The Gray code corresponding to the binary number 101101_2 is:
Binary to Gray: G5=B5=1, G4=B5^B4=$1^0$=1, G3=B4^B3=$0^1$=1, G2=B3^B2=$1^1$=0, G1=B2^B1=$1^0$=1, G0=B1^B0=$0^1$=1. Result: 111011.
The binary equivalent corresponding to the Gray code 110101_2 is:
Gray to Binary: B5=G5=1; B4=B5^G4=$1^1$=0; B3=B4^G3=$0^0$=0; B2=B3^G2=$0^1$=1; B1=B2^G1=$1^0$=1; B0=B1^G0=$1^1$=0. So Binary is 100110_2 = 100110.
In IEEE 754 single-precision floating-point format, the binary word C0000000_16 represents which decimal value?
$Hex C0000000 = 1100 0000 0000 0000 0000 0000 0000 0000. Sign S=1 (-), Exponent = 10000000_2 = 128. Actual exponent = 128 - 127 = 1. Mantissa = 0. Value = -1 * 2^1 * (1.0) = -2.0.$
How many bits are used for the biased exponent in IEEE 754 double-precision floating-point standard?
IEEE 754 double precision uses 64 bits total: 1 sign bit, 11 exponent bits (bias 1023), and 52 fraction/mantissa bits.
Which of the following statements regarding number systems and digital codes is/are CORRECT?
Excess-3 is self-complementing and unweighted. 2's complement has only 0000 for zero. Gray code is unweighted. For Hamming code: $2^p$ >= d + p + 1 => $2^3$ = 8 >= 4 + 3 + 1 = 8.