參考教案數(shù)電

1、DigitalDesign Chapter 22.1Performthefollowingnumbersystem2.2ConvertDigitalDesign Chapter 22.1Performthefollowingnumbersystem2.2Convertthefollowingoctalnumbersintobinaryand2.3ConvertthefollowinghexadecimalnumbersintobinaryandConvertthefollowingnumbersinto（e）10100.11012=24+22+2-1+2-2+2-（j）15C.3816=116

2、2+516+12+316-1+816-Performthefollowingnumber-system Addthefollowingpairsofbinarynumbers,showingallC2.8Subtractingandshowingthe2.9Addthefollowingpairsofoctal011110577341066610222.10AddthefollowingpairsofhexadecimalF 8-bitssigned- 2.12 Indicate whether or not overflow occurs when adding the following

3、8-bit twos- S2.12 Indicate whether or not overflow occurs when adding the following 8-bit twos- S符號(hào)位 COUTCIN，溢S2.19Assumingthatthesolutioniscorrectin somebase,wecan writethefollowingequationusing base-10arithmetic:Solvingthequadratic,wegetb=13orb=25.Next, wetrythesolutionforbothpossiblebases,and tha

4、titworksonlyforb=13;theMartianshad132.24 P41. An addition overflows if the addendssigns are the same but the sums sign is fromthe n 位二進(jìn)制數(shù)補(bǔ)碼數(shù)，當(dāng)mn 時(shí)，舍去xd 位（d=n-m）得到的m 位二進(jìn)制補(bǔ)y x y 的符號(hào)位。m n 位二Dn=dn-1 dn-2 dm-1 dm-2 d1 Dm=dm-1 dm-2 d1 Dn=dn-1 dn-2 dm-1 dm-2 d1 d0=dm-1 dm-2 d1 （1-若即顯然表達(dá)式相等則符號(hào)位相同，即 dn-1=dm- -2n-1 +dn-22n-2+2m-1+dm-22m-2+d121 +d0 =-2m-1+dm-22m-2+dm-32m-3 +d121 -2n-1 +dn-22n-2+dn-32n-3+dm-12m-1=-2m-dm- dm- 2d=dn-22d-dn-2dn-3dn-4dm dm-1=1 得證：對(duì)給定的n 位二進(jìn)制數(shù)補(bǔ)碼mn x最左d 位（d=n-m）得到的m 1。（2） 該數(shù)為正dn-1=0dm-0+dn-22n-2+0+dm-22m-2+d121+d0=0+dm-22m-2+dm-32m-3+d121dn-22n-2+dn-32n-dn-2dn-3dn-4dm