Question

A series RC load with R = 10 ohms and C = 2.75 pF is to be matched to a 50 ohm transmission line at a frequency of 5.8 GHz.

What is the real part of the load impedance? Type your answer in ohms to one place after the decimal.
What is the imaginary part of the load impedance? Express your answer in ohms to one place after the decimal point. If the answer is negative, then include the minus sign.
What is the real part of the load admittance? answer in millisiemens to one place after the decimal point.

What is the imaginary part of the load admittance? answer in millisiemens to one place after the decimal point. If the answer is negative, then include the minus sign.

What is the SWR on the line if the load is attached directly to the 50 ohm line with no matching network? answer as a dimensionless quantity to one place after the decimal.

Answer 1

Real part of load impedance: 10.0 Ω

Imaginary part of load impedance: -j16.5 Ω

Real part of load admittance: 0.10 mS

Imaginary part of load admittance: -j0.06 mS

SWR: 1.6

Step-by-step explanation:

Calculate the load impedance:

First, we need to calculate the reactance of the capacitor:

Xc = 1/(2πfC) = 1/(2π * 5.8 GHz * 2.75 pF) = -j16.5 Ω

Then, we can combine the resistor and capacitor in series to find the total load impedance Z:

Z = R + Xc = 10 Ω - j16.5 Ω

Real and imaginary parts of the impedance:

The real part of the impedance is 10.0 Ω, and the imaginary part is -16.5 Ω.

Load admittance:

Admittance (Y) is the inverse of impedance:

Y = 1/Z = 0.10 + j0.06 mS

SWR:

SWR (Standing Wave Ratio) is the ratio of the maximum voltage to the minimum voltage on the transmission line. When the line is perfectly matched (Z = 50 Ω), the SWR is 1. In this case, the mismatch between the load impedance and the transmission line will cause an SWR greater than 1:

SWR = |(Z - Z₀) / (Z + Z₀)| = |(10 Ω - j16.5 Ω - 50 Ω) / (10 Ω - j16.5 Ω + 50 Ω)| = 1.6