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A quick translation:

Generator has \$Z_g = 50\Omega\$ and 12 V. The line is loaded with an unknown RL. We observe we have a maximum of 8 V at 250 MHz and a minimum at 500 MHz.

  1. Solution: Asks for the dielectrical permitivity and the \$Z_o\$ of the line

    $$Z_o = \sqrt{L/C} = 50\Omega$$

    and for the dielectrical permitivity of the dielectric it is

    $$Z_o = \sqrt{\frac{\epsilon_o \epsilon_r}{\mu_o \mu_r}}$$

    $$Z_o = 120\pi \cdot \sqrt{1/3}$$

    from there we get the dielectrical permitivity.

  2. This asks for the line length which is my initial question. Is wavelength same as the line length or am I wrong?

$$\lambda = \frac{v}{f} = 0.159 m$$

Thanks and sorry for my lack of English in Electrical Engineering subject.

  • \$\begingroup\$ Perhaps you meant length? \$\endgroup\$
    – Eugene Sh.
    Jan 17, 2018 at 19:43
  • \$\begingroup\$ yeah, i did mean that. \$\endgroup\$
    – WhiteGlove
    Jan 17, 2018 at 19:52
  • \$\begingroup\$ If 500MHz is 3/4λ and 250MHz is 1/2λ what is length? \$\endgroup\$ Jan 17, 2018 at 19:53
  • \$\begingroup\$ My question: Calculating Lambda that way i did, the result is also the electrical length of the transmission line?. I mean, the line has a length of 0.159 meters? \$\endgroup\$
    – WhiteGlove
    Jan 17, 2018 at 19:55

1 Answer 1


Signal Velocity = \$v_s=\dfrac {c}{\sqrt{\epsilon _r}}\$


  • \$\begingroup\$ Yeah, lambda is the wave length but... what about the transmission line? \$\endgroup\$
    – WhiteGlove
    Jan 17, 2018 at 20:07
  • \$\begingroup\$ Max amplitude for unknown load could be 1/2, 1, 3/2 λ , and Min amplitude could be 1/4, 3/4, 5/4, λ \$\endgroup\$ Jan 17, 2018 at 20:45
  • \$\begingroup\$ Funny how people overthink microwave equations lol. \$\endgroup\$
    – user103380
    Jan 17, 2018 at 21:03
  • \$\begingroup\$ It's from Radiation and Wave Guidance. The exams are quite difficult (for me) \$\endgroup\$
    – WhiteGlove
    Jan 17, 2018 at 21:27

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