Problem 2.66 A 0-Ω transmission line is to be matched to a computer terminal with Z L = ( j25) Ω by inserting an appropriate reactance in parallel with the line. If f = 800 MHz and ε r = 4, determine the location nearest to the load at which inserting: (a) A capacitor can achieve the required matching, and the value of the capacitor. (b) An inductor can achieve the required matching, and the value of the inductor. Solution: (a) After entering the specified values for Z L and Z 0 into Module 2.6, we have z L represented by the red dot in Fig. P2.66(a), and y L represented by the blue dot. By moving the cursor a distance d = 0.093λ, the blue dot arrives at the intersection point between the SWR circle and the S = 1 circle. At that point y(d) = 26126 j1.54026. To cancel the imaginary part, we need to add a reactive element whose admittance is positive, such as a capacitor. That is: ωc = (1.546) Y 0 = 1.546 = 1.546 = 7.71 3, Z 0 0 which leads to C = 7.71 3 2π 8 8 = 1.53 12 F.
Figure P2.66(a) (b) Repeating the procedure for the second intersection point [Fig. P2.66(b)] leads to y(d) = 00001 + j1.5691, at d2 = 47806λ. To cancel the imaginary part, we add an inductor in parallel such that 1 1.5691 =, ωl 0 from which we obtain L= 0 = 2.618 8 H. 1.52 2π 8 8
Figure P2.66(b)
0.9 7 3 Problem 2.68 A -Ω lossless line is to be matched to an antenna with Z L = (75 j) Ω using a shorted stub. Use the Smith chart to determine the stub length and distance between the antenna and stub. 9 8 9 8 170 ± 180-170 7 0.04 > WAVELENGTHS TOWARD GENERATOR > 0.0 0.0 < WAVELENGTHS TOWARD LOAD < 7 160-160 6 6 0.04 0.05 5 1-1 5 0.05 0.9 4 0.06-140 140 0.06 4 3 0.07 0.07 3-130 130 0.08 2 2 0.08 1-1 04 λ 0.09 1 INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) 0.09 1-1 Y-STUB-IN-1 1 1 9 1 9 0-0 0.9 2 8 Y-LOAD 2 8 90-90 3 7-80 80 4 6 4 5 5 5 70 Y-LOAD-IN-1 6-70 Z-LOAD 5 6 4 6 4 60-60 7 3 3 7 73 λ - 8 2 8 2 1 40 9-40 9 1 30-30 9 1 1-9 8 ANGLE OF REFLECTION COEFFICIENT IN DEGREES 2 2 7 3 8 4 6 3 Y-SHT 5 5 6 4 7 Figure P2.68: (a) First solution to Problem 2.68. Solution: Refer to Fig. P2.68(a) and Fig. P2.68(b), which represent two different solutions. z L = Z L (75 j) Ω = = 1.5 j Z 0 Ω and is located at point Z-LOAD in both figures. Since it is advantageous to work in admittance coordinates, y L is plotted as point Y -LOAD in both figures. Y -LOAD is at 0.041λ on the WTG scale.
0.9 7 3 For the first solution in Fig. P2.68(a), point Y -LOAD-IN-1 represents the point at which g = 1 on the SWR circle of the load. Y -LOAD-IN-1 is at 45λ on the WTG scale, so the stub should be located at 45λ 0.041λ = 04λ from the load (or some multiple of a half wavelength further). At Y -LOAD-IN-1, b = 2, so a stub with an input admittance of y stub = 0 j2 is required. This point is Y -STUB-IN-1 and is at 23λ on the WTG scale. The short circuit admittance is denoted by point Y -SHT, located at λ. Therefore, the short stub must be 23λ λ = 73λ long (or some multiple of a half wavelength longer). 9 8 9 8 170 ± 180-170 7 0.04 > WAVELENGTHS TOWARD GENERATOR > 0.0 0.0 < WAVELENGTHS TOWARD LOAD < 7 160-160 6 6 0.04 0.05 5 1-1 5 0.05 0.9 4 0.06-140 140 0.06 4 3 0.07 0.07 3-130 130 0.08 2 2 0.08 1-1 0.09 1 0.09 1-1 Y-STUB-IN-2 INDUCTIVE REACTANCE COMPONENT (+jx/zo), OR CAPACITIVE SUSCEPTANCE (+jb/yo) RESISTANCE COMPONENT (R/Zo), OR CONDUCTANCE COMPONENT (G/Yo) CAPACITIVE REACTANCE COMPONENT (-jx/zo), OR INDUCTIVE SUSCEPTANCE (-jb/yo) 27 λ 1 1 9 1 9 0-0 0.9 2 8 Y-LOAD 2 8 90-90 3 7-80 80 4 6 4 6 5 5 5 70-70 Z-LOAD Y-LOAD-IN-2 5 6 4 4 6 60-60 7 3 3 7-8 2 8 2 1 40 9-40 9 1 14 λ 30-30 9 1 1-9 8 ANGLE OF REFLECTION COEFFICIENT IN DEGREES 2 2 7 3 8 4 6 3 Y-SHT 5 5 6 4 7 Figure P2.68: (b) Second solution to Problem 2.68. For the second solution in Fig. P2.68(b), point Y -LOAD-IN-2 represents the point at which g = 1 on the SWR circle of the load. Y -LOAD-IN-2 is at 55λ on the WTG scale, so the stub should be located at 55λ 0.041λ = 14λ from the
load (or some multiple of a half wavelength further). At Y -LOAD-IN-2, b = 2, so a stub with an input admittance of y stub = 0 + j2 is required. This point is Y -STUB-IN-2 and is at 0.077λ on the WTG scale. The short circuit admittance is denoted by point Y -SHT, located at λ. Therefore, the short stub must be 0.077λ λ + 0.0λ = 27λ long (or some multiple of a half wavelength longer).
Problem 3.12 Two lines in the x y plane are described by the expressions: Line 1 x+2y = 6, Line 2 3x+4y = 8. Use vector algebra to find the smaller angle between the lines at their intersection point. 30 25 15 (0, 2) -35-30 -25 - -15 - (0, -3) - -15 A 15 25 30 35 B (, -13) - -25-30 θ AB Figure P3.12: Lines 1 and 2. Solution: Intersection point is found by solving the two equations simultaneously: 2x 4y = 12, 3x+4y = 8. The sum gives x =, which, when used in the first equation, gives y = 13. Hence, intersection point is (, 13). Another point on line 1 is x = 0, y = 3. Vector A from (0, 3) to (, 13) is A = ˆx()+ŷ( 13+3) = ˆx ŷ, A = 2 + 2 = 0. A point on line 2 is x = 0, y = 2. Vector B from (0,2) to (, 13) is B = ˆx()+ŷ( 13 2) = ˆx ŷ15,
B = 2 + 15 2 = 625. Angle between A and B is ( ) ( ) A B 400+1 θ AB = cos 1 = cos 1 = 1. A B 0 625