qr code vb.net free Fig. 196 in Visual Studio .NET

Maker Code 128 Code Set B in Visual Studio .NET Fig. 196

For convenience in the above, let d 1 h2 ). Then, by inspection, we have tan  B=A h=d = h2 =d 1=h hence,  arctan 1=h or, since h !=!1 and !1 1=RC,  arctan !1 =! arctan 1=!RC 331 or, since !1 1=RC is constant (in any given case), let us, for convenience here, set !1 1. Then eq. (331) becomes  arctan 1=! 332

We have said, correctly, that if a network produces phase shift, then, in order that there be zero phase-shift distortion, the phase shift must be proportional to frequency !. Investigation of eq. (332) shows, however, that phase shift is not proportional to frequency, neither at low frequencies nor at high frequencies. O hand, this would seem to mean that the high-pass lter of Fig. 194 would produce an unacceptable amount of phase distortion, even at the desired higher frequencies. The point to notice, however, is that for high frequencies the phase shift approaches zero degrees, and thus the time delay approaches zero as we go to the higher frequencies (eq. (13-A), note 20, Appendix). Thus the criterion for zero phase-shift distortion is that either the network should produce a negligibly small amount of phase shift or, if a sizeable amount of phase shift is produced, then such phase shift must be proportional to frequency.

In the gures, R is resistance in ohms and L is inductance in henrys, so that XL !L is " the inductive reactance in ohms. We ll take Vi to be the input voltage reference vector and " " write that Vi Vi =08 Vi . The amplitude of Vi is to remain xed in value while we allow its frequency ! to increase slowly from a very low value to a very high value. The vector " " output voltage is denoted by Vo , as shown. As previously, we ll let G denote the ratio of " " output to input voltage, G Vo =Vi . The operation of both circuits depends upon the fact that the value of the reactance !L, increases as the frequency increases, and thus (since Vi is constant) the voltage drop across R decreases as the frequency increases. Thus, in Fig. 197 the higher the frequency the lower " " the value of Vo , while in Fig. 198 the higher the frequency the greater the value of Vo . Note that, in both cases, we have, by Ohm s law, that " I Vi R j!L 333

and upon making use of this fact we nd, for the LOW-PASS case of Fig. 197, that R " jGj q 2 !L 2 R and  arctan !L=R 335 334

and for the HIGH-PASS case of Fig. 198, that !L " jGj q R2 !L 2 and  arctan R=!L Problem 171 Prove that eqs. (334) and (335) are correct. Problem 172 Prove that eqs. (336) and (337) are correct. Problem 173 Show that the half-power frequency in Fig. 197 is R=L radians/second. 337 336

The RC and RL lters of section 9.5 have the advantage of simplicity, but they have the two disadvantages of having high loss of signal power and poor frequency response characteristics. This is illustrated in Fig. 199, which shows the general form of the curve of " the relative magnitude of G versus frequency for the basic RC and RL networks of Figs. " " 190 and 197 in section 9.5. As already de ned, jGj jVo =Vi j, and !0 is the half-power " frequency, that is, the frequency at which the voltage ratio jGj is reduced to 70.7% of its "j is at ! 0). maximum value* (here the maximum value of jG