Tile y parameters of “two-port network (Figure 9- t) are derived from the following equations: Note that each term in each equation must represent a current, since the left side of each equation is- a current. Therefore. each of the parameters y Y12, Y210 and must be an admittance; Y parameters are also called admittance parameters. The defining equation Cor each Y parameter is obtained from the network equations (9-56) or (9-57) following a procedure similar to that used to obtain II-parameter definitions. Setting V2 = 0 in equation 9-56 gives is called the short-circuit input (off’ col since it is the admittance at the input terminals when the output is short-circuited. With Vl = O .Yll can be obtained (rom equation 9-57:.”21 is called the short-circuit forward transfer ad». at once. Setting VI = 0 ill equation <)-56 gives )’12 is called the short-circuit reverse transfer admittance. With VI = 0 in equation 9-57, we can solve for Y22 is called the short-circuit output admittance. We see that all of the y-parameter definitions arc obtained by shorting either the input (VI = 0) or output (v~ ::: 0) terminals. For that reason, y parameters arc formally known as short-circuit admittance parameters.Find the real and imaginary parts of conch of the y parameters of the circuit shown in Figure 9-25 when the frequency of operation is 7.l)5~ kHz. Solution”. We must first find the capacities reactant of the 2-pF capacitor (lit 7.958 kHz: Figure 9-26 shows the network with its output short-circuited, as necessary to compute YII and Y21′ As can be seen in the figure. the impedance z at the input terminals when the output is short-circuited is the parallel combination of the active reactant and the resistance:Thus, the real part of designated is 1 X 10-7 S, and the imaginary part, . ), is 1 X to-7 S. Recall that is the conductance component of the admittance, and .f is the acceptance component. (Acceptance is the reciprocal of reactant.) Also recall that the total admittance of components in parallel is the sum of the individual admittance, so the result YII = 10-7 + j 10-7 S could have been written immediately. The most direct way to find is to assume a value for VI in Figure 9-26 and then compute the current Assuming that VI = we find Therefore (Y21) = -10-7 Sand 1m(Y21) = O.Notice that it is possible to have negative admittance values. To find Yl2 and it is necessary to short-circuit the input terminals of the network. as shown in Figure 9-27. Assume that V2 = lLQ: V. Then .Like” parameters. ” parameters are given the letter subscripts i. r, [, and 0 in place of the double-number subscripts 11. 12. 21. and 22. respectively. Following is a summary of the y Norplant~l designations. Which We will use hereafter.
