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barcode in vb.net 2010 Copyright 2006, 2002, 1997, 1993 by The McGrawHill Companies, Inc. Click here for terms of use. in Software
Copyright 2006, 2002, 1997, 1993 by The McGrawHill Companies, Inc. Click here for terms of use. Code 3/9 Recognizer In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Code39 Printer In None Using Barcode creator for Software Control to generate, create ANSI/AIM Code 39 image in Software applications. 380 Amplifiers and Oscillators
Reading Code39 In None Using Barcode recognizer for Software Control to read, scan read, scan image in Software applications. Code 39 Printer In C# Using Barcode creator for Visual Studio .NET Control to generate, create Code 3/9 image in .NET framework applications. 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: USS Code 39 Maker In .NET Framework Using Barcode maker for ASP.NET Control to generate, create Code39 image in ASP.NET applications. Code39 Printer In .NET Using Barcode drawer for .NET framework Control to generate, create Code 3/9 image in .NET applications. Gain (dB) = 20 log(Eout /Ein ) The base10 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 base10 logarithm unless otherwise specified. Draw Code 39 Full ASCII In VB.NET Using Barcode creator for .NET framework Control to generate, create Code 39 image in .NET framework applications. Print UCC  12 In None Using Barcode creation for Software Control to generate, create EAN / UCC  14 image in Software applications. Problem 241 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 242 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: Bar Code Encoder In None Using Barcode maker for Software Control to generate, create bar code image in Software applications. Painting UPCA In None Using Barcode maker for Software Control to generate, create Universal Product Code version A image in Software applications. Gain (dB) = 20 log(Iout /Iin ) Making Code 128B In None Using Barcode creation for Software Control to generate, create Code 128B image in Software applications. Barcode Encoder In None Using Barcode printer for Software Control to generate, create bar code image in Software applications. 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: Encode USPS Intelligent Mail In None Using Barcode printer for Software Control to generate, create Intelligent Mail image in Software applications. Printing Barcode In Visual Basic .NET Using Barcode drawer for .NET Control to generate, create barcode image in .NET applications. 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. Drawing Bar Code In None Using Barcode drawer for Font Control to generate, create barcode image in Font applications. Create 1D Barcode In .NET Using Barcode maker for ASP.NET Control to generate, create Linear Barcode image in ASP.NET applications. Problem 243 Suppose a power amplifier has an input of 5.03 W and an output of 125 W. What is the gain in decibels Making Code 128 In None Using Barcode generation for Font Control to generate, create Code 128 Code Set C image in Font applications. GS1 128 Encoder In VB.NET Using Barcode maker for VS .NET Control to generate, create UCC  12 image in Visual Studio .NET applications. Basic Bipolar Transistor Amplifier 381
Read ANSI/AIM Code 128 In VB.NET Using Barcode scanner for .NET framework Control to read, scan read, scan image in Visual Studio .NET applications. Code 128 Code Set A Encoder In .NET Using Barcode maker for .NET framework Control to generate, create ANSI/AIM Code 128 image in .NET applications. 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 244 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 base10 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 stepup transformer, in theory, has voltage gain, but this alone doesn t make a signal more powerful. A stepdown 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 realworld increase in signal strength. Similarly, negative power gain (or power loss) always represents a true decrease in signal strength.

