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Creator USS Code 128 in .NET framework Reactance and Impedance

CHAPTER 8 Reactance and Impedance
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eq. (254)). Thus, upon multiplying both sides of eq. (251) by 1=R0 , we have that   R !C L j !2 L2 R2 " Zp R0 R0 C 2 2 2 2 R0 R ! C ! LC 1 2
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Actually, the above equation can be expressed in much better form if certain practical approximations are made. To show how this can be done, let us rst denote the ratio of the coil reactance at resonance to its resistance by Q ; thus !0 L Q R Next square both sides of eq. (252), thus getting !2 0 But note that   R2 !2 R2 R 2 0 2 !2 !2 =Q2 0 0 !0 L L2 !2 L 0 and upon making this substitution into eq. (257) we have that   1 1 2 !0 1 2 LC Q R2 1 2 LC L 257 256
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At this point it should be noted that all of the foregoing equations concerning Fig. 167, are exact equations; that is, no simplifying assumptions have been made. Let us now, however, take into account the fact that in most practical applications of Fig. 167 the value of Q, as de ned by eq. (256), will be EQUAL TO OR GREATER THAN 10; that is, in most practical work it will be true that Q ! 10. Let us therefore base the rest of our discussion of Fig. 167 on the assumption that Q will be equal to or greater than 10. Thus, for practical purposes we can write that 1 1 1 Q2
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and hence, for practical purposes, eq. (258) becomes !2 0 thus !0 L 1=!0 C 260 1 LC 259
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showing that, for practical purposes, in Fig. 167 it can be taken that XL XC at the same " frequency, !0 , that Zp R0 . Our goal, now, is to express eq. (255) in dimensionless form (similar to eqs. (246) and (248) in section 8.6), making use of the assumption that 1 1=Q2 1. To do this requires a certain amount of trial and error; let us suppose, after a few trials, we try writing the denominator of eq. (255) in the form R2 !2 !=!0 2 C 2 !=!0 2 !2 LC 1 2 0 0 261
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CHAPTER 8 Reactance and Impedance
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Now let (see eq. (247) in section 8.6) d ! !0 262
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Using this notation, and also noting, by eq. (259), that !2 LC 1, eq. (261) becomes 0 R2 d 2 !0 C 2 d 2 1 2 263 Next, by eq. (260), !0 C 1=!0 L, and using this relationship, and also the de nition of eq. (256), we can now write eq. (255) with a new, dimensionless denominator; thus   R !C L 2 2 2 j ! L R " Zp R0 R0 C 264 2 R0 d 2 2 d 1 Q2 Now let us work on the numerator on the right-hand side of the above equation, as follows. First, making use of eqs. (254), (260), and (256), we have that   R R2 C R2 !0 C R 2 1 2 265 R0 L !0 L !0 L Q Next note that the quantity inside the parentheses in the imaginary part of eq. (264) can be written as   L !2 !2 L2 !0 L !2 L2 0 2 2 2 ! L R 02 C !0 C !2 Q 0 !0 L 2 d 2 1 1=Q2 266 where we used the relationships !0 L 1=!0 C and R !0 L=Q. Thus, substituting the results of eqs. (265) and (266) into eq. (264), we have that   1 j !C !0 L 2 1 2 d 1 2 " Zp Q2 R0 Q 267 R0 d2 2 2 d 1 Q2 Now, for the last step (remembering that !0 L 1=!0 C), note that !C !0 L 2 !C !0 L 2 RC ! !0 C !0 L 2 R !0 C L !0 !0 L R0 ! R d !0 !0 L Q
and upon making this substitution into eq. (267), then multiplying the numerator and denominator by Q2 , we get the nal desired result " Zp 1 jdQ d 2 1 1=Q2 R0 d 2 Q2 d 2 1 2 also, therefore, "  Z p    R  0 q 1 d 2 Q2 d 2 1 1=Q2 2 d 2 Q2 d 2 1 2 268
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