New PDF release: Introduction to Digital Electronics (Essential Electronics

By John Crowe

This article takes the coed from the very fundamentals of electronic electronics to an advent of cutting-edge strategies utilized in the sphere. it truly is perfect for any engineering or technological know-how scholar who needs to review the topic from its easy rules in addition to serving as a consultant to extra complex subject matters for readers already acquainted with the topic. The assurance is satisfactorily in-depth to permit the reader to growth easily onto larger point texts.

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Extra resources for Introduction to Digital Electronics (Essential Electronics Series)

Example text

8 UNIVERSAL GATES Universal gates, as the name suggests, are gates from which any digital circuit can be built. There are two such gates, but far from being more complex than anything considered so far they are in fact the N A N D and NOR gates. The reason they are universal is that because any circuit can be dualled (by complementing all of the variables and swapping operators) then any gate capable of being used (either singly or in combinations of itself) to implement either the AND or OR operation and the NOT operator must be universal.

27 Subtract-5 from-8 using binary notation. Solution -8 01000 two's comp. 11000 - 5 -00101 ~ +11011 - 13 110011 The sign (fifth bit) indicates that the result is negative. The two's complement of 10011 is 01101 = 13~0 giving the result as-13. 3 Multiplication Long multiplication in binary is performed in exactly the same way as in decimal arithmetic. However, in binary arithmetic digits can only be multiplied by 0 or 1. g. 22= 1002, simply results in the binary number being shifted n digits to the left and n zeros being added as the LSBs.

C =A + B + C which can be implemented using three-input A N D and OR gates respectively. l~. C A ,. , ~ y C Fig. 25 Using Boolean algebra, since (B+ C)=(B- C) then Y = A - B . C, the A N D gate implementation. To obtain the sum expression (OR gate implementation): Y= A.. (B + C) inverting both sides = A + (B + C) de Morgan's theorem =A+B+C Hence Y = A + B + C as above. 6. Note that the A N D gate produces a 1 when A = 1 and B= C=0, which specifies a single row in the truth table. The OR operator can be considered in a slightly different way.

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