For impedance matching between a 50 Ω source and a 25 Ω load at a single frequency, which approach is correct and what Z0 is required for a quarter-wave transformer?

For impedance matching between a 50 Ω source and a 25 Ω load at a single frequency, which approach is correct and what Z0 is required for a quarter-wave transformer?

• A. Quarter-wave transformer with Z0 ≈ 35 Ω; length λ/4
• B. L-section transformer with Z0 = 50 Ω
• C. Stub tuner with Z0 = 25 Ω
• D. RC network with Z0 = 12.5 Ω

Answer: (A)

Impedance matching at a single frequency can be done with a quarter-wave transformer, which uses a short section of transmission line whose length is exactly one-quarter of a wavelength. The key relation is that the input impedance seen into a λ/4 line terminated by the load is Z_in = Z0^2 / Z_L. To match a source of 50 Ω to a load of 25 Ω, you set Z_in equal to 50 Ω. So solve Z0^2 / 25 = 50, giving Z0^2 = 1250 and Z0 ≈ 35.4 Ω. Therefore, a quarter-wave transformer with a characteristic impedance of about 35 Ω and a length of λ/4 will transform the 25 Ω load to appear as 50 Ω to the source at that frequency. This approach is the cleanest single-frequency solution because it provides a direct, lossless impedance transformation without extra reactive tuning. The other approaches wouldn’t yield the exact 50 Ω match at that frequency with a single, passive element: they either require different impedance values, additional components, or would introduce reactance or frequency-dependent behavior that prevents a perfect match at the designated frequency.

Impedance matching at a single frequency can be done with a quarter-wave transformer, which uses a short section of transmission line whose length is exactly one-quarter of a wavelength. The key relation is that the input impedance seen into a λ/4 line terminated by the load is Z_in = Z0^2 / Z_L. To match a source of 50 Ω to a load of 25 Ω, you set Z_in equal to 50 Ω.

So solve Z0^2 / 25 = 50, giving Z0^2 = 1250 and Z0 ≈ 35.4 Ω. Therefore, a quarter-wave transformer with a characteristic impedance of about 35 Ω and a length of λ/4 will transform the 25 Ω load to appear as 50 Ω to the source at that frequency. This approach is the cleanest single-frequency solution because it provides a direct, lossless impedance transformation without extra reactive tuning.

The other approaches wouldn’t yield the exact 50 Ω match at that frequency with a single, passive element: they either require different impedance values, additional components, or would introduce reactance or frequency-dependent behavior that prevents a perfect match at the designated frequency.