Question

An amplifier with an output impedance of 75 ? is connected to a filter with a characteristic impedance of 60 – 20j ?. Design a single shunt stub matching network using an open circuit for a frequency of 500 MHz. Show ALL calculations. a) Calculate the normalized load admittance. b) Determine the distance (in ?) to the stub in the direction of the generator. c) Determine the normalized input admittance of the stub. d) Calculate the susceptance of the stub. e) Calculate the length of the short circuit stub. f) Determine the standing wave ratio (SWR) for the normalized admittance point. g) Sketch the layout of the system, indicating the distance to and the length of the stub

Answer 1

Answer :- a) Normalized load impedace = 75 / (60 - 20j) , hence normalized load admittance = 0.8 - 0.267j

b) First we need to draw the constant SWR circle (taking OA as radius) on smith chart as shown below-

Here A is the point of normalized load admittance and S is SWR point i.e. SWR = 1.5. Now see that point B and C are the intersection of SWR and unity circles. In below figure-

At B, y1 = 1 + 0.38j and at C, y2 = 1 - 0.38j. Moving from point A clockwise (towards generator), point B is nearer than point C. The disatnce from A to B = ((0.5 - 0.412) + 0.139)*lambda = 0.227*lambda. Here lambda = 3000/500 = 6 m.

For point B, susceptace of the stub = -0.38j. The point D shows the point on smith chart as shown below-

SC is the short circuit point on the smith chart (admittance form). So the length of stub = distance between SC and point B = (0.442 - 0.25)*lambda = 0.192*lambda.

The system can be draw as -