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Main points of this past exam are: Network Connecting, Network Connecting, Control Signal, Signed Operations, Complement Numbers, Subtractor, Multiplexors,, Basic Gates, One-Bit Registers, Truth Table
Typology: Exams
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A B Ci S Co 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1
(1100.0011) 2 = 14.14 (^8) (178) 16 = 1(16) 2 + 7(16) + 8(1) = 376 10 (46.5625) 10 = 101110.1001 (^2)
b. Convert 46.25 to IEEE 754 single precision format and provide the hexadecimal value of the encoding.
First convert 46.25 to binary = 101110. Since the number is +ve sign bit is 0. Rewriting the number 1.0111001 x 2^5. Expanding the fraction to 23 bits 01110010000000000000000. The exponent is 5. Biasing 5 + 127 = 132. In binary this is 10000100. Putting this all together 01000010001110010000000000000000 0x
Two’s Complement -2 n-1^2 n-1^ - 1
Sign Magnitude -2n-1^ - 1 2 n-1^ - 1
Ans