2-Port Quarter Wave Transmission Line Impedance Matching

Question

Design a quarter wave transformer to match a load impedance of 25 Ohms at a frequency of 2.4 GHz.

Assume a 50 Ohm transmission line with a permittivity of 2.7.

Neatly sketch the transmission line solution (provide all lengths in m).

Plot Zin (magnitude and phase) from DC to 5 X the design frequency.

Plot the s-parameters: S11 and S21 in dB from DC to 5 X the design frequency (lossless networks).

Discuss the quarter-wave transformer to match the purely real load across the frequency band of step 1.2 above. How do you know it's matched?

Answer 1

The characteristic impedance of the quarter-wave transformer, ZTcap Z sub cap T

𝑍𝑇

, is calculated using the formula:

ZT=Z0ZLcap Z sub cap T equals the square root of cap Z sub 0 cap Z sub cap L end-root

𝑍𝑇=𝑍0𝑍𝐿√

, where

Z0cap Z sub 0

𝑍0

is the characteristic impedance of the transmission line and

ZLcap Z sub cap L

𝑍𝐿

is the load impedance.

In this case, ZT=50 Ω×25 Ω=1250 Ω2≈35.36 Ωcap Z sub cap T equals the square root of 50 cap omega cross 25 cap omega end-root equals the square root of 1250 cap omega squared end-root is approximately equal to 35.36 cap omega

𝑍𝑇=50 Ω×25 Ω√=1250 Ω2√≈35.36 Ω

.The wavelength, λlambda

𝜆

, in the transmission line is determined by the speed of light in the medium and the design frequency.

The speed of light in the medium, vpv sub p

𝑣𝑝

, is calculated as vp=cϵrv sub p equals the fraction with numerator c and denominator the square root of epsilon sub r end-root end-fraction

𝑣𝑝=𝑐𝜖𝑟√

, where

cc

𝑐

is the speed of light in vacuum (

3×108 m/s3 cross 10 to the eighth power m/s

3×108 m/s

) and

ϵrepsilon sub r

𝜖𝑟

is the relative permittivity.

Therefore, vp=3×108 m/s2.7≈1.82×108 m/sv sub p equals the fraction with numerator 3 cross 10 to the eighth power m/s and denominator the square root of 2.7 end-root end-fraction is approximately equal to 1.82 cross 10 to the eighth power m/s

𝑣𝑝=3×108 m/s2.7√≈1.82×108 m/s

.

The wavelength, λlambda

𝜆

, is then calculated as λ=vpflambda equals the fraction with numerator v sub p and denominator f end-fraction

𝜆=𝑣𝑝𝑓

, where

ff

𝑓

is the design frequency (

2.4 GHz2.4 GHz

2.4 GHz

).

So, λ=1.82×108 m/s2.4×109 Hz≈0.0758 mlambda equals the fraction with numerator 1.82 cross 10 to the eighth power m/s and denominator 2.4 cross 10 to the nineth power Hz end-fraction is approximately equal to 0.0758 m

𝜆=1.82×108 m/s2.4×109 Hz≈0.0758 m

.

The length of the quarter-wave transformer, Lcap L

𝐿

, is λ4the fraction with numerator lambda and denominator 4 end-fraction

𝜆4

.

Thus, L=0.0758 m4≈0.01895 mcap L equals the fraction with numerator 0.0758 m and denominator 4 end-fraction is approximately equal to 0.01895 m

𝐿=0.0758 m4≈0.01895 m

.