# Simple Transmission Line Question

#### xz4chx

Joined Sep 17, 2012
71
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

$$\lambda= \frac {\lambda_0}{\sqrt {\epsilon_r}}$$

$$\lambda_0 = c/f$$

$$\lambda_0 = \frac {3*10^8} {1*10^9} = .3 meters$$

$$\lambda = \delta V_{minima} = 6 centimeters * 10^{-2} = .06 meters$$

$$(\frac {\lambda_0}{\lambda})^2 = \epsilon_r$$

$$\epsilon_r = 25$$

$$\epsilon_r = 6.25$$

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.

#### w2aew

Joined Jan 3, 2012
219
Think about this. For a standing wave, is the distance between the maximum and the closest minimum measured in wavelengths. Or, what is the distance in wavelengths between two consecutive nodes (minimums)? Did you factor this into your calculations?

• anhnha

#### KL7AJ

Joined Nov 4, 2008
2,229
Here's how you can check your work, using the Velocity Factor. At 1GHz, the wavelength in free space is 30cm. The distance between two nulls, per your measurement is -34-(-28), which is 6 cm, which is 1/2 wavelength. A full wavelength (as measured) is 12 cm. Your velocity factor is 12cm/30cm, which is .4.

The relative permittivity is the reciprocal of the square of the velocity factor. So 1/.4^2 is indeed 6.25.

Your calculations are correct, I believe your actual measurements are in error. The VF should be on the order of 80-90% for air dielectric.

Eric

• anhnha

#### xz4chx

Joined Sep 17, 2012
71
If there is a voltage minima at -34 [cm] and -28 [cm], i don't understand how that is only .5 wavelengths? There is a maximum between them at -31 [cm]. So if the two minima (through) of the wave are at -34 [cm] and -28 [cm], isn't one wavelength from through to through or in this case minima to minima?

#### xz4chx

Joined Sep 17, 2012
71
For all my smith chart calculations, my teacher uses a wavelength of 12 [cm].

#### KL7AJ

Joined Nov 4, 2008
2,229
If there is a voltage minima at -34 [cm] and -28 [cm], i don't understand how that is only .5 wavelengths? There is a maximum between them at -31 [cm]. So if the two minima (through) of the wave are at -34 [cm] and -28 [cm], isn't one wavelength from through to through or in this case minima to minima?
No...minima to minima (or maximum to maximum is 1/2 wave.

Watch this!

Eric

#### xz4chx

Joined Sep 17, 2012
71
If I have the function sin(x)
Is the minima at 0 or - 1?

#### anhnha

Joined Apr 19, 2012
881
If there is a voltage minima at -34 [cm] and -28 [cm], i don't understand how that is only .5 wavelengths? There is a maximum between them at -31 [cm]. So if the two minima (through) of the wave are at -34 [cm] and -28 [cm], isn't one wavelength from through to through or in this case minima to minima?
I think this may help. #### Attachments

• xz4chx
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