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6 Pages

### wilkinson

Course: EE 5347, Fall 2008
School: UT Arlington
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Word Count: 842

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Wilkinson The Power Divider Reference: E. J. Wilkinson, An N-Way Hybrid Power Divider, IRE Trans. on Microwave Theory and Techniques, Vol. MTT-8, pp. 116 - 118, Jan. 1960. The divider must isolate the output lines and be matched. Isolation is shown by putting a voltage on one of the output lines and see where the signal goes. Put an applied voltage of value V on spline # 1 s and see if isolation occurs. In spline...

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Wilkinson The Power Divider Reference: E. J. Wilkinson, An N-Way Hybrid Power Divider, IRE Trans. on Microwave Theory and Techniques, Vol. MTT-8, pp. 116 - 118, Jan. 1960. The divider must isolate the output lines and be matched. Isolation is shown by putting a voltage on one of the output lines and see where the signal goes. Put an applied voltage of value V on spline # 1 s and see if isolation occurs. In spline 1 shown in the lower diagram, V1 = V + ej + V ej Note the reverse of the usual sign convention for the exponentials because the right hand 1 side is at z = 0 and the left hand side is at z = . At = 0, Va = V + + V I1a = I + I = V+V Zo then I1a = V + j V j e e Zo Zo This voltage expression is converted into sine and cosine form. V1 = V + (cos + j sin ) + V (cos j sin ) V1 = cos (V + + V ) + j sin (V + V ) V1 = Va cos + jI1a Zo sin V1 = jI1a Zo When = /2, (1) The expression for the current, I1a is also put in the form of sines and cosines. I1a = V+ V (cos + j sin ) (cos j sin ) Zo Zo V++V V+V + j sin = cos Zo Zo Va = I1a cos + j sin Zo Va Zo (2) where for = /2 I1a = j For any of the other splines, say the nth one, the following diagram would be appropriated. 2 By analogy with the previous analysis, Va = Vn cos + jIna Zo sin Va = jZo Ina Similarly Ina = Ina cos + j where for = /2 (3) Vn sin Zo (4) where for = /2 Ina = j Vn Zo (4) By symmetry, all the rest of the n 1 currents are the same. At node a the sum of the currents is zero at this input. I1a = (n 1)Ina + Va Ro (5) At the nth output the current can be found by symmetry. Inb = I1b n1 (6) Therefore the node equation at the nth output is Ina = I1b Vn Ro n1 At the rst output, the partial circuit is given below. (I1a + I1b )Ro = V V1 The voltage drop through the node from output 1 to output n is: I1b + so that from (6) I1b n1 Rx = V1 Vn (7) (8) 3 Ina = and from (3) I1b Vn Ro n1 Vn I1b Va = jZo Ro n1 Vn jVa I1b = (n 1) + Ro Zo Equation (8) is put in the form n Rx = V1 Vn I1b n1 and from (8) jVa n Vn + Rx V1 Vn = (n 1) Ro Zo n1 nRx Vn jVa nRx + V1 Vn = Zo Ro jVa Ro Ro Ro 0= + Vn + 1 V1 Zo nRx nRx Then from V (7) = I1b Ro + Ro I1a + V1 From (9) V1 Vn = I1b or I1b = n n1 Rx (9) (10) V1 Vn (n 1) nRx This can be put in for I1b above and using (2) to replace I1a , so that V = Ro or V = V1 1 + n1 Ro Vn nRx n1 nRx Ro + jVa Ro Zo (11) V1 Vn (n 1) + RO nRx jVa Zo + V1 For perfect isolation, Vn = 0. In this case (10) becomes, 0= and (11) becomes V = V1 1 + Now substitute for jVa Ro /Zo 4 n1 Va Ro Ro + j nRx Zo V1 Ro jVa Ro Zo nRx Ro n1 Ro + nRx nRx Ro V = V1 1 + Rx 1 V1 = V 1 + Ro /Rx V = V1 1 + Use (4) to replace Ina in ((5) so that (12) V1 Va jVn = + (n 1) jZo Ro Zo V1 Ro Ro +J 0 = Va + j(n 1)Vn Zo Zo For Vn = 0, (13) becomes jVa = V1 and (10) becomes Zo nRx so that by equating these two equations Zo Ro = V1 V1 Zo nRx 2 Ro Ro = 2 Zo nRx jVa = V1 Ro Zo (13) (14) For a matched circuit, the load impedance must match the source resistance Ro . The load is carrying I1a + I1b through a voltage V1 / Making use of (2), the matched load must be Zmatch = From (7) = V1 I1a + I1b V1 Ro V V1 = Ro This implies that V1 = V V1 V = 2V1 Then from (12) 1 1 = 2 1 + Ro /Rx or 5 Ro = Rx and from (14) 2 1 Ro = 2 Zo n (15) whihc gives Zo = nRo (16) Now that Zo is known for each of the lines, the input impedance at node a is found. There are n input lines, each of which is in parallel. There...

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