Convert the following: (a) (1001)8 = ( )2 (b) (.45)10 = ( )8 (c) (2CC)16 = ( )5. Explain non-inverting amplifier. Define canonical form of Boolean expressions.

Steps to Solve

1. Convert (1001)₈ to Binary (Base 2)

Each octal digit corresponds to a 3-bit binary number:

• (1 = 001)
• (0 = 000)
• (0 = 000)
• (1 = 001)

So, combining them gives:
[ (1001)₈ = 001000000001 = (10010001)₂ ]

2. Convert (.45)₁₀ to Octal (Base 8)

To convert a decimal fraction to octal:

• Multiply by 8:
  (0.45 \times 8 = 3.6) → Take (3) as the first digit.
• Take the decimal part (0.6) and multiply again by 8:
  (0.6 \times 8 = 4.8) → Take (4) as the next digit.
• Continue with (0.8):
  (0.8 \times 8 = 6.4) → Take (6).
• Continue with (0.4):
  (0.4 \times 8 = 3.2) → Take (3).
• Continue with (0.2):
  (0.2 \times 8 = 1.6) → Take (1).
• Continue with (0.6) for further digits if needed.

So we have:
[ (.45)_{10} \approx (0.34631)₈ ]

3. Convert (2CC)₁₆ to Base 5

First, convert hexadecimal to decimal:
[ (2CC)_{16} = 2 \times 16^2 + 12 \times 16^1 + 12 \times 16^0 = 2 \times 256 + 12 \times 16 + 12 \times 1 = 512 + 192 + 12 = 716 ]

Next, convert decimal to base 5:

• Divide by 5, keeping track of the remainders:
  (716 \div 5 = 143) remainder (1)
  (143 \div 5 = 28) remainder (3)
  (28 \div 5 = 5) remainder (3)
  (5 \div 5 = 1) remainder (0)
  (1 \div 5 = 0) remainder (1)

Reading remainders from bottom to top gives:
[ (716){10} = (10331){5} ]