Question

The state of a 12 bit register is 100010010111. What is its content if it represents the three decimal digits in the excess-3 code and in bcd?

Answer 1

Answer:

for the excess-3 code we have divided it into 3 parts  now converting each part into decimal and then subtract (-3) from each as it is excess.

1000 1001 0111

0111 -> 2^3∗0+2^2∗1+2^1∗1+1∗1=0+4+2+1=7

for excess 3

(7−3)=4

1001-> 2^3∗1+2^2∗0+2^1∗0+1∗1=8+0+0+1=9

for excess 3

(9−3)=6

1000-> 2^3∗1+2^2∗0+2^1∗0+1∗0=8+0+0+0=8

for excess 3

(8−3)=5

Hence  1000 1001 0111 ->564

Now when it represents the three decimal digits in the bcdwe have divided it into 3 parts  now converting each part into decimal

1000 1001 0111

Now converting this into decimal

0111 -> 2^3∗0+2^2∗1+2^1∗1+1∗1=0+4+2+1=7

1001-> 2^3∗1+2^2∗0+2^1∗0+1∗1=8+0+0+1=9

1000-> 2^3∗1+2^2∗0+2^1∗0+1∗0=8+0+0+0=8

Answer 2

Answer:

hi 1234

Explanation: