Worked example 7.7.1: Feeding a 2.45 GHz patch antenna

${\lambda }_0=c/f=122.4\mathrm{mm}$ at $2.45\mathrm{GHz}$.

$Z_T=\sqrt{50\times 200}=100\Omega$. From Problem 7.7.1: $w=3.06\mathrm{mm}$ for $50\Omega$ and $w=0.71\mathrm{mm}$ for $100\Omega$.

The $100\Omega$ line has ${\epsilon }_{\mathrm{eff}}=3.042$, so ${\lambda }_g=122.4/1.744=70.2\mathrm{mm}$ and

With the ${\epsilon }_{\mathrm{eff}}$ of the $50\Omega$ line one would get $16.8\mathrm{mm}$, $4\%$ too short.

$\beta \ell =90^{\circ }\times f/(2.45\mathrm{GHz})$ (dispersion neglected): $88.2^{\circ }$ at $2.40\mathrm{GHz}$ and $91.8^{\circ }$ at $2.50\mathrm{GHz}$. Then

a return loss of $32\mathrm{dB}$ at both band edges.

At $2.45\mathrm{GHz}$ the $50\Omega$ line loses ${\alpha }_c+{\alpha }_d=0.73+7.37\approx 8.1dB/m$, so $40\mathrm{mm}$ cost $0.32\mathrm{dB}$. The $100\Omega$ transformer loses $1.59+6.75=8.34dB/m$, over $17.6\mathrm{mm}$ $0.15\mathrm{dB}$. In total $0.47\mathrm{dB}$, so $1-10^{-0.047}=10\%$ of the power heats the board.