ECE 6130 Pi-Matching NetworksPortfolio Question: How do you design a pi-matching network.Reference: Bowick, RF Circuits, pp. 70-71The basic Pi-Network is shown below:This network will be NARROWER BAND than the L-network (will have a higher Q), and the Q will be adjustable.This is designed as a set of Two L-Matching Networks:An artificial impedance R is placed between them. R< Rg,RLChoice of R defines Q.Now we design the TWO L-Networks:QRg RLRORRRg RLQ max( , )max( , )112First find R:Use same example as for single L-network:RL = 100 ohms, Rg = 50 ohmsDefine Q = 15 (given)Then: R = max (100,500 / (152 +1) = 0.44 ohmsNow design Network 2:Xp2 = Rp2/Q = 100 / 15 = 6.67 ohmsXs2 = Q Rs2 = (15) (0.44) = 6.6 ohmsNext design Network 1:Q1 = sqrt (Rp1/Rs1 -1) = sqrt (50 / 0.44 -1) = 10.6Xp1 = Rp1/Q1 = 50 / 10.6 = 4.72 ohmsXs1 = Q1 Rs1 = (10.6) (0.44) = 4.664 ohmsNow define elements to be used in the design:Xs1,Xp1 must be opposite typesXs2,Xp2 must be opposite typesRemember X_capacitor = 1/ CX_inductor = LThis gives four choices: The series elements can be combined:What if the Load has an imaginary part?Either ABSORB it into the matching network If the imaginary part of the load is smaller than the associated value of the matching network,Use the imaginary part of the load as PART of the matching X.OR Resonate it Choose  = sqrt(LC) to define L or C.ADD this value of L or C to the matching network to determine the necessary value of installed component.The T-Network:Design is the same as for the Pi-network, except the artificial impedance at the center of the arms( R) will be less than either Rg or RL:Then design two L-networks:QRRg RL min( , )1Note different placement of Xs and Xp than before.After the Xs1,Xp1,Xs2,Xp2 have been defined,Remove R (it was artificial anyway), combine Xp1 and Xp2.If ZL has an imaginary part, either absorb it into the matching network or resonate