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Problem 4-15 Consider five resistors in parallel. Call them R1 through R5, and call the total resistance R as shown in Fig. 4-9. Let the resistance values be as follows: R1 = 100 , R2 = 200 , R3 = 300 , R 4 = 400 , and R5 = 500 . What is the total resistance, R, of this parallel combination

Suppose the following resistances are hooked up in series with each other: 112 , and 680 . What is the total resistance of the series combination (Fig. 4-8) , 470

inherent in human speech to reduce the amount of information required. The technique still relies on PCM encoding, but adds an additional step to carry out its task. The 64-Kbps PCM-encoded signal is fed into an ADPCM transcoder, which considers the prior behavior of the incoming stream to predict the behavior of the next sample. Here s where the magic happens: instead of transmitting the actual value of the predicted sample, it encodes in 4 bits and transmits the difference between the actual and predicted samples. Since the difference from sample to sample is typically quite small, the results are generally considered to be very close to toll-quality. This 4-bit transcoding process, which is based on the known behavior characteristics of human voice, enables the system to transmit 8,000 4-bit samples per second, thus reducing the overall bandwidth requirement from 64 to 32 Kbps. It should be noted that ADPCM works well for voice, because the encoding and predictive algorithms are based upon its behavior characteristics. It does not, however, work as well for higher bit rate data (above 4800 bps), which has an entirely different set of behavior characteristics.

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4-9 Five resistors of values R1 through R5, connected in parallel,

4-8 Three resistors in series (Problem 4-14).

Just add the values, getting a total of 112 + 470 + 680 1262 . You might round this off to 1260 . It depends on the tolerances of the components how precise their actual values are to the ones specified by the manufacturer.

produce a net resistance R. Illustration for Problems 4-15 and 4-16.

Continuously Variable Slope Delta (CVSD)

When resistances are placed in parallel, they behave differently than they do in series. In general, if you have a resistor of a certain value and you place other resistors in parallel with it, the overall resistance will decrease. One way to look at resistances in parallel is to consider them as conductances instead. In parallel, conductances add, just as resistances add in series. If you change all the ohmic values to siemens, you can add these figures up and convert the final answer back to ohms. The symbol for conductance is G. This figure, in siemens, is related to the resistance R, in ohms, by the formulas: G R 1/R, and 1/G

Converting the resistances to conductance values, you get: G1 = 1/100 = 0.01 S, G2 = 1/200 = 0.005 S, G3 = 1/300 = 0.00333 S, G4 = 1/400 = 0.0025 S, and G5 = 1/500 = 0.002 S. Adding these gives G = 0.01 + 0.005 + 0.00333 + 0.0025 + 0.002 = 0.0228 S. The total resistance is therefore R = 1/G = 1/0.0228 = 43.8 .

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