# Why a wave does not enter a cable? Understanding TDR (Time Domain Reflectometry)

I have been doing some TDR (Time Domain Reflectometry) measures which involves sending gaussian pulses through a cable and measuring the s11 (reflection) parameter.

In this measure, I could see that from some frequency on, most of the pulse was reflected in the beginning of the cable and it didn't (or it didn't seem to) enter the cable. This reflection happens due to the difference of the impedance of the cable and the impedance of the source but it seemed to be more important at higher frequencies.

The effect of the wave not entering the cable, was also partially because of the higher attenuation at higher frequencies. Anyway, the peak of the reflection in the beginnig was also much bigger in higher frequencies.

So...

Is it because the wave cannot be difracted (or refracted?) into the cable for its small wavelenght?

Or is it because the difference between the impedances of the source and the cable has gotten bigger, so the reflection coefficient has grown bigger? If both impedances depend on frequency, the difference shouldn't grow that much, or what?

25MHz

250MHz

1GHz

• Could you please provide more details such as frequency range, type of cable (e.g. coax), source impedance, length of the cable? Aug 23, 2013 at 14:07
• Did you keep the source impedance of the pulse constant over the whole range of frequencies? Aug 23, 2013 at 14:11
• Sorry. I really don't 'get' these tdr plots. What is the -1 to 1 axis? Aug 23, 2013 at 17:26
• What are the units of the X axis? Aug 23, 2013 at 17:48
• The question would be a much better resource for future readers if you left the plots in. Aug 25, 2013 at 23:34

I have been doing some TDR (Time Domain Reflectometry) measures which involves sending gaussian pulses through a cable and measuring the s11 (reflection) parameter.

But in comments you say the source of your signal is a VNA.

A VNA can not do the measurement you describe. It cannot produce a gaussian pulse.

What it can do is produce sinusoidal stimulus at different frequencies. Then sweep the frequency to obtain the S11 scattering parameter.

Then it can do some mathematical analysis to tell you what the time domain reflection would look like, assuming the system is linear.

I could see that from some frequency on, most of the pulse was reflected in the beginning of the cable and it didn't (or it didn't seem to) enter the cable. This reflection happens due to the difference of the impedance of the cable and the impedance of the source but it seemed to be more important at higher frequencies.

If you are seeing a frequency dependent reflection, then something in your system doesn't have consistent behavior across frequencies. Some possibilites are

• Your transmission line is not maintaining its characteristic impedance at high frequencies. It may also become a multimode transmission line at high enough frequencies.

• Your connectors are not ideal at higher frequencies. They may have some excess capacitance or inductance that causes a reflection. Excess shunt capacitance would cause a negative reflection at high frequencies and excess series inductance would cause a positive reflection at high frequencies, when viewed in a TDR. However its possible the parasitics of a connector might not be simple enough to model with a single parasitic element.

• It's unlikely if you're testing with a VNA that the source would not perform consistently across frequencies that it's able to produce. However if you did not calibrate the system correctly (An Open-Short-Load cal is best, but even a Short response cal is probably okay if you don't need perfect accuracy) you could see some issues.

Edit

Thanks for including the plots.

What you're seeing is not that the reflection is different at different frequencies, but that the way you are measuring, you are effectively filtering the response with a low pass filter. As you reduce the cut-off frequency of this filter you're smearing out the reflection in time. But you're not adding energy to the reflection, so the peak amplitude has to decrease as the pulse width increases.

If you have a VNA, and you want to see how the reflection depends on frequency, it would be more clear to just plot |S11| vs. frequency.

• It is true that the VNA doesn't send diretly the gaussian pulse, but it does an intern calcule in order to produce the same effect. With the VNA you get the H(w) of the cable then this is multiplied by the gaussian in frequency domain (the X(w) signal) and what you get is the Y(w). With an IFFT you get your y(t) which is you TDR refletogram. So do you consider that is has nothing to do with the geometry of the connector? What you are saying is that a parasitic L is created in the connector that rises the impedance of the connector?
– Xabi
Aug 23, 2013 at 16:52
• @Xabi, can you share your TDR results? How do you know the reflection happens mostly at high frequencies? Aug 23, 2013 at 16:59
• I don't get why do you say I am filtering the signal. What happenned to the other answer? He was saying that the skin effect may change the impedance of the cable... Is that true? So in your opinion the same proportion of energy is reflected but as the peak in high frequency is thinner it is also higher?
– Xabi
Aug 23, 2013 at 17:34
• The guy who did the other answer has a history of abusing the site, so his answers (and accounts) get removed whenever he shows up. Aug 23, 2013 at 17:38
• As for why you are filtering...Your plots don't show signs that the reflection is different at different frequencies. They show that when you change the frequency range of the measurement used (with all that math) to estimate the reflection response, you measure a different result. If there was a selective reflection at, say, 500 MHz, you would see almost no reflection when you measure with 25 MHz bandwidth, and you would see a ringing response when you measure with 1 GHz bandwidth. Aug 23, 2013 at 17:42

I am observing similar things with my FDR. At higher sweep frequencies, impedance discontinuities are more apparent. I think it's because lower frequencies are more forgiving because the wavelength is much larger than the feature causing the discontinuity.