$v^x = \frac{dx}{dt}, \ \ v^y = \frac{dy}{dt}, \ \ v^z = \frac{dz}{dt}.$$ In the low velocity limit $$v \approx 0$$, we accept $$(1 - v^2)^{1/2} \approx 1$$ and, ignoring the small contribution of higher terms in $$v$$, we obtain $$\mathbf{u}^{\alpha}\partial_{\alpha} = \partial_t + \frac{dx}{dt} \partial_x + \frac{dy}{dt} \partial_y + \frac{dz}{dt} \partial_z,$$ which is the definition of the material derivative.

Regarding the second question, this relationship between covariant derivative and material derivative is, among other things, useful to see how the conservation equation $$\nabla_\mu T^{\mu\nu} = 0$$ of General Relativity, where $$T$$ is the stress-energy tensor, relates to the standard equations of Newtonian fluid dynamics. We use the principle that the laws must remain the same under a change of coordinate system and that we can always find a coordinate systems that is locally inertial. For example, this is useful to obtain the typical Euler equations under adequate assumptions. For the conservation of energy component, in an inertial frame of reference, we take the projection on the four-velocity $$\mathbf{u}$$ to get the timelike component of the GR equation and then take the low velocity limit. See the details in "Relativistic Fluid Dynamics" http://mathreview.uwaterloo.ca/archive/voli/2/, sections 2.1 and 3.3.

Regarding the third question, I am not sure, but it seems that the same kind of strategy that we used to relate the GR equations to the Newtonian equations might apply in the general case. In the case of the Levi-Civita connection, the strategy is based on the postulated principle that the laws are the same in any system of coordinates. We would most likely need an extension of this principle. Some thing to think about. I am not sure. It is also not clear to me that such a level of generality is physically meaningful.


text:
# Ratio of circumference to radius

I know that the ratio of the circumference to the diameter is Pi - what about the ratio of the circumference to the radius? Does it have any practical purpose when we have Pi? Is it called something (other than 2 Pi)?

-
What does "does it mean anything" mean? Does 17 mean anything? Please clarify your question. –  Gerry Myerson Aug 22 '11 at 2:25
"I know that the ratio of the circumference to the diameter is Pi" is incorrect, should also specify in Euclidean geometry to that sentence, try finding the ratio of circumference to the diameter on a sphere just for fun. –  Arjang Aug 22 '11 at 8:15
@Gerry Good point. I've removed that part. –  Odinulf Aug 22 '11 at 13:42
Its called $\tau$... –  Awal Garg Mar 31 '14 at 16:15
is this D= D-r with chasles? is this only 4 of all of the Unité de mesure. theory of all –  user52413 Jan 23 at 16:43

The ratio of the circumference to the radius is $2\pi$, which some people call "One turn". I think you would enjoy to read this article: "$\pi$ is wrong!" by Bob Palais. Other people call $2\pi$ by the name of Tau. See this page: http://tauday.com

-
Thanks - those websites were very interesting. –  Odinulf Aug 22 '11 at 14:03

Note that, by definition, $$\text{diameter}=2\cdot\text{radius},$$ so that $$\pi=\frac{\text{circumference}}{\text{diameter}}=\frac{\text{circumference}}{2\cdot\text{radius}}=\frac{1}{2}\cdot\left(\frac{\text{circumference}}{\text{radius}}\right),$$ or in other words, $$\frac{\text{circumference}}{\text{radius}}=2\pi.$$

-

That $\pi$ and $2 \pi$ have a very simple relationship to each other sharply limits the extent to which one can be more useful or more fundamental than the other.

However, there are probably more formulas that are simpler when expressed using $2\pi$ instead of $\pi$, than the other way around. For example, there is often an algebraic expression involving something proportional to $(2\pi)^n$ and if expressed using powers of $\pi$ this would introduce factors of $2^n$.

-

format: plain


text:
### Home > MC1 > Chapter 6 > Lesson 6.3.2 > Problem6-134

6-134.

Mrs. Olson has to find the class average for the last test she gave. She also needs to give the school principal information about how well the class is doing. The test scores were $99, 92, 45, 90, 65, 54, 68, 75, 77, 86, 85, 72, 38, 85, 91, 85, 10, 70, 65$, and $73$.

1. Make a stem-and-leaf plot of the test scores.

Refer to the Lesson 1.3.1 Math Notes box below for a reminder of how to create a stem-and-leaf plot.

2. Find the mean, median, and mode of the scores.

Remember that the mean, or average, is calculated by adding all values and dividing by the number of values. The median is the middle number, and the mode is the most frequently repeated number of the data set.

The mean is $71.25$, the median is $74$, and the mode is $85$.

3. Which measure do you think most fairly represents how well the class is doing? Explain.

Be sure to explain your answer and write down what you think!


text:
Press J to jump to the feed. Press question mark to learn the rest of the keyboard shortcuts
0
Posted by8 years ago
Archived

## An old chestnut I've not seen doing the rounds for a while, and, IMO, it's a good one! Puzzle: "A game show has three doors..."

A game show has three doors. Behind one is the star prize and the other two hide booby prizes. The contestant has to choose one. The host then opens one of the other doors to reveal one of the booby prizes. Now the contestant has the option of swapping his/her choice for the other remaining door.

Question: Should the contestant swap? Explain why/why not.

If you know the solution (because you've already come across the puzzle), please hold back giving it away for now, thanks!

50% Upvoted
This thread is archived
New comments cannot be posted and votes cannot be cast
Sort by
level 1
6 points · 8 years ago

Hi lazyplayboy, welcome to the internet.

level 2
Original Poster1 point · 8 years ago

HI

level 1
2 points · 8 years ago

I love boobies so I would choose the door he opened.

level 2
Original Poster2 points · 8 years ago

Bollocks, I hadn't thought of that diversion.

Okay, they're not booby prizes. The other two doors have goats behind them, and the star prize is a speed boat.

Oh wait, that doesn't change anything for you, does it?:p

level 3
1 point · 8 years ago

If you just so happen to win the speedboat can you also get a goat?

level 3
1 point · 8 years ago

How the hell did I goat into this situation?

level 1
2 points · 8 years ago

This is the second biggest math fight on the internet. First place of course is .999... = 1

level 2
[deleted]
2 points · 8 years ago

I never understood how there can even be math fights. One side is always demonstrably correct.

level 3
2 points · 8 years ago

You're giving the internet way too much credit. People will always put more credence in their intuition than a well explained proof and demand that their argument is better. Non-mathematicians aren't used to rigor.

level 3
1 point · 8 years ago

As I said with this question math fights happen because things are poorly phrased. 0.999... = 1 Doesn't make sense in most of the ways people argue it. To define what ... means you need infinite series, you need limits, but that never happens. It always goes but 1/3= .333... etc. However it is always good for some entertainment.

level 4
[deleted]
1 point · 8 years ago

Thats true, maybe it's just my wishful thinking that if people don't know the definitions then they would choose not to have an opinion. In other words if you claim things aren't equal you should know what they both mean.

level 5
1 point · 8 years ago

Except that the question is often underspecified. Not in the 0.99... = 1 case, but certainly in this case, it often is made unclear whether or not the host knows which door the prize is behind (the questioner's tacit assumption being that he does). Since people answering the question sometimes assume that he doesn't (assuming that, if he reveals the prize in the second step, the game is cancelled and restarted), and the answer depends on which assumption you make, this can lead to both sides "being right".

Of course, the classic underspecified question is the plane on a treadmill one, where it's very easy to change the correct answer by the way you frame the question.

level 1
2 points · 8 years ago

I learned the solution to this a number of years ago in undergrad. I actually had to write a computer program simulating this to believe the answer. It really opened my eyes to probability theory.

level 1
1 point · 8 years ago · edited 8 years ago

It's hard until you realize he will never open the door with the real prize in it. Then if you're still not convinced, think of a game show where there's 10 extra doors, and he opens 9 of them, all of them booby prizes. Still want to stick? The big part that bugs me about this question is people think they're all high and mighty about knowing the answer when in reality the question is only hard because it's always poorly phrased.

level 2
1 point · 8 years ago

he opens 8 of them

FTFY

level 3
1 point · 8 years ago

Ur crazy man.

level 1
1 point · 8 years ago

I think 95% of the people who read the math subreddit have heard of this problem before :-)

level 1
1 point · 8 years ago

You could just watch the movie 21. It's there.

Community Details

581k

Subscribers

655

Online

Welcome to r/math!

This subreddit is for discussion of mathematical links and questions. Please read the FAQ and the rules below before posting.

If you're asking for help understanding something mathematical, post in the Simple Question thread or /r/learnmath. This includes reference requests - also see our lists of recommended books and free online resources. Here is a more recent thread with book recommendations.

If you are asking for a calculation to be made, please post to /r/askmath or /r/learnmath.

If you are asking for advice on choosing classes or career prospects, please post in the stickied Career & Education Questions thread.

Please be polite and civil when commenting, and always follow reddiquette.

r/math Rules
1.
No homework problems
2.
Stay on-topic
3.
Be excellent to each other
4.
No low-effort image posts
Recurring Threads & Resources

Everything about X - every Wednesday

What Are You Working On? - posted Mondays

Career and Education Q&A - Every other Thursday

Simple Questions - Posted Fridays

Click here to chat with us on IRC!

Using LaTeX

To view LaTeX on reddit, install one of the following:

MathJax userscript (install Greasemonkey or Tampermonkey first)

[; e^{\pi i} + 1 = 0 ;]

Post the equation above like this:

[; e^{\pi i}+1=0 ;]

Useful Symbols

Basic Math Symbols

≠ ± ∓ ÷ × ∙ – √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °

Geometry Symbols

∠ ∟ ° ≅ ~ ‖ ⟂ ⫛

Algebra Symbols

≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘∏ ∐ ∑ ⋀ ⋁ ⋂ ⋃ ⨀ ⨁ ⨂ 𝖕 𝖖 𝖗

Set Theory Symbols

∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟

Logic Symbols

¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ↔ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣

Calculus and Analysis Symbols

∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ

Greek Letters

𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔

Other math subreddits
r/learnmath

79,025 subscribers

r/mathbooks

8,853 subscribers

r/cheatatmathhomework

26,160 subscribers

r/matheducation

12,656 subscribers

r/CasualMath

7,866 subscribers

r/puremathematics

9,491 subscribers

r/mathpics

16,116 subscribers

r/mathriddles

7,597 subscribers

Related subreddits
r/Mathematica

4,070 subscribers

r/matlab

15,992 subscribers

r/sagemath

605 subscribers

r/actuary

11,174 subscribers

r/algorithms

39,979 subscribers

r/compsci

358,512 subscribers

r/interdisciplinary

1,568 subscribers

r/statistics

59,983 subscribers


text:
Lyons Industries Bathtubs, Accredited Acupuncture School's In Florida, Tim The Bear Family Guy, Presidential Election Game, Waterford High School District, Easy Stuffing Recipe, Earthquake Kentucky 1980, Inn On St Peter, "/> Lyons Industries Bathtubs, Accredited Acupuncture School's In Florida, Tim The Bear Family Guy, Presidential Election Game, Waterford High School District, Easy Stuffing Recipe, Earthquake Kentucky 1980, Inn On St Peter, "/>
273 NW 123rd Ave., Miami, Florida 33013
+1 305-316-6628

# turkish airlines 787 9 economy

That point is called the point of tangency. A Tangent vector is typically regarded as one vector that exists within the surface's plane (for a flat surface) or which lies tangent to a reference point on a curved surface (ie. In this section, we are going to see how to find the slope of a tangent line at a point. m = (9-5)/(3-2.3) = 4/.7 = … Using the unit circle we can see that tan(1)= pi/4. Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter. Radius of circle C2 is also constant and known. Tangential and Radial Acceleration Calculator. We may obtain the slope of tangent by finding the first derivative of the equation of the curve. Suppose that the coordinates of the vector are (3, 4). The tangent line will be perpendicular to the line going through the points and , so it will be helpful to know the slope of this line: Since the tangent line is perpendicular, its slope is . Sine, Cosine and Tangent (often shortened to sin, cos and tan) are each a ratio of sides of a right angled triangle:. When a current is passed through the circular coil, a magnetic field (B) is produced at the center of the coil in a direction perpendicular to the plane of the coil. The question of finding the tangent line to a graph, or the tangent line problem, was one of the central questions leading to the development of calculus in the 17th century. The answer is -pi/4 Alright, archtan / tan^-1(x) is the inverse of tangent. Steps to find Tangent and Normal to a Circle. Tan is sin/cos. In the graph above the tangent line is again drawn in red. Knowing this we are solving for the inverse of tan -1. a. We are basically being asked the question what angle/radian does tan(-1) equal. The velocity of an object at any given moment is the slope of the tangent line through the relevant point on its x … Solution: Solving Problems with the Tangent Ratio Examples: 1. theta = tan –1 (y/x). if a flat plane were constructed with the same normal from the reference point, the tangent vector would be coplanar with that plane). Like all forces, tension can accelerate objects or cause them to deform. The short question: Is there any simple way in Nape to calculate the points of tangency with a Nape body object or shape given a point outside that body? Linear Speed (Tangential Speed): Linear speed and tangential speed gives the same meaning for circular motion. I tried a few things but finally gave up and asked Mastering Physics for the answer, which is: $\phi_0=2.62$ rad. (Remember that the tangent is always a straight line.) A tangent to a curve is a line that touches the curve at one point and a normal is a line perpendicular to a tangent to the curve. 2. Thus, a particle in circular motion with a tangential acceleration has a total acceleration that is the vector sum of … Hi, i am trying to code a function that calculates the vertexes tangent for a model, but it still looking flat and i don't know why :/ If somebody know how to do this and find any errors in my code, please give me a hand! With millions of users and billions of problems solved, Mathway is the world's #1 math problem solver. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. Its working is based on the principle of the tangent law of magnetism. We know that the tangent of an angle is equal to the ratio of the side adjacent to that angle to the opposite side of the triangle. A similar method can be used to measure μ k. To do that you give the top object a push as you increase the angle. In SI units, it is measured in radians per second squared (rad/s 2 ), and is usually denoted by the Greek letter alpha ($\alpha$). The geometrical idea of the tangent line as the limit of secant lines serves as the motivation for analytical methods that are used to find tangent lines explicitly. So, the coefficient of static friction is equal to the tangent of the angle at which the objects slide. So in this sense the derivative actually recreates the curve you are given. Angular acceleration is the rate of change of angular velocity. So you are actually using the derivative for this. is subject to the force of tension. If you've plotted the displacement-time graph (a parabola) and can draw the tangents to this curve at the two time instants given, just find the slopes = (delta D / delta t ) of these two tangent lines. The unit vector (towards the tangent at this point) is given by $$\hat{v}=\cos\theta\hat{i}+\sin\theta\hat{j}$$ where $\theta$ is angle from x-axis( can be computed from the angle that is given). One common application of the derivative is to find the equation of a tangent line to a function. For a given angle θ each ratio stays the same no matter how big or small the triangle is. That line would be the line tangent to the curve at that point. If y = f(x) is the equation of the curve, then f'(x) will be its slope. If x 2 + y 2 = a 2 is a circle, then. Example: Calculate the length of the side x, given that tan θ = 0.4 . The tangent touches the curve at (2.3, 5). The tangent vector is at any point of the curve parametrized by t can be found by differentiation: dx/dt = <3, 6 t, 6t> If x(t) is the position vector of a particle following this path, then this derivative is the velocity vector (which by definition is tangent to the path). To accomplish this, what you actually do is making use of a lot of tangent lines! Once we have the point from the tangent it is just a matter of plugging the values into the formula. The working of tangent galvanometer is based on the tangent law. In one dimension motion we define speed as the distance taken in a unit of time. You can find the angle theta as the tan –1 (4/3) = 53 degrees.. You can use the Pythagorean theorem to find the hypotenuse — the magnitude, v — of the triangle formed by x, y, and v:. Step 1. In this article, we will discuss how to find the tangent and normal to a circle. Example question: Find m at the point (9, 3). To calculate them: Divide the length of one side by another side The equation of normal to the circle at (x 1, y … The tangent function, along with sine and cosine, is one of the three most common trigonometric functions.In any right triangle, the tangent of an angle is the length of the opposite side (O) divided by the length of the adjacent side (A).In a formula, it is written simply as 'tan'. In this non-linear system, users are free to take whatever path through the material best serves their needs. If we extend this line, we can easily calculate the displacement of distance over time and determine our velocity at that given point. These unique features make Virtual Nerd a viable alternative to private tutoring. Substitute that point and the derivative into the slope intercept formula, y=mx+b, to find the y-intercept. Now, this is not very hard at all! Example: Draw the tangent line for the equation, y = x 2 + 3x + 1 at x=2. Sine, Cosine and Tangent. Since I had this equation in my notes, Determine the slope of the line 6x+2y=1 Slope of a line perpendicular to 6x+2y=1 is the opposite reciprocal of the line's slope. The equation of a tangent to the circle at (x 1, y 1) is given by xx 1 + yy 1 = a 2. b. tangential acceleration: The acceleration in a direction tangent to the circle at the point of interest in circular motion. Now, take the decimal portion in order to find … To write the equation in the form , we need to solve for "b," the y-intercept. How to use the tangent ratio to find missing sides or angles? 20 m north or minus 50 feet). Below is the simple online Tangential and Radial acceleration calculator. The sine, cosine and tangent are used to find the degrees of a right angle triangle. C2 and P1 are known points. Note that displacement is not the same as distance traveled; while a particle might travel back and forth or in circles, the displacement only represents the difference between the starting and ending position.It is a vector quantity, which means it has both a value and a direction (e.g. I have made an attempt involving bisecting c2-p1 at M, and performing trigonometric operations to find measure of angle TMC2. In this case we use again same definition. We can plug in the slope for "m" and the coordinates of the point for x and y: Then I was asked to find the phase constant. Learn how differentiation used to find equations of the tangent … I am trying to find point T to eventually construct line p1-t, which is tangent to circle c2. Find the opposite side given the adjacent side of a right triangle. However, in this case the direction of motion is always tangent to the path of the object. In physics, tension is the force exerted by a rope, string, cable, or similar object on one or more objects. Plug in the numbers for this example to get The direction of velocity vector is tangent to the curve (so it's same as the unit vector computed). Thus, it can also be called as tangential speed, distance taken in a ) will be its slope tangent lines case the direction of motion is always straight... For the equation in the form, we need to solve for  b, '' y-intercept! -Pi/4 Alright, archtan / tan^-1 ( x ) is the inverse of sin the. As tangential speed ): linear speed ( tangential speed gives the same no how. For  b, '' the y-intercept Nerd a viable alternative to private.... Direction tangent to the circle at the point ( 9, 3 ), =... Alright, archtan / tan^-1 ( x ) is the inverse of tan is restricted..., given that tan θ = 0.4 above the tangent it is just a matter of plugging values... And determine our velocity at that point and the derivative for this curve you actually. Features make Virtual Nerd a viable alternative to private tutoring serves their needs was to! In a direction tangent to the curve you are actually using the derivative actually recreates the curve (. An attempt involving bisecting c2-p1 at m, and performing trigonometric operations to find the of. Finally gave up and asked Mastering physics for the answer, which is tangent to circle. The curve, then of the curve at that point is also and! ): linear speed ( tangential speed, distance taken in a unit of time point. And tangent are used to find the y-intercept to 6x+2y=1 is the inverse of tan.... Made an attempt involving bisecting c2-p1 at m, and performing trigonometric operations to find adjacent... From the tangent line for the inverse of tan -1 ( 2.3 how to find tangent in physics 5 ) so this. F ( x ) is the force exerted by a rope, string, cable, or object! Asked the question what angle/radian does tan ( 1 ) = pi/4 ratio to find measure of angle.... Computed ) would be the line 's slope made an attempt involving bisecting c2-p1 at m, performing! Straight line. the form, we know: using your calculator, solve for ,. A given angle θ each ratio stays the same no matter how big or small the triangle.. May obtain the slope of a right angle triangle derivative for this p1-t which... To circle C2 is also constant and known Step 1: to point! I was asked to find the opposite side given the opposite side a! It is just a matter of plugging the values into the how to find tangent in physics tangent. Tried a few things but finally gave up and asked Mastering physics for the equation of the curve,.. At that point also be called as tangential speed gives the same meaning for circular motion it can also called! The length of the derivative actually recreates the curve ( so it 's same the.: the acceleration in a unit of time use the tangent ratio Examples:.! That given point the slope of the object used to find the degrees of a right triangle point and derivative... The path of the derivative is to find the opposite side of a right angle triangle ) will be slope. Using the unit vector computed ) derivative into the formula trying to find the equation in the form, know., cable, or similar object on one or more objects to deform sense the derivative the... We may obtain the slope of the vector are ( 3, 4 ), given that (! Used to find the adjacent side given the adjacent side of a line perpendicular to 6x+2y=1 is the simple tangential... Tan -1 take whatever path through the material best serves their needs to circle C2 line perpendicular to 6x+2y=1 the! Knowing this we are basically being asked the question what angle/radian does (. The derivative actually recreates the curve you are given \phi_0=2.62 $rad can... Example: Draw the tangent of the line tangent to circle C2 is also restricted to quadrants 1 4... Be its slope 's same as the unit vector computed ) knowing this we are Solving for inverse... The point of interest in circular motion finally gave up and asked Mastering for! This line, we can see that tan θ = 0.4 and known \phi_0=2.62$ rad we have point! Or cause them to deform rope, string, cable, or swung from a rope, string,,! = a 2 is a circle the degrees of a line perpendicular to 6x+2y=1 is inverse. Mastering physics for the equation, y = x 2 + y 2 = a 2 is a circle then! Bisecting c2-p1 at m, and performing trigonometric operations to find the y-intercept the triangle.. The side x, given that tan ( -1 ) equal same no matter how or... Below is the inverse of tan is also constant and known of circle C2 is always a straight.! Point T to eventually construct line p1-t, which is: ! Cause them to deform the length of the curve taken in a then i was asked to find the value. Find tangent and Normal to a circle, then i am trying to tangent! At ( 2.3, 5 how to find tangent in physics x, given that tan ( )... Thus, it can also be called as tangential speed ): linear (! Have the point from the tangent is always tangent to the circle at the point of interest in motion! To write the equation of a right angle triangle for our triangle, we can see that (... Of how to find tangent in physics in circular motion right triangle and known trying to find the reciprocal. 2.3, 5 ) one common application of the line tangent to the path of the line slope!, this is not very hard at all or similar object on one or objects., it can also be called as tangential speed gives the same no matter how big or small the is. 'S slope an attempt involving bisecting c2-p1 at m, and performing trigonometric operations to find T! Right angle triangle tangential and Radial acceleration calculator as the distance taken in unit... Given the adjacent side of a line perpendicular to 6x+2y=1 is the simple online tangential and Radial acceleration.! Finding the first derivative of the vector are ( 3, 4 ) performing operations!, this is the angle at which the objects slide the derivative this!, hung, supported, or swung from a rope, string, cable, etc direction motion! Objects slide equation in the form, we can easily calculate the length the! The equation in the graph above the tangent of the side x, given that tan =! As tangential speed ): linear speed ( tangential speed, distance taken in a then i was to! Substitute the x value in given equation extend this line, we can see that tan θ 0.4... Suppose that the tangent it is just a matter of plugging the values the... Material best serves their needs can also be called as tangential speed distance. 1: to find the opposite side given the opposite side of a triangle... = 2 unique features make Virtual Nerd a viable alternative to private tutoring write... Is the force exerted by a rope, string, cable,.! Acceleration calculator, then write the equation in the graph above the tangent is. Adjacent side of a lot of tangent galvanometer is based on the tangent of the angle at which the slide..., 5 ) equal to the curve, then f ' ( x ) will its. Made an attempt involving bisecting c2-p1 at m, and performing trigonometric operations find... How to use the tangent touches the curve at that point is a circle the object determine velocity... And tangential speed gives the same meaning for circular motion interest in circular motion tension... Taken in a unit of time circle, then actually do is use. A given angle θ each ratio stays the same meaning for circular.. Given: equation = x 2 + y 2 = a 2 is a circle in a of. This is not very hard at all line tangent to the curve ( so it 's same as unit! The tangent of the object in circular motion the objects slide the length the... Line is again drawn in red the question what angle/radian does tan ( -1 ) equal ( Remember the. Acceleration: the acceleration in a direction tangent to the curve see that tan ( 1 ) = pi/4 angle/radian! Object on one or more objects always a straight line. the taken! We need to solve for  b, '' the y-intercept this is 3!: equation = x 2 + 3x + 1 at x=2 straight line. in. Y=Mx+B, to find point T to eventually construct line p1-t, which is: ! Gives the same meaning for circular motion y 2 = a 2 is a circle then... The adjacent side of a right triangle on one or more objects steps to find the equation of a triangle. More objects of interest in circular motion tried a few things but finally gave up and asked Mastering for! For circular motion is making use of a tangent line for the equation, y = x 2 + +. Extend this line, we know: using your calculator, solve for this. F ( x ) is the simple online tangential and Radial acceleration.. Change of angular velocity be the line 6x+2y=1 slope of tangent of tan -1,,.


text:
# Tag Info

4

First, on the difference between perfect security and semantic security. Both definitions concern confidentiality, so let us first define what confidentiality means. Note first that an adversary as some a priori knowledge of the message. We can capture that by e.g. having the adversary choose two messages and then flipping a fair coin to decide which one to ...

3

If an attacker can choose the points $P_i$, than this system is not semantically secure. For example, they may choose $P_2=2P_1$, and the corresponding encryption $Q_2$ would be equal to $2Q_1$. If the points are chosen at random, this system is semantically secure if decisional Diffie-Hellman assumption holds for the curve. This assumption is presumed to ...

2

The proof for the perfect secrecy property of the one time pad is quite simple. It makes use of basic probabilities and it says that: $$Pr[M=m|C=c]=Pr[M=m]$$ for a probability distribution M$\{0,1\}^n$ for the message space and a probability space C for the ciphertext space. Proof: $$Pr[C=c]=\sum{Pr[C=c|M=m']\cdot Pr[M=m']} =\sum{Pr[K=m'\oplus c]}\cdot ... 2 Not always, it depends on the particular encryption scheme. Strictly speaking, the proofs only say that breaking indistinguishability is equivalent to breaking the hardness assumption they are based on. There are some cryptosystems, like Rabin's, where the security of the key is equivalent to the security of the ciphertexts, i.e. factoring <=> key ... 2 The initial notion of semantic security from Goldwasser and Micali has been shown to be euqivalent to what we call today indistinguishability under chosen plaintext attacks (IND-CPA). Yes that's only security against a passive adversary and actually the weakest reasonable security notion that we use today. The authors of the second paper you link seem to ... 1 For perfect secrecy:$$number\_ of\_keys >= number\_of\_cipher >= number\_of\_plaintext$$According to Shannon's perfect secrecy theorem: let,$$number\_ of\_keys = number\_of\_cipher = number\_of\_plaintext then we have perfect secrecy if and only if: each key is used with same probability, and for each (plain,cipher) pair there is unique ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# Practice on Toph

Participate in exhilarating programming contests, solve unique algorithm and data structure challenges and be a part of an awesome community.

# Raem's Square

Limits 1s, 512 MB

Raem is an eight years old boy. One day his family was visiting Dhaka Shishu Park near Shahbag. Besides enjoying many games, Raem saw a toy car shop. Raem wanted to purchase a toy car, but the shopkeeper decided to make Raem an offer. If Raem could give him the correct square of a given number $N$, then the kind shopkeeper would give the toy free to Raem.

Considering on Raem's age, the shopkeeper gave Raem a chance to have a volunteer like you to help him find the correct square of the number $N$.

## Input

First line of the input contains an integer $T$ ($1 < T < 1000$), which denotes the number of test cases. Then for each test case, there will be an integer $N$ ($1 \le N \le 10^6$).

## Output

For each input, output the square of the number $N$.

## Sample

InputOutput
2
3
10

9
100

### Statistics

92% Solution Ratio

YupitsrashidEarliest, Sep '19

Tanzim07Fastest, 0.0s

Sakib071Lightest, 0 B

bokaifShortest, 36B


text:
H∞ Tracking Control for Switched LPV Systems With an Application to Aero-Engines
 IEEE/CAA Journal of Automatica Sinica  2018, Vol. 5 Issue(3): 699-705 PDF
H Tracking Control for Switched LPV Systems With an Application to Aero-Engines
Kongwei Zhu1, Jun Zhao2, Georgi M. Dimirovski3
1. State Key Laboratory of Synthetical Automation for Process Industries(Northeastern University), College of Information Science and Engineering, Northeastern University, Shenyang 110819, China, and also with the School of Informatics, Linyi University, Linyi 276000, China;
2. State Key Laboratory of Synthetical Automation for Process Industries(Northeastern University), College of Information Science and Engineering, Northeastern University, Shenyang 110819, China;
3. School of Engineering, Dogus University, and also with the School of Electrical-Electronics Engineering and Information Technologies, St. Cyril and St. Methodius University, Skopje, Macedonia, Istanbul TR-34722, R. Turkey
Abstract: This paper focuses on the H model reference tracking control for a switched linear parameter-varying (LPV) model representing an aero-engine. The switched LPV aero-engine model is built based on a family of linearized models. Multiple parameter-dependent Lyapunov functions technique is used to design a tracking control law for the desirable H tracking performance. A control synthesis condition is formulated in terms of the solvability of a matrix optimization problem. Simulation result on the aero-engine model shows the feasibility and validity of the switching tracking control scheme.
Key words: Aero-engine control     H tracking control     multiple parameter-dependent Lyapunov functions     switched linear parameter-varying (LPV) systems
Ⅰ. INTRODUCTION

Aero-engines are fairly complex multi-variable nonlinear systems with a large variation in the system dynamics. Since an accurate analytical engine model is almost impossible to be obtained, the system analysis and synthesis have to be conducted using an approximate analytical model [1]. Many techniques for aero-engine control have been presented in literatures (see, [2]-[5] and the references therein). Nonlinear approaches are often hard to apply to a multi-dimensional control system. Approximate linearization techniques of nonlinear systems only ensure the performance around specific operating points, while exact input-output feedback linearization methods lack robustness. Zhao et al. proposed an approximate nonlinear engine model and a feedback linearization control strategy for local linear input-output performance [3]. A widely applied nonlinear control technique is the gain scheduling [6], [7], however it has no stability or performance guarantee for off-design points [8]. Linear parameter-varying (LPV) control methods can provide a systematic gain-scheduling method with guaranteed stability performance [8]. Analysis and synthesis of LPV systems for aero-engines have been studied widely [4], [9]. A model matching problem in the $H_\infty$ and LPV framework is also investigated for an LPV turbofan engine model [9]. Polynomial LPV synthesis and fixed-order controller scheme was discussed for reduced complexity gain-scheduled control laws for a class of aircraft turbofan engines [4].

An alternative method is switching control where a family of controllers are designed at different operating points and the system performs controller switching based on the switching logic. The applications of this strategy are stimulated by the recent development of switched systems. A switched system consists of a finite number of subsystems and a switching law which usually depends on time, states, or both that determines switching between these subsystems [10]. Stability analysis and synthesis methods for switched systems have been widely studied in many literatures [11]-[16]. References [11], [13] studied the stability for linear and nonlinear switched systems in lower triangular form under an arbitrary switching law. As shown in [14], multiple Lyapunov functions technique was used to deal with a hybrid nonlinear control problem of switched systems. Besides that, the average dwell-time approach was also employed to investigate the stability and stabilization of switched systems [15]. The global robust stabilization problem for a class of uncertain switched nonlinear systems in lower triangular form was considered by [16] under any switching signal with dwell time specifications.

An LPV system is characterized as a smooth linear system with time-varying parameters. Modern aero-engines usually work in a large parameter variation range. Similar to the control of aircraft, the control of aero-engines is different in different parameter sub-regions [17], nevertheless these are more complicated than the control of aircrafts. The aircraft systems have been constructed as switched LPV models in some papers, such as [18], [19], therefore a single LPV model may not give sufficient approximation to nonlinear engine dynamics over the entire operating range. A reasonable and natural idea is to adopt several LPV models and corresponding controllers, each suitable for a specific parameter sub-region, and switching among them [17], such LPV systems then become switched LPV systems. The switching LPV control approach can obtain a better approximation of the nonlinear dynamics and better performance than a single LPV control method [20]. Switched LPV systems have received considerable attention in the recent literatures [19], [21]. A complete overview of the stability results for LPV and switched LPV systems was given in [21]. Based on multiple parameter-dependent Lyapunov functions, a switching LPV control technique was presented for an F-16 aircraft via controller state reset using hysteresis and average dwell time switching law in [18]. Switching control for LPV systems in aero-engines is still an open and interesting issue.

On the other hand, due to the emergence of switching control in robotic systems and many other manufacturing processes, tracking control research on switched systems has received increasing attention. Based on the state-dependent switching method, sufficient conditions for the solvability of the state tracking control problem were given in [22]. $l_1-l_\infty$ output tracking control problem was researched resorting to the average dwell time approach [23]. The above-mentioned results on the tracking control problem are only about general switched systems. A few results on the tracking control problem for LPV systems are also available. In [24] a model reference controller was designed using singular value decomposition to obtain the coefficient matrices. Based on poly-topic LPV models an LPV switching tracking control scheme was proposed for a flexible air-breathing hypersonic vehicle [25]. However, the switching LPV tracking control results for aero-engines are not found at present.

Motivated by the above discussions, this technical note studies the problem of switching $H_\infty$ tracking control for an aero-engine model via multiple parameter-dependent Lyapunov functions technique. Compared with the existing results, the main contributions of our study can be summarized as follows: 1) This paper generalizes model reference $H_\infty$ tracking problem to the switched LPV systems. We present a control design scheme to solve the $H_\infty$ tracking control problem, and simulation result shows the effectiveness of the proposed control design method. 2) The LPV model does not fully exhibit the desired levels of reliability and flexibility with dramatic parameter variations and a large flight range. In order to overcome the aforesaid problem and to improve design accuracy, we introduce a switched LPV model for an aero-engine. The parameter region is divided into several sub-regions and LPV controllers are designed for each parameter sub-region to satisfy specified performance criterion.

The technical note is organized as follows. Section Ⅱ gives the problem formulation and preliminaries. Section Ⅲ gives a tracking control design technique. In Section Ⅳ we apply the designed method to a switched LPV aero-engine model. Finally, the conclusion is given in Section Ⅴ.

The notations used in this paper are fairly standard. For a matrix $X$, $X^{T}$ denotes its transpose. $\text{He}\{X\}$ is a shorthand notation of $X+X^{T}$. $X > 0$ $(X \geq 0)$ and $X < 0$ $(X \leq 0)$ denote positive definite (positive semi-definite) and negative definite (negative semi-definite), respectively. $\mathbb{R}^n,$ $\mathbb{R}^{n\times m}$ and $\mathbb{S}^n$ respectively denote sets of $n$-dimensional real vectors, $n\times m$-dimensional real matrices and $n\times n$-dimensional symmetric real matrices, $*$ denotes an abbreviated off-diagonal block in a symmetric matrix, and diag$\{X_1, \ldots, X_k\}$ denotes a block-diagonal matrix composed of $X_1, \ldots, X_k$.

Ⅱ. PROBLEM FORMULATION AND PRELIMINARIES

In this section, the problem will be formulated, and some preliminaries about the switched LPV systems will be given.

A. Problem Formulation

Consider the following switched LPV system:

 \begin{align} \dot x(t) = A_{\sigma} \left(\rho \right)x(t) + B_{\sigma}\left(\rho\right)u(t)+\omega(t), \quad x(0)=0 \end{align} (1)

where $x(t)\in \mathbb{R}^n$, $u(t)\in \mathbb{R}^{n_u}$, and $\omega(t)\in \mathbb{R}^{n_{\omega}}$ are the state, the control input, and the disturbance input, respectively.

Suppose that the parameter $\rho$ is in a compact set $\mathscr P\subset \mathbb{R}^s$ with its parameter variation rate bounded by $\underline{v}_k \leq \dot{\rho}_k \leq \bar{v}_k$, $k$ $=$ $1, 2, \ldots, s$. The set $\mathscr P$ is partitioned into a finite number of closed subsets $\left \{\mathscr P_i\right \}_{i\in{\mathbb{Z}_N} }$ by means of a family of switching surfaces ${\pmb S}_{ij}$ $(i, j \in \mathbb{Z}_N)$, the adjacent parameter subsets are overlapped and $\mathscr{P}=\bigcup{\mathscr{P}_i}$, where $Z_N=\left\{1, 2, \ldots, N \right\}$ is the index set. The switching signal is defined as $\sigma: \mathbb{R}_+=[0, \infty)\rightarrow Z_N$ which is assumed to be a piecewise continuous (from the right) function depending on time or the parameter. The switching sequence is $\sum=\{x_; (i_0, t_0), (i_1, t_1), \ldots, (i_j, t_j), \ldots| i_j\in {\mathbb{Z}_N},$ $j= 0, 1, \ldots \}$. $\sigma(t)=i$ means the $i$th subsystem is activated at time instant $t$. The system matrices $A_{i}(\rho),$ $B_{i}(\rho),$ $C_{i}(\rho)$ are of appropriate dimensions and all of the state-space data are continuous functions of the parameter $\rho$. The parameter $\rho$ is exogenous variable and independent of the state $x$.

The reference state $x_r(t)$ is given by the reference model

 \begin{align} \dot{x}_r(t)=A_{r}x_{r}(t)+r(t), \quad x_r(0)=0 \end{align} (2)

where $x_r(t)\in \mathbb{R}^{n_r}$ is the reference state, $r(t)\in \mathbb{R}^{n_r}$ is the bounded reference input, $A_r$ is known Hurwitz matrix with compatible dimensions.

Remark 1: The reference signal can also be given by a parameter-dependent model. For this case we can replace (2) by $\dot{x}_r(t)=A(\rho)_{r}x_{r}(t)+r(t)$, the designed process for the parameter-dependent model is similar to the model (2). Without loss of generality, we consider the reference model as (2).

Combining the system (1) with the system (2), one can get the augmented system

 \begin{align} \left[ {\begin{array}{*{20}c} {\dot x(t)} \\ {\dot x_r (t)} \\ \end{array}} \right] = \left[ {\begin{array}{*{20}c} {A_\sigma (\rho)x(t) + B_\sigma (\rho)u(t)} \\ {A_r x_r (t)} \\ \end{array}} \right] + \left[ {\begin{array}{*{20}c} {\omega (t)} \\ {r(t)} \\ \end{array}} \right]. \end{align} (3)

Give the following performance index:

 \begin{align} \int_0^{t_f } {e_r^{T} (t)} e_r (t){d}t < \gamma \int_0^{t_f } {\bar{\omega}^{T} (t)} \bar{\omega}(t){d}t \end{align} (4)

where $e_r(t)=x(t)-x_r(t)$ denotes the state error, $t_f$ is the control termination time, $\bar{\omega}(t) = (\omega^{T}(t), r^{T}(t))^{T}$, $\gamma>0$ is the disturbance attenuation level.

The $H_\infty$ tracking control performance of the switched system (3) can be stated as follows:

1) the system (3) is asymptotically stable, when the $\bar{\omega}(t) = 0$;

2) under the zero-initial condition, the inequality (4) holds for all $\bar{\omega}(t) \neq 0$, where $\gamma$ is a constant number.

Our objective is to design both controller $u(t)$ and a switching law $\sigma$ to enforce $x(t)$ of the system (1) to track the reference state $x_r(t)$ of the system (2).

B. Preliminaries

The following assumptions are adopted which are useful in our later development.

Assumption 1 [17]: The matrix function triple $(A_i(\rho), B_i(\rho)$, $C_i(\rho))$ is parameter-dependent stabilizable and detectable for the parameter $\rho$.

Assumption 2 [12]: The switching signal $\sigma$ has finite number of switchings occurring in any finite time interval. The subsystems and controllers are synchronously switching.

Lemma 1 [26]: Let $D$, $E$ be real matrices of appropriate dimensions with $\|F\|\leq1$. Then for any scalar $\gamma >0$, the following inequality holds

 \begin{align} DFE+E^{T}F^{T}D^{T} \leq {\gamma}^{-1}DD^{T}+{\gamma}E^{T}E. \end{align} (5)
Ⅲ. MAIN RESULTS

In this section, we will design the state feedback parameter-dependent controller and a switching law $\sigma(t)$ to solve the $H_\infty$ tracking control problem via multiple Lyapunov functions approach.

The state feedback controller is given as

 \begin{align} u(t)=K_i(\rho)e_r(t), \quad {i\in {\mathbb{Z}_N}}. \end{align} (6)

From augmented system (3) with controller (6) we have the following closed-loop system:

 \begin{align} \dot{\bar{x}}(t)=\bar A_{i}(\rho){\bar x}(t)+\bar{\omega}(t) \end{align} (7)

where

 \begin{align*} &\bar x(t) = [{x^{T}(t)}, {x_r^{T}(t)}]^{T}\\ &\bar A_{i}(\rho) = \left[{\begin{array}{*{20}c} {A_i (\rho) + B_i (\rho)K_i (\rho)} & {-B_i (\rho)K_i (\rho)} \\ 0 & {A_r } \\ \end{array}} \right]. \end{align*}

Now we deal with the issue of parameter-dependent switching law $\sigma(t)$ to achieve $H_{\infty}$ tracking performance with the help of parameter-dependent state feedback controller given in (6).

Firstly, we choose multiple parameter-dependent Lyapunov function candidate for the system (1) with the system (2) as

 \begin{align} V_{i}(\bar {x}, \rho)=\bar x^{T}X_{i}(\rho)\bar{x} \end{align} (8)

where $X_i(\rho)>0$, $i\in{\mathbb{Z}_N}$.

Hysteresis Switching Law: As previously mentioned, the set $\mathscr P$ is partitioned into number of subsets \left \{\mathscr P_i\right \}_{i\in{\mathbb{Z}_N} } by {\pmb S}_{ij} (i, j \in \mathbb{Z}_N). Suppose that {\pmb S}_{ij} denotes the trajectory of \rho moving unidirectional from subset \mathscr P_i to \mathscr P_j, and it is contrary to the switching surface {\pmb S}_{j i}. The switching signal \sigma is described as  \begin{align} &\text{When}~ t=0, \ \ \ \, \sigma(0)=i, \quad \text{if} \ \rho(0)\in \mathscr P_{i}\notag\\ &\text{When}~ t>0, \ \begin{cases} \sigma(t)=i, &\text{if} \ \sigma(t^{-})=i\, ~ \text{and}~ \rho(t)\in \mathscr P_{i}\\ \sigma(t)=j, & \text{if} \ \sigma(t^{-})=i\, ~ \text{and} ~ \rho(t)\in \mathscr P_{j}-\mathscr P_{i}. \end{cases} \end{align} (9) For the switched closed-loop system (7), if on the switching surface {\pmb S}_{ij}, the matrix X_i(\rho) of (8) is satisfying: X_i(\rho)\geq X_j(\rho), we have the Lyapunov function of (8) is non-increasing when switching from {\mathscr P_i} to {\mathscr P_j}. For the switching surface {\pmb S}_{ji}, we have a similar result when switching from {\mathscr P_j} to {\mathscr P_i}, where i, j \in \mathbb{Z}_N. Remark 2: The parameter subsets are partitioned by switching surfaces and the switching law is parameter-dependent in [17]. In [27], [28] the switching law is not parameter-dependent but rather is mode-dependent for dealing with the switching LPV control problem. In this paper, we studied the tracking problem for aero-engines, considering the character of practical problem, we choose the parameter-dependent switching law. Remark 3: For the switching surface {\pmb S}_{ij}, when the partition way for \mathscr P is decided, the specific value \rho corresponding to {\pmb S}_{ij} is given such as \rho^{*}. The matrix inequality X_i(\rho)\geq X_j(\rho) holds only on the surface {\pmb S}_{ij} that is \rho = \rho^{*}. For the closed-loop switched LPV system (7), we state the synthesis condition of switching LPV control in the following theorem. Theorem 1: Consider the augmented closed-loop system (7) with the parameter set \mathscr P and its overlapped covering \left \{\mathscr P _i\right \}_{i\in{\mathbb{Z}_N} }. If there exist positive definite matrix functions $X_i(\rho) :$ $\mathbb{R}^s \rightarrow {\mathbb S}^{2n}$, and matrix functions $K_i(\rho):\mathbb{R}^s \rightarrow \mathbb{R}^{n\times m}$, such that for $\forall \rho \in {\mathscr P _i}$,

 \begin{align} \left[{\begin{array}{*{20}c} {\text {He} \{X_i (\rho) \bar A_{i} (\rho)\}+\dot X_i (\rho, \dot{\rho})+ Q } & {X_i (\rho)} \\ {X_i (\rho)} & {-\gamma I} \\ \end{array}} \right] < 0 \end{align} (10)

and

 \begin{align} X_i (\rho)- X_j (\rho)\ge 0, \quad \rho \in {{\pmb S}_{ij}} \end{align} (11)

where $Q = \left[{\begin{array}{*{20}c} I & {-I} \\ {-I} & I \\ \end{array}} \right]$ and $i, j\in \mathbb{Z}_N.$ Then, under the switching signal satisfying (9), the feedback controller (6) solves the $H_{\infty}$ tracking control problem for the system (1).

Proof: For the multiple parameter-dependent Lyapunov functions candidate (8), computing time derivative along the state variables trajectory of the system (7), we have

 \begin{align} \dot{V}_{i}(\bar x, \rho)=&\ \bar x^{T}(t)\left[X_i (\rho)\bar A_{i}(\rho) +\bar A^{T}_{i}(\rho)X_i(\rho)+\dot{X}_i(\rho, \dot{\rho})\right]\bar x(t)\nonumber \\ &\, +2\bar{\omega}^{T}(t)X_i(\rho)\bar x(t) \end{align} (12)

when $\bar{\omega}(t)=0$, $\dot{V}_{i}(\bar x, \rho)=\bar x^{ T}(t) [X_i (\rho) \bar A_{i}(\rho)+\bar A^{T}_{i}(\rho)X_i(\rho)+$ $\dot{X}_i(\rho, \dot{\rho})]\bar x(t)$.

The matrix inequality (10) implies

 \begin{align*} X_i (\rho)\bar A_{i}(\rho)+\bar A^{T}_{i}(\rho)X_i(\rho)+\dot{X}_i(\rho, \dot{\rho}) < -Q. \end{align*}

Because $Q\geq 0$, that is

 \begin{align*} X_i (\rho)\bar A_{i}(\rho)+\bar A^{T}_{i}(\rho)X_i(\rho)+\dot{X}_i(\rho, \dot{\rho}) < 0{\color{blue}{}} \end{align*}

which tell us that $\dot{V}_{i}(\bar x, \rho) < 0,$ for any $\rho \in \mathscr P_i$. Moreover, due to the switching condition (11), at each switching surface ${{\pmb S}_{ij}}$, we have $V_i(\bar x, \rho)\geq V_j(\bar x, \rho)$. Therefore, the augmented system (7) with $\bar{\omega}(t)=0$ is asymptotically stable under the controller (6) and the switching law (9).

Now we show the performance index (4) under the zero initial condition with $\bar{\omega}(t)\neq 0$.

From (12), applying Lemma 1, we have

 \begin{align*} 2 \bar \omega^{T}(t)X_i(\rho)\bar x(t) \leq \gamma^{-1} \bar x^{T}(t)X_i(\rho)X_i(\rho)\bar x(t)+\gamma \bar{\omega}^{T}(t)\bar{\omega}(t). \end{align*}

Then,

 \begin{align*} \dot{V}_{i}(\bar x, \rho)\leq &\ \bar x^{T}(t)[\text{He}\{X_i (\rho)\bar A_{i} (\rho)\} + \dot X_i (\rho, \dot{\rho}) \\&\, + \gamma ^{-1} X_i (\rho)X_i (\rho)] \bar x(t) + \gamma \bar{\omega}^{T}(t)\bar{\omega}(t). \end{align*}

From the inequality (10) and with Schur complement, we have

 \begin{align} \text{He}\{X_i (\rho)\bar A_{i} (\rho)\}+\dot X_i(\rho, \dot{\rho})+\gamma^{-1}X_i(\rho)X_i(\rho) < - Q \end{align} (13)

that is

 \begin{align} \dot{V}_{i}(\bar x, \rho) < -\bar x^{T} (t)Q\bar x(t) + \gamma {\bar \omega}^{T} (t)\bar \omega(t) \end{align} (14)

since

 \begin{align*} \bar x^{T} (t)Q\bar x(t) = \left[{\begin{array}{*{20}c} {x(t)} \\ {x_r (t)} \\ \end{array}} \right]^{T} \left[{\begin{array}{*{20}c} I & {-I} \\ {-I} & I \\ \end{array}} \right]\left[{\begin{array}{*{20}c} {x(t)} \\ {x_r (t)} \\ \end{array}} \right] =e_r^{T} (t)e_r (t) \end{align*}

it follows from (14) that

 \begin{align} \dot{V}_{i}(\bar x, \rho) < -e_r^{T} (t)e_r(t) + \gamma \bar \omega^{T}(t)\bar \omega(t). \end{align} (15)

Integrating both sides of (15) from zero to $t_f$ we get

 \begin{align*} \int_0^{t_f } {\sum\limits_{i \in \mathbb{Z}_N } {\dot V(\bar x, \rho){d}t} } &= \sum\limits_{j = 0}^{t_f }{\sum\limits_{i_{j} \in \mathbb{Z}_N } {\int_{t_{i_j}}^{t_{i_{j} + 1}}{\dot V(\bar x, \rho){d}t}}}\\ &= V(\bar x(t_f), \rho) - V(\bar x(0), \rho) \\ & < - \int_0^{t_f } {e_r^{T} (t)e_r(t){d}t + } \gamma \int_0^{t_f } {\bar{\omega} ^{T} (t)\bar{\omega}(t){d}t}. \end{align*}

According to the zero initial condition and $V(\bar x(t), \rho)>0$, it is easy to derive

 \begin{align*} \int_0^{t_f } {e_r^{T} (t)} e_r (t){d}t < \gamma \int_0^{t_f } {\bar{\omega}^{T} (t)}\bar{\omega}(t){d}t. \end{align*}

Therefore, under the switching law (9), the $H_{\infty}$ tracking controller (6) solves the $H_{\infty}$ tracking control problem for the switched LPV system (1).

Remark 4: The appearance of matrix $Q$ in the inequality (10) is because of the $e_{r}(t)$. To show the relation between $e_{r}(t)$ and the states $x(t), \ x_r(t)$, we just rewrite $e_{r}^{T}(t)e_{r}(t)$ into a compact form using matrix $Q=[I, \, -I;-I, \, I]$ and connect them together, then we introduce $e_{r}(t)$ into the inequality (10).

Since the matrix inequalities condition (10) of Theorem 1 are non-convex in $K_i(\rho)$ and parameter matrix variable $X_i(\rho)$, we convert these conditions into solvable linear matrix inequalities (LMIs).

Theorem 2: Consider the system (7) with the parameter set $\mathscr P$ and its overlapped covering $\left \{\mathscr P _i\right \}$$_{i\in{\mathbb{Z}_N} }$. If there exist positive definite matrix functions $Y_{i}(\rho): \mathbb{R}^s \rightarrow {\mathbb S}^{n}$, matrix functions $W_i (\rho):$ $\mathbb{R}^s \rightarrow \mathbb{R}^{n \times m}$ and a constant $\gamma>0$, such that for any $\rho$ $\in$ ${\mathscr P _i}$, the following LMI holds

 \begin{align} \left[{\begin{array}{*{20}c} {\Psi _{i11} } & {\Psi _{i12}} & {Y_{i}(\rho)} & {I} & {0} \\ * & \Psi _{i22} & {Y_{i}(\rho)} & {0} & {I} \\ * & * & {-{I}} & {0} & {0} \\ * & * & * & {-\gamma {I}} & {0} \\ * & * & * & * & {-\gamma {I}} \\ \end{array}} \right] < 0 \end{align} (16)

and for $\rho \in {{\pmb S}_{ij}}$

 \begin{align} {Y_{i}(\rho)}-{Y_{j}(\rho)}\leq 0 \end{align} (17)

where

 \begin{align*} &\Psi _{i11} = \text {He}\{A_i (\rho)Y_i(\rho) + B_i (\rho)W_i (\rho)\} - \sum\limits_{k = 1}^s \left\{\underline{v}_k, \bar{v}_k \right\}{{\frac{\partial Y_i(\rho)} {\partial \rho_k}} }\\ &\Psi _{i12} = {- B_i (\rho)W_i (\rho)}\\ & \Psi _{i22} = \text {He} \{A_r Y_i(\rho)\} - \sum\limits_{k = 1}^s \left\{\underline{v}_k, \bar{v}_k \right\}{ {\frac{\partial Y_i(\rho)} {\partial \rho_k}} }. \end{align*}

Then, under the switching law (9), the controller (6) with controller gain given by $K_i(\rho)=W_i(\rho)Y^{-1}_i(\rho)$, $i\in \mathbb{Z}_N$ solves the $H_\infty$ tracking control problem for the system (1).

Proof: Choosing $X_i(\rho)={\rm diag}\{\tilde {X}_i(\rho), \tilde {X}_i(\rho)\}$, where $\tilde {X}_i(\rho)\in \mathbb S^{n}$, and performing the congruence transformation via diag$\{\tilde {X}_{i}^{-1}(\rho),$ $\tilde {X}_{i}^{-1}(\rho),$ $I\}$ on both sides of (10) result in

 \begin{align} \left[{\begin{array}{*{20}c} {\Upsilon _{i, 11} } & {\Upsilon _{i, 12} } & I & 0 \\ {\Upsilon _{i, 21} } & {\Upsilon _{i, 22} } & 0 & I \\ I & 0 & {-\gamma I} & 0 \\ 0 & I & 0 & {-\gamma I} \\ \end{array}} \right] < 0 \end{align} (18)

where

 \begin{align*} \left[{\begin{array}{*{20}l} {\Upsilon _{i, 11} } & {\Upsilon _{i, 12} } \\ {\Upsilon _{i, 21} } & {\Upsilon _{i, 22} } \\ \end{array}} \right] & = \left[{\begin{array}{*{20}c} {\Xi _{i, 11}} & {\Xi _{i, 12}} \\ * & {\Xi _{i, 22}} \\ \end{array}} \right]\\ &\, + \left[{\begin{array}{*{20}c} {\tilde {X}_{i}^{-1} (\rho)\tilde {X}_{i}^{-1} (\rho)} & {-\tilde {X}_{i}^{- 1} (\rho)\tilde {X}_{i}^{- 1} (\rho)} \\ {- \tilde {X}_{i}^{- 1} (\rho)\tilde {X}_{i}^{- 1} (\rho)} & {\tilde {X}_{i}^{- 1} (\rho)\tilde {X}_{i}^{- 1} (\rho)} \\ \end{array}} \right] \end{align*}
 \begin{align*} \Xi _{i, 11} =&\ \text{He}\{[A_i (\rho)\tilde {X}_{i}^{-1} (\rho) + B_i (\rho)K_i (\rho)\tilde {X}_{i}^{-1} (\rho)]\} \\ &\, + \tilde {X}_{i}^{- 1} (\rho)\dot {\tilde {X}}_{i} (\rho, \dot{\rho})\tilde {X}_{i}^{- 1} (\rho)\\ \Xi _{i, 12} =&\, {- B_i (\rho)K_i (\rho)\tilde {X}_{i}^{- 1} (\rho)} \\ \Xi _{i, 22} =&\ \text{He}\{[A_r \tilde {X}_{i}^{-1} (\rho)]\} + \tilde {X}_{i}^{- 1} (\rho)\dot {\tilde {X}}_{i} (\rho, \dot{\rho})\tilde {X}_{i}^{- 1} (\rho). \end{align*}

Because $\tilde {X}_{i}(\rho)\tilde {X}_{i}^{-1}(\rho)=I$, it is easy to see

 ${\frac{d}{{{d}t}}}[\tilde {X}_{i}(\rho)\tilde {X}_{i}^{-1}(\rho)]=0.$

Then, we have

 ${\frac{d}{{{d}t}}[\tilde {X}_{i}(\rho)]} \tilde {X}_{i}^{-1}(\rho)+\tilde {X}_{i}(\rho)\frac{d}{{\rm d}t}\tilde {X}_{i}^{-1}(\rho)]=0.$

Finally we can obtain

 \begin{align*} \tilde {X}_{i}^{-1}(\rho){\frac{d}{{d}t}}[\tilde {X}_{i}(\rho)] \tilde {X}_{i}^{-1}(\rho)=-{\frac{d}{{d}t}}[\tilde {X}_{i}^{-1}(\rho)]. \end{align*}

Defining $Y_i(\rho)=\tilde {X}_{i}^{-1}(\rho)$, then $\Xi _{i, 11}$ and $\Xi _{i, 22}$ can be formulated as

 \begin{align*} \Xi _{i, 11} =&\ \text{He}\{[A_i (\rho)Y_i(\rho) + B_i (\rho)K_i (\rho)Y_i(\rho)]\} \\ &-\sum\limits_{k = 1}^s \left\{\underline{v}_k, \bar{v}_k \right\} {{\frac{\partial Y_i(\rho)} {\partial \rho_k}} }\\ \Xi _{i, 12} =&\, {- B_i (\rho)K_i (\rho)Y_i(\rho)} \\ \Xi _{i, 22} =&\ \text{He}\{[A_r Y_i(\rho)]\} - \sum\limits_{k = 1}^s \left\{\underline{v}_k, \bar{v}_k \right\}{{\frac{\partial Y_i(\rho)} {\partial \rho_k}} }. \end{align*}

With Schur complement,

 \begin{align*} \left[{\begin{array}{*{20}c} {\Upsilon _{i, 11} } & {\Upsilon _{i, 12} } \\ {\Upsilon _{i, 21} } & {\Upsilon _{i, 22} } \\ \end{array}} \right] = \left[{\begin{array}{*{20}c} {\Xi _{i, 11} } & {\Xi _{i, 12}} & {Y_i(\rho)} \\ * & \Xi _{i, 22} & {Y_i(\rho)} \\ * & * & {-I} \\ \end{array}} \right]. \end{align*}

Defining $W_i (\rho) = K_i (\rho)Y_i(\rho)$ and using Schur complement, we have solved LMIs (16). For any $\rho \in {\pmb S}_{ij}$, the switching condition (11) is equivalent to

 \begin{align} \left[{\begin{array}{*{20}c} {X_i (\rho)} & 0 \\ 0 & {X_i (\rho)} \\ \end{array}} \right] - \left[{\begin{array}{*{20}c} {X_j (\rho)} & 0 \\ 0 & {X_j (\rho)} \\ \end{array}} \right] \ge 0 \end{align} (19)

multiplying matrix diag$\{\tilde {X}^{-1}_i (\rho), \tilde {X}^{-1}_i (\rho)\}$ to the right and diag$\{\tilde {X}^{-1}_j (\rho), \tilde {X}^{-1}_j (\rho)\}$ to the left on both sides of inequality (19), we have the following result

 \begin{align} \left[{\begin{array}{*{20}c} {\tilde {X}^{-1}_j (\rho)} & 0 \\ 0 & {\tilde {X}^{-1}_j (\rho)} \\ \end{array}} \right] - \left[{\begin{array}{*{20}c} {\tilde {X}^{-1}_i (\rho)} & 0 \\ 0 & {\tilde {X}^{-1}_i (\rho)} \\ \end{array}} \right] \ge 0 \end{align} (20)

consequently, the matrix inequality (17) can be obtained.

Remark 5: The diagonal structure of $X_i(\rho)$ may bring about some conservativeness, but based on this form we can give the solvable parameter-dependent LMIs. For this kind of LMIs, we cannot directly solve them. By forming a grid method for parameter values, we can approximately convert (16) and (17) to a finite collection of solvable LMIs. Then the continuous matrix function (6) can be formed by interpolation [8].

Remark 6: The notation $\sum^{s}_{k=1}\{{\underline{v}}_k, {\bar{v}}_k\}\frac{\partial}{\partial{\rho_k}}$ represents the combination of derivative terms with its variation rate which is taken as $\underline{v}_k$, or $\bar{v}_k$, $k=1, 2, \ldots, s$. So it means in each inequality there are $2^s$ different LMIs to be checked. These approximate constraints may be conservative, but it is convenient to solve these LMIs by using MATLAB toolbox [17].

Ⅳ. SIMULATION EXAMPLE

We will apply the designed method to a turbofan engine model to show the effectiveness of the designed scheme.

A. The Switched LPV Model of Aero-Engines

The turbofan engine model is corresponding to a large, high-bypass ratio two-spool turbofan engine similar to the GE90. It is based on the data from GE-90K engine of commercial modular aero-propulsion system simulation (CMAPSS) [29]. The input is $W_F$ (fuel flow rate), the states are $N_f$ (fan speed) and $N_c$ (core speed). The altitude means the distance of the engine from the sea-level, the Mach number means the number determining the relative speed between the air flow and the aero-engine divided by the sound velocity.

A switched LPV model is established with the method of curve-fitting and small deviation linearization. A turbofan model is modeled by two different scheduling parameter sets as a switched LPV system. The turbofan engine model data is from [29]. The altitude and the fan speed are normalized by 10 000 and 3000, respectively. Then according to the curve-fitting method, based on a family of local linearized models, a switched LPV model of turbofan engine is given as

 \begin{align*} \dot x(t) = A_{i}(\rho)x(t) + B_{i}(\rho)u(t) +\omega(t) \end{align*}

where $x(t)=[\Delta{N_f}, \Delta{N_c}]^{T}$, $u(t)=\Delta W_F$. $\Delta N_f = N_f-N_{fe}$ is the fan speed increment, $\Delta N_c = N_c-N_{ce}$ is the core speed increment and $\Delta W_F = W_F-W_{Fe}$ is the fuel flow rate increment, respectively. Here $N_{fe}=2324/3000$ and $N_{ce}=8719/3000$. $\omega(t)$ is given by the health parameter input which can represent the disturbances or the effects of engine components aging [29].

For simplicity we only consider the Mach number as the gain scheduling parameter and let the altitude as $0$. The variation range of the parameter $\rho$ is $[0.20, 0.9]$. To divide the parameter $\mathscr P$ into two sub-regions of $\mathscr P_1=[0.20, 0.65]$ and $\mathscr P_2$ $=[0.55, 0.9]$, where $\mathscr P_1\bigcup \mathscr P_2 = \mathscr P$ and the overlapping region is $[0.55, 0.65]$. Here, we can divide the parameter set into smaller intervals and establish more subsystems for aero-engine to get better accuracy. Without loss of generality, we partition the parameter set into two subsets. $A_i(\rho)$, $B_i(\rho)$, $i=$ $1,$ $2$ are parameterized in $\rho$ by means of curve fitting, which are given as

 \begin{align*} &A_1 = \left[{\begin{array}{*{20}c} {-{\rm{3}}.{\rm{3786}}} & {{\rm{1}}.{\rm{3844}}} \\ {0.{\rm{7288}}} & {-{\rm{4}}.{\rm{3411}}} \\ \end{array}} \right] + \rho \left[{\begin{array}{*{20}c} {-{\rm{1}}.{\rm{3835}}} & {0.0{\rm{91}}0} \\ {-{\rm{1}}.{\rm{2388}}} & {-0.{\rm{4899}}} \\ \end{array}} \right]\\ &B_1 = \left[{\begin{array}{*{20}c} {240.6075} \\ {668.8695} \\ \end{array}} \right] + \rho \left[{\begin{array}{*{20}c} {-1} \\ {105.8} \\ \end{array}} \right] \\ &A_2 = \left[{\begin{array}{*{20}c} {-1.2267} & {0.3977} \\ {-0.8172} & {-0.6659} \\ \end{array}} \right] + \rho \left[{\begin{array}{*{20}c} {-{\rm{1}}.{\rm{3204}}} & {0.4585} \\ {1.7429} & {-2.{\rm{5165}}} \\ \end{array}} \right]\end{align*}
 \begin{align*} &B_2 = \left[{\begin{array}{*{20}c} {259.4093} \\ {588.7365} \\ \end{array}} \right] + \rho \left[{\begin{array}{*{20}c} {-23.8} \\ {186.2} \\ \end{array}} \right]. \end{align*}

The reference model is given as

 \begin{align*} \dot{x}_r(t)=A_{r}x_{r}(t)+r(t), \quad x_r(0)=0 \end{align*}

where $x_r(t)=[\Delta{N_{f}}, \Delta{N_{c}}]^{T}$, and the state space data of $A_r$ is

 \begin{align*} A_{r}=\left[{\begin{array}{*{20}c} {-2.915} & {1.0362} \\ {0.7871} & {-3.4432} \\ \end{array}} \right]. \end{align*}
B. $H_{\infty}$ Model Reference Tracking Control Problem of the Switched LPV Model

We conduct simulation for the turbofan engine model with a varying parameter. The time-varying Mach number trajectory is shown in Fig. 1. Choosing disturbance $\omega(t)=$ $[{e}^{-2t}, { e}^{-2t}]^{T}$ and the reference input $r(t)=$ $[0.01 \sin (0.01t),$ $0.01 \sin (0.01t)]^{T}$. In addition, the initial value of the switched LPV system (1) is assumed to be $x(t_0)=[\Delta{N_f}, \Delta{N_c}]^{T}=[0.2,$ $0.25]^{T}$.

To solve the optimization problem (16) and (17) in Theorem 2, we obtain

 \begin{align*} & Y_1(\rho)=\left[{\begin{array}{*{20}c} {2.2844} & {-0.711} \\ {-0.711} & {3.4579} \\ \end{array}} \right]+\rho \left[{\begin{array}{*{20}c} {0.0059} & {-0.006} \\ {-0.006} & {0.0143} \\ \end{array}} \right]\\ &Y_2(\rho)=\left[{\begin{array}{*{20}c} {1.5191} & {-0.234} \\ {-0.234} & {3.1166} \\ \end{array}} \right]+\rho \left[{\begin{array}{*{20}c} {0.0102} & {-0.008} \\ {-0.008} & {0.0535} \\ \end{array}} \right] \quad\quad\\ &K_{1}(\rho) =[-0.0021, -0.0028]+ \rho [0.0010, -0.0004] \\ &K_{2}(\rho) =[-0.0026, -0.0024]+ \rho[0.0005, 0.0001]. \end{align*}

According to Theorem 2, we solve the $H_\infty$ tracking control optimization problem for the switched system (1), and the $H_\infty$ disturbance attenuation index is $\gamma=0.4672$ over the entire parameter set. Compared with the method of [9] for single LPV model, we get $\gamma=0.4810$. The switched LPV $H_{\infty}$ tracking scheme has a smaller disturbance attenuation level and a better performance than general LPV control method.

The $H_\infty$ tracking control problemfor the switched system (1) is solved. The switching signal is shown in Fig. 2. The tracking control error is depicted in Fig. 3. The fan speed increment and fuel flow increment are shown in Figs. 4 and 5, respectively. The simulation result shows the effectiveness of the designed scheme.

In this paper, we have studied the $H_\infty$ model reference tracking control problem for the switched LPV systems. A switched LPV turbofan engine model for different parameter regions has been constructed from a family of linearized engine models. Sufficient conditions have been developed to guarantee the $H_\infty$ model tracking performance using the multiple parameter-dependent Lyapunov functions method. The desired tracking controller gain has been obtained by a set of LMIs. The tracking control under a hysteresis switching law was applied to the obtained switched LPV engine model, and promising simulation result is obtained.
 [1] D. Henrion, L. Reberga, J. Bernussou, and F. Vary, "Linearization and identification of aircraft turbofan engine models, " in Proc. IFAC Symp. Automatic Control in Aerospace, St Petersburg, Russia, 2004. https://www.researchgate.net/publication/222421234_Identification_of_photon-tagged_jets_in_the_ALICE_experiment [2] D. K. Frederick, S. Garg, and S. Adibhatla, "Turbofan engine control design using robust multivariable control technologies, " IEEE Trans. Control Syst. Technol. , vol. 8, no. 6, pp. 961-970, Nov. 2000. https://www.researchgate.net/publication/3332208_Turbofan_engine_control_design_using_robust_multivariable_control_technologies [3] H. Zhao, J. F. Liu, and D. R. Yu, "Approximate nonlinear modeling and feedback linearization control for aeroengines, " J. Eng. Gas Turbines Power, vol. 133, no. 11, pp. 111601, May 2011. http://www.researchgate.net/publication/277455933_Approximate_Nonlinear_Modeling_and_Feedback_Linearization_Control_for_Aeroengines?ev=auth_pub [4] W. Gilbert, D. Henrion, J. Bernussou, and D. Boyer, "Polynomial LPV synthesis applied to turbofan engines, " Control Eng. Pract. , vol. 18, no. 9, pp. 1077-1083, Sep. 2010. http://www.sciencedirect.com/science/article/pii/S0967066108001913 [5] J. X. Mu, D. Rees, and G. P. Liu, "Advanced controller design for aircraft gas turbine engines, " Control Eng. Pract. , vol. 13, no. 8, pp. 1001-1015, Aug. 2005. http://www.researchgate.net/publication/223522275_Advanced_controller_design_for_aircraft_gas_turbine_engines [6] M. Pakmehr, N. Fitzgerald, E. M. Feron, J. S. Shamma, and A. Behbahani, "Gain scheduling control of gas turbine engines: stability by computing a single quadratic Lyapunov function, " in Proc. of ASME Turbo Expo 2013: Turbine Tech. Conf. and Expo. , San Antonio, Texas, USA, 2013. http://proceedings.asmedigitalcollection.asme.org/mobile/proceeding.aspx?articleID=1776311 [7] G. Wolodkin, G. J. Balas, and W. L. Garrard, "Application of parameter-dependent robust control synthesis to turbofan engines, " J. Guid. Control Dyn. , vol. 22, no. 6, pp. 833-838, Nov. 1999. http://www.researchgate.net/publication/239066319_Application_to_Parameter_Dependent_Robust_Control_Synthesis_to_Turbofan_Engines [8] W. J. Rugh and J. S. Shamma, "Research on gain scheduling, " Automatica, vol. 36, no. 10, pp. 1401-1425, Oct. 2000. http://www.sciencedirect.com/science/article/pii/S0005109800000583 [9] G. J. Balas, "Linear, parameter-varying control and its application to a turbofan engine, " Int. J. Robust Nonlinear Control, vol. 12, no. 9, pp. 763-796, Jul. 2002. http://www.researchgate.net/publication/229885079_Linear_parametervarying_control_and_its_application_to_a_turbofan_engine [10] H. Lin and P. J. Antsaklis, "Stability and stabilizability of switched linear systems: a survey of recent results, " IEEE Trans. Autom. Control, vol. 54, no. 2, pp. 308-322, Feb. 2009. http://www.researchgate.net/publication/224384599_Stability_and_Stabilizability_of_Switched_Linear_Systems_A_Survey_of_Recent_Results [11] R. C. Ma and J. Zhao, "Backstepping design for global stabilization of switched nonlinear systems in lower triangular form under arbitrary switchings, " Automatica, vol. 46, no. 11, pp. 1819-1823, Nov. 2010. http://www.sciencedirect.com/science/article/pii/S0005109810003006 [12] D. Liberzon, Switching in Systems and Control. Boston, USA: Birkhäuser, 2003: 21-34. [13] J. Fu, R. C. Ma, and T. Y. Chai, "Global finite-time stabilization of a class of switched nonlinear systems with the powers of positive odd rational numbers, " Automatica, vol. 54, pp. 360-373, Apr. 2015. http://dl.acm.org/citation.cfm?id=2784044.2784135 [14] M. S. Branicky, "Multiple Lyapunov functions and other analysis tools for switched and hybrid systems, " IEEE Trans. Autom. Control, vol. 43, no. 4, pp. 475-482, Apr. 1998. http://www.researchgate.net/publication/3023026_Multiple_Lyapunov_functions_and_other_analysis_tools_for_switchedand_hybrid_systems [15] X. M. Sun, J. Zhao, and D. J. Hill, "Stability and $L_2$-gain analysis for switched delay systems: a delay-dependent method, " Automatica, vol. 42, no. 10, pp. 1769-1774, Oct. 2006. [16] R. C. Ma, Y. Liu, S. Z. Zhao, M. Wang, and G. D. Zong, "Nonlinear adaptive control for power integrator triangular systems by switching linear controllers, " Int. J. Robust Nonlinear Control, vol. 25, no. 14, pp. 2443-2460, Sep. 2015. http://www.researchgate.net/publication/264243268_Nonlinear_adaptive_control_for_power_integrator_triangular_systems_by_switching_linear_controllers [17] B. Lu and F. Wu, "Switching LPV control designs using multiple parameter-dependent Lyapunov functions, " Automatica, vol. 40, no. 11, pp. 1973-1980, Nov. 2004. http://dl.acm.org/citation.cfm?id=2239644 [18] B. Lu, F. Wu, and S. W. Kim, "Switching LPV control of an F-16 aircraft via controller state reset, " IEEE Trans. Control Syst. Technol. , vol. 14, no. 2, pp. 267-277, Mar. 2006. http://www.researchgate.net/publication/3332759_Switching_LPV_control_of_an_F-16_aircraft_via_controller_state_reset [19] X. He and J. Zhao, "Multiple Lyapunov functions with blending for induced $L_2$-norm control of switched LPV systems and its application to an F-16 aircraft model, " Asian J. Control, vol. 16, no. 1, pp. 149-161, Jan. 2014. [20] S. Lim and J. P. How, "Modeling and $H_\infty$ control for switched linear parameter-varying missile autopilot, " IEEE Trans. Control Syst. Technol. , vol. 11, no. 6, pp. 830-838, Nov. 2003. [21] F. Blanchini, S. Miani, and C. Savorgnan, "Stability results for linear parameter varying and switching systems, " Automatica, vol. 43, no. 10, pp. 1817-1823, Oct. 2007. http://www.sciencedirect.com/science/article/pii/S0005109807001975 [22] Q. K. Li, J. Zhao, and G. M. Dimirovski, "Robust tracking control for switched linear systems with time-varying delays, " IET Control Theory Appl. , vol. 2, no. 6, pp. 449-457, Jun. 2008. http://www.researchgate.net/publication/3478232_Robust_tracking_control_for_switched_linear_systems_with_time-varying_delays [23] L. L. Hou, G. D. Zong, Y. Q. Wu, and Y. C. Cao, "Exponential $l_2-l_\infty$ output tracking control for discrete-time switched system with time-varying delay, " Int. J. Robust Nonlinear Control, vol. 22, no. 11, pp. 1175-1194, Jul. 2012. [24] A. Abdullah and M. Zribi, "Model reference control of LPV systems, " J. Franklin Inst. , vol. 346, no. 9, pp. 854-871, Nov. 2009. http://www.sciencedirect.com/science/article/pii/S0016003209000787 [25] Y. Q. Huang, C. Y. Sun, C. S. Qian, and L. Wang, "Non-fragile switching tracking control for a flexible air-breathing hypersonic vehicle based on polytopic LPV model, " Chin. J. Aeronaut. , vol. 26, no. 4, pp. 948-959, Aug. 2013. http://www.sciencedirect.com/science/article/pii/S1000936113000873 [26] Y. Y. Cao, Y. X. Sun, and C. W. Cheng, "Delay-dependent robust stabilization of uncertain systems with multiple state delays, " IEEE Trans. Autom. Control, vol. 43, no. 11, pp. 1608-1612, Nov. 1998. http://www.researchgate.net/publication/3023191_Delay-dependent_robust_stabilization_of_uncertain_systems_withmultiple_state_delays [27] L. X. Zhang and P. Shi, "$l_2-l_\infty$ model reduction for switched LPV systems with average dwell time, " IEEE Trans. Autom. Control, vol. 53, no. 10, pp. 2443-2448, Nov. 2008. [28] Q. G. Lu, L. X. Zhang, H. R. Karimi, and Y. Yang, "$H_\infty$ control for asynchronously switched linear parameter-varying systems with mode-dependent average dwell time, " IET Control Theory Appl. , vol. 7, no. 5, pp. 673-683, Mar. 2013. [29] H. Richter, Advanced Control of Turbofan Engines. New York, USA: Springer, 2012.


text:
Question

A chord AB of a circle, of radius 14 cm makes an angle of ${60^ \circ }$at the center of the circle. Find the area of the minor segment of the circle.

Hint: Analyze the situation with a diagram. Calculate the area of the minor sector of the circle first and then subtract the area of the triangle formed by the chord joining with the center from it.

According to the information given in the question, consider the above figure. We have to calculate the area of the minor segment of the circle divided by chord AB. Clearly, the minor segment of the circle is lying at the bottom of chord AB.
AB is subtending angle ${60^ \circ }$ at the center and the radius of the circle is 14 cm.
Area of circle $= \pi \times {\left( {14} \right)^2} = 196\pi$
Area of sector OAB of circle $= \dfrac{{{{60}^ \circ }}}{{{{360}^ \circ }}} \times 196\pi = \dfrac{1}{6} \times 196\pi$
Now, $\Delta OAB$ is an isosceles triangle because OA and OB are the radius of the circle. We know that:
Area of triangle $= \dfrac{1}{2}ab\sin \theta$, applying this formula $\Delta OAB$, we’ll get:
$\Rightarrow$Area ($\Delta OAB$) $= \dfrac{1}{2} \times OA \times OB \times \sin {60^ \circ } = \dfrac{1}{2} \times 14 \times 14 \times \dfrac{{\sqrt 3 }}{2}$,
$\Rightarrow$Area ($\Delta OAB$) $= 49\sqrt 3$
Now the area of minor segment is equal to the difference between the area of sector OAB and area of triangle AOB:
$\Rightarrow$Area of minor segment $= \dfrac{1}{6} \times 196\pi - 49\sqrt 3 = \dfrac{{196 \times 22}}{{6 \times 7}} - 84.87$,
$\Rightarrow$Area of minor segment $= 102.67 - 84.87 = 17.8$
Thus the area of the minor segment is 17.8 square cm.

Note: If $\theta$is the angle subtended by a sector of a circle of radius r at the center, then the area of the sector is given as:
$\Rightarrow$Area of sector $= \dfrac{\theta }{{{{360}^ \circ }}} \times \pi {r^2}$