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For Voltage Consider a circuit with an rms ac input voltage of Ein and an rms ac output voltage of Eout, specified in the same units as Ein. Then the voltage gain of the circuit, in decibels, is given by this formula:
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Gain (dB) = 20 log(Eout /Ein ) The base-10 logarithm of a value x is written log x. Log functions of base 10 and base e (another type of logarithm sometimes used in physics and engineering) can be determined easily using scientific calculators. From now on, when we say logarithm, we mean the base-10 logarithm unless otherwise specified.
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Problem 24-1 Suppose a circuit has an rms ac input of 1.00 V and an rms ac output of 14.0 V. What is the gain in decibels First, find the ratio Eout/Ein. Because Eout = 14.0 V rms and Ein = 1.00 V rms, the ratio is 14.0/1.00, or 14.0. Next, find the logarithm of 14.0. This is about 1.146128036. Finally, multiply this number by 20, getting something like 22.92256071. Round this off to three significant figures, because that s all you re entitled to, getting 22.9 dB. Problem 24-2 Suppose a circuit has an rms ac input voltage of 24.2 V and an rms ac output voltage of 19.9 V. What is the gain in decibels Find the ratio Eout/Ein = 19.9/24.2 = 0.822 . . . . (The three dots indicate extra digits introduced by the calculator. You can leave them in until the final roundoff.) Find the logarithm of this: log 0.822 . . . = 0.0849 . . . . The gain is 20 times this, or 1.699 . . . dB, which rounds off to 1.70 dB. For Current Current gain or loss is calculated in the same way as voltage gain or loss. If Iin is the rms ac input current and Iout is the rms ac output current specified in the same units as Iin, then:
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Gain (dB) = 20 log(Iout /Iin )
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For Power The power gain of a circuit, in decibels, is calculated according to a slightly different formula. If Pin is the input signal power and Pout is the output signal power expressed in the same units as Pin, then:
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Gain (dB) = 10 log (Pout /Pin ) The coefficient (that is, the factor by which the logarithm is to be multiplied) in the formula for power gain is 10, whereas for voltage and current gain it is 20.
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Problem 24-3 Suppose a power amplifier has an input of 5.03 W and an output of 125 W. What is the gain in decibels
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Find the ratio Pout/Pin = 125/5.03 = 24.85 . . . . Then find the logarithm: log 24.85 . . . = 1.395 . . . . Finally, multiply by 10 and round off. The gain is thus 10 1.395 . . . = 14.0 dB.
Problem 24-4 Suppose an attenuator (a circuit designed deliberately to produce power loss) provides 10 dB power reduction. The input power is 94 W. What is the output power An attenuation of 10 dB represents a gain of 10 dB. We know that Pin = 94 W, so the unknown in the power gain formula is Pout. We must solve for Pout in this formula:
10 = 10 log (Pout /94) First, divide each side by 10, getting: 1 = log (Pout /94) To solve this, we must take the base-10 antilogarithm, also known as the antilog, or the inverse log, of each side. The antilog function undoes the log function. The antilog of a value x is written antilog x. It can also be denoted as log 1x or 10x. Antilogarithms can be determined with any good scientific calculator. The solution process goes like this: antilog ( 1) = antilog [log (Pout /94)] 0.1 = Pout /94 94 0.1 = Pout Pout = 9.4 W
Decibels and Impedance When determining the voltage gain (or loss) and the current gain (or loss) for a circuit in decibels, you should expect to get the same figure for both parameters only when the input impedance is identical to the output impedance. If the input and output impedances differ, the voltage gain or loss is generally not the same as the current gain or loss. Consider how transformers work. A step-up transformer, in theory, has voltage gain, but this alone doesn t make a signal more powerful. A step-down transformer can exhibit theoretical current gain, but again, this alone does not make a signal more powerful. In order to make a signal more powerful, a circuit must increase the signal power the product of the voltage and the current! When determining power gain (or loss) for a particular circuit in decibels, the input and output impedances don t matter. In this sense, positive power gain always represents a real-world increase in signal strength. Similarly, negative power gain (or power loss) always represents a true decrease in signal strength.