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From now on:

From now on:. Combinatorial Circuits: Study analyses and design of circuits made up of gates without memory Design components needed for a simple “Computer” Sequential Circtuits: Study basic memory elements Study Ckts that can be designed using memory elements and gates

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  1. From now on: • Combinatorial Circuits: • Study analyses and design of circuits made up of gates without memory • Design components needed for a simple “Computer” • Sequential Circtuits: • Study basic memory elements • Study Ckts that can be designed using memory elements and gates • Design a computer (and understand design of computers in general) CS231 Boolean Algebra

  2. Combinatorial Circuits • Analysis of Circuits: • Given a circuit with gates (but no feedback : i.e. no memory), figure out what it does • Express it as “boolean expressions” • Methods for simplifying circuits • Boolean algebra • Karnaugh Maps (K-maps) CS231 Boolean Algebra

  3. An expression is finite but not unique A function table is unique but infinite f(x,y) = 2x + y = x + x + y = 2(x + y/2) = ... Functions • Computers take inputs and produce outputs, just like functions in math! • Mathematical functions can be expressed in two ways: • We can represent logical functions in two analogous ways too: • A finite, but non-unique Boolean expression. • A truth table, which will turn out to be unique and finite. CS231 Boolean Algebra

  4. Basic Boolean operations • There are three basic operations for logical values. NOT (complement) on one input AND (product) of two inputs OR (sum) of two inputs Operation: Expression: xy, or xy x + y x’ Truth table: CS231 Boolean Algebra

  5. Boolean expressions • We can use these basic operations to form more complex expressions: f(x,y,z) = (x + y’)z + x’ • Some terminology and notation: • f is the name of the function. • (x,y,z) are the input variables, each representing 1 or 0. Listing the inputs is optional, but sometimes helpful. • A literal is any occurrence of an input variable or its complement. The function above has four literals: x, y’, z, and x’. • Precedences are important, but not too difficult. • NOT has the highest precedence, followed by AND, and then OR. • Fully parenthesized, the function above would be kind of messy: f(x,y,z) = (((x +(y’))z) + x’) CS231 Boolean Algebra

  6. f(x,y,z) = (x + y’)z + x’ f(0,0,0) = (0 + 1)0 + 1 = 1 f(0,0,1) = (0 + 1)1 + 1 = 1 f(0,1,0) = (0 + 0)0 + 1 = 1 f(0,1,1) = (0 + 0)1 + 1 = 1 f(1,0,0) = (1 + 1)0 + 0 = 0 f(1,0,1) = (1 + 1)1 + 0 = 1 f(1,1,0) = (1 + 0)0 + 0 = 0 f(1,1,1) = (1 + 0)1 + 0 = 1 Truth tables • A truth table shows all possible inputs and outputs of a function. • Remember that each input variable represents either 1 or 0. • Because there are only a finite number of values (1 and 0), truth tables themselves are finite. • A function with n variables has 2n possible combinations of inputs. • Inputs are listed in binary order—in this example, from 000 to 111. CS231 Boolean Algebra

  7. Primitive logic gates • Each of our basic operations can be implemented in hardware using a primitive logic gate. • Symbols for each of the logic gates are shown below. • These gates output the product, sum or complement of their inputs. NOT (complement) on one input AND (product) of two inputs OR (sum) of two inputs Operation: Expression: xy, or xy x + y x’ Logic gate: CS231 Boolean Algebra

  8. Expressions and circuits • Any Boolean expression can be converted into a circuit by combining basic gates in a relatively straightforward way. • The diagram below shows the inputs and outputs of each gate. • The precedences are explicit in a circuit. Clearly, we have to make sure that the hardware does operations in the right order! (x + y’)z + x’ CS231 Boolean Algebra

  9. Circuit analysis • Circuit analysis involves figuring out what some circuit does. • Every circuit computes some function, which can be described with Boolean expressions or truth tables. • So, the goal is to find an expression or truth table for the circuit. • The first thing to do is figure out what the inputs and outputs of the overall circuit are. • This step is often overlooked! • The example circuit here has three inputs x, y, z and one output f. CS231 Boolean Algebra

  10. Write algebraic expressions... • Next, write expressions for the outputs of each individual gate, based on that gate’s inputs. • Start from the inputs and work towards the outputs. • It might help to do some algebraic simplification along the way. • Here is the example again. • We did a little simplification for the top AND gate. • You can see the circuit computes f(x,y,z) = xz + y’z + x’yz’ CS231 Boolean Algebra

  11. ...or make a truth table • It’s also possible to find a truth table directly from the circuit. • Once you know the number of inputs and outputs, list all the possible input combinations in your truth table. • A circuit with n inputs should have a truth table with 2n rows. • Our example has three inputs, so the truth table will have 23 = 8 rows. All the possible input combinations are shown. CS231 Boolean Algebra

  12. 1 1 1 0 1 1 1 0 0 0 Simulating the circuit • Then you can simulate the circuit, either by hand or with a program like LogicWorks, to find the output for each possible combination of inputs. • For example, when xyz = 101, the gate outputs would be as shown below. • Use truth tables for AND, OR and NOT to find the gate outputs. • For the final output, we find that f(1,0,1) = 1. CS231 Boolean Algebra

  13. Finishing the truth table • Doing the same thing for all the other input combinations yields the complete truth table. • This is simple, but tedious. CS231 Boolean Algebra

  14. Expressions and truth tables • Remember that if you already have a Boolean expression, you can use that to easily make a truth table. • For example, since we already found that the circuit computes the function f(x,y,z) = xz + y’z + x’yz’,we can use that to fill in a table: • We show intermediate columns for the terms xz, y’z and x’yz’. • Then, f is obtained by just OR’ing the intermediate columns. CS231 Boolean Algebra

  15. Truth tables and expressions • The opposite is also true: it’s easy to come up with an expression if you already have a truth table. • We saw that you can quickly convert a truth table into a sum of minterms expression. The minterms correspond to the truth table rows where the output is 1. • You can then simplify this sum of minterms if desired—using a K-map, for example. f(x,y,z) = x’y’z+ x’yz’ + xy’z +xyz = m1+ m2 + m5 +m7 CS231 Boolean Algebra

  16. Circuit analysis summary • After finding the circuit inputs and outputs, you can come up with either an expression or a truth table to describe what the circuit does. • You can easily convert between expressions and truth tables. Find the circuit’s inputs and outputs Find a Boolean expression for the circuit Find a truth table for the circuit CS231 Boolean Algebra

  17. Boolean operations summary • We can interpret high or low voltage as representing true or false. • A variable whose value can be either 1 or 0 is called a Boolean variable. • AND, OR, and NOT are the basic Boolean operations. • We can express Boolean functions with either an expression or a truth table. • Every Boolean expression can be converted to a circuit. • Next time, we’ll look at how Boolean algebra can help simplify expressions, which in turn will lead to simpler circuits. CS231 Boolean Algebra

  18. Expression simplification • Normal mathematical expressions can be simplified using the laws of algebra • For binary systems, we can use Boolean algebra, which is superficially similar to regular algebra • There are many differences, due to • having only two values (0 and 1) to work with • having a complement operation • the OR operation is not the same as addition CS231 Boolean Algebra

  19. Formal definition of Boolean algebra • A Boolean algebra requires • A set of elements B, which needs at least two elements (0 and 1) • Two binary (two-argument) operations OR and AND • A unary (one-argument) operation NOT • The axioms below must always be true (textbook, p. 33) • The magentaaxioms deal with the complement operation • Blueaxioms (especially 15) are different from regular algebra CS231 Boolean Algebra

  20. Comments on the axioms • The associative laws show that there is no ambiguity about a term such as x + y + z or xyz, so we can introduce multiple-input primitive gates: • The left and right columns of axioms are duals • exchange all ANDs with ORs, and 0s with 1s • The dual of any equation is always true CS231 Boolean Algebra

  21. Are these axioms for real? • We can show that these axioms are true, given the definitions of AND, OR and NOT • The first 11 axioms are easy to see from these truth tables alone. For example, x + x’ = 1 because of the middle two lines below (where y = x’) CS231 Boolean Algebra

  22. Proving the rest of the axioms • We can make up truth tables to prove (both parts of) DeMorgan’s law • For (x + y)’ = x’y’, we can make truth tables for (x + y)’ and for x’y’ • In each table, the columns on the left (x and y) are the inputs. The columns on the right are outputs. • In this case, we only care about the columns in blue. The other “outputs” are just to help us find the blue columns. • Since both of the columns in blue are the same, this shows that (x + y)’ and x’y’ are equivalent CS231 Boolean Algebra

  23. Simplification with axioms • We can now start doing some simplifications x’y’ + xyz + x’y = x’(y’ + y) + xyz [ Distributive; x’y’ + x’y = x’(y’ + y) ] = x’1 + xyz [ Axiom 7; y’ + y = 1 ] = x’ + xyz [ Axiom 2; x’1 = x’ ] = (x’ + x)(x’ + yz) [ Distributive ] = 1  (x’ + yz) [ Axiom 7; x’ + x = 1 ] = x’ + yz [ Axiom 2 ] CS231 Boolean Algebra

  24. Let’s compare the resulting circuits • Here are two different but equivalent circuits. • In general the one with fewer gates is “better”: • It costs less to build • It requires less power • But we had to do some work to find the second form CS231 Boolean Algebra

  25. Some more laws • Here are some more useful laws (p. 37). Notice the duals again! • We can prove these laws by either • Making truth tables: • Using the axioms: x + x’y = (x + x’)(x + y) [ Distributive ] = 1  (x + y) [ x + x’ = 1 ] = x + y [ Axiom 2 ] CS231 Boolean Algebra

  26. The complement of a function • The complement of a function always outputs 0 where the original function outputted 1, and 1 where the original produced 0. • In a truth table, we can just exchange 0s and 1s in the output column(s) f(x,y,z) = x(y’z’ + yz) CS231 Boolean Algebra

  27. Complementing a function algebraically • You can use DeMorgan’s law to keep “pushing” the complements inwards • You can also take the dual of the function, and then complement each literal • If f(x,y,z) = x(y’z’ + yz)… • …the dual of f is x + (y’ + z’)(y + z)… • …then complementing each literal gives x’ + (y + z)(y’ + z’)… • …so f’(x,y,z) = x’ + (y + z)(y’ + z’) f(x,y,z) = x(y’z’ + yz) f’(x,y,z) = ( x(y’z’ + yz) )’ [ complement both sides ] = x’ + (y’z’ + yz)’ [ because (xy)’ = x’ + y’ ] = x’ + (y’z’)’ (yz)’ [ because (x + y)’ = x’ y’ ] = x’ + (y + z)(y’ + z’) [ because (xy)’ = x’ + y’, twice] CS231 Boolean Algebra

  28. Standard forms of expressions • We can write expressions in many ways, but some ways are more useful than others • A sum of products (SOP) expression contains: • Only OR (sum) operations at the “outermost” level • Each term that is summed must be a product of literals • The advantage is that any sum of products expression can be implemented using a two-level circuit • literals and their complements at the “0th” level • AND gates at the first level • a single OR gate at the second level • This diagram uses some shorthands… • NOT gates are implicit • literals are reused • this is not okay in LogicWorks! f(x,y,z) = y’ + x’yz’ + xz CS231 Boolean Algebra

  29. Minterms • A minterm is a special product of literals, in which each input variable appears exactly once. • A function with n variables has 2n minterms (since each variable can appear complemented or not) • A three-variable function, such as f(x,y,z), has 23 = 8 minterms: • Each minterm is true for exactly one combination of inputs: x’y’z’ x’y’z x’yz’ x’yz xy’z’ xy’z xyz’ xyz Minterm Is true when… Shorthand x’y’z’ x=0, y=0, z=0 m0 x’y’z x=0, y=0, z=1 m1 x’yz’ x=0, y=1, z=0 m2 x’yz x=0, y=1, z=1 m3 xy’z’ x=1, y=0, z=0 m4 xy’z x=1, y=0, z=1 m5 xyz’ x=1, y=1, z=0 m6 xyz x=1, y=1, z=1 m7 CS231 Boolean Algebra

  30. Sum of minterms form • Every function can be written as a sum of minterms, which is a special kind of sum of products form • The sum of minterms form for any function is unique • If you have a truth table for a function, you can write a sum of minterms expression just by picking out the rows of the table where the function output is 1. f = x’y’z’ + x’y’z + x’yz’ + x’yz + xyz’ = m0 + m1 + m2 + m3 + m6 = m(0,1,2,3,6) f’ = xy’z’ + xy’z + xyz = m4 + m5 + m7 = m(4,5,7) f’ contains all the minterms not in f CS231 Boolean Algebra

  31. The dual idea: products of sums • Just to keep you on your toes... • A product of sums (POS) expression contains: • Only AND (product) operations at the “outermost” level • Each term must be a sum of literals • Product of sums expressions can be implemented with two-level circuits • literals and their complements at the “0th” level • OR gates at the first level • a single AND gate at the second level • Compare this with sums of products f(x,y,z) = y’ (x’ + y + z’) (x + z) CS231 Boolean Algebra

  32. Maxterms • A maxterm is a sum of literals, in which each input variable appears exactly once. • A function with n variables has 2n maxterms • The maxterms for a three-variable function f(x,y,z): • Each maxterm is false for exactly one combination of inputs: x’ + y’ + z’ x’ + y’ + z x’ + y + z’ x’+ y + z x + y’ + z’ x + y’ + z x + y + z’ x + y + z Maxterm Is false when… Shorthand x + y + z x=0, y=0, z=0 M0 x + y + z’ x=0, y=0, z=1 M1 x + y’ + z x=0, y=1, z=0 M2 x + y’ + z’ x=0, y=1, z=1 M3 x’ + y + z x=1, y=0, z=0 M4 x’ + y + z’ x=1, y=0, z=1 M5 x’ + y’ + z x=1, y=1, z=0 M6 x’ + y’ + z’ x=1, y=1, z=1 M7 CS231 Boolean Algebra

  33. Product of maxterms form • Every function can be written as a uniqueproduct of maxterms • If you have a truth table for a function, you can write a product of maxterms expression by picking out the rows of the table where the function output is 0. (Be careful if you’re writing the actual literals!) f = (x’ + y + z)(x’ + y + z’)(x’ + y’ + z’) = M4 M5 M7 = M(4,5,7) f’ = (x + y + z)(x + y + z’)(x + y’ + z) (x + y’ + z’)(x’ + y’ + z) = M0 M1 M2 M3 M6 = M(0,1,2,3,6) f’ contains all the maxterms not in f CS231 Boolean Algebra

  34. Minterms and maxterms are related • Any minterm mi is the complement of the corresponding maxterm Mi • For example, m4’ = M4 because (xy’z’)’ = x’ + y + z Minterm Shorthand x’y’z’ m0 x’y’z m1 x’yz’ m2 x’yz m3 xy’z’ m4 xy’z m5 xyz’ m6 xyz m7 Maxterm Shorthand x + y + z M0 x + y + z’ M1 x + y’ + z M2 x + y’ + z’ M3 x’ + y + z M4 x’ + y + z’ M5 x’ + y’ + z M6 x’ + y’ + z’ M7 CS231 Boolean Algebra

  35. Converting between standard forms • We can convert a sum of minterms to a product of maxterms • In general, just replace the minterms with maxterms, using maxterm numbers that don’t appear in the sum of minterms: • The same thing works for converting from a product of maxterms to a sum of minterms From before f = m(0,1,2,3,6) and f’ = m(4,5,7) = m4 + m5 + m7 complementing (f’)’ = (m4 + m5 + m7)’ so f = m4’ m5’ m7’ [ DeMorgan’s law ] = M4 M5 M7 [ By the previous page ] = M(4,5,7) f = m(0,1,2,3,6) = M(4,5,7) CS231 Boolean Algebra

  36. Summary • So far: • A bunch of Boolean algebra trickery for simplifying expressions and circuits • The algebra guarantees us that the simplified circuit is equivalent to the original one • Introducing some standard forms and terminology • Next: • An alternative simplification method • We’ll start using all this stuff to build and analyze bigger, more useful, circuits CS231 Boolean Algebra

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