Receiver With Adaptive Strobe Offset Adjustment
    1.
    发明申请
    Receiver With Adaptive Strobe Offset Adjustment 有权
    接收机采用自适应频闪偏移调整

    公开(公告)号:US20080205565A1

    公开(公告)日:2008-08-28

    申请号:US11913211

    申请日:2006-04-25

    IPC分类号: H04L7/00

    CPC分类号: H04L7/048 H04L7/0337

    摘要: Receiver for receiving a data stream via a data bus, which receiver samples the bits of the data stream in an over-sampling process, in which n bit strobe offsets are used and n data sets with i bits are sampled,—applies a decision criterion for identifying those data sets with correct bit values. This decision uses checksum CRC,—selects one of the identified data sets with correct bit values and—uses the bit strobe offset, which was used for receiving the selected data streams, for receiving the data stream. In this way the multiphase clock with optimal phase shifts is selected.

    摘要翻译: 用于经由数据总线接收数据流的接收机,哪个接收机在过采样过程中对数据流的比特进行采样,其中使用n比特选通偏移量,并对n比特的n个数据集进行采样, 用于识别具有正确位值的那些数据集。 该决定使用校验和CRC, - 选择具有正确位值的所识别数据集之一,并使用用于接收所选数据流的位选通偏移量来接收数据流。 以这种方式选择具有最佳相移的多相时钟。

    Timeslot sharing over different cycles in tdma bus
    2.
    发明申请
    Timeslot sharing over different cycles in tdma bus 审中-公开
    在tdma总线上的不同周期的时隙共享

    公开(公告)号:US20060224394A1

    公开(公告)日:2006-10-05

    申请号:US10555259

    申请日:2004-04-26

    IPC分类号: G06Q99/00 G05B19/418 G06F9/46

    摘要: The invention relates to a method of transmitting user data via a communications medium (2) between subscribers (3) connected to the communications medium (2), wherein the data are transmitted in recurrent cycles (8) and at least one slot (9, 10) in each cycle (8) is intended for the user data of at least one subscriber (3). In order to permit a particularly efficient data transmission via the communications medium (2) at least one of the slots (10) is used to transmit the user data of different subscribers (3) (A, C, F) in different cycles (8). In addition, a bus guardian (6) of one subscriber (3) determines whether the subscriber (3) may transmit user data in the current slot (9, 10) of the current cycle (8), the bus guardian (6) having at least indirect access to a universal condition available throughout the entire communications system (1). In particular, an internal counter of the bus guardian (6) is synchronized to a universal cycle counter.

    摘要翻译: 本发明涉及一种通过连接到通信介质(2)的用户(3)之间的通信介质(2)发送用户数据的方法,其中数据以循环周期(8)和至少一个时隙(9, 每个周期(8)中的10)用于至少一个用户(3)的用户数据。 为了允许经由通信介质(2)的特别有效的数据传输,使用时隙(10)中的至少一个用于以不同的周期(8)发送不同用户(3)(A,C,F)的用户数据 )。 另外,一个用户(3)的总线监控器(6)确定用户(3)是否可以在当前周期(8)的当前时隙(9,10)中发送用户数据,总线监控器(6)具有 至少间接访问整个通信系统(1)中可用的通用条件。 特别地,总线监控器(6)的内部计数器与通用周期计数器同步。

    Method for temporal synchronization of clocks
    4.
    发明申请
    Method for temporal synchronization of clocks 有权
    时钟同步时钟的方法

    公开(公告)号:US20070033294A1

    公开(公告)日:2007-02-08

    申请号:US10555258

    申请日:2004-04-26

    IPC分类号: G06F15/16

    CPC分类号: H04J3/0655 G06F1/12 H04L7/08

    摘要: In order to carry out in a communication system (1) a temporal synchronization of clocks in a particularly rapid and efficient manner, a method is proposed which has the following steps: acquiring state values which are dependent on a time base (10); filing each acquired state value at a position in a first list L comprising (k+1) positions, if the acquired state value is smaller than or equal to the (k+1) smallest element of the list L, where k is a predefinable error tolerance; filing the acquired state value at a position in a second list H comprising (k+1) positions, if the acquired state value is greater than or equal to the (k+1) greatest element of the list H; forming a mean value from the (k+1) smallest element of the list L and the (k+1) greatest element of the list H, if the number of acquired state values is greater than or equal to (2k+2); determining a correction value as a function of the mean value; and correcting a current state value of the clocks that are to be synchronized.

    摘要翻译: 为了以特别快速和有效的方式在通信系统(1)中执行时钟的时间同步,提出了一种方法,其具有以下步骤:获取取决于时基(10)的状态值; 如果所获取的状态值小于或等于列表L的(k + 1)最小元素,则在包括(k + 1)个位置的第一列表L中的位置处归档每个获取的状态值,其中k是可预定义的 误差容差 如果所获取的状态值大于或等于列表H的(k + 1)个最大元素,则在包括(k + 1)个位置的第二列表H中的位置处归档所获取的状态值; 如果所获取的状态值的数量大于或等于(2k + 2),则从列表L的第(k + 1)个最小元素和列表H的(k + 1)最大元素形成平均值; 确定作为所述平均值的函数的校正值; 并且校正要同步的时钟的当前状态值。