Chapter 3 Combinational Logic Circuits
Chapter 3Combinational Logic Circuits12 Hours24 Marks3.1 Standard representation for logical functionsBoolean expressions / logic expressions / logical functions are expressed interms of logical variables. Logical variables can have value either ‘0’ or ‘1’. Thelogical functions can be represented in one of the following forms.- Sum of Products form (SOP form)- Product of Sums form (POS form)3.1.1 Sum of Products (SOP) formIn this form, Boolean expression is defined by sum of product terms.Various AND terms are ORed together. Each AND term may be a singlevariable or a product of multiple variables (each variable may be either incomplemented or un-complemented form).Example 1:𝑌 𝐴𝐶 𝐴̅𝐵𝐶 𝐵Here, Y is represented in SOP form with three terms. First AND termcontains two variables, second contains three variables and third contains asingle variable.Example 2:We can obtain sum of products expression from a truth table byconsidering the input combinations those produce logic ‘1’ output as shownbelow.Table 3.1: Example Truth st logic ‘1’ output in the truth table is for A 0, B 0 and C 1. This ispossible when 𝐴̅, 𝐵̅ and C are ANDed. So, first term is 𝐴̅. 𝐵̅ . 𝐶 and other terms canbe obtained similarly to get the expression for the truth table 3.1 in sum ofproducts form as,3-1
𝑌 𝐴̅. 𝐵̅ . 𝐶 𝐴̅. 𝐵. 𝐶̅ 𝐴̅. 𝐵. 𝐶 𝐴. 𝐵. 𝐶̅As this expression is a function of three variables, it can be expressed as𝑓(𝐴, 𝐵, 𝐶). It cans also be represented as 𝑌(𝐴, 𝐵, 𝐶).𝑌 𝑓(𝐴, 𝐵, 𝐶) 𝐴̅. 𝐵̅ . 𝐶 𝐴̅. 𝐵. 𝐶̅ 𝐴̅. 𝐵. 𝐶 𝐴. 𝐵. 𝐶̅Both the examples shown above are in sum of products form. But in firstexample, even though the expression is function of three variables, first term iscontaining only two variables whereas third term is containing only onevariable.𝑌 𝐴. 𝐶 𝐴̅. 𝐵. 𝐶 𝐵We may write each such incomplete term of the above expression inexpanded form as,𝐴. 𝐶 𝐴. 𝐶. (𝐵 𝐵̅ ) 𝐴. 𝐵. 𝐶 𝐴. 𝐵̅ . 𝐶𝐵 𝐵. (𝐴 𝐴̅) 𝐴. 𝐵 𝐴̅. 𝐵 𝐴. 𝐵. (𝐶 𝐶̅ ) 𝐴̅. 𝐵. (𝐶 𝐶̅ ) 𝐴. 𝐵. 𝐶 𝐴. 𝐵. 𝐶̅ 𝐴̅. 𝐵. 𝐶 𝐴̅. 𝐵. 𝐶̅Therefore, the expanded sum of product expression can be written as,𝑌 𝐴. 𝐵. 𝐶 𝐴. 𝐵̅ . 𝐶 𝐴̅. 𝐵. 𝐶 𝐴. 𝐵. 𝐶 𝐴. 𝐵. 𝐶̅ 𝐴̅. 𝐵. 𝐶 𝐴̅. 𝐵. 𝐶̅𝑌 𝐴. 𝐵. 𝐶 𝐴. 𝐵̅ . 𝐶 𝐴̅. 𝐵. 𝐶 𝐴. 𝐵. 𝐶̅ 𝐴̅. 𝐵. 𝐶̅Expanded form of Boolean expression in which each term contains all theBoolean variables in it is called as canonical form. It is also called asstandard sum of products form.Example 1:Convert following Boolean expression into standard SOP form.𝐴𝐶̅ 𝐵𝐷The above expression is a function of four variables A, B, C and D.As each term is not containing all the variables, it is not standard SOP. Theterms can be expanded to make it standard SOP as,𝑌(𝐴, 𝐵, 𝐶, 𝐷)𝑌(𝐴, 𝐵, 𝐶, 𝐷) 𝐴. 𝐶̅ 𝐵. 𝐷 𝐴. (𝐵 𝐵̅ ). 𝐶̅ 𝐵. 𝐷 𝐴. 𝐵. 𝐶̅ 𝐴. 𝐵̅ . 𝐶̅ 𝐵. 𝐷̅ ) 𝐴. 𝐵̅ . 𝐶̅ . (𝐷 𝐷̅ ) 𝐵. 𝐷 𝐴. 𝐵. 𝐶̅ . (𝐷 𝐷̅ 𝐴. 𝐵̅ . 𝐶̅ . 𝐷 𝐴. 𝐵̅ . 𝐶̅ . 𝐷̅ 𝐵. 𝐷 𝐴. 𝐵. 𝐶̅ . 𝐷 𝐴. 𝐵. 𝐶̅ . 𝐷̅̅̅̅̅̅̅̅ (𝐴 𝐴̅). 𝐵. 𝐷 𝐴. 𝐵. 𝐶 . 𝐷 𝐴. 𝐵. 𝐶 . 𝐷 𝐴. 𝐵 . 𝐶 . 𝐷 𝐴. 𝐵 . 𝐶 . 𝐷̅ 𝐴. 𝐵̅ . 𝐶̅ . 𝐷 𝐴. 𝐵̅ . 𝐶̅ . 𝐷̅ 𝐴. 𝐵. 𝐷 𝐴. 𝐵. 𝐷̅ 𝐴. 𝐵. 𝐶̅ . 𝐷 𝐴. 𝐵. 𝐶̅ . 𝐷̅̅̅̅̅̅ 𝐴. 𝐵̅ . 𝐶 . 𝐷 𝐴. 𝐵̅ . 𝐶 . 𝐷̅ 𝐴. 𝐵. (𝐶 𝐶 ). 𝐷 𝐴. 𝐵. 𝐶 . 𝐷 𝐴. 𝐵. 𝐶 . 𝐷̅̅ 𝐴. 𝐵. (𝐶 𝐶 ). 𝐷̅ 𝐴. 𝐵̅ . 𝐶̅ . 𝐷 𝐴. 𝐵̅ . 𝐶̅ . 𝐷̅ 𝐴. 𝐵. 𝐶. 𝐷 𝐴. 𝐵. 𝐶̅ . 𝐷 𝐴. 𝐵. 𝐶̅ . 𝐷 𝐴. 𝐵. 𝐶̅ . 𝐷̅̅̅ 𝐴. 𝐵. 𝐶. 𝐷 𝐴. 𝐵. 𝐶 . 𝐷As, 𝐴 𝐴 𝐴̅ 𝐴. 𝐵̅ . 𝐶̅ . 𝐷 𝐴. 𝐵̅ . 𝐶̅ . 𝐷̅ 𝐴. 𝐵. 𝐶. 𝐷 𝐴. 𝐵. 𝐶. 𝐷̅ 𝐴. 𝐵. 𝐶̅ . 𝐷 𝐴. 𝐵. 𝐶̅ . 𝐷Example 2:Standardize following Boolean expression.𝑌 𝐴𝐶 𝐵̅ 𝐶3-2
The above expression is a function of three variables A, B and C. Aseach term is not containing all the variables, it is not standard SOP. The termscan be expanded to make it standard SOP as,𝑌(𝐴, 𝐵, 𝐶)𝑌(𝐴, 𝐵, 𝐶) 𝐴. 𝐶 𝐵̅ . 𝐶 𝐴. (𝐵 𝐵̅ ). 𝐶 𝐵̅ . 𝐶 𝐴. 𝐵. 𝐶 𝐴. 𝐵̅ . 𝐶 𝐵̅ . 𝐶 𝐴. 𝐵. 𝐶 𝐴. 𝐵̅ . 𝐶 (𝐴 𝐴̅). 𝐵̅ . 𝐶 𝐴. 𝐵. 𝐶 𝐴. 𝐵̅ . 𝐶 𝐴. 𝐵̅ . 𝐶 𝐴̅. 𝐵̅ . 𝐶As, 𝐴 𝐴 𝐴 𝐴. 𝐵. 𝐶 𝐴. 𝐵̅ . 𝐶 𝐴̅. 𝐵̅ . 𝐶Single term in standard sum of products is called as minterm. Eachexpression can be represented using minterms. Some examples are shownbelow.Example 1:𝑌 𝐴. 𝐵. 𝐶 𝐴. 𝐵̅ . 𝐶 𝐴̅. 𝐵. 𝐶 𝐴. 𝐵. 𝐶̅ 𝐴̅. 𝐵. 𝐶̅As the above expression in standard sum of products, all terms in it areminterms. They are,𝐴. 𝐵. 𝐶111m7𝐴. 𝐵̅ . 𝐶101m5𝐴. 𝐵. 𝐶̅110m6𝐴̅. 𝐵. 𝐶̅010m2𝐴̅. 𝐵. 𝐶011m3Therefore the expression can be represented as,𝑌(𝐴, 𝐵, 𝐶) 𝑚(2,3,5,6,7)Example 2:̅ 𝐴. 𝐵̅ . 𝐶̅ . 𝐷 𝐴. 𝐵̅ . 𝐶̅ . 𝐷̅ 𝐴. 𝐵. 𝐶. 𝐷 𝐴. 𝐵. 𝐶. 𝐷̅𝑌 𝐴. 𝐵. 𝐶̅ . 𝐷 𝐴. 𝐵. 𝐶̅ . 𝐷As the above expression in standard sum of products, all terms in it areminterms. They are,𝐴. 𝐵. 𝐶̅ . 𝐷m131101̅𝐴. 𝐵. 𝐶̅ . 𝐷m121100𝐴. 𝐵̅ . 𝐶̅ . 𝐷m91001̅𝐴. 𝐵̅ . 𝐶̅ . 𝐷m81000𝐴. 𝐵. 𝐶. 𝐷m151111̅𝐴. 𝐵. 𝐶. 𝐷m141110Therefore the expression can be represented as,𝑌(𝐴, 𝐵, 𝐶, 𝐷) 𝑚(8,9,12,13,14,15)3.1.2 Product of Sums (POS) formIn this form, Boolean expression is defined by product of sum terms.Various OR terms are ANDed together. Each OR term may be a single variable3-3
or a sum of multiple variables (each variable may be either in complemented orun-complemented form).Example 1:𝑌 (𝐴 𝐶). (𝐵̅ 𝐶). 𝐴Here, Y is represented in POS form with three terms. First OR termcontains two variables, second contains two variables and third contains a singlevariable.Example 2:We can obtain product of sums expression from a truth table byconsidering the input combinations those produce logic ‘0’ output as shownbelow.Table 3.2: Example Truth st logic ‘0’ output in the truth table is for A 0, B 0 and C 0. This ispossible when 𝐴, 𝐵 and C are ORed. So, first term is (𝐴 𝐵 𝐶) and other termscan be obtained similarly to get the expression for the truth table 3.2 in productof sums form as,𝑌 (𝐴 𝐵 𝐶). (𝐴̅ 𝐵 𝐶). (𝐴̅ 𝐵 𝐶̅ ). (𝐴̅ 𝐵̅ 𝐶̅ )As this expression is a function of three variables, it can be expressed as𝑓(𝐴, 𝐵, 𝐶). It cans also be represented as 𝑌(𝐴, 𝐵, 𝐶).𝑌 𝑓(𝐴, 𝐵, 𝐶) (𝐴 𝐵 𝐶). (𝐴̅ 𝐵 𝐶). (𝐴̅ 𝐵 𝐶̅ ). (𝐴̅ 𝐵̅ 𝐶̅ )Both the examples shown above are in product of sums form. But in firstexample, even though the expression is function of three variables, first term iscontaining only two variables, second term is also containing only two variablesand third term is containing only one variable.𝑌 (𝐴 𝐶). (𝐵̅ 𝐶). 𝐴We may write each such incomplete term of the above expression inexpanded form as,(𝐴 𝐶) (𝐴 𝐵. 𝐵̅ 𝐶) (𝐴 𝐵 𝐶). (𝐴 𝐵̅ 𝐶)(𝐵̅ 𝐶) (𝐴. 𝐴̅ 𝐵̅ 𝐶) (𝐴 𝐵̅ 𝐶). (𝐴̅ 𝐵̅ 𝐶)𝐴 (𝐴 𝐵. 𝐵̅ ) (𝐴 𝐵). (𝐴 𝐵̅ ) (𝐴 𝐵 𝐶. 𝐶̅ ). (𝐴 𝐵̅ 𝐶. 𝐶̅ ) (𝐴 𝐵 𝐶). (𝐴 𝐵 𝐶̅ ). (𝐴 𝐵̅ 𝐶). (𝐴 𝐵̅ 𝐶̅ )3-4
Therefore, the expanded sum of product expression can be written as,𝑌 (𝐴 𝐵 𝐶). (𝐴 𝐵̅ 𝐶). (𝐴 𝐵̅ 𝐶). (𝐴̅ 𝐵̅ 𝐶). (𝐴 𝐵 𝐶). (𝐴 𝐵 𝐶̅ ). (𝐴 𝐵̅ 𝐶). (𝐴 𝐵̅ 𝐶̅ )𝑌 (𝐴 𝐵 𝐶). (𝐴 𝐵̅ 𝐶). (𝐴̅ 𝐵̅ 𝐶). (𝐴 𝐵 𝐶̅ ). (𝐴 𝐵̅ 𝐶̅ )Expanded form of Boolean expression in which each term contains all theBoolean variables in it is called as canonical form. It is also called asstandard product of sums form.Example 1:Convert following Boolean expression into standard POS form.(𝐴 𝐶̅ ). (𝐵 𝐷)The above expression is a function of four variables A, B, C and D.As each term is not containing all the variables, it is not standard POS. Theterms can be expanded to make it standard POS as,𝑌(𝐴, 𝐵, 𝐶, 𝐷)𝑌(𝐴, 𝐵, 𝐶, 𝐷) (𝐴 𝐶̅ ). (𝐵 𝐷) (𝐴 𝐵. 𝐵̅ 𝐶̅ ). (𝐵 𝐷) (𝐴 𝐵 𝐶̅ ). (𝐴 𝐵̅ 𝐶̅ ). (𝐵 𝐷)̅ ). (𝐴 𝐵̅ 𝐶̅ 𝐷. 𝐷̅ ). (𝐵 𝐷) (𝐴 𝐵 𝐶̅ 𝐷. 𝐷̅ ). (𝐴 𝐵̅ 𝐶̅ 𝐷). (𝐴 𝐵̅ 𝐶̅ (𝐴 𝐵 𝐶̅ 𝐷). (𝐴 𝐵 𝐶̅ 𝐷̅ 𝐷). (𝐵 𝐷)̅ ). (𝐴 𝐵̅ 𝐶̅ 𝐷). (𝐴 𝐵̅ 𝐶̅ (𝐴 𝐵 𝐶̅ 𝐷). (𝐴 𝐵 𝐶̅ 𝐷̅̅ 𝐷). (𝐴. 𝐴 𝐵 𝐷)̅̅ ). (𝐴 𝐵̅ 𝐶̅ 𝐷). (𝐴 𝐵̅ 𝐶̅(𝐴 𝐵 𝐶 𝐷). (𝐴 𝐵 𝐶̅ 𝐷̅ ). (𝐴 𝐵 𝐷). (𝐴̅ 𝐵 𝐷) 𝐷̅̅ ). (𝐴 𝐵̅ 𝐶̅ 𝐷). (𝐴 𝐵̅ 𝐶̅ (𝐴 𝐵 𝐶 𝐷). (𝐴 𝐵 𝐶̅ 𝐷̅ ). (𝐴 𝐵 𝐶. 𝐶̅ 𝐷). (𝐴̅ 𝐵 𝐶. 𝐶̅ 𝐷) 𝐷̅̅ ). (𝐴 𝐵̅ 𝐶̅ 𝐷). (𝐴 𝐵̅ 𝐶̅ (𝐴 𝐵 𝐶 𝐷). (𝐴 𝐵 𝐶̅ 𝐷̅ ). (𝐴 𝐵 𝐶 𝐷). (𝐴 𝐵 𝐶̅ 𝐷). (𝐴̅ 𝐵 𝐶 𝐷 𝐷). (𝐴̅ 𝐵 𝐶̅ 𝐷)As, 𝐴. 𝐴 𝐴̅ ). (𝐴 𝐵̅ 𝐶̅ 𝐷). (𝐴 𝐵̅ 𝐶̅ (𝐴 𝐵 𝐶̅ 𝐷). (𝐴 𝐵 𝐶̅ 𝐷̅ ). (𝐴 𝐵 𝐶 𝐷). (𝐴̅ 𝐵 𝐶 𝐷). (𝐴̅ 𝐵 𝐶̅ 𝐷) 𝐷Example 2:Standardize following Boolean expression.𝑌 (𝐴 𝐵)(𝐴̅ 𝐶)The above expression is a function of three variables A, B and C. Aseach term is not containing all the variables, it is not standard POS. The termscan be expanded to make it standard POS as,𝑌(𝐴, 𝐵, 𝐶)𝑌(𝐴, 𝐵, 𝐶) (𝐴 𝐵). (𝐴̅ 𝐶) (𝐴 𝐵 𝐶. 𝐶̅ ). (𝐴̅ 𝐶) (𝐴 𝐵 𝐶). (𝐴 𝐵 𝐶̅ ). (𝐴̅ 𝐶) (𝐴 𝐵 𝐶). (𝐴 𝐵 𝐶̅ ). (𝐴̅ 𝐵. 𝐵̅ 𝐶) (𝐴 𝐵 𝐶). (𝐴 𝐵 𝐶̅ ). (𝐴̅ 𝐵 𝐶). (𝐴̅ 𝐵̅ 𝐶)As, 𝐴 𝐴 𝐴 (𝐴 𝐵 𝐶). (𝐴 𝐵 𝐶̅ ). (𝐴̅ 𝐵 𝐶). (𝐴̅ 𝐵̅ 𝐶)3-5
Single term in standard product of sumss is called as maxterm. Eachexpression can be represented using maxterms. Some examples are shownbelow.Example 1:𝑌 (𝐴 𝐵 𝐶). (𝐴 𝐵̅ 𝐶). (𝐴̅ 𝐵 𝐶). (𝐴 𝐵 𝐶̅ ). (𝐴̅ 𝐵 𝐶̅ )As the above expression in standard product of sums form, all terms in itare maxterms. They are,𝐴 𝐵 𝐶0 0 0M0𝐴 𝐵̅ 𝐶0 1 0M2𝐴 𝐵 𝐶̅0 0 1M1𝐴̅ 𝐵 𝐶̅1 0 1M5𝐴̅ 𝐵 𝐶1 0 0M4Therefore the expression can be represented as,𝑌(𝐴, 𝐵, 𝐶) 𝑀(0,1,2,4,5)Example 2:̅ ). (𝐴 𝐵̅ 𝐶̅ 𝐷)𝑌 (𝐴 𝐵 𝐶̅ 𝐷). (𝐴 𝐵 𝐶̅ 𝐷As the above expression in standard product of sums form, all terms in itare maxterms. They are,𝐴 𝐵 𝐶̅ 𝐷0 0 1 0M2̅𝐴 𝐵 𝐶̅ 𝐷0 0 1 1M3𝐴 𝐵̅ 𝐶̅ 𝐷M60 1 1 0Therefore the expression can be represented as,𝑌(𝐴, 𝐵, 𝐶, 𝐷) 𝑀(2,3,6)3.2 Karnaugh MapIn the last chapter, we have simplified Boolean expression using Booleanlaws. That was algebraic simplification method. In that method, it is hard toguess whether the simplified expression is in its simplest form or not (i.e.whether simplification is to be continued or not).This drawback has beenovercome in Karnaugh map simplification.Karnaugh map (also called K-map) is graphical representation of a logicsystem. It can be drawn from minterm or maxterm Boolean expressions.Number of cells in a K-map depends on number of variables in the Booleanexpression. Karnaugh map of a Boolean expression with n variables contain 2ncells (squares). Each cell in the K-map corresponds with one input combination.Each row and each column of K-map is assigned by 0s and 1s.A two-variable K-map can be drawn with various possibilities. Twopossibilities are shown in figure 3.1. In these notes we will use pattern shown infigure 3.1a.3-6
B0A01A010123B0110213a.b.Figure 3.1: Two ways of representing a 2-variable K-mapAs discussed previously, each square (or cell) in K-map corresponds to oneinput combination. In figure 3.1a, cell 0 corresponds to input combination ‘00’(i.e. 𝐴̅. 𝐵̅), cell 1 corresponds to input combination ‘01’ (i.e. 𝐴̅. 𝐵), cell 2corresponds to input combination ‘10’ (i.e. 𝐴. 𝐵̅) and cell 3 corresponds to inputcombination ‘11’ (i.e. 𝐴. 𝐵).A three-variable K-map can be drawn with various possibilities. Fourpossibilities are shown in figure 3.2. In these notes we will use pattern shown infigure 236745c.112a.AB 00011110010BC 00011110104153726d.Figure 3.2: Four ways of representing a 3-variable K-mapA four-variable K-map can be drawn with various possibilities. Twopossibilities are shown in figure 3.3. In these notes we will use pattern shown infigure 3.3a.3-7
CD00AB 000111100111AB00100132457612131514891110CD 000111100111100412815139371511261410a.b.Figure 3.3: Two ways of representing a 4-variable K-mapWe can represent any expression which is in standard SOP (minterms)form or standard POS (maxterms) form in a K-map.3.2.1 Representation SOP on K-mapIf the given Boolean expression is not in standard SOP form, it should befirst converted to standard SOP form. Then its minterms are written. For eachminterm in the expression, ‘1’ is written in the corresponding cell in the K-mapand the remaining cells are marked as ‘0’.Example 1:Represent following Boolean expression by K-map.𝑌 𝐵𝐶 𝐴̅𝐵̅ 𝐶̅ 𝐴𝐵𝐶̅The above expression is in SOP form. It is a function of threevariables A, B and C. As each term is not containing all the variables, it is not instandard SOP form. So, converting it into standard SOP form,𝑌(𝐴, 𝐵, 𝐶) 𝐵. 𝐶 𝐴̅. 𝐵̅ . 𝐶̅ 𝐴. 𝐵. 𝐶̅ (𝐴 𝐴̅). 𝐵. 𝐶 𝐴̅. 𝐵̅ . 𝐶̅ 𝐴. 𝐵. 𝐶̅ 𝐴. 𝐵. 𝐶 𝐴̅. 𝐵. 𝐶 𝐴̅. 𝐵̅ . 𝐶̅ 𝐴. 𝐵. 𝐶̅111 011000110m7m3m0m6𝑌(𝐴, 𝐵, 𝐶) 𝑚(0,3,6,7)𝑌(𝐴, 𝐵, 𝐶)The above minterms can be represented in K-map. All the present termsare marked as ‘1’ and remaining cells are marked as ‘0’. As the function is ofthree variables, a three-variable K-map is used for representation.ABC000 0 11 4 0150100371111261001Example 2:Represent following expression using K-map.𝑓(𝐴, 𝐵, 𝐶, 𝐷) 𝑚(0,1,2,5,13,15)The above minterms can be represented in K-map. All the presentterms are marked as ‘1’ and remaining cells are marked as ‘0’. As the function isof four-variables, a four-variable K-map is used for representation.3-8
AB 00011110CD000 14 012 08 0011 15 113 19 0113 07 015 111 0102 16 014 010 0Example 3:Represent following expression using K-map.𝑌 𝑚(0,1,3)The above minterms can be represented in K-map. All the presentterms are marked as ‘1’ and remaining cells are marked as ‘0’. As the function isof two-variables, a two-variable K-map is used for representation.A0102B010131113.2.2 Representation POS on K-mapIf the given Boolean expression is not in standard POS form, it should befirst converted to standard POS form. Then its maxterms are written. For eachmaxterm in the expression, ‘0’ is written in the corresponding cell in the K-mapand the remaining cells are marked as ‘1’.Example 1:Represent following Boolean expression by K-map.𝑌 (𝐴 𝐵) (𝐴̅ 𝐵̅ 𝐶̅ ) (𝐴 𝐵 𝐶̅ )The above expression is in POS form. It is a function of threevariables A, B and C. As each term is not containing all the variables, it is not instandard POS form. So, converting it into standard POS form,𝑌(𝐴, 𝐵, 𝐶)𝑌(𝐴, 𝐵, 𝐶)𝑌(𝐴, 𝐵, 𝐶) (𝐴 𝐵) (𝐴̅ 𝐵̅ 𝐶̅ ) (𝐴 𝐵 𝐶̅ ) (𝐴 𝐵 𝐶. 𝐶̅ ) (𝐴̅ 𝐵̅ 𝐶̅ ) (𝐴 𝐵 𝐶̅ ) (𝐴 𝐵 𝐶) (𝐴 𝐵 𝐶̅ ) (𝐴̅ 𝐵̅ 𝐶̅ ) (𝐴 𝐵 𝐶̅ ) (𝐴 𝐵 𝐶) (𝐴 𝐵 𝐶̅ ) (𝐴̅ 𝐵̅ 𝐶̅ )0 0 00 0 11 1 1M0M1M7 𝑀(0,1,7)The above maxterms can be represented in K-map. All the present termsare marked as ‘0’ and remaining cells are marked as ‘1’. As the function is ofthree variables, a three-variable K-map is used for representation.ABC000 0 01 4 11501013711102610113-9
Example 2:Represent following expression using K-map.𝑓(𝐴, 𝐵, 𝐶, 𝐷) 𝑀(1,3,5,7,9,11,13,15)The above maxterms can be represented in K-map. All the presentterms are marked as ‘0’ and remaining cells are marked as ‘1’. As the function isof four-variables, a four-variable K-map is used for representation.AB 00011110CD000 14 112 18 10105 013 09 01107 015 011 0131016 114 110 12Example 3:Represent following expression using K-map.𝑌 𝑀(0,2)The above maxterms can be represented in K-map. All the presentterms are marked as ‘0’ and remaining cells are marked as ‘1’. As the function isof two-variables, a two-variable K-map is used for representation.A0102B000131113.2.3 Grouping of adjacent cellsK-maps are used for simplification of Boolean expressions. Thesimplification is based on grouping of the terms in adjacent cells. Importantaspects of grouping are discussed below.- Number of cells in a group must always be a power of 2. (i.e. 1, 2, 4, 8,16 etc.)- Two cells are said to be adjacent if they differ in only one variable.Four cells are said to be adjacent if they differ in two variables. Eightcells are said to be adjacent if they differ in three variables and so on.- If K-map is represented for a SOP, grouping is done for adjacent 1’sand if K-map is represented for a POS, grouping is done for adjacent0’s.- While grouping, each cell containing ‘1’ in case of SOP (‘0’ in case ofPOS) must be considered at least once.- We should try to make minimum number of groups (i.e. each group ofmaximum possible size).- ‘Don’t care entries’ may be considered in any group.3-10
Combinational Logic Circuits 12 Hours 24 Marks 3.1 Standard representation for logical functions Boolean expressions / logic expressions / logical functions are expressed in terms of logical variables. Logical variables can have value either ‘0
3.5 Combinational Circuits 138 Digital logic chips are combined to give us useful circuits. These logic circuits can be categorized as either combinational logic (Section 3.5) or sequential logic (Sec. 3.6). 3.5.1 Basic Concepts 138 The key concept in recognizing a combinational circu
Apr 03, 2020 · Combinational logic circuits 2. Sequential logic circuit. Compare combinational and sequential circuits (four points). Standard representation for logical functions: Boolean expressions / logic expressions / logical functions are expressed in terms of logical variables. Log
The University of Texas at Arlington Sequential Logic - Intro CSE 2340/2140 – Introduction to Digital Logic Dr. Gergely Záruba The Sequential Circuit Model x 1 Combinational z1 x n zm (a) y y Y Y Combinational logic logic x1 z1 x n z m Combinational logic with n inputs and m switching functions: Sequential logic with n inputs, m outputs, r .
To implement simple logical operations using combinational logic circuits 4 To design combinational logic circuits, sequential logic circuits 5 To impart to student the concepts of sequential circuits, enabling them to analyze sequential systems interms of state machines. 6
Ch. 5 Control Task Basics 1 Chapter 5 Control Task Basics In his chapter, we begin programming a process. The first task is to conceptualize the problem, then move to describe a solution with combinational and then sequential logic. Combinational logic is simply And/Or logic. This chapter will discuss combinational logic and use the
3 Introduction Logic circuits for digital systems may be – Combinational – Sequential A combinational circuit consists of logic gates whose outputs at any time are determined by the current input values, i.e., it has no memory elements A sequential circuit consists of logic gates whose outputs at any time are determi
design combinational logic circuits Combinational logic circuits do not have an internal stored state, i.e., they have no memory. Consequently the output is solely a function of the current inputs. Later, we will study circuits having a stored internal
combinational logic. This is an ideal location to introduce the language because the reader has just learned about combinational logic theory inChap. 4. This allows the student to begin gainingexperience using the VHDL simulation tools on basic combinational
