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Theorems Table 26-1B shows several logic equations Such facts are called theorems Statements on either side of the equals sign in each case are logically equivalent When two statements are logically equivalent, it means that one is true if and only if (iff ) the other is true For example, the statement X = Y means that X implies Y, and also that Y implies X Logical equivalence is sometimes symbolized by a double arrow with one or two shafts ( or ) Boolean theorems are used to analyze and simplify complicated logic functions This makes it possible to build a circuit to perform a specific digital function, using the smallest possible number of logic switches

Even if all the pulses have the same amplitude and the same duration, modulation can still be accomplished by varying how often they occur In PAM and PDM, the pulses are always sent at the same time interval, such as 00001 second But in pulse interval modulation (PIM), also called pulse frequency modulation (PFM), pulses might occur more or less frequently than the zero-modulation interval This is shown in Fig 26-12C Every pulse has the same amplitude and the same duration; it is their frequency that changes When there is no modulation, the pulses are evenly spaced with respect to time An increase in the instantaneous data amplitude might cause pulses to be sent more often, as is the case in Fig 26-12C Or, an increase in instantaneous data level might slow down the rate at which the pulses are sent Either scheme will work equally well

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and x 3=0 giving us x = 5 or x = 3 We can substitute these two values for x into either of the original functions to get the y-values Let s use the first one: y = 2x 2 3x 4 For x = 5, we have y = 2 ( 5)2 3 ( 5) 4 = 2 25 ( 15) 4 = 50 + 15 4 = 35 4 = 39 For x = 3, we have y = 2 32 3 3 4 = 2 9 9 4 = 18 9 4 = 27 4 = 31 The two solutions of the system are therefore (x, y) = ( 5, 39) and (x, y) = (3, 31)

In recent years, the transmission of data has been done more and more by digital means In digital communications, the modulating data attains only certain defined states, rather than continuously varying in an analog way Digital transmission offers better efficiency than analog transmission With digital modes, the signal-to-noise ratio is better, the bandwidth is narrower, and there are fewer errors Teleprinter data is always sent digitally, as is Morse code But voices and video can be sent digitally, too The only drawback that early digital experimenters faced was somewhat degraded fidelity But today, that has been overcome to the extent that digitized music recordings and transmissions actually sound better than the best analog reproductions In pulse-code modulation (PCM), any of the above aspects amplitude, duration, or frequency of a pulse train can be varied But rather than having infinitely many possible states, there are finitely many The greater the number of states, the better the fidelity But the transmitting and receiving equipment must be more and more sophisticated as the

Finally, we check Even though it may seem redundant, we should check both solutions in the original equations to remove all doubt about their correctness Let s begin by plugging ( 5, 39) into the first equation, paying careful attention to the signs:

All binary digital devices and systems employ high-speed electronic switches that perform Boolean operations These switches are called logic gates By combining logic gates, sophisticated digital systems can be built up Even the most advanced computers are, at the basic level, comprised of logic gates

4x 2 + 6x + 2y + 8 = 0 4 ( 5)2 + 6 ( 5) + 2 ( 39) + 8 = 0 4 25 + ( 30) + ( 78) + 8 = 0 100 + ( 30) + ( 78) + 8 = 0 70 + ( 78) + 8 = 0 8 + 8 = 0 0=0 Next, we check (3, 31) in that equation: 4x 2 + 6x + 2y + 8 = 0 4 3 + 6 3 + 2 ( 31) + 8 = 0

Image transmission 487 number of digital states increases An example of eight-level PAM/PCM is shown in Fig 26-12D

4 9 + 18 + ( 62) + 8 = 0 36 + 18 + ( 62) + 8 = 0 54 + ( 62) + 8 = 0 8 + 8 = 0 0=0 Next, we check ( 5, 39) in the second original equation: 3x 2 + y + 5x 11 = 0 3 ( 5) + ( 39) + 5 ( 5) 11 = 0 3 25 + ( 39) + ( 25) 11 = 0 75 + ( 39) + ( 25) 11 = 0 36 + ( 25) 11 = 0 11 11 = 0 0=0

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