Q1-ANS)
for -117d:
−117d=−(01110101)b
−(01110101)b=10001010(oc8)+1=10001011(tc8)
for -127d:
−127d=−(01111111)b
−(01111111)b=10000000(oc8)+1=10000001(tc8)
for 127d:
127d=01111111b=01111111(tc8)
for 0d:
0d=00000000b
Two′scomplement=one′scomplement+1=11111111(oc8)+1=00000000(tc8)
due to overflow the 9th digit is removed
Q2-ANS)
1.
10010001(tc8)=−(01101110+1)d=−(01101111)b=−((0×27)+(1×26)+(1×25)+(0×24)+(1×23)+(1×22)+(1×21)+(1×20))d=−111d
2.
00010001(tc8)=((0×26)+(0×25)+(1×24)+(0×23)+(0×22)+(0×21)+(1×20))d=17d
3.
01111111(tc8)=((1×26)+(1×25)+(1×24)+(1×23)+(1×22)+(1×21)+(1×20))d=127d
4.
0000000(tc8)=((0×26)+(0×25)+(0×24)+(0×23)+(0×22)+(0×21)+(0×20))d=0d
5.
11111111(tc8)=−(00000000+1)b=−(00000001)=−1d
Answer submitted by Amirhossein Azizafshari from Thursday 1 PM class.