Matching Examples - ~, 9 .. , n \Viouhciru‘vxc‘s...

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Unformatted text preview: ~, 9 .. , n \Viouhciru‘vxc‘s LTXOJMQKCS \‘lreth KY‘CLvS Oct/U1 i‘ietcci’t 154 CHAPTER} TRANSMISSION LINES Example 3-18. 4-patch antenna array. An array of four patch antennas is mounted on a a. = 4 substrate. The patches are in a linear array with l/\ spacing between them as in Fig. 3-21. Each patch has a 50-0 terminal resistance. The four patches are to be fed in-phase from a lO-GHZ transmitter via a single 50-0 microstrip line. Design the feed system. Solution. To feed all four patches in-phase requires that the line lengths be identical from each patch to the feed line connection. A popular method is with a “corporate feed." The term alludes to a business organization with a president, two vice presidents. and four assistant vice presidents. Here the patches are the assistant vice presidents and the president is at the feed point C. First. we connect the left-hand pair ofpatches in parallel at pointA and the right- hand pair at point B. Two 50-!) lines in parallel combine as 25 Q atA and B. With M4 sections this is transformed to 50 Q on the lines to C. At C the same procedure could be repeated but instead a /\/4 section transforms each 50-!) line to 100 Q. These then combine to 50 Q on the line to the transmitter. We still need to calculate the resistance of the M4 sections and specify the phys- ical dimensions of the feed network. From (20), when [3x = M4, Z(/\/4) = \/R(in)R(0ut) = J25 X 50 = 35.4 D atA and B Ans. Z(A/4) = J50 x 100 = 70.7 n at C Ans. At 10 GHz, c 3 X 108 ms‘1 A0 _ ‘ m — 30 mm (free space) and A0 _ I — 7.5 mm (free space) Patch antennas F— 20 = 30 mm —>l To transmitter FIGL‘RE 3-21 Four-patch antenna array. 3-5 BAND’WIDTH 155 Assuming that all of the energy is transmitted in the substrate (which it is not), the length of the M4 sections should be about A0 __ 7.5mm 4 VIE = 3.75 mm Ans. The length of the y sections is arbitrarily made )to/4 = 7.5 mm. For simplicity, we have neglected mutual impedances or the effect of one patch on the impedences of the others. Problem 3-4-14. Design a feed system for the four-patch array of Example 3-18 with feed lines from all four patches connected directly to the main feed point C. This is a “Deming feed” alluding to a Deming-type organization with direct lines of com- munication from bottom to top. To keep all lines equal in length this will require zigzagging of the lines from the center two patches. Match in two ways: (a) with M4 sections in the four patch lines and (b) with a M4 section in the line to the transmitter. Ans. (a) Z(/\/4) = 100 D; (b) Z()t/4) = 25 Q. I;)Qubi€— Stub Tuning \: x amp ie Ifa transmission line has two wires as in Fig. 3—18, Example 3— 14. it is easy to move the stub along the line to the proper position. But if the line is a ceaxial cable, it requires a special telescoping section called a “line stretcher” to do this. It is often more. convenient to have two stub lines attached at fixed positions and achieve a match by adjusting the lengths of the two stubs. Such double—stub tuners are very common. See Figure 3—20 on page 152. Example 3-16. Double-stub matching. A 50-!) coaxial line is terminated in a load impedance ZL = 100 + jSO Q with two stub tuners at MS spacings as shown in Fig. 3-20. Each stub is “tuned” with a sliding short—circuit plug. All lines and stubs are 50 .0. Find: (a) length an of stub l, (b) length d; of stub 2, (c) VSWR on )t/8 line between the load and stub ], (d) VSWR on /\/8 line between the stubs, (e) VSWR on stub l, and (f) VSWR on stub 2. Solution Step I: Normalize the load impedance by dividing by the line impedance. Thus, 100 + j50 _ =———-———-—2+' Z” 50 11 and enter the chart at this value (point A). Step 2: Draw a VSWR = 2.4 circle [Ans. to (c)] through pointA and go halfway around the chart to point B, converting Z,, to the admittance Y,, = 0.4 - j0.2. We are still at the load. FIGURE 3-21) Double- Step7 StUb (11: 0.1951 ——-Ans. (a) tuner. Step 3: We now move away from the load M8 to the first stub which is at point C on the chart. To reach the center of the chart from C, we need to think ahead. Thus. let us rotate the G,. = l circle (shown dashed) IVS (90°) toward the load or to the “up” position as shown in Fig. 3-20 because we want to be on this circle when we arrive at stub 2. To do this, we move on a constant 0,, circle to point D on the chart. 7 Step 4: To go from C(Yn = 0.5 + j0.5) to point D(Y,, = 0.5 + jO. 14) requires that stub 1 present a susceptance 8,, = —O.5 + 0.14 = ~O.36 to the main line. The total admittance at point D is then Y” = 0.5 + jO.5 — 10.36 = 0.5 + j0.l4 Points C and D are both at stub 1. Step 5: Now we move A/8 from stub 1 to stub 2, or from point D on the con— stant VSWR = 2.0 circle [Ans to (d)]. to point E on the G,, = l circle where the line admittance is Y,, = 1+ j0.73 Step 6: To achieve a match, the length of stub 2 is adjusted so as to present. a susceptance of — jO.73 to the main line taking us from E to the center of the chart (Home) where Y,, ——- 1 + jO. When G,l = l, R” = 1 and the impedance presented to the main line at stub 2 is equal to -R,, x R0 = 1 X 50 = 50 Q for a match. Step 7: Reading off the stub lengths along the edge of the chart by moving away, or clockwise, from the short~circuit plug (G,1 = I) to — j0.36 for stub l and to — jO.73 for stub 2, we obtain (11 = 0.195/\ [An5. to (a)] and d3 = 0.15/\ [Ans to (17)]. Also VSWR = x [Ans to (e) and (f)]. ...
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Matching Examples - ~, 9 .. , n \Viouhciru‘vxc‘s...

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