0
\$\begingroup\$

I am currently looking for a term for a lossless line with a wave resistance Zw which is open ended and I am not sure if I have calculated correctly.

My thoughts about the calculation: If the line is open at the end, then the reflection r at the end is 1 $$r_{end} = 1$$. Because it is a lossless line, the reflection factor is only turned in phase, so the reflection at the beginning of the line is $$ r_{begin} = r_{end}e^{-j2\beta l}$$

the reflection coefficient r can be calculated from the impedances in general: $$r = \frac{Z - Z{w}}{Z+Z_{w}}$$

so that $$ r_{begin} = \frac{Z_{w} - Z_{begin}}{Z_{w} + Z_{begin} } $$

This is my solution for the input impedance Z_begin:

$$r_{begin} = r_{end}*e^{-j2bl} \Leftrightarrow \frac{Z_{w} - Z_{begin}}{Z_{w} + Z_{begin}} = r_{end}*e^{-j2bl} \Leftrightarrow _{r_{end} = 1} Z_{begin} = \frac{Z_{w}(1-e^{-j2bl})}{1+e^{-j2bl}}$$

Is it correct what I have calculated, or do I have a big mistake somewhere?

EDIT:

many thanks @user287001 I Would like to add my comment to the question here, so that it is easier to read with my new solution.

I can see the following from your note:

$$Z_{begin} = U_{begin}/I_{begin}$$

$$U_{begin} = U_{beginfront} + U_{beginreflected}$$

$$I_{begin} = I_{beginfront} + I_{beginreflected}$$

the following applies: $$U_{endfront} = U_{beginfront} * e ^{(-\gamma l)}$$ $$U_{beginreflected} = U_{endreflected} * e^{(-\gamma l)}$$

Besides that: $$U_{endreflected} = r_{end} * U_{endfront}$$ so that : $$U_{beginreflected} = U_{beginfront} * r_{end} * e^{(-2 \gamma l)}$$

Equivalently, this also applies to the current I_begin

results from this : $$Z_{begin} = \frac{U_{beginfront} + U_{beginfront} * r_{end}* e ^{-2 \gamma l}}{I_{beginfront} - I_{beginfront} * r_{end} * e^{-2 \gamma l}}$$

with $$I_{beginfront} = U_{beginfront}/Z_{w}$$ I then get $$Z_{begin} = Z_{w} \frac{1 + r_{end} e^{-2 \gamma l} }{1 - r_{end} e^{-2\gamma l}}$$

finally I have to use for r_end = 1 and for gamma = (- j b)

is that correct ?

\$\endgroup\$
4
  • \$\begingroup\$ Are you trying to calculate the impedance looking into a t-line with an open circuit at the far end? \$\endgroup\$
    – Andy aka
    Sep 11, 2020 at 8:20
  • \$\begingroup\$ Yes, exactly. I would like to know what the input impedance of such a line would be \$\endgroup\$
    – H4x9r
    Sep 11, 2020 at 8:23
  • \$\begingroup\$ Have you googled the answer? It's quite a common formula and resides on several internet sites. \$\endgroup\$
    – Andy aka
    Sep 11, 2020 at 9:04
  • 1
    \$\begingroup\$ Just for your info: The gamma in your last formula should be j*2*Pi/wavelength where j=the imaginary unit. I do not know what your b exactly is. \$\endgroup\$
    – user136077
    Sep 11, 2020 at 17:59

1 Answer 1

1
\$\begingroup\$

EDIT this is valid only for the original question. The edited version of the question has simpler calculation principle and the result is right.

===

Your result is wrong. It resembles the right formula, but there's wrong signs. Your extra trip via the equation between impedance and reflection factor creates numerous possibilities to make errors.

You should calculate the impedance as Ut/It where Ut is the total voltage at the beginning of the line and It is the total current at the beginning of the line.

Assume there's at the beginning of the line an arbitary incident wave with voltage amplitude Uo and summed with it there's the reflected wave Uo*r where r is the reflection factor with the delay exponential term.

Calculate the incident and reflected current components with the wave impedance. Keep in mind that for the reflected current component you must take it's direction as reversed, so there's not Io+Ir, but Io-Ir as the total current. That can be shown with the energy flow direction rule of the general circuit theory.

\$\endgroup\$
4
  • \$\begingroup\$ Thanks for the answer.I think I am now on the right track: \$\endgroup\$
    – H4x9r
    Sep 11, 2020 at 11:31
  • \$\begingroup\$ Z_b = U_b/I_b U_b = U_bf + U_br I_b = I_bf + I_br the following applies: U_ef = U_bf * e ^(-g * l) U_br = U_er * e^(-g * l) Besides that: U_er = r_e * U_ef so that : U_br = U_bf * r_e * e^(-2 * g * l) Equivalently, this also applies to the current I_b results from this : Z_b = (U_bf + U_f * r_e* e ^(-2 * g * l)) / (I_bf - I_bf * r_e * e^(-2 * g * l) with I_bf = U_bf/Z_w I then get Z_b = Z_w (1 + r_e^(-2*gl)) / (1 - r_e^(-2*gl)) b stands for begin, e for end , r for reflected and f for front e.g. U_bf means U_begin_front and U_br means U_begin_reflected. \$\endgroup\$
    – H4x9r
    Sep 11, 2020 at 11:44
  • \$\begingroup\$ Sorry, but the comment is a little too complex to be red without errors. The question has now a correct calculation. \$\endgroup\$
    – user136077
    Sep 11, 2020 at 12:46
  • \$\begingroup\$ Thanks a lot for your hint, my understanding of the topic has increased significantly, also a thanks for that =) \$\endgroup\$
    – H4x9r
    Sep 11, 2020 at 12:49

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.