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Fast Boolean Algebra ELEC 2607 notes with the overburden removed A fast way to learn enough to get the prelab done honorably Printed; 03/01/05 Modified; January 3, 2005 Slide 1 Department of Electronics, Carleton University © John Knight Digital Circuits p. 1 Fast Boolean Algebra Fast Boolean Algebra These are the lecture notes, with details left out that are not needed to prepare for the lab. The example design of a 2-bit comparator, is not in the lecture notes. The design of a full-adder is used in the second lab. Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight Comment on Slide 1 Digital Circuits p. 2 Two Valued Algebra Boolean Algebra The algebra of 1 and 0 Two Valued Algebra Values are called “1” and “0” or True and False or High and Low. Variables X=2 X = -1 X = 3+j6 X=1 X=0 A variable X can have only two values, 0 or 1. Symbol Operations Complement, Not, or Inverse X Schematic Truth Table X X X is the opposite of X AND a•b ab X is 1 if a AND b are both 1. OR X-OR, XOR, Exclusive OR X is 1 if exactly one of a or b is 1. a⊕b X a b a+b X is 1 if a OR b or both are 1. Printed; 03/01/05 Modified; January 3, 2005 a b a b Department of Electronics, Carleton University © John Knight a 0 0 1 1 b 0 1 0 1 a 0 0 1 1 b 0 1 0 1 X X X 0 0 0 1 a 0 0 1 1 X 0 1 1 0 X X 0 1 1 0 b 0 1 0 1 X 0 1 1 1 Slide 2 onSlide63 Comment Digital Circuits p. 3 Basic Boolean Algebra Operations Basic Boolean Algebra Operations Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight Comment on Slide 2 Digital Circuits p. 4 Operations (continued) Symbol Schematic Truth Table Operations (continued) NAND a b a•b ab NOT-AND X a 0 0 1 1 NOR NOT-OR a+b a b X X-NOR Exclusive-NOR, NOT-XOR. (Equivalence) a⊕b Multiple Input Operations AND “-” means either 1 or 0, (don’t care) OR a b a•b•c abc a + b+ c a bc a bc b 0 1 0 1 X a 0 1 X b c - 0- 0 11 a 0 0 1 X 1 1 0 0 0 a 0 0 1 1 X 0 0 0 1 X Number of “1” X-OR inputs Exclusive-OR, XOR. Printed; 03/01/05 Modified; January 3, 2005 a⊕b⊕c a bc X 0 1 2 3 a 0 1 - b 0 1 0 1 X 1 1 1 0 b 0 1 0 1 X 1 0 0 1 b 0 1 - c 0 1 X 0 1 1 1 X 0 1 0 1 Slide 3 Department of Electronics, Carleton University © John Knight Digital Circuits p. 5 Boolean Operations Boolean Operations Common Mistake a• b is not the same as a •b 1.• PROBLEM Prove the above statement by evaluating both expressions for a=1, b=0. Multiple Input Operations N-input Gates In theory a gate can have any number of inputs. For transistor circuits, the gates become very slow with more than four inputs. However theoretical circuits will often have more than four inputs. With transistors it is easier to build NAND and NOR than AND and OR. However people make fewer mistakes if the design is done with AND and OR. 2.• PROBLEM Is a+b is the same as a + b. If not, give a counter example. Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight Comment on Slide 3 Digital Circuits p. 6 Multiple Input Operations Basic Laws Laws with Zeros and Ones X+0=X (Z1) X+1=1 ( O 1) or just (Z) or just (Z) X•0 = 0 (Z2) X•1 = X (O 2) (I1) X•X = X (I2) or (I) for (N2) No 2nd form (N1) X•X = 0 (N2) or (N) (C1) X•Y = Y•X (C2) or (C) (A2) or (A) Indempotent or just Z for any of these four A variable is unchanged by operating with itself. X+X=X Double Negative (X) =X Negation Laws X+X=1 Commutative Laws X+Y=Y+X Associative Laws X + (Y + Z) = (X + Y) + Z (A1) X•(Y•Z) = (X•Y)•Z Distributive Laws (X + Y) Z = XZ + YZ (D1) (D2) X Y + Z = (X + Z)(Y + Z) Remember This Printed; 03/01/05 Modified; January 3, 2005 Slide 4 Department of Electronics, Carleton University © John Knight Basic Laws Digital Circuits p. 7 Multiple Input Operations Basic Laws Numbering System for the Laws In this course one frequently needs to refer to these laws, hence they have been given numbers. (Z1) stands for zero law number 1, (Z2) for zero law number 2. (O 1) for one law number 1, etc., where O is a script letter “Oh”. If you do not want to remember these numbers, most of the laws on these pages are simple enough to just write a small example instead. On the next page both the number and an example are given. The above laws will be taken as axioms These laws are fairly obvious from the description of the gates (operators). The only strange on is the second distributive law (D2). This one has to be memorized because it is very useful as well as strange. We will treat all these laws as axioms except (D2). This will be proven very shortly. Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight Comment on Slide 4 Digital Circuits p. 8 Basic Laws Multiple Input Operations Using Boolean Algebra Prove: AB + B = B Simplification law (S) Proof: AB +B = AB + B·1 = ( A + 1)B = (1)B =B Prove: from (O 2) from (D1) from (O 1) from (O 2) AB + C = (A + C)(B + C) X •1 = X (X + Y)Z = XZ + YZ X+1=1 X •1 = X Second distributive law (D2) Proof: (A +C)(B + C) = Q(B + C) = QB + QC = (A + C)B + (A + C)C = AB + CB + AC + CC = AB + CB + AC + C = AB + CB + C + AC + C = AB + C +C = AB + C Simplify: let Q = (A + C) from (D1) (X + Y)Z = XZ + YZ since Q = (A + C) from (D1) from ( I ) XX=X from ( I ) X+ X=X use (S) twice XY + X = X from ( I ) AB + AB + AB AB +AB + AB = AB + AB + AB +AB = A( B + B)+(A + A)B = A(1) + (1)B =A+B Printed; 03/01/05 Modified; January 3, 2005 from ( I ) from (D1) from (N1) from (O 1) X+ X=X (X + Y)Z = XZ + YZ X+X =1 X+1=1 Department of Electronics, Carleton University © John Knight Basic Laws Slide 5 Digital Circuits p. 9 Useful Rule Useful Rule Simplification or Absorption Rule x + xy = x is one of the most useful rules, and fairly easy to remember. Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight Comment on Slide 5 Digital Circuits p. 10 Basic Laws Boolean Algebra: Exhaustive Proofs Boolean Algebra: Exhaustive Proofs Good for a small number of inputs (up to about 4) Make a table a) List all possible inputs b) Calculate all values of the left hand side c) Calculate all values of the right hand side d) See if they agree Alternate Proof Second distributive law: (A + C)(B + C) = AB + C Proof: Use a truth table and prove for all cases LHS RHS ABC (A + C) (B + C)(A+C)(B+C) AB AB + C 000 001 011 010 100 101 111 110 0 1 1 0 1 1 1 1 0 1 1 1 0 1 1 1 0 1 1 0 0 1 1 1 Printed; 03/01/05 Modified; January 3, 2005 0 0 0 0 0 0 1 1 0 1 1 0 0 1 1 1 The two sides are equal for all combinations of ABC. Department of Electronics, Carleton University © John Knight Basic Laws Slide 6 Digital Circuits p. 11 Useful Rule Trying All Combinations The method is good for proofs, with not very many variables. It is not good for simplification. Exhaustive Problems 3.• PROBLEM WITH XORS Prove that A⊕B = A ⊕B = Α⊕Β 4.• PROBLEM Prove that C + BC = C + B Prove it (a) algebraically and (b) with a truth table. 5.• PROVE Ab + bc + cA = Ab + cA Consensus theorem This is hard to prove algebraically.1 6.• PROVE xA + xB = (x + A)(x +B) Swap rule This is easy to prove algebraically after you have proven the consensus theorem. 7.• PROVE THAT: A ⊕B = A⊕B 1. Hint bc = bcA + bcA. Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight Comment on Slide 6 Digital Circuits p. 12 Basic Laws Boolean Algebra: Exhaustive Checking Boolean Algebra: Exhaustive Checking Checking An Answer You think the following simplification is true: (A + BC)(B + C)(B + C) = AB + C Check it out. ABC 000 001 011 010 100 101 111 110 (A + BC) (B + C) (B + C) (A+BC)(A+C)(B+C) AB AB + C 0 0 0 0 1 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 0 1 0 0 0 0 0 1 1 1 0 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 Not the same With 3 inputs table has 8 entries With 4 inputs table has 16 entries With 5 inputs table has 32 entries With 6 inputs table has 64 entries With 7 inputs, algebra starts to look a lot easier. Printed; 03/01/05 Modified; January 3, 2005 Slide 7 Department of Electronics, Carleton University © John Knight Digital Circuits p. 13 Basic Laws Useful Rule Circuit Problems 8.• PROBLEM: FOR CIRCUIT SHOWN Simplify the circuit. A A AA=A F= 9.• PROBLEM Simplify to 4 literals (letters)AD +AE + AF B 10.• PROBLEM Implement:ABCD + ABCE + ABCF with one multi-input OR and one multi-input AND. 11.• PROBLEM Simplify (A + B)(A + C)(A + D) 12.• PROBLEM Implement:(A+B+C+D)(A+B+C+E)(A+B+C+F) with one multi-input OR and one multi-input AND1 1. Hint; try Problem 11.• first. Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight Comment on Slide 7 Digital Circuits p. 14 Basic Laws Deriving Formulas From Truth Tables Deriving Formulas From Truth Tables A formulas for XOR using AND, OR and NOT Truth Table a 0 0 1 1 b 0 1 0 1 (a⊕b) Equation for when (a⊕b) is “1” 0 1 a•b 1 a•b 0 (a⊕b) = a•b + a•b (a⊕b) is “0” for all other combinations XOR FORMULA 1. Write out the Truth Table 2. Look for where the result is “1”. 3. Write down the letters corresponding to the inputs (a⊕b) = a•b + a•b Thus a “0” in the “a” column is written as “a”. a “1” in the “a” column is written as “a”. 4. AND the letters to make a term like a•b. 5. The equation is the OR of all the terms where the output is “1”. Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight Basic Laws Slide 8 Digital Circuits p. 15 Obtaining Formulas from Truth Tables Obtaining Formulas from Truth Tables This method will give formulas in what is called OR of ANDs or (SUM of PRODUCTs) form. These formulas are letters or their inverses ANDed together and the resulting terms ORed together1, like abc + acd + cf + ace + .... We need only find all the terms that make the final answer “1”. If no term make the formula “1”, then it defaults to “0”. To Find the Formula Go through the truth table and writes down each terms that make the result “1”. Then OR all these terms together That is the desired formula. XOR and XNOR gates These are more complex to implement than most other two-input gates. They are often implemented using formula like the one derived above. 1. SUM of PRODUCTS form has no bracketed terms like ab(f + d) or the brackets implied by long inverting bars like ab = (ab) Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight Comment on Slide 8 Digital Circuits p. 16 Basic Laws Deriving Circuits From Truth Tables Deriving Circuits From Truth Tables Truth Table The Comparator A1A0 B1B0 This compares two, 2-bit numbers A1 A0 z=1 B1 B0 z=1 If: A1A0 >B1B0 z is “1” if any of the terms in the table are “1”. z = A 1 •A 0 • B 1 •B 0 + A 1 •A 0 •B 1 •B 0 + A 1 •A 0 •B 1 •B 0 + A 1 •A 0 •B 1 •B 0 + A 1 •A 0 •B 1 •B 0 + A 1 •A 0 •B 1 •B 0 (1) To simplify z, look for the terms with the most common factors A 1 •A 0 •B 1 •B 0 + A 1 •A 0 •B 1 •B 0 + A 1 •A 0 •B 1 •B 0 + A 1 •A 0 •B 1 •B 0 0 0 0 0 0 1 0 1 0 1 0 1 0 0 1 1 1 1 0 0 1 0 0 1 A 1 •A 0 •B 1 •B 0 1 0 0 0 1 A 1 •A 0 •B 1 •B 0 1 0 0 1 1 A 1 •A 0 •B 1 •B 0 1 1 0 0 1 A 1 •A 0 •B 1 •B 0 1 1 0 1 1 A 1 •A 0 •B 1 •B 0 1 1 1 0 1 A 1 •A 0 •B 1 •B 0 0 0 0 1 0 0 = A1•B1•(A0•B0 + A0•B0 + A0•B0 + A0•B0) (D1) 0 0 1 0 = A1•B1•(A0•(B0 + B0) + A0•(B0 + B0)) (D1) 0 1 1 0 0 X+X=1 0 0 1 1 0 = A1•B1•(A0 + A0) X •1 = X 0 1 1 1 0 = A1•B1•(1) X+X=1 1 0 1 1 0 = A1•B1•(A0•(1) + A0•(1)) = A 1 •B 1 Printed; 03/01/05 Modified; January 3, 2005 Boolean Formula z 0 X •1 = X (2) Slide 9 Department of Electronics, Carleton University © John Knight Digital Circuits p. 17 Basic Laws Compare Compare Comparator circuits compare numbers. Here we only consider non-negative binary numbers, further the examples are restricted to two bits. Decimal A1A0 value 0 0 1 1 The “(1)” and “(2)” are equation numbers. 13.• PROBLEM ON COMPARE Design a circuit which compares two 2-bit numbers A1A0 and B1B0. It gives out Z=1 if the two numbers are equal. In this problem please do not use XOR or XNOR gates. 0 1 0 1 0 1 2 3 A1 A0 B1 B0 z=1 z=1 If: A 0=B0 , and A1 =B1 Deriving Circuits From Truth Tables Recall we only need consider cases which evaluate to “1”. Boolean formulas default to “0” if they are not “1”. If the output is “1” when A1, A0, B1,B0 = 1,0,1,0, this gives the term A1A 0B1B0 OR together all the terms that give an output of “1”, to get the complete formula. The formulas so derived will not be in a reduced form. Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight Comment on Slide 9 Digital Circuits p. 18 Basic Laws Deriving Circuits From Truth Tables Deriving Circuits From Truth Tables The Comparator, reducing the equation.) A1 A0 z=1 B1 B0 z=1 If: A1A0 >B1B0 z = A1•A0•B1•B0 + A1•A0•B1•B0 + A1•A0•B1•B0 + A1•A0•B1•B0 + A1•A0•B1•B0 + A1•A0•B1•B0 (1) Substitute equation (2) into (1) = A 1 •A 0 •B 1 •B 0 + A 1 •B 1 + A 1 •A 0 •B 1 •B 0 (3) Look for term(s) having the most common factors with terms A1•A0•B1•B0 and A1•A0•B1•B0 A1 •A0 •B1•B0 has 3 with the 1st, and gain A 1•A0 •B1 •B0 has 3 with the 2nd. In (3), OR in 2 extra copies of A1•A 0•B 1•B0. z = A 1 •A 0 •B 1 •B 0 + A 1 •B 1 + A 1 •A 0 •B 1 •B 0 + A 1 •A 0 •B 1 •B 0 + A 1 •A 0 •B 1 •B 0 (X = X + X) = A 1 •B 1 + A 1 •A 0 •B 1 •B 0 + A 1 •A 0 •B 1 •B 0 + A 1 •A 0 •B 1 •B 0 + A 1 •A 0 •B 1 •B 0 (X +Y = Y + X) = A 1 •B 1 + A 0 •B 1 •B 0 •A 1 + A 0 •B 1 •B 0 •A 1 + A 1 •A 0 •B 0 •B 1 + A 1 •B 0 •A 0 •B 1 (X•Y=Y •X) = A1•B1 + A0•B1•B0•(A1 + A1) + A1•A0•B0•(B1 + B1) = A1•B1 + A0•B1•B0•(1) + A1•A0•B0•(1) = A 1 •B 1 + A 0 •B 1 •B 0 + A 1 •A 0 •B 0 (D1) (X+X=1) (X•1=X) Final reduced answer z = A 1 •B 1 + A 0 •B 1 •B 0 + A 1 •A 0 •B 0 ) Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight Basic Laws Slide 10 Digital Circuits p. 19 The Comparator The Comparator The final formula is easy to understand. Use “-” to mean either “1” or “0”. The first term (A1B1) says A1A0 > B1B0 when A1A0 = 1 - and B1B0 = 0 -. That is 1- > 0- . 1 - is either decimal 3 or decimal 2. 0 - is either decimal 1 or decimal 0. Thus (A1B1) covers 3 > 1 or 0 and 2 > 1 or 0. The second term (A0B1B0) says A1A0 > B1B0 when A1A0 = - 1 and B1B0 = 0 0. In decimal 3 > 0 and 1 > 0. Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight Comment on Slide 10 Digital Circuits p. 20 Basic Laws Deriving Circuits From Truth Tables Deriving Circuits From Truth Tables (More) a 0 0 0 0 1 1 1 1 The Full Adder The full-adder adds 3 bits A B Cy (out) Cy(in) Full Adder Result, Σ Decimal b Cy(in) Cy(out) 0 0 0 0 1 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 0 1 0 1 0 1 1 0 1 0 0 1 2 1 1 2 3 2 Σ (sum) a 0 0 0 0 1 1 1 1 b Cy(in) Cy(out) 0 0 0 0 1 0 1 1 1 a•b•Cy 1 0 0 0 0 0 0 1 1 a•b•Cy 1 1 1 a•b•Cy 1 0 1 a•b•Cy Printed; 03/01/05 Modified; January 3, 2005 Σ Equation for when Cy(out) is “1” 0 1 a•b•Cy 0 1 a•b•Cy 1 a•b•Cy 0 1 a•b•Cy 0 Cy(out) = a•b•Cy + a•b•Cy + a•b•Cy + a•b•Cy Cy(out) is “0” for all other combinations Equation for when Σ is “1” Σ = a•b•Cy + a•b•Cy + a•b•Cy + a•b•Cy Department of Electronics, Carleton University © John Knight Basic Laws Slide 11 Digital Circuits p. 21 The Comparator The Adder1 0 0 00 0 1 01 1 0 01 A0 B0 The four results from adding two bits Cy This upper circuit is called a half adder because it adds only two bits. A full adder (lower box) can add three bits. The third bit is the carry input from a previous stage. Σ For adding one-bit numbers, or the first bits of a multibit number, a half added is all right. Otherwise a full adder is needed. Half Adder Cy (out) CY(IN) A B 1. We will use which. 1 1 10 Full Adder Σ (sum) for addition here, since + was used for OR. In many places + is used for both, and you have to figure out which is Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight Comment on Slide 11 Digital Circuits p. 22 Basic Laws Deriving Circuits From Truth Tables The Full Adder (Cont) Simplify the full adder equations O Cy(out) = a•b•Cy + a•b•Cy + a•b•Cy + a•b•Cy = a•b•Cy + a•b•Cy + a•b•Cy + a•b•Cy + a•b•Cy+ a•b•Cy = (a + a)b•Cy + ( b + b)a•Cy + a•b(Cy+ Cy) (D1) ay + by = (a + b)y (N) x + x = 1 = (1)b•Cy + ( 1)a•Cy + a•b(1) = b•Cy + a•Cy + a•b = a majority gate (I) x + x = x (O ) x•1 = x ≥2 a b Cy(out) Cy Σ = a•b•Cy + a•b•Cy + a•b•Cy + a•b•Cy = a•b•Cy + a•b•C y + a•b•C y + a•b•C y (C) x+y=y+x = (a•b + a•b)•C y + (a•b + a•b)C y (D1) ay + by = (a + b)y = (a•b + a•b)•C y + (a•b + a•b)C y = (a ⊕ b)•Cy + (a ⊕ b)•Cy = (a ⊕ b)⊕Cy a b Cy (Formula for XOR) a•b + a•b = (a ⊕ b) Σ Printed; 03/01/05 Modified; January 3, 2005 (Formula for XNOR) a•b + a•b = a•b + a•b (Formula for XOR) u•Cy + u•Cy = (u ⊕ Cy) u = (a ⊕ b) Slide 12 Department of Electronics, Carleton University © John Knight Basic Laws Digital Circuits p. 23 Simplifying the Full Adder Simplifying the Full Adder Carry Term This is straightforward Sum Term This requires being aware of the various forms of XOR and XNOR. A⊕B = A⊕B = Α⊕Β a⊕b = ab+ab a⊕ b = a•b + ab (a⊕b) c + (a⊕b)c = a⊕b⊕c 14.• PROBLEM Starting on the second line of the simplification of the carry, Cy(out) = a•b•Cy + a•b•Cy + a•b•Cy + a•b•Cy + a•b•Cy+ a•b•Cy use the simplification rule to reduce the amount of writing to get the final answer. Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight Comment on Slide 12 Digital Circuits p. 24 Basic Laws Deriving Circuits From Truth Tables Common Mistakes 1. Saying a•b is the same as a•b 2. Not using AB + A =A to simplify expressions before using more more complex rules. Also not reducing using A + AE = A + E. Simplifying and reducing first may save many lines of algebra. 3. Saying X + 1 = X Everybody knows better than this, but they still do it. Printed; 03/01/05 Modified; January 3, 2005 Slide 13 Department of Electronics, Carleton University © John Knight Digital Circuits p. 25 Basic Laws Larger Adders Larger Adders The full adder only adds one-bit numbers. To add multibit numbers, one needs several full adders. A 4-bit adder which adds two 4-bit numbers is shown. 4-Bit Adder made from four full-adders A3 B3 C4 Full adder A2 B2 C3 C3 Σ3 A3+B3+C3 Σ3 Printed; 03/01/05 Modified; January 3, 2005 Full adder Σ2 A3 B3 C4 Full adder A1 B1 C2 C2 A2 B 2 C3 A2+B2+C2 A0 B0 C1 C1 Σ1 C2 Σ2 Department of Electronics, Carleton University © John Knight Σ1 C0 Σ0 A1 B1 A1+B1+C1 Full adder A0 B0 C1 A0+B0+C0 C0 0 Σ0 Comment on Slide 13 Digital Circuits p. 26 Basic Laws DeMorgan’s Theorems, Simple Forms DeMorgan’s Theorem DeMorgan’s Theorems, Simple Forms DeMorgan The 2nd DeMorgan A + B = A·B (DeM1) D+E Inverse The 2nd inverse A + B = A·B A 0 0 1 1 B 0 1 0 1 1 0 1 0 D· E = D + E AB AB A B A+B 1 1 0 0 (DeM2) D·E = 1 1 1 0 0 0 0 1 D 0 0 1 1 1 1 1 0 E D+E 0 1 1 0 0 0 1 0 D E D·E 1 1 0 0 1 0 1 0 1 0 0 0 Equivalent graphical forms: A·B = K A K B NAND A·B A B AND = A+B A B NAND = C C Printed; 03/01/05 Modified; January 3, 2005 = A B D+E D E K NOR = G G = D E D·E G NOR A+B AND AND D+E C D E OR = F F = D E Department of Electronics, Carleton University © John Knight Basic Laws D·E F OR Slide 14 Digital Circuits p. 27 DeMorgan’s Theorems DeMorgan’s Theorems A theorem relating NANDs and NORs. • An OR gate with inverted inputs is equivalent to an AND gate with an inverted output. • An AND gate with inverted inputs is equivalent to an OR gate with an inverted output. • Inverting inputs and outputs of an OR makes it an AND. • Inverting inputs and outputs of an AND makes it an OR. EXAMPLE Convert (a + b)(a + c) to an expression with 3 letters and inversion bars only over single letters. (a + b)(a + c) = (a + b) + (a + c) = (a·b) + (a·c) = a ·b + a ·c = a(b + c) 41.• PROBLEM (a+b) + a·b Reduce (DeM1) (DeM2) (Clear brackets) (D1) xy + xz = x(y+z) to four letters with inversion bars over single letters only. 42.• PROBLEM d(de) + (de)e to four letters with inversion bars over single letters only. Reduce Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight Comment on Slide 14 Digital Circuits p. 28 Basic Laws DeMorgan’s Theorems, Simple Forms Using DeMorgan’s Theorem Equivalent Gate Symbols AND NOR OR NOR OR NAND AND AND NAND Real Gates are NAND and NOR Easy to Understand Gates are AND and OR NOR NAND OR AND One cannot make AND and OR gates directly. Circuit with real gates a b NAND c d NAND NAND Circuit with simple gates a b AND F c d AND F = (ab) • (cd) OR F F = ab + cd Which one is easier for you to understand? Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight DeMorgan’s Theorem Slide 15 Digital Circuits p. 29 DeMorgan’s Theorems DeMorgan’s Theorem The two expressions for the circuit with real gates and the circuit with simple gates, are equivalent ab + cd = F = (ab) • (cd) Use High-True And-Or Signals for Thinking Thinking • Thinking in nand-nor logic is difficult. Just look at any industrial schematic used extensively for maintenance. The margin will be full of “1”s and “0”s pencilled in by users. These Notes • If the logic is important ANDs and ORs will be used. • If the gate design is important, as when we talk about CMOS gates, NANDs and NORs will be used. 43.• PROBLEM Prove, using DeMorgan’s Theroem(s), that Printed; 03/01/05 Modified; January 3, 2005 ab + cd = (ab) • (cd) Department of Electronics, Carleton University © John Knight Comment on Slide 15 Digital Circuits p. 30 DeMorgan’s Theorem DeMorgan Transfers Real Gates Into DeMorgan Transfers Real Gates Into Simple Gates Circuit with real gates a b NAND c d NAND NAND NAND NAND NAND NAND a b AND NAND c d AND NAND Use DeMorgan’s NAND gate symbol at output F = (ab) • (cd) • F Circuit with simple gates Inverting circles cancel each other OR F F = ab + cd Circuits meant for understanding the logic use ANDs and ORs. - Draw your circuits with ANDs and ORs. • Circuits for construction are drawn with NANDs and NORs. - Lab circuits to be wired are drawn with NANDs and NORs Draw construction diagrams by: - transforming the understandable circuit into the real circuit - using the DeMorgan alternate symbols. NAND • NAND Thus Compromise drawings have NANDS and NORs with the circles are arranged to cancel each other. Printed; 03/01/05 Modified; January 3, 2005 NAND Department of Electronics, Carleton University © John Knight DeMorgan’s Theorem alternate NAND symbol Slide 16 Digital Circuits p. 31 Transforming NAND-NOR Diagrams into Transforming NAND-NOR Diagrams into AND-OR Diagrams • • • Diagrams in this course will be drawn with ANDs and ORs as much as possible. Diagrams for construction or maintainance, that want to show exactly what gates were used, will be drawn with NANDs and NORs. This is particularly true of older diagrams. A compromise method, which is almost as easy to follow, but shows NOR NAND the real gates as used, is to make alternate gates with the alternate symbols, the ones with the inverting circles on the inputs. NOTE: One output circle cancels all the input circles it feeds. 44.• PROBLEM Transform this circuit into simple gates. a) Printed; 03/01/05 Modified; January 3, 2005 Department of Electronics, Carleton University © John Knight Comment on Slide 16 Digital Circuits p. 32 DeMorgan’s Theorem Printed; 03/01/05 Modified; January 3, 2005 DeMorgan’s Theorem Printed; 03/01/05 Modified; January 3, 2005 DeMorgan Transfers Real Gates Into Department of Electronics, Carleton University © John Knight Slide 17 Digital Circuits p. 33 Transforming NAND-NOR Diagrams into Department of Electronics, Carleton University © John Knight Comment on Slide 17 Digital Circuits p. 34