E) Solving-11 +(-2) will yield which two\'s-complement answer? a. 1110 1101 c. 1

Question

E) Solving-11 +(-2) will yield which two's-complement answer? a. 1110 1101 c. 1111 0011 d. 1110 1001 F) Add the following BCD numbers. (Note that you have 3 independent BCD addition problems here) 0110 0111 1001 0101 1000 1000 a. 0000 1011 0000 1111 0001 0001 b. 0001 0001 0001 0101 0001 0001 c. 0000 1011 0000 1111 0001 0111 d. 0001 0001 0001 0101 0001 0111 G) The 2's-complement system is to be used to add the signed binary numbers 11110010 and 11110011. Determine, in decimal, the sign and value of each number and their sum. a. -113 and -114,-227 b. -14 and-13,-27 c. -11 and-16,-27 d. -27 and-13,-40 H) Which is the decimal equivalent for the BCD number, 10110110? a. 182 b. 36 c. 116 d. 10110110 is not a valid BCD number. I) Why is the Gray code more practical to use when coding the position of a rotating shaft? a. All digits change between counts. b. Two digits change between counts. c. Only one digit changes between counts 2 of 10

Explanation / Answer

E.) -11 + (-2)

Both are negative. So taking 2's compliment for both numbers and adding them :

1. Binary equivalent for 11 = 00001011

1's compliment = 11110100

Adding 1 +1

2's compliment(-11) = 11110101

Similarly,

Binary equivalent for 2 = 00000010

1's compliment = 11111101

Adding 1 +1

2's compliment(-2) = 11111110

Adding both, 11110101

+ 11111110

= 1)  11110011, ignoring 1

So result is option c.

You can verify the same as:

-11+(-2) = -13 and 11110011 is 2's compliment of 00001101 which is 13

F) BCD Addition:

In BCD Addition, whenever the sum is more than 9 and carry is 0 or sum is less than or equal to 9 and carry is 1, an additional 6 (0110) is added to the sum to get the correct result. This is because each digit is considered separate BCD number.

For your problem, 0110 0111 1001

0101 1000 1000

Considering them as three different additions, starting from leftmost:

0110 (6)

+ 0101 (5)

1011, which is 11 an incorrect BCD number as explained above(more than 9 and carry 0).

So, adding 6

1011

+ 0110

0001 0001 (11)   

Second Addition: 0111 (7)

1000 (8)

1111, which again is an incorrect BCD number as it is 15, more than 9 and carry 0.

So, adding 6:

1111

+ 0101

0001 0101 (15)

Third addition: 1001 (9)

+ 1000 (8)

1 0001, which again is an incorrect BCD number as per the rule, less than 9 and carry 1.

So adding 6: 0001 0001

+ 0000 0110

0001 0111 (17)

Thus, the correct result is (d) 0001 0001 0001 0101 00010111

(G) 11110010 is signed representation of -14.

Explanation: Subtract 1 from 11110010 = 11110001

Taking 1's complement, 00001110 which is 14

11110011 is signed representation of -13

Explanaton: Subtract 1 from 11110011 = 11110010

Taking 1's complement, 00001101 which is 13

Adding both, 11110010

11110011

= 1) 11100101, which is signed representation of -27

Explanation: Subtract 1 from 11100101 = 11100100

Taking 1's complement, 00011011, which is 27

Thus, the correct answer is b) -14 and -13, -27

H) 10110110 is not a valid BCD number

Explanation: In the BCD numbering system, a decimal number is separated into four bits for each decimal digit within the number. Each decimal digit is represented by its weighted binary value performing a direct translation of the number. So a 4-bit group represents each displayed decimal digit from 0000 for a zero to 1001 for a nine. The remaining six binary code combinations of: 1010 (decimal 10), 1011 (decimal 11), 1100 (decimal 12), 1101 (decimal 13), 1110 (decimal 14), and 1111 (decimal 15) are classed as forbidden numbers and can not be used.

Thus 1011 0110 is an invalid BCD number, making d) the correct answer.

I) c.) Only one digit changes between counts

When coding the position of a rotating shaft, Gray code is considered more practical than binary numbers because the main problem with binary systems is that using binary code in tracks, there are many positions where several tracks change state at same time. This may result in error. Actually in Gray code only one track can change at same time during rotation. So then if error occurs, the resulting error will be only one bit. Gray code facilitates an easy correction through a microcontroller using a lookup table.

So that's all. Please rate if your question is solved and ask in case of queries :)

Thanks :)

  

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