0
\$\begingroup\$

enter image description here

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.

\$\endgroup\$
4
  • \$\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

4
\$\begingroup\$

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

\$\lambda=\dfrac{v_s}{f}\$

\$\endgroup\$
4
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.