1Circuits• and gate• or gate• not gateCombining & Determining I/O Relationship• P ∨∨∨∨ ~ (Q ^ R)PQR2Draw the Circuit for:P,Q&RareinputsSimplifybefore buildingthe circuit00000100001001101001010110111111OutputRQPNumber Conversions• Base of the Number System– 10 (decimal), 2 (binary), 8 (octal), 16 (hexadecimal)– tells how many different numerals are used– determines the value of each place• Conversions from anything to Base 10– use the definition of the number system• Conversions from Base 10 to anything– use repeated integer division3Addition of Binary Numbers• Carry if the number would be too large for the number system -- if it is greater than 11001 1001 1011 1101+ 10 + 11 + 11 +111------- ------- ------ ------Addition of Oct and Hex Numbers• Carry if the number would be too large for the number system ( larger than 7 or 15 )72382658ABC16CDE16+128+ 338+ 1216+ED16------- ------- ------ ------4Using a Circuit for Addition• Write as input outputa Logic Expression • Translate toCircuits0111100110100000SumCarryQPHalf AdderQP and Q are binary values (1 bit each)sum = (P v Q) ^ ~(P ^ Q)carry = (P ^ Q)Psumcarry5Full Adder• P Q and R are binary digits• P + Q + R gives sum value and carry valuehalf-adder #1half-adder #2PQRS1C1CSC2Parallel Adders• Chain these half adders and full adders together for multi-bit addition• X1X2X3+ Y1Y2Y3 = CA1A2A3 half-adderfull-adderfull-adderX3Y3X2Y2X1Y1A3carrycarrycarryA2A162's ComplimentTo Represent Negative Values using Binary1. Find the binary equivalent of the absolute value.2. Pad on the left to completely fill the bits.3. Switch all of the 1's to 0's and 0's to 1's.4. Add 1 to the result.i.e.: Find the 8-bit 2's compliment representation of -43.1. 4310= 10101122. 0010101123. 1101010024. 110101012 =
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