Single parallel stub matching at an antenna the smith an antenna operates at a wavelength of 2 m and is designed with an impedance of 75 n. however, because of mistakes in design, the antenna is badly mismatched. the measured impedance after installation is 15 +j60 . the antenna is connected to a 75 line as shown in figure 15.14. calculaie (a) the required shorted stub and its location on the line to match the antenna to the line. the line and stub have the same characieristic impedance (b) the shoresi required open circuit stub that will accomplish the same purpose as the shor circuit stub in (a). figure 15.14 mismatched antenna connected to a line and a stub designed to match z75 z 15+j60 z 75 the antenna to the line solution: first, we find a location on the line at which the real part of the line admitance is equal to the characteristic admitance of the line; that is, find zd) such that y (d) = y jb (d). now, we connect a shorted stub in parallel with the line at this point and of a length such that the imaginary part of the line admittance is canceled. the open circuit stub in (b) is placed at the same location and its length is that of the shor circuit stub 4 (a) in this case, it is simpler to use the smith chart as an admittance char. to do so, we first calculate the normalized load impedance 15+60 = 0.2 +08. 75 (1) we mark this point as p an the smith chart in figure 15.15, using the chart as an impedance chari. the reflection coefficient circle is now drawn around the center of the chart, with the radius equal to the disance between p and p (2) to find the load admittance, we draw a straight line from p through p and extend this line to the periphery of the char. the line intersects the reflection coefficient circle at point p. this point is the normalized load admittance y 0.294-. 76 (3) as we move around the reflection coefficient cirele, the line admittance changes. to match the load, we must find the location at which the real part of the line admittance equals the characteristic admittance. since we are working with normalized admittances, this happens when rely 1. this happens a the locations al which the reflection coefficient circle intersects the circle g= 1. the two possible points are p and ps. the line admittance at these points is at p y1+2.53 at ps 1-/2.53 each one of these points provides one possible solution exercise 15.1 suppose that in example 15.3, part (a), it is not physically possible location found. the nearest location at which a stub may be connected is 1 m from the load: to connect the stub at either (a) what are the solutions for di and ? (b) are these solutions unique?