单输入延迟系统中的标准H无穷问题

infinite
problem
需积分: 3 1 下载量 160 浏览量 更新于2024-07-25 收藏 478KB PDF 举报
"标准H无限问题在具有单个输入延迟的系统中的研究" 本文主要探讨了在具有单个纯输入延迟的线性时不变系统（LTI系统）中解决标准H无限问题的方法。H无限控制理论是现代控制理论的一个重要分支，它关注的是系统在无限频率范围内的性能，特别是其在最大增益指标下的稳定性与性能优化。 Gilead Tadmor，一位IEEE的资深成员，提出了基于状态空间分析的解决方案。这个方案结合了一个有限维模型和一个抽象的演化模型，利用这类分布式系统的相对简单结构，将相关的Riccati微分方程简化为两个代数Riccati方程和一个在延迟区间上的微分Riccati方程的组合。这种简化使得计算变得更加可行。 在介绍部分，作者指出这篇论文旨在处理控制端存在单个纯输入延迟的系统中的标准H无限问题。具有控制延迟或观测延迟的系统构成了分布参数系统中最简单的类别。这类问题的解决对于理解和设计这类系统中的控制器至关重要，因为它们广泛存在于各种工程应用中，如航空航天、过程控制和通信网络。 文章的核心是将原本复杂的延迟系统转化为更易于处理的形式。通过代数和微分Riccati方程的转换，可以更有效地求解控制器设计的关键参数，如反馈增益矩阵。这些方程的解提供了系统性能的定量评估，包括稳定性、鲁棒性和性能指标。 此外，文章还强调，这些结果可以扩展到有限时间和时变问题中。在这种情况下，代数Riccati方程被相应的微分Riccati方程替代，以覆盖整个过程时间。这种方法的灵活性允许对不断变化的环境或参数进行动态控制设计。 关键词包括：H无限控制，输入延迟，矩阵Riccati方程。 这篇文章为处理具有单个输入延迟的线性系统中的标准H无限问题提供了一种新的、实用的分析方法。通过巧妙地将问题转化为更简单的代数和微分形式，该方法不仅解决了静态问题，还能够适应动态变化的系统条件，从而为实际工程应用提供了有力的理论工具。

Force at a point in the interior of a semi-infinite solid

2010-07-23 上传
根据提供的标题“Force at a point in the interior of a semi-infinite solid”以及描述中的信息“经典力学解答文献，1936，弹性均质提中，作用集中力的解答！”可以推断出本文档主要讨论了在半无限大的弹性固体...

## Problem 5: Remainder Generator Like functions, generators can also be higher-order. For this problem, we will be writing `remainders_generator`, which yields a series of generator objects. `remainders_generator` takes in an integer `m`, and yields `m` different generators. The first generator is a generator of multiples of `m`, i.e. numbers where the remainder is 0. The second is a generator of natural numbers with remainder 1 when divided by `m`. The last generator yields natural numbers with remainder `m - 1` when divided by `m`. Note that different generators should not influence each other. > Hint: Consider defining an inner generator function. Each yielded generator varies only in that the elements of each generator have a particular remainder when divided by m. What does that tell you about the argument(s) that the inner function should take in? ```python def remainders_generator(m): """ Yields m generators. The ith yielded generator yields natural numbers whose remainder is i when divided by m. >>> import types >>> [isinstance(gen, types.GeneratorType) for gen in remainders_generator(5)] [True, True, True, True, True] >>> remainders_four = remainders_generator(4) >>> for i in range(4): ... print("First 3 natural numbers with remainder {0} when divided by 4:".format(i)) ... gen = next(remainders_four) ... for _ in range(3): ... print(next(gen)) First 3 natural numbers with remainder 0 when divided by 4: 4 8 12 First 3 natural numbers with remainder 1 when divided by 4: 1 5 9 First 3 natural numbers with remainder 2 when divided by 4: 2 6 10 First 3 natural numbers with remainder 3 when divided by 4: 3 7 11 """ "*** YOUR CODE HERE ***" ``` Note that if you have implemented this correctly, each of the generators yielded by `remainder_generator` will be infinite - you can keep calling next on them forever without running into a `StopIteration` exception.

2023-05-31 上传
In the example usage, we create a `remainders_generator` object with `m=4`, and then for each remainder value `i` from 0 to 3, we get the corresponding generator and print the first 3 numbers it ...

Enter your answers into the text box below. a) Given the following:   float data[54];      Write a line of C++ code that causes an “out of bounds” problem. [1 marks] b) Write the function prototype for a function with two string parameters that returns a boolean. [2 marks] c) Write a few lines of C++ code to show an example of an infinite loop.      Explain why it is an infinite loop [2 marks] d) Given the following:   string s; s = "Good evening";      What is the integer value stored in s[3] ? [2 marks]

2023-06-12 上传
` will cause an out of bounds problem as the array `data` has a size of 54, meaning the last index is 53. Accessing index 54 will result in undefined behavior. b) `bool functionName(string param1, ...

Write a generator function merge that takes in two infinite generators a and b that are in increasing order without duplicates and returns a generator that has all the elements of both generators, in increasing order, without duplicates.

2023-06-01 上传
The `merge` function is then used to merge these two generators into a single generator `merged` that generates all the numbers in increasing order, without duplicates. Finally, the list ...

用c++写Bob has a magic computer which has infinite USB ports. But there is a problem with this computer——it can only read and write the two most recently inserted USB disks. Initially, Bob has n USB disks and every USB disk has one unique file. He wants to insert all the n USB disks into his computer. Since his computer is old, he must insert USB disk in the following way: Select 2 unplugged USB disks, insert them into computer. The two USB disks inserted in the previous step will share the files with each other, which means after that both two USB disks will have all files either in the first one or in the second one. If all the n USB disks are in the computer, the task is finished. Otherwise, choose one USB disk in the computer, unplug it and go to step 1 . Bob wants to know the minimum number of n such that there is a sequence of operations that ends up with at least k different files in each USB disk. Input Only one number,k (2≤k≤100000 ),meaning that finally each USB disk owns at least k different files. Output You need to output one number n , which means the minimum number of USB disks in module 998244353.

2023-05-26 上传
a = 1ll * a * a % MOD; b >>= 1; } return res; } int main() { int k; cin >> k; int ans = power(2, k - 1); cout ; return 0; } ``` 我们可以发现，每次插入两个 USB 磁盘，它们的文件集合会并集，也...

Like functions, generators can also be higher-order. For this problem, we will be writing remainders_generator, which yields a series of generator objects. remainders_generator takes in an integer m, and yields m different generators. The first generator is a generator of multiples of m, i.e. numbers where the remainder is 0. The second is a generator of natural numbers with remainder 1 when divided by m. The last generator yields natural numbers with remainder m - 1 when divided by m. Note that different generators should not influence each other.

2023-06-02 上传
This function takes an integer m as input and generates m different generators using a for loop. Each generator is created using a generator expression that yields numbers with the corresponding ...

: Use this idea to write a procedure invert-unit-series that computes 1/S for a unit power series S. You will need to use mul-series. To test your procedure, invert the series S1 (from problem 1) and show that you get the series 1 - x. (Convince yourself that this is the correct answer.) Turn in a listing of your procedure. This is a very short procedure, but it is very clever. In fact, to someone looking at it for the first time, it may seem that it can’t work-that it must go into an infinite loop. Write a few sentences of explanation explaining why the procedure does in fact work, and does not go into a loop

2023-06-07 上传
The procedure uses a recursive helper function `iter`, which takes two series `S1` and `S2` and computes the next coefficient of the reciprocal series using the formula `a_n = -a_0*S_1[n] - a_1*S_2[n]...

Use this idea to write a procedure invert-unit-series that computes 1/S for a unit power series S. You will need to use mul-series. To test your procedure, invert the series S1 (from problem 1) and show that you get the series 1 - x. (Convince yourself that this is the correct answer.) Turn in a listing of your procedure. This is a very short procedure, but it is very clever. In fact, to someone looking at it for the first time, it may seem that it can’t work-that it must go into an infinite loop. Write a few sentences of explanation explaining why the procedure does in fact work, and does not go into a loop（please use racket）

2023-06-07 上传
- The procedure recurses on `add-series one (invert-unit-series s)`, which is a unit power series with a non-zero constant coefficient. - Therefore, the recursion will eventually reach the base case ...

infinite or NaN

2023-05-17 上传
In mathematics, infinity is often used to represent the limit of a sequence or function as its input approaches a certain value. "NaN", on the other hand, stands for "Not a Number". This is a ...

animation: move 15s ease 1s infinite

2023-04-04 上传
This is a CSS animation shorthand property that sets ...Overall, this animation will move an element in a certain way for 15 seconds, with a 1-second delay before it starts, and will repeat indefinitely.
四旋翼小菜一枚
  • 粉丝: 0
  • 资源: 2
上传资源 快速赚钱

最新资源