let's consider a coaxial cable of length L and see what happens at port 1 when port 2 is left open. Let's consider what is written on these slides:

enter image description here

1) I do not understand the reason of this graph. In fact I thought that the absolute value of input reflection coefficient should be always equal to 1 (or 0 in dB) since for a transmission line we have total reflection if at the end it is left open. Why is there this curve that decreases with frequency?

2) Now let's consider the following slide:

enter image description here

It is written that in resonance the absolute value of the input reflection coefficient shows a periodic dependence of frequency. I do not understand:

  • what is exactly the resonance condition

  • why is the curve oscillating

  • why is the curve different from that seen in 1) and in general from |ᴦ|dB = 0

Reference Link

  • \$\begingroup\$ It's the same reason why when you blow over the end of a pipe it makes a tuneful sound. Length of the pipe changes the pitch of the tone. Standing waves! \$\endgroup\$ – Andy aka Nov 1 '19 at 12:53
  • \$\begingroup\$ Put a link to the slides in your question, this is missing a lot of context that could help to answer the question. \$\endgroup\$ – Captainj2001 Nov 1 '19 at 14:34
  • \$\begingroup\$ Perfect, link added \$\endgroup\$ – Kinka-Byo Nov 1 '19 at 14:38

Due to reflection from the open circuit end, the impedance seen at the sending end can have an array of different values: -

enter image description here

And this is all purely due to the phase angle of the returning wave when it hits the sending end. For instance, if the applied frequency is such that the line is one complete wavelength, the line behaves like an open circuit at the sending end (depicted as a tuned parallel LC). This is also the case at half a wavelength of cable.

For a quarter wave line, the input impedance becomes a short circuit (depicted by a series L and C). Ditto at 3/4 of a wavelength. In between those notable points the line can look inductive or capacitive. Purely down to how the returning wave interacts (on a phase angle basis) with the sending end voltage.

| improve this answer | |
  • \$\begingroup\$ But why if I try to calculate the input reflection coefficient I get that its absolute value is always 1? For an open transmission line Zin = -j Z0 cot(kl) and so the input reflection coefficient will be: ᴦ = (Zin - Z0)/(Zin + Z0) = [-j Z0 cot(kl) - Z0]/[-j Z0 cot(kl) + Z0]. But numerator and denominator are complex conjugate numbers, their absolute values are the same and so |ᴦ| = 1 at any frequency \$\endgroup\$ – Kinka-Byo Nov 1 '19 at 13:39
  • \$\begingroup\$ Where did you get the formula? Zin and how it relates with Zo has nothing to do with reflections coming back up the line from an unterminated far end. \$\endgroup\$ – Andy aka Nov 1 '19 at 14:26
  • \$\begingroup\$ well: Zin = Z0 * [ZL + jZ0tan(kl] / [Z0 + jZLtan(kl]. If ZL is very high (open circuit), we get: Zin = Z0 * ZL / [jZLtan(kl] = -j Z0 cot (kl) \$\endgroup\$ – Kinka-Byo Nov 1 '19 at 14:29
  • \$\begingroup\$ I don't know where you are going with this but the math indicates a changing reactance with variations in length (or frequency) as per the graph in my answer. \$\endgroup\$ – Andy aka Nov 1 '19 at 14:34

The resonance condition is when the cable is an integer multiple of half a wavelength in the cable, at the frequency of interest. $$ L = m\lambda/2 $$

The first graph is likely measured from a real cable, hence includes effects of power dissipation in the cable. The last graph is actually quite confusing and you should ignore it.


Since you are curious about the derivation of the input impedance I will show it for a lossless transmission line of system impedance \$Z_0\$ terminated in an open circuit (\$Z_L = \infty\$): $$ Z_{in} = \frac{Z_L + jZ_0 \tan(\beta L)}{Z_0 + jZ_L \tan(\beta L)} = -jZ_0\cot(\beta L) $$ $$ \beta = 2\pi/\lambda = 2\pi f/c $$ $$ Z_{in} = -jZ_0\cot(2\pi fL/c) $$ At this point it is fairly clear that the input impedance is a periodic function of frequency or wavelength.

You can calculate the resonance condition based on when the input impedance goes to infinity.

| improve this answer | |
  • \$\begingroup\$ so is it not correct to say that |ᴦ|dB = 0 for an open transmission line? That is what I have always read on transmission lines theory books \$\endgroup\$ – Kinka-Byo Nov 1 '19 at 12:59
  • \$\begingroup\$ There are two steps to calculating the input reflection coefficient into a lossless or lossy transmission line. First, calculate the input impedance looking into the cable based on the load condition. Second calculate the reflection coefficient using the system impedance and your calculated input inpedance. \$\endgroup\$ – Captainj2001 Nov 1 '19 at 13:05
  • \$\begingroup\$ and the absolute value of the reflection coefficient is not always 1? \$\endgroup\$ – Kinka-Byo Nov 1 '19 at 13:18
  • \$\begingroup\$ @Kinka-Byo The absolute value of the reflection coefficient at the end of the transmission line at the load is equal to 1. However "looking into" the transmission line from the source you will see a different impedance based on the length of the transmission line. This is an effect often used to transform load impedances into matched impedances with the source (think quarter wavelength transformer). \$\endgroup\$ – Captainj2001 Nov 1 '19 at 13:25
  • \$\begingroup\$ what is wrong in this computation? For an open transmission line Zin = -j Z0 cot(kl). Therefore the input reflection coefficient will be: ᴦ = (Zin - Z0)/(Zin + Z0) = [-j Z0 cot(kl) - Z0]/[-j Z0 cot(kl) + Z0]. Since numerator and denominator are complex conjugate numbers, their absolute values are the same and so |ᴦ| = 1 \$\endgroup\$ – Kinka-Byo Nov 1 '19 at 13:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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