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I was trying to look for the answer of what I think it should be quite a simple question but I haven't found any nice answer (at least any that has convinced me). Here it is:

Through a cable (reflectometry) or even a in a RADAR, in order to detect small defects/objects, you usually use high frequencies, because the resolution will be better; my questions is why? I have always assumed instinctively that the object must be bigger than the wavelength in order to be detected but I would like to know why... Does the reflection coefficient get bigger with frequency? Is the reflected energy the same but as the reflected pulse would be wider with low frequencies, it is just more difficult to detect?

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Small objects (small in comparison with wavelength) diffract the wave around them, rather than reflect the wave back, they are effectively invisible (unresolved). By decreasing the wavelength (increasing frequency) the objects diffract less of the waves energy and reflect more of it making them detectable.

Sharp edges, small holes etc. do the same. By decreasing the wavelength this enables such features to be resolved.

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  • \$\begingroup\$ In this case I was thinking of a cable, so I calculate the reflection coefficient of a default in the cable: gamma=(Zdefault-Zo)/(Zdefault+Zo) Can you apply the effect of the frequency to this equation? How can you calculate the difraction effect in this case? \$\endgroup\$
    – Xabi
    Commented Aug 22, 2013 at 16:37
  • \$\begingroup\$ If you search for the wikipedia definition of characteristic impedance, you will see the frequency component. The real answer is not that simple, however. Due to imperfections in a build, real-world loss and other things, the reflection (sometimes known as return loss or VSWR) will vary wildly over frequency (most times worse at higher frequencies. Not always). \$\endgroup\$
    – scld
    Commented Aug 22, 2013 at 16:54
  • \$\begingroup\$ If we're talking an abrupt change in transmission medium such as light passing from air to water or an RF signal in a transmission line interrupted by a short or open circuit that's a different matter and you need to specifically look at the effects on the electric and magnetic fields at the transition. \$\endgroup\$ Commented Aug 22, 2013 at 17:13
  • \$\begingroup\$ I'll put an example. During some TDR (time domain reflectometry) measures which involves sending gaussian pulses through a cable, I could see that from some frequency on, most of the pulse was reflected in the beginning of the cable and it didn't enter the cable. Of course it was also partially beacuse of the higher attenuation in higher frequencies that the wave couldn't pass through. But at the same time the peak of the reflection in the beginnig was also much bigger in higher frequencies. Is it because the wave cannot be difracted (or refracted?) into the cable for its small wavelenght? \$\endgroup\$
    – Xabi
    Commented Aug 22, 2013 at 21:32
  • \$\begingroup\$ 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? \$\endgroup\$
    – Xabi
    Commented Aug 22, 2013 at 21:37

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