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After completing a calculation, you should always round the answer off to the least number of digits given in the input data numbers If you follow this rule in Problem 4-6, you must round off the answer to two significant digits, getting 140 V, because the resistance (99 ) is only specified to that level of accuracy If the resistance were given as 990 , then you would round off the answer to 143 V If the resistance were given as 9900 , then you could state the answer as 1431 V However, any further precision in the resistance value would not entitle you to go to any more digits in your answer, unless the current were specified to more than four significant figures This rule takes some getting used to if you haven t known about it or practiced it before.

You can calculate the power, in watts, in a dc circuit such as that shown in Fig. 4-7, by the formula P EI or the product of the voltage in volts and the current in amperes. You might not be given the voltage directly, but can calculate it if you know the current and the resistance. Remember the Ohm s Law formula for obtaining voltage: E IR. If you know I and R, but don t know E, you can get the power P by means of the formula P (IR)I I 2R. That is, you take the current in amperes, multiply this figure by itself, and then multiply the result by the resistance in ohms. You can also get the power if you aren t given the current directly. Suppose you re given only the voltage and the resistance. Remember the Ohm s Law formula for obtaining current: I E /R. Therefore, P E(E/R) E 2/R. Take the voltage, multiply it by itself, and divide by the resistance. Stated all together, these power formulas are: P EI I 2R E 2/R

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Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

Now you are ready to do some problems in power calculations. Refer once again to Fig. 4-7.

But after a while, it will become a habit..

noise and is inevitable when linear measurement systems, such as the one suggested by Figure 3-75, are employed in CODECs. Needless to say, design engineers recognized this problem rather quickly, and equally quickly came up with an adequate solution. It is a fairly well-known fact among psycholinguists and speech therapists that the human ear is far more sensitive to discrete changes in amplitude at low volume levels than it is at high volume levels, a fact not missed by the network designers tasked with optimizing the performance of digital carrier systems intended for voice transport. Instead of using a linear scale for digitally encoding the PAM samples, they designed and employed a nonlinear scale that is weighted with much more granularity at low volume levels that is, close to the zero line than at the higher amplitude levels. In other words, the values are extremely close together near the x-axis, and get farther and farther apart as they travel up and down the y-axis. This nonlinear approach keeps the quantizing noise to a minimum at the low amplitude levels where hearing sensitivity is the highest and enables it to creep up at the higher amplitudes, where the human ear is less sensitive to its presence. It turns out that this is not a problem, because the inherent shortcomings of the mechanical equipment (microphones, speakers, the circuit itself) introduce slight distortions at high amplitude levels that hide the effect of the nonlinear quantizing scale. This technique of compressing the values of the PAM samples to make them fit the nonlinear quantizing scale results in a bandwidth savings of more than 30 percent. In fact, the actual process is called companding, because the sample is first compressed for transmission, and then expanded for reception at the far end, hence the term.

Problem 4-10

Ohms Law can be used to find a resistance between two points in a dc circuit when the voltage and the current are known.

Suppose that the voltmeter reads 12 V and the ammeter shows 50 mA. What is the power dissipated by the potentiometer Use the formula P EI. First, convert the current to amperes, getting I 0.050 A. (Note that the zero counts as a significant digit.) Then P EI 12 0.050 0.60 W. You might say that this is 600 mW, although that is to three significant figures. It s not easy to specify the number 600 to two significant digits without using a means of writing numbers called scientific notation. That subject is beyond the scope of this discussion, so for now, you might want to say 600 milliwatts, accurate to two significant figures. (You can probably get away with 600 milliwatts and nobody will call you on the number of significant digits.)

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