x??0.01110,余数??0.10111*2?5 y9．(1) x = 2-011*0.100101, y = 2-010*(-0.011110)

[x]浮 = 11101,0.100101 [y]浮 = 11110,-0.011110 Ex-Ey = 11101+00010=11111

[x]浮 = 11110,0.010010(1) x+y 0 0. 0 1 0 0 1 0 (1) + 1 1. 1 0 0 0 1 0 1 1. 1 1 0 1 0 0 (1) 规格化处理: 1.010010 阶码 11100 x+y= 1.010010*2-4 = 2-4*-0.101110 x-y 0 0. 0 1 0 0 1 0 (1) + 0 0. 0 1 1 1 1 0 0 0 1 1 0 0 0 0 (1)

规格化处理: 0.110000 阶码 11110

x-y=2-2*0.110001

(2) x = 2-101*(-0.010110), y = 2-100*0.010110 [x]浮= 11011,-0.010110 [y]浮= 11100,0.010110

Ex-Ey = 11011+00100 = 11111

[x]浮= 11100,1.110101(0) x+y 1 1. 1 1 0 1 0 1 + 0 0. 0 1 0 1 1 0 0 0. 0 0 1 0 1 1 规格化处理: 0.101100 阶码 11010 x+y= 0.101100*2-6 x-y 1 1.1 1 0 1 0 1 + 1 1.1 0 1 0 1 0 1 1.0 1 1 1 1 1 规格化处理: 1.011111 阶码 11100 x-y=-0.100001*2-4

10．(1) Ex = 0011, Mx = 0.110100

Ey = 0100, My = 0.100100 Ez = Ex+Ey = 0111 Mx*My 0. 1 1 0 1 * 0.1 0 0 1 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 1 1 1 0 1 0 1 规格化： 26*0.111011

(2) Ex = 1110, Mx = 0.011010

Ey = 0011, My = 0.111100 Ez = Ex-Ey = 1110+1101 = 1011 [Mx]补 = 00.011010

[My]补 = 00.111100, [-My]补 = 11.000100 0 0 0 1 1 0 1 0 +[-My] 1 1 0 0 0 1 0 0 1 1 0 1 1 1 1 0 0 1 0 1 1 1 1 0 0 +[My] 0 0 1 1 1 1 0 0 1 1 1 1 1 0 0 0 0.0 1 1 1 1 0 0 0 0 +[My] 0 0 1 1 1 1 0 0 0 0 1 0 1 1 0 0 0.01 0 1 0 1 1 0 0 0 +[-My] 1 1 0 0 0 1 0 0 0 0 0 1 1 1 0 0 0.011 0 0 1 1 1 0 0 0 +[-My] 1 1 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0.0110 1 1 1 1 1 0 0 0 +[My] 0 0 1 1 1 1 0 0 0 0 1 1 0 1 0 0 0.01101 0 1 1 0 1 0 0 0 +[-My] 1 1 0 00 1 0 0 0 0 1 0 1 10 0 0.01101 商 = 0.110110*2-6, 余数=0.101100*2-6 11．

4位加法器如上图，

Ci?AiBi?AiCi?1?BiCi?1?AiBi?(Ai?Bi)Ci?1?AiBi?(Ai?Bi)Ci?1(1)串行进位方式

C1 = G1+P1C0 其中：G1 = A1B1 P1 = A1⊕B1（A1＋B1也对） C2 = G2+P2C1 G2 = A2B2 P2 = A2⊕B2 C3 = G3+P3C2 G3 = A3B3 P3 = A3⊕B3 C4 = G4+P4C3 G4 = A4B4 P4 = A4⊕B4 (2)并行进位方式 C1 = G1+P1C0

C2 = G2+P2G1+P2P1C0

C3 = G3+P3G2+P3P2G1+P3P2P1C0

C4 = G4+P4G3+P4P3G2+P4P3P2G1+P4P3P2P1C0 12．(1)组成最低四位的74181进位输出为：

C4 = Cn+4 = G+PCn = G+PC0， C0为向第0位进位

其中，G = y3+y2x3+y1x2x3+y0x1x2x3，P = x0x1x2x3，所以 C5 = y4+x4C4

C6 = y5+x5C5 = y5+x5y4+x5x4C4 (2)设标准门延迟时间为T，“与或非”门延迟时间为1.5T，则进位信号C0，由最低位传送至C6需经一个反相器、两级“与或非”门，故产生C0的最长延迟时间为 T+2*1.5T = 4T

(3)最长求和时间应从施加操作数到ALU算起：第一片74181有3级“与或非”门（产生控制参数x0, y0, Cn+4），第二、三片74181共2级反相器和2级“与或非”门（进位链），第四片74181求和逻辑（1级与或非门和1级半加器，设其延迟时间为3T），故总的加法时间为：

t0 = 3*1.5T+2T+2*1.5T+1.5T+3T = 14T 13．串行状态下： C1 = G1+P1CO