# Wavelength and 300/f what is the deference ?

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#### votientu

##### Member level 2
Hi everyone, I have a question about wavelength on transmission line,
According to Microwave Engineering Book of Pozar, $\lambda$ = 2*pi / beta (2.10)
And beta is calculated from R,L,G,C on the transmission line.

Elec.length = 90 means this is mode quater wavelength, so it means wavelength is 8*4=32 mm, if I change the frequency to 2.4, I give the wavelength is 17*4=68 mm.

So, my question is if $\lambda$ is dependent on beta, then what is the frequency for ?

The frequency is a measure of oscillating charges; it is defined by signal source.
Wave length is a parameter of electromagnetic wave with a given frequency, measured along the line of propagation. If the wave propagates in air or vacuum, the equation c = f. lambda is valid , where c is the speed of light in vacuum.
If you propagate the wave in another medium or in a transmission line, c is different by that medium (permittivity) and/or by the structure of the transmission line.

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### votientu

Points: 2
From the real model of transmission line it compose of inductor, capacitor whose Reactance changes with frequency which effects lambda..

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### votientu

Points: 2
Oki, I get it. Thank all of you !

300/f is a crude approximation to the real equation λ = C/f. C is the speed of light in whatever medium you happen to be in.

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### votientu

Points: 2
300/f is a crude approximation to the real equation λ = C/f. C is the speed of light in whatever medium you happen to be in.

actually 300/f is a quite accurate lambda in vacuum, where f is in Mhz and result is in Meters. It differs from the real value by less than 0.07%

The factor of 300 will vary quite a lot depending on which medium the EM is passing through, and in fact this ratio ( c / Vp) is called the the refractive index of that medium, which is always greater than 1, and in fact can be quite large for some materials. Alternatively it is the ratio of permitivity or permeability of the material to that of vacuum for opaque materials and electrical signals

In effect, both frequency and c are "hidden" inside the other parameters in the asker's refered equations.

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### votientu

Points: 2
So if the ratio c/Vp is constant for one medium, lamda is constant for one medium at one frequency beacause lamda = Vp/f (2.11 in Microwave Engineering of Pozar).
Please refer this link https://www.emtalk.com/mscalc.php?e...ts_list=Wmm&La=8.21969195722&L_units_list=Lmm and only change R to 60, you will give another lamda, lamda = L * 4 at mode quater wave length.
Thanks!

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So, it leads to the conclution that L is not equal to lamda/4. Why ? Electric length is 90, it is mode quater wave length, isn't it ?

I understand that lamda just depends on the medium the EM passes through and frequecy.
L at mode quater wave length will vary depending on Zo and of course lamda.
What do you think about it ?

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