# Calculating wavelength

Given is fundamental frequency $50\,\text{MHz}$, $50\%$ Duty Cycle, Rise and fall time is $5\,\text{ns}$, wire length is $30\, \text m$, $R_i=50\,\Omega$ and $c$.

How do I calculate wavelength?

I'm confused which formula I should use.

Should I simply use $$\lambda=\frac{c}{f}$$

or

$$f_{sig}=\frac{1}{t_r\pi}$$

$$\lambda_{sig}=\frac{c}{f_{sig}}$$

$t_r$ is rising/fall time

How should I calculate it?

• Wavelength of what? cable? you need to change c with $\epsilon$ and length of line to determine standing wave length. Otherwise bandwidth for rise time 10~90% , BW f=0.35/t – Tony Stewart Sunnyskyguy EE75 Nov 26 '17 at 21:41
• Of cable. No I don't know to because with $c$ I meant $3*10^8m/s$. @TonyStewart.EEsince'75 – Alena Nov 26 '17 at 21:44
• Most coax use c = 2e8 [m/s] due to dielectric constant. – Tony Stewart Sunnyskyguy EE75 Nov 26 '17 at 21:45
• This is just random example @TonyStewart.EEsince'75, but can we switch to telling me why and which formula should be used :( – Alena Nov 26 '17 at 21:46
• In a coaxial cable, the classic speed of light (299 m/us) is not used, because the dielectric material in the cable slows down the wave. The actual speed depends on the dielectric constant of the material. What Tony is telling you is that a typical speed is around 200 m/us. But if your problem is telling you to use 300m/us, so be it. Then lambda = c/f. – mkeith Nov 27 '17 at 4:22

wave velocity $v_c=\dfrac{c}{\sqrt{\epsilon}}$ thus
$\lambda(f)=\dfrac{c}{\sqrt{\epsilon}f}$
When f < 10% $\lambda$, the rise time is limited by RC time constant of cable C (100pF/m typ.) and total R = Rs+Rload (neglecting cable loss)