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Basic Gates An inverter or NOT gate has one input and one output It reverses the state of the input An OR gate can have two or more inputs If both, or all, of the inputs are 0 (low), then the output is 0 If any of the inputs is 1 (high), then the output is 1 An AND gate can have two or more inputs If both, or all, of the inputs are 1, then the output is 1 Otherwise the output is 0 Other Gates Sometimes an inverter and an OR gate are combined This produces a NOR gate If an inverter and an AND gate are combined, the result is a NAND gate An exclusive OR gate, also called an XOR gate, has two inputs and one output If the two inputs are the same (either both 1 or both 0), then the output is 0 If the two inputs are different, then the output is 1 The functions of logic gates are summarized in Table 26-2 Their schematic symbols are shown in Fig 26-2 Black Boxes Logic gates can be combined to form circuits with many inputs and outputs When two or more logic gates are combined, the outputs are always specific logical functions of the inputs A complex combination of logic gates is sometimes called a black box The functions of a black box can always be determined using Boolean algebra, if the gates inside, and the way they are interconnected, is known Conversely, if a certain complex logical function is needed for an application, a black box can be designed to perform that function by using Boolean algebra to break the function down into components of NOT, OR, AND, NOR, NAND, and XOR

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The modulation techniques used for image transmission are similar to those employed for sending voices Nonmoving pictures can be sent within the same bandwidth as a voice For high-resolution, fast-scan moving images, the necessary bandwidth is greater A thorough discussion of image transmission is beyond the scope of this book The basics of the three most common video communications modes are discussed here For further detail, a text on video communications is recommended

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Both of these equations are presented as functions of x , so we can mix the right sides directly without having to manipulate When we do that, we get a1x 2 + b1x + c1 = a2x 2 + b2x + c2 The next several steps are tricky It s easy to make mistakes with the signs and the grouping To minimize the risk of error, let s do negative additions, rearrange and regroup the symbols, and change the negative additions to subtractions after we ve put everything in the proper order We can add the negatives of a2x2, b2x, and c2 to each side, getting a1x 2 + b1x + c1 + ( a2x 2) + ( b2x) + ( c2) = 0 Using the commutative law for addition, we can rewrite this as a1x 2 + ( a2x 2) + b1x + ( b2x) + c1 + ( c2) = 0 The distributive law of multiplication over addition allows us to group the coefficients and the constants to get [a1 + ( a2)]x 2 + [b1 + ( b2)]x + [c1 + ( c2)] = 0 Changing the negative additions back to subtractions, we obtain (a1 a2)x 2 + (b1 b2)x + (c1 c2) = 0 We can solve this equation with the quadratic formula For reference, here it is, yet one more time, in its classical form (Have you memorized it yet ) When we see the quadratic equation ax 2 + bx + c = 0 the roots are given by x = [ b (b2 4ac)1/2] / (2a) To solve the equation at hand, we can make these substitutions in the classical version of the quadratic formula: Write (a1 a2) in place of a Write (b1 b2) in place of b Write (c1 c2) in place of c When we make these changes in copy-and-paste fashion, we get x = { (b1 b2) [(b1 b2)2 4(a1 a2)(c1 c2)]1/2} / [2(a1 a2)] We can simplify this slightly by getting rid of the minus sign in the first expression on the right side of the equals sign, and then reversing the positions of b1 and b2 inside the parentheses We can also multiply out.

gate (A), an AND gate (B), an OR gate (C), an XOR gate (D), a NAND gate (E), and a NOR gate (F)

the denominator But there s no good reason to expand the products of the binomials That would produce a formula with fewer grouping symbols, but the arithmetic would be more cumbersome in practical use Let s express the formula as x = {(b2 b1) [(b1 b2)2 4(a1 a2)(c1 c2)]1/2} / (2a1 2a2) After we ve found the x-values of the solutions by plugging in the coefficients, plugging in the constants, and going through the arithmetic, we can put those x-values into either of the original two functions and calculate the y-values If we call the x-values x1 and x2, and if we use the first of the original functions, we have y1 = a1x12+ b1x1 + c1 and y2 = a1x22 + b1x2 + c1 The solutions of the whole system can be written as ordered pairs in the form (x1, y1) and (x2, y2)

Still images are transmitted by facsimile ( fax) If data is sent slowly enough, any amount of detail can be transmitted within a voice band This is how telephone fax works A high-resolution commercial fax image has upwards of 1000 lines of data The image is scanned from left to right and from top to bottom like reading a book The complete image takes several minutes to send Many of the black-and-white photographs in the daily newspaper are sent via fax Practically all weather satellite images are faxed

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