定点补码一位乘法器的设计.rar

上传者: u012429555 | 上传时间: 2021-06-17 09:54:44 | 文件大小: 252KB | 文件类型: RAR
(1)用[X]补×[Y]补直接求[X×Y]补 讨论当相乘的两个数中有一个或二个为负数的情况,在讨论补码乘法运算时,对被乘数或部分积的处理上与原码乘法有某些类似,差别仅表现在被乘数和部分积的符号位要和数值一起参加运算。 若[Y]补=Y0Y1Y2…Yn 当Y0为1时,则有Y=-1+Yi×2-i 故有 X×Y=X×Yi×2-1-X当Y为负值时,用补码乘计算[X×Y]补,是用[X]补乘上[Y]补的数值位,而不理[Y]补符号位上的1,乘完之后,在所得的乘积中再减X,即加-[X]补。实现补码乘法的另一个方案是比较法,是由BOOTH最早提出的,这一方法的出发点是避免区分乘数符号的正负,而且让乘数符号位也参加运算。技巧上表现在分解乘数的每一位上的1为高一位的一个+1和本位上的一个-1:X×Y=X×(-1+Yi×2i) (逐项展开则得)=X×[-Y0+Y1×2-1+Y2×2-2+…+Yn×2-n]=X×[-Y0+(Y1-Y1×2-1)+(Y2×2-1-Y2×2-2)+…+(Yn×2-(n-1)-Yn×2-n)](合并相同幂次项得) =X×[(Y1-Y0)+(Y2-Y1) ×2-1+…+(Yn-Yn-1) ×2-(n-1)+(0-Yn) ×2-n]=X×(Yi+1-Yi)×2-i(写成累加求和的形式,得到实现补码乘运算的算法)将上述公式展开,则每一次的部分积为: P1=[2-1(Yn+1-Yn) ×X]补 P2=[2-1(P1+(Yn-Yn-1) ×X)]补 … Pi=[2-1(Pn-i+(Yn-I+2-Yn-I+1) ×X)]补 … Pn=[2-1(Pn-1+(Y2-Y1) ×X)]补 Pn+1=[ (Pn+(Y1-Y0) ×X)]补 则最终补码乘积为[X*Y]补=[Pn+1]补 由上述公式可以看出,比较法是用乘数中每相邻的两位判断如何求得每次的相加数。每两位Yi和Yi+1的取值有00,01,10,11四种组合,则它们的差值分别为0,1,-1和0,非最后一次的部分积,分别为上一次部分积的1/2(右移一位)的值Rj,Rj+[X]补,Rj-[X]补(即Rj+[-X]补)和Rj,但一定要注意:最后一次求出的部分积即为最终乘积,不执行右移操作。用此法计算乘积,需要乘数寄存器的最低一位之后再补充一位Yn+1,并使其初值为0,再增加对Yn和Yn+1两位进行译码的线路,以区分出Yn+1-Yn 4种不同的差值。对N位的数(不含符号位)相乘,要计算N+1次部分积,并且不对最后一次部分积执行右移操作。此时的加法器最好采用双符号位方案。

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