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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

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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

As Figs. 301 and 302 show, Fig. 300 can be considered to consist of two two-ports in series; in Fig. 302 note that N2 denotes the single shunt impedance Z and N1 denotes the transistor.

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From our work in section 11.9 concerning series-connected two-ports, we know that the general matrix representation for Fig. 302 will be equal to the SUM of the matrix representations of N1 and N2 . Thus, since we ll be dealing with the sum of two matrices, and since the matrix representation for N2 is in terms of impedance (eq. (543)), it follows that the matrix representation for N1 (the transistor) must also be expressed in terms of impedance. Therefore, upon making use of eq. (543) and the impedance matrix representation of a transistor (note 30 in Appendix), we have that the matrix equation for Fig. 300 is ! ! ! V1 Z11 Z Z12 Z I1 545 Z21 Z Z22 Z I2 V2 To continue with another example, let us nd the matrix representation for a network consisting of a transistor, in the CE mode, using a collector-to-base feedback impedance of Z ohms (or, if we wish, Y 1=Z mhos), as shown in Fig. 303.

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