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本义介绍一个带校验操作的浮点二进制运算器的逻辑设计。这个运算器是作为一个研究报告提出来的,并不打算具体应用在那一台计算机里。校验操作是这样来进行的:将原来参加运算的数加以修正,重复一次运算以得到与原来结果相补的数;然后将两者相加,校验是否等于零。运算器的长度为64位,假定采用6相的1兆赫时钟脈冲,几个主要运算的速度如下(包括校验及规格化时间,但不包括访问存储器的时间): 代数加法……9微砂乘法……23微秒除法……23微秒开平方……23微秒本文所讨论的基本方法是在作加法时将部份和及进位分别用两个寄存器存貯起来,接着逐次对进位进行清除,得到最终的结果。对两个64位数的进位清除工作可以在1微秒内完成。同样,规格化及移位也可以在1微秒内完成。这个运算器大约需要用到1,500个触发器及25,000个二极管。与工作周期为6微秒的存储器相配合,每秒能进行40,000多次带校验的浮点运算。如果将存储器的工作周期缩短到2微秒,则每秒能进行80,000次运算。本文假定计算机为4地址。 ,
The original meaning of a floating-point binary calculator with a verify operation of the logic design. This calculator is presented as a research report and does not intend to be specific to that computer. The verification operation is performed by correcting the number of the original operation, repeating the operation to get the number that complements the original result, and then adding the two to check whether the value equals zero. The length of the calculator is 64 bits, assuming 1-phase 6-phase clocks and the speed of several main operations is as follows (including checksum normalization time but excluding memory access time): Algebraic addition ... 9 micro Sand multiplication ... 23 microseconds division ... 23 microseconds square ... 23 microseconds The basic method discussed in this article is to add parts and carry to two registers separately for summing, Clear, get the final result. The carry-out of two 64-bit numbers can be completed in 1 microsecond. Also, normalization and shifting can be done in 1 microsecond. This calculator needs about 1,500 flip-flops and 25,000 diodes. Combined with a memory operating at 6 microseconds, it performs over 40,000 verified floating-point operations per second. If you shorten the memory cycle to 2 microseconds, you can do 80,000 operations per second. This article assumes that the computer is 4 addresses. ,