By Wai-Kai Chen

Lively community and suggestions Amplifier conception

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If 2 , Eq. 94) can be violated for any value of ω on the real-frequency 4G 1 G 2 < gm axis, indicating that Hh ( j ω) is not nonnegative definite and H(s) is not positive real. 3, it is suggested that the reader use the definition to verify the above results. 6. The first two conditions are clearly satisfied by the matrix Y(s) of Eq. 86). To verify condition 3, we compute the associated residue matrix. The admittance matrix Y(s) has a pole at the infinity, which is customarily considered to be on September 1, 2016 10:33 Active Network Analysis: Feedback …– 9in x 6in b2428-ch01 page 33 CHARACTERIZATIONS OF NETWORKS 33 the real-frequency axis.

Likewise, we compute h 12 (s) and h 22 (s). The resulting hybrid matrix is given by (see Prob. 67) we let u(s) ˜ = [I1 (s), I2 (s)] and y˜ (s) = [V1 (s), V2 (s)] From Eq. 68b) which are obtained from the network of Fig. 17 by open-circuiting the output port. In a similar manner, we can compute z 12 (s) and z 22 (s). 70) we let u(s) ˜ = [V1 (s), V2 (s)] and y˜ (s) = [I1 (s), I2 (s)] . From Eq. 71b) September 1, 2016 10:33 Active Network Analysis: Feedback …– 9in x 6in b2428-ch01 page 26 26 ACTIVE NETWORK ANALYSIS They are obtained by short-circuiting the output port of the two-port network of Fig.

94) can be violated for any value of ω on the real-frequency 4G 1 G 2 < gm axis, indicating that Hh ( j ω) is not nonnegative definite and H(s) is not positive real. 3, it is suggested that the reader use the definition to verify the above results. 6. The first two conditions are clearly satisfied by the matrix Y(s) of Eq. 86). To verify condition 3, we compute the associated residue matrix. The admittance matrix Y(s) has a pole at the infinity, which is customarily considered to be on September 1, 2016 10:33 Active Network Analysis: Feedback …– 9in x 6in b2428-ch01 page 33 CHARACTERIZATIONS OF NETWORKS 33 the real-frequency axis.

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