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2 RXL jR2 XL XL " jXL Zp R R2 in VS .NET
2 2 RXL jR2 XL XL " jXL Zp R R2 Recognize Code 128 Code Set A In Visual Studio .NET Using Barcode Control SDK for .NET framework Control to generate, create, read, scan barcode image in .NET applications. ANSI/AIM Code 128 Generator In .NET Framework Using Barcode printer for VS .NET Control to generate, create Code128 image in Visual Studio .NET applications. 395 396 Scan Code128 In Visual Studio .NET Using Barcode recognizer for Visual Studio .NET Control to read, scan read, scan image in Visual Studio .NET applications. Bar Code Generation In .NET Using Barcode printer for .NET Control to generate, create bar code image in .NET framework applications. that is, " Zp r jXL
Bar Code Scanner In Visual Studio .NET Using Barcode reader for .NET Control to read, scan read, scan image in .NET applications. Code 128 Code Set B Creator In C#.NET Using Barcode creator for VS .NET Control to generate, create Code 128B image in Visual Studio .NET applications. 2 XL =R.
Making USS Code 128 In .NET Using Barcode generation for ASP.NET Control to generate, create Code 128B image in ASP.NET applications. Create Code 128 Code Set A In Visual Basic .NET Using Barcode generator for Visual Studio .NET Control to generate, create ANSI/AIM Code 128 image in Visual Studio .NET applications. where r Note that eq. (396) represents a resistance of r ohms in series with the reactance jXL . Thus the parallel circuit of Fig. 235 can be replaced by the series circuit of Fig. 236. Make Barcode In .NET Framework Using Barcode printer for .NET Control to generate, create bar code image in Visual Studio .NET applications. Data Matrix ECC200 Creation In VS .NET Using Barcode creator for .NET framework Control to generate, create Data Matrix ECC200 image in Visual Studio .NET applications. Fig. 236
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Barcode Printer In Java Using Barcode maker for Eclipse BIRT Control to generate, create bar code image in Eclipse BIRT applications. Printing Barcode In VS .NET Using Barcode generation for ASP.NET Control to generate, create barcode image in ASP.NET applications. With all the foregoing in mind, note that the original circuit of Fig. 232 can now be drawn in the form of Fig. 237. In this gure, r can include any resistance the primary and secondary windings may have. Now, for the nal step, replace the transformer with its T equivalent (Fig. 231). Doing " " this, and also adding the two loop currents I1 and I2 , Fig. 237 becomes Fig. 238. " We now wish to nd the value of G in eq. (393). Since, by inspection of Fig. 238, " " Vt jXC I2 , eq. (393) becomes " " G jXC I2 =Vi 397 EAN13 Printer In None Using Barcode creation for Online Control to generate, create EAN 13 image in Online applications. EAN13 Supplement 5 Printer In None Using Barcode creator for Font Control to generate, create GS1  13 image in Font applications. " " Thus, to nd G we must nd the value of I2 . To do this, let us begin by writing the two loopvoltage equations for Fig. 238, which, as you can verify, are " " r j XL XC I1 jXm I2 jgm Vi XC " " jXm I1 r j XL XC I2 0 398 399 Generating Bar Code In Java Using Barcode generation for Java Control to generate, create bar code image in Java applications. Code 128 Reader In Java Using Barcode reader for Java Control to read, scan read, scan image in Java applications. CHAPTER 10 Magnetic Coupling. Transformers
Reading Code 128 Code Set B In None Using Barcode decoder for Software Control to read, scan read, scan image in Software applications. GS1 128 Printer In None Using Barcode encoder for Font Control to generate, create GTIN  128 image in Font applications. Fig. 238
" One way to solve the above two equations for I2 is to use the method of determinants. To do this, as you ll recall, the rst step is to nd the value of D (delta), where D, here, is the value of the determinant formed from the coe cients of the two unknown currents, which, as you should verify, gives the value 2 D r2 Xm XL XC 2 j2r XL XC
and thus, continuing with determinant procedure, we have g VX X " I2 m i m C D and hence, by eq. (397), 2 2 jgm Xm XC " jgm Xm XC G 2 D r2 Xm XL XC 2 j2r XL XC
400 The di culty now is that it s hard to get a handle on eq. (400), to see what it really means. Let us therefore work on the equation and try to get it in a di erent form, better suited to our needs. This can be done by a combination of algebraic manipulation and making use of certain relationships that are known to exist in practical applications of Fig. 232. One such procedure is as follows. We begin by de ning that, in our work here, the resonant frequency will be the frequency at which XL XC . In regard to Fig. 232, since the primary and secondary circuits separately consist of equal parallel values of R, L, and C, it follows that the primary and secondary circuits will separately have the same value of resonant frequency, which we ll denote by !0 radians/second. (Our equations will appear less cluttered up if we use rad/sec instead of cycles/sec, where, as always, ! 2pf .) Thus, to begin, we de ne that, in Fig. 232, the resonant frequency is de ned by the equation !0 L 1 !0 C 401 For frequencies other than the resonant frequency we write ! instead of !0 . Thus, XL !L and XC 1=!C will denote reactance values at any frequency !, while the notations X0L !0 L and X0C 1=!0 C will denote reactance values at the resonant frequency. Next, after a certain amount of trial and error, we nd that the following algebraic manipulations produce some very useful relationships. Let us begin by writing XL XC !L 1 !!0 L !0 !!0 C !C !0 CHAPTER 10 Magnetic Coupling. Transformers
and thus, by eq. (401), we have that ! !0 XL XC !0 L !0 ! ! !0 ! !0 !0 ! 402 Next note that ! !0 !2 !2 0 !0 ! !0 ! In the above equation let d
403 ! !0 !0 404 in which ! !0 is the frequency di erence between any frequency ! and the resonant frequency !0 , as illustrated in the gure below. Thus, in eq. (404) we see that d is the FRACTIONAL DEVIATION FROM RESONANCE for any given frequency !. (For instance, if ! 1:02!0 , then d 0:02; that is, ! is two percent greater than !0 .) In our work here, !0 will be the frequency of the carrier wave, while ! will be the highest sideband frequency of importance in the modulated wave. Thus, to satisfactorily amplify a given modulated wave, the circuit of Fig. 232 should pass all frequencies in the range !0 !. Returning now to eqs. (402) through (404), we see that ! !0 XL XC !0 Ld ! or, since (by eq. (404)) ! d!0 !0 , the above equation becomes d 2 XL XC !0 Ld d 1 405 At this point let s pause to consider what actual values of d we might expect to encounter in practical work. In doing this, it should be noted that the doubletuned circuit of Fig. 232 is especially suited for use as an intermediate frequency (IF) ampli er in AM and FM receivers. In this regard consider, for example, the standard broadcastband FM receiver. Here f0 (the IF) is generally selected to be 10,700 kHz (10.7 MHz), with side bands extending to 100 kHz either side of f0 . Thus in this case the value of d is, by eq. (404),* equal to d 10,800 10,700 =10,700 0:01, approx: This illustrates that the value of d will normally be very much less than 1, and thus in practical engineering work eq. (405) can be written as XL XC 2!0 Ld

