qr code vb.net free * See note 23 in Appendix. in VS .NET

Printing Code-128 in VS .NET * See note 23 in Appendix.

345

" " You ll note that the last four equations give the value of the CURRENT ratio I1 =I2 . If, however, we wish to work in terms of VOLTAGE ratio, this can be done by noting that " " " " " " " V1 I1 Zin and V2 I2 ZL ; hence for any value of ZL , " " " V1 Zin I1 346 "2 ZL I2 " " V " " " or, for Zin ZL Z0 ; " " V 1 I1 "2 I2 " V 347

Thus, as eq. (347) shows, at the particular frequency for which a symmetrical T network is terminated in its characteristic impedance, the voltage ratio is EQUAL to the current ratio.

" " In this section we study a form of symmetrical T network in which Z1 and Z2 will be given to be PURE REACTANCES of opposite sign, where (using the terminology of Fig. 202) " " Z1 jXL j!L and Z2 jXC 1=j!C " and where the network is terminated in a pure resistance of ZL RL , as shown in Fig. 208.

Note that at VERY LOW frequencies XL is VERY LOW while XC is VERY GREAT, " " " " so that V2 is practically equal to V1 at such frequencies (V2 is actually EQUAL to V1 at ! 0). On the other hand, at very HIGH frequencies XL is VERY GREAT and XC is very " LOW, so that V2 is practically equal to zero at such frequencies. Thus, in just a general way, we see that Fig. 208 constitutes a low-pass type of network. Such a general observation is, of course, not su cient for engineering purposes; to get speci c information let us now apply the algebra of complex numbers to Fig. 208, as follows (using the notation of Fig. 202). From note 23 in the Appendix, " " " " V Z V1 Z2 " I2 1 2 2   "1 " D Z "1 Z2 Z1 Z2 ZL " " " Z 4 2

hence, by Ohm s law,

" " " Now make the substitutions Z1 jXL , Z2 jXC , and ZL RL ; upon doing this, then multiplying the numerator and denominator by j, you should nd that (taking " V1 V1 =08 V1 as reference vector), " X R V2  C L    X X V1 XC L RL jXL XC L 2 4 Now, on the right-hand of the above equation, multiply the numerator and denominator by 1=XC RL ; doing this, then making the substitutions XL !L and XC 1=!C, you should nd that " V2 V1 ! LC 1 2

348

At this point let s pause and try to decide upon a reasonable value for RL in Fig. 208. To do this, let us begin by noting that at ZERO frequency ! 0 Fig. 208 would become

" " Thus, for ! 0 the generator would see a pure resistance of RL ohms, with V2 =V1 1 " "2 and V1 , which would be the desired condition here and with zero phase shift between V because Fig. 208 is to be a low-pass lter passing, as uniformly as possible, all frequencies from ! 0 to whatever the cut-o frequency is to be. With this in mind, and upon setting " " Z1 j!L and Z2 1=j!C in eq. (340), we have that the CHARACTERISTIC IMPEDANCE of the T-network in Fig. 208 is equal to s L !2 L2 " Z0 C 4 349

We have agreed, however, that the generator will see the actual value of RL for ! 0; thus, upon setting ! 0 in eq. 349 we have that " Z0 RL r L C 350

which is the resistance the generator will see only at ! 0, and is thus the actual value of RL that will be used in Fig. 208.