Active Network Analysis. Feedback Amplifier Theory by Wai-Kai Chen

By Wai-Kai Chen

Energetic community and suggestions Amplifier thought

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69), but the amount of work involved is considerable (see Prob. 10). Thus, to test for passivity or activity of a device, the choice of an appropriate matrix is as important as the test itself. The reader is urged to compare the amount of labor involved in the above three cases. 4 Let y(t) be the response of an excitation u(t) of an n-port network N. 97) t0 for all initial time t0 and for all excitations u(t). The condition is clearly necessary, because Eq. 10) must be true for all time t and, in particular, for t = ∞, and because u (t)y(t) = v (t)i(t).

Likewise, the unbiased tunnel diode of Fig. 11 is a causal one-port under voltage excitation-current response, and a noncausal one-port under current excitation-voltage response (see Prob. 3). 3 Consider a two-port ideal transformer of turns ratio k:1, whose port voltages and currents are related by the equation (Fig. 35) i 2 (t) i 1 (t) 0 −k If u(t) = [v1 (t), i 2 (t)] is taken as the excitation, then the response y(t) = [i 1 (t), v2 (t)] is uniquely determined through Eq. 35). Consequently, for any two identical excitations we obtain identical responses, meaning that the transformer is causal under the chosen excitation and response.

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. 95) which is hermitian and nonnegative definite. Thus, condition 3 is satisfied. 96) Yh ( j ω) =   1 gm G 2 2 2 , confirming Eq. 90). which is nonnegative definite if and only if 4G 1 G 2 gm 2.

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