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The second use of Ohm s Law is to find unknown voltages when the current and the resistance are known. For the following problems, uncover the ammeter and cover the voltmeter scale instead in your mind.

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125 microseconds; companding, during which the values are weighted toward those most receptive to the human ear; quantization, in which the weighted samples are given values on a nonlinear scale; and finally encoding, during which each value is assigned a distinct binary value. Each of the stages of Pulse Code Modulation (PCM) will now be discussed in detail.

Problem 4-4

1. 6. 11. 16. b a c a 2. 7. 12. 17. a b b d 3. 8. 13. 18. b d d c 4. 9. 14. 19. d c a c 5. 10. 15. 20. c b b d

Pulse Code Modulation (PCM)

Suppose the potentiometer (Fig. 4-7) is set to 100 ohms, and the measured current is 10 mA. What is the dc voltage Use the formula E I R. First, convert the current to amperes: 10 mA = 0. 01 A. Then multiply: E 0. 01 100 1 V. That s a low, safe voltage, a little less than what is produced by a flashlight cell.

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Thanks to the work performed by Harry Nyquist at Bell Laboratories in the 1920s, we know that to optimally represent an analog signal as a digitally encoded bitstream, the analog signal must be sampled at a rate that is equal to twice the bandwidth of the channel over which the signal is to be transmitted. Since each analog voice channel is allocated 4 KHz of bandwidth, it follows that each voice signal must be sampled at twice that rate, or 8,000 samples per second. In fact, that is precisely what happens in T-Carrier systems, which we now use to illustrate our example. The standard T-Carrier multiplexer accepts inputs from 24 analog channels, as shown in Figure 3-73. Each channel is sampled in turn every one-eight thousandth of a second in a round-robin fashion, resulting in the generation of 8,000 pulse amplitude samples from each channel every second. The sampling rate is important. If the sampling rate is too high, too much information is transmitted, and bandwidth is wasted; if the sampling rate is too low, then we run the risk of aliasing. Aliasing is the interpretation of the sample points as a false waveform, due to the paucity of samples.

Problem 4-5

1. 6. 11. 16. c d a a 2. 7. 12. 17. c c c c 3. 8. 13. 18. a a b d 4. 9. 14. 19. d d a d 5. 10. 15. 20. b c c b

Adjust the potentiometer (Fig. 4-7) to a value of 157 K , and let the current reading be 17 mA. What is the voltage of the source Now you have to convert both the resistance and the current values to their proper units. A resistance of 157 K is 157,000 ; a current of 17 mA is 0. 017 A. Then E IR 0.017 157,000 2669 V 2.669 kV. You might want to round this off to 2.67 kV. This is a dangerous voltage. If you touch the terminals you ll get clobbered.

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.

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Problem 4-6

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You set the potentiometer (Fig. 4-7) so that the meter reads 1.445 A, and you observe that the potentiometer scale shows 99 ohms. What is the voltage These units are both in their proper form. Therefore, you can plug them right in and use your calculator: E IR 1. 445 99 143.055 V. This can, and should, be rounded off to 143 V. A purist would go further and round it to the nearest 10 volts, to 140 V. It s never a good idea to specify your answer to a problem with more significant figures than you re given. The best engineers and scientists go by the rule of significant figures: keep to the least number of digits given in the data. If you follow this rule in Problem 4-6, you must round off the answer to two significant figures, getting 140 V, because the resistance specified (99 ) is only accurate to two digits.

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