High accuracy amplitude and phase measurements based on a double heterodyne architecture

In the digital low level RF (LLRF) system of a circular (particle) accelerator, the RF field signal is usually down converted to a fixed intermediate frequency (IF). The ratio of IF and sampling frequency determines the processing required, and differs in various LLRF systems. It is generally desirable to design a universally compatible architecture for different IFs with no change to the sampling frequency and algorithm. A new RF detection method based on a double heterodyne architecture for wide IF range has been developed, which achieves the high accuracy requirement of modern LLRF. In this paper, the relation of IF and phase error is systematically analyzed for the first time and verified by experiments. The effects of temperature drift for 16 h IF detection are inhibited by the amplitude and phase calibrations.     

国务院再发文加快发展第三方检验检测认证服务

<正>不久前,国务院发布《国务院关于加快科技服务业发展的若干意见》(简称《意见》)。《意见》提到,重点发展研究开发、技术转移、检验检测认证、创业孵化、知识产权、科技咨询、科技金融、科学技术普及等专业科技服务和综合科技服务,提升科技服务业对科技创新和产业发展的支撑能力。检验检测认证作为科技服务业的重要组成部分,受到国务院关注。《意见》提到,加快发展第三方检验检测认证服务,鼓励不同所有制检验检测认证机构平等参与市场竞争。加     

Efficient scheme of quantum SWAP gate and multi-atom cluster state via cavity QED

In this paper, we propose a physical scheme to realize quantum SWAP gate by using a large-detuned single-mode cavity field and two identical Rydberg atoms. It is shown that the scheme can also be used to create multi-atom cluster state. During the interaction between atom and cavity, the cavity is only virtually excited and thus the scheme is insensitive to the cavity field states and cavity decay. With the help of our scheme it is very simple to prepare the N-atom cluster state with perfect fidelity and probability. The practical feasibility of this method is also discussed.     

Quantum entanglement in the mode-mode competition system

Considering a two-level atom interacting with the competing two-mode field, this paper investigates the entanglement between the two-level atom and the two-mode field by using the quantum reduced entropy, and that between the two-mode field by using the quantum relative entropy of entanglement. It shows that the two kinds of entanglement are dependent on the relative coupling strength of atom-field and the atomic distribution, and exhibit the periodical evolution. The maximal atom-field entanglement state can be prepared via the appropriate selection of system parameters and interaction time.     

Atom-photon entanglement in the system with competing k-photon and l-photon transitions

We have investigated the evolution of the atomic quantum entropy and the entanglement of atom-photon in the system with competing k-photon and l-photon transitions by means of fully quantum theory, and examined the effects of competing photon numbers （k and l）, the relative coupling strength between the atom and the two-mode field （A/g）, and the initial photon number of the field on the atomic quantum entropy and the entanglement of atom-photon. The results show that the multiphoton competing transitions or the large relative coupling strength can lead to the strong entanglement between atoms and photons. The maximal atom-photon entanglement can be prepared via the appropriate selection of system parameters and interaction time.     

走进国际级统计研究团队——介绍吴建福教授及其工程统计研究组

吴建福(C.F.Jeff Wu)教授是美国工程院院士、台湾中研院院士、国际工程统计领域的领军性人物。自2003年加盟佐治亚理工学院工业和系统工程系(ISyE)担任工程统计讲座教授以来,短短几年间将ISyE系的工程统计方向提升到了国际级水准,其证明是,经过吴建福教授聘任到这个小组的前五位教员全部拿到了美国国家科学基金青年学者奖(EARLY CAREER AWARD)。     

关于n~(1/k)的不等式猜想的研讨

文 [1 ]给出了一个关于kn的不等式猜想 ,猜想的右侧不等式是 :正整数n ,k >1 ,则nk 2时 ,( 1 )式成立 .为证明上述结论 ,先给出两个引理引理 1　 [贝努利 (Bernoulli)不等式 ]若x >- 1且k是正整数 ,则 ( 1 +x) k≥ 1 +kx .等号当且仅当x =0时成立 .利用二项式定理易证引理 1 .引理 2 [2 ] 　若 - 1 相似文献   

余弦定理在四面体的一个推广

余弦定理　在△ABC中 ,设内角A ,B ,C的对边分别为a ,b ,c,则b2 =a2 c2 - 2accosB .( 1 )文 [1 ]给出了余弦定理在四面体的一个推广如下 :定理 1 　在任意四面体中 ,它的一个面的面积的平方 ,等于其他三个面的面积的平方和 ,减去这三个面中每个面的面积与它们所夹二面角的余弦的积的和的两部 .文 [2 ]给出了余弦定理在四边形的一个推广如下 :定理 2　设凸四边形ABCD的四边长依次为AB=a ,BC=b ,CD=c,DA =d ,两对角线长AC =p,BD =q ,则(pq) 2 =(ac) 2 (bd) 2 -2abcdcos(B D)(2 )本文给出余弦定理在四面体的一个有别于定理 1的推…     

数学问题解答

2006年1月号问题解答(解答由问题提供人给出)1591如图,△ABC中,CD⊥AB于D,△ACD、△BCD的内切圆分别切AC,BC于E,F.求证:(1)若∠ACB=90°,则∠EDF=90°.(2)若∠EDF=90°,则∠ACB=90°.证明(1)因为∠ACB=90°,CD⊥AB,所以△ACD∽△CDB,所以ACBC=BCDD=CADD=BACC CADD--BCDD.因为A     

一个无理不等式的类比

文[1]给出了如下不等式： 设a，b〉0，0〈λ≤2，则 √a/a＋λb＋√b/λa＋b≤2/√1＋λ（1）