A measurement of a transmission line at 1 [GHz] has voltage maximum at z= -31 [cm] with a magnitude of 1 [V]. The closest voltage minima (the minima that are the closest to the indicated voltage maximum) are at z = -34 [cm] and -28 [cm] with a magnitude of 0 [V]. Characteristic impedance is 50 [Ω] and place at z=0.
a) What is the relative permittivity of the line?
Formulas that could help
\( \lambda= \frac {\lambda_0}{\sqrt {\epsilon_r}} \)
\( \lambda_0 = c/f \)
\( \lambda_0 = \frac {3*10^8} {1*10^9} = .3 meters \)
\( \lambda = \delta V_{minima} = 6 centimeters * 10^{-2} = .06 meters \)
\( (\frac {\lambda_0}{\lambda})^2 = \epsilon_r \)
\( \epsilon_r = 25 \)
My solutions answer is:
\( \epsilon_r = 6.25 \)
which means the ratio of \( \frac {\lambda_0}{\lambda} \) needs to be half of what i got, and i don't understand what i did wrong.
a) What is the relative permittivity of the line?
Formulas that could help
\( \lambda= \frac {\lambda_0}{\sqrt {\epsilon_r}} \)
\( \lambda_0 = c/f \)
\( \lambda_0 = \frac {3*10^8} {1*10^9} = .3 meters \)
\( \lambda = \delta V_{minima} = 6 centimeters * 10^{-2} = .06 meters \)
\( (\frac {\lambda_0}{\lambda})^2 = \epsilon_r \)
\( \epsilon_r = 25 \)
My solutions answer is:
\( \epsilon_r = 6.25 \)
which means the ratio of \( \frac {\lambda_0}{\lambda} \) needs to be half of what i got, and i don't understand what i did wrong.