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Optimal-control-applied-basic
最优控制是现代控制理论的核心,它研究的主要问题是:在满足一定约束条件下,寻求最优控制策略,使得性能指标取极大值或极小值(The optimal control is the core of modern control theory it is the main problem is: conditions meet certain constraints, finding the optimal control strategy, and performance to take the maximum or minimum value)
- 2020-10-23 16:37:22下载
- 积分:1
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Multi-Instance
Multi-Instance Learning matlab code
- 2013-09-24 15:00:08下载
- 积分:1
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audio_combine
利用发生机理进行语音 的合成,并且形成变速不变调和变调不变速的处理方法。(Use of the mechanism of occurrence of the synthesized speech, and forming a shift processing method does not change harmonizing shift.)
- 2014-11-16 10:23:12下载
- 积分:1
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PUMA-nonlinear-control
MATLAB/SIMULINK BASED
NON Linear Control of PUMA 560
- 2014-12-12 04:04:09下载
- 积分:1
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epaneth
用MATLAB调用EPA的几个例子,实现EPA中的管网模拟(Calling EPA with MATLAB)
- 2021-01-21 22:58:46下载
- 积分:1
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Matlab
matlab 经典教程 从入门到精通 实例讲解 非常经典 值得收藏(matlab classic tutorial on the very classic examples from entry to the proficient worth collecting)
- 2013-12-16 15:12:02下载
- 积分:1
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fdmtm
finite difference method有限差分法求波导的TM模式(finite difference method finite difference method for the TM waveguide mode)
- 2007-04-25 10:11:51下载
- 积分:1
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MATLAB
QPSK调制全称Quadrature Phase Shift Keying ,意为正交相移键控,是一种数字调制方式。数字信号的四相相移键控调制与解调技术以其灵活性和通用性而得到广泛的应用,符合未来数字通信技术发展的方向。随着移动通信技术的发展,以前在数字通信系统中采用FSK、ASK、PSK等调制方式,逐渐被许多优秀的调制技术所替代。(QPSK modulation partial Quadrature Phase Shift Keying, Quadrature Phase Shift Keying means, is a digital modulation. Digital signal quadrature phase shift keying modulation and demodulation technology for its flexibility and versatility which has been widely used, in line with the future direction of development of digital communication technologies. With the development of mobile communication technology, previously used in digital communication systems FSK, ASK, PSK modulation, etc., many excellent modulation gradually being replaced by technology.)
- 2013-05-22 17:54:51下载
- 积分:1
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3
说明: 求三角对角方程组Tx=f的解,其中T=[1 1 0 0 0,1 2 1 0 0,1 1 3 1 0,0 0 1 4 1, 0 0 0 1 5],x
=[x1,x2,x3,...x5],f=[3,8,15,24,29](Diagonal triangle demand equations Tx = f the solution, in which T = [1 1 0 0 0,1 2 1 0 0,1 1 3 1 0,0 0 1 4 1, 0 0 0 1 5], x = [ x1, x2, x3, ... x5], f = [3,8,15,24,29])
- 2010-08-26 21:09:57下载
- 积分:1
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04
说明: thanks mes chers amis
- 2011-02-14 21:22:50下载
- 积分:1