Question

Find all two-component matching networks that provide impedance matching from a 50-ohm source to a load impedance, ZL= 30 - j40 ohm. Find the values of all inductors and capacitors if a perfect match is desired at f₀= 450 MHz.

Answer 1

To achieve impedance matching at 450 MHz from a 50-ohm source to a load impedance of 30 - j40 ohm, a 0.0359 μH inductor and a 7.05 pF capacitor should be used.

Step-by-step explanation:

To find the two-component matching networks that provide impedance matching from a 50-ohm source to a load impedance of 30 - j40 ohm at 450 MHz, we can use the formulas for the reactances of inductors and capacitors. The reactance for an inductor, XL, is given by XL = 2πfL, where f is the frequency and L is the inductance. The reactance for a capacitor, XC, is given by XC = 1/(2πfC), where C is the capacitance. By equating XL and XC to the imaginary part of the load impedance, we can solve for the values of L and C.

For the given load impedance, the imaginary part is -40 ohm, which indicates an inductive reactance. Using the formula XL = 2πfL and plugging in the values, we can solve for L:

XL = 2π * 450e6 * L

-40 = 2π * 450e6 * L

L ≈ -0.0359 μH

Now, using the formula XC = 1/(2πfC), we can solve for C:

XC = 1/(2π * 450e6 * C)

-40 = 1/(2π * 450e6 * C)

C ≈ 7.05 pF.

Therefore, to achieve impedance matching at 450 MHz, a 0.0359 μH inductor and a 7.05 pF capacitor should be used.