What is meant by the E and H being approximately equal in the far field? I thought the ratio of E/H was 377:1? (ie the characteristic impedance of free space)
Good question. In the far field the E and H fields have approximately the same wave impedance (377 ohms). This is often defined as the distance (wavelength / 2 * pi) from the radiator. So for a signal at 100 MHz (lamba = 3 meters), the transition from near field to far field occurs at a distance of about 0.5 m. Prior to this point, the wave impedances will be unequal, and the nature of the radiator determines which impedance dominates: this includes things like whether the antenna has a dipole versus a loop type structure, whether the source is high or low current, etc. Although this is an important topic (and one which I'm planning to cover in a separate future video), as a practical matter, successful EMI debugging with near field probes generally does not require knowledge of wave impedances, etc. :) In the case of debugging a PCB, for example, it's simply important to important to know that the probe should be held close (1-2 cm) to the board, different probe types (E vs. H) should be tried, and especially in the case of loop-style H field probes, the probe should be rotated and/or have it's orientation changed. The low amplitudes of some emissions and the need for high spatial resolution (i.e. which specific trace, pin, component, etc. is causing the emission), often require the probe to be held much closer to the EUT than the mathematical demarcation point between near and far field, which is why I left the math out of this presentation :) Hope some of that is helpful. Thanks again for the question!
You explained very well at the beginning that the E field and the H field move 90 degrees relative to each other and later in the video that the H field is a round loop surrounding the conductor. For the E field I understand well perpendicular to the conductor, ok for the E field but how can the H field be at 90 degrees from E if it is a round loop?
Thanks UA-cam for recommending me this excellent video! This answers pretty much all of my questions.
Thanks for watching!
This series of videos is great - thank you!
Thank you!
Educational video! Please make more so that we could know more about R&S products.
This was crystal clear. Thank you very much.
Thanks for watching!
What is meant by the E and H being approximately equal in the far field? I thought the ratio of E/H was 377:1? (ie the characteristic impedance of free space)
Good question. In the far field the E and H fields have approximately the same wave impedance (377 ohms). This is often defined as the distance (wavelength / 2 * pi) from the radiator. So for a signal at 100 MHz (lamba = 3 meters), the transition from near field to far field occurs at a distance of about 0.5 m. Prior to this point, the wave impedances will be unequal, and the nature of the radiator determines which impedance dominates: this includes things like whether the antenna has a dipole versus a loop type structure, whether the source is high or low current, etc.
Although this is an important topic (and one which I'm planning to cover in a separate future video), as a practical matter, successful EMI debugging with near field probes generally does not require knowledge of wave impedances, etc. :) In the case of debugging a PCB, for example, it's simply important to important to know that the probe should be held close (1-2 cm) to the board, different probe types (E vs. H) should be tried, and especially in the case of loop-style H field probes, the probe should be rotated and/or have it's orientation changed. The low amplitudes of some emissions and the need for high spatial resolution (i.e. which specific trace, pin, component, etc. is causing the emission), often require the probe to be held much closer to the EUT than the mathematical demarcation point between near and far field, which is why I left the math out of this presentation :)
Hope some of that is helpful. Thanks again for the question!
You explained very well at the beginning that the E field and the H field move 90 degrees relative to each other and later in the video that the H field is a round loop surrounding the conductor. For the E field I understand well perpendicular to the conductor, ok for the E field but how can the H field be at 90 degrees from E if it is a round loop?
great presentation
Thank you!
Very helpful!
Thanks for the feedback!
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