[SI-LIST] Re: Units Conversion Question
- From: Rob Hinz <rob@xxxxxxxxxx>
- To: cpad@xxxxxxxxx
- Date: Mon, 13 Aug 2001 13:38:49 -0700
Hi Chris,
I am not sure I completely followed your derivation. The free space
impedance relates the magnitude of the E-field and H-field in a TEM (plane)
wave propagating in free space. So for an X-polarized plane wave
propagating in the Z direction we get:
Ex (V/m) / Hy (A/m) = sqrt(u/eps) = 377 (Ohms)
So, just convert dBuV/m back to V/m and divide by 377 to get A/m
For example, if you have X (dBuV/m):
convert to V/m by
10^(X/20) uV/m = 10^(X/20)*10^(-6) V/m
Now you have E-field strength, so convert to H-field by dividing E-field by
free space impedance
(10^(X/20)*10^-6)/377 A/m
That should do it. As you can see from above, V/m is E-field, and A/m is
H-field. So you actually ARE doing it with E's and H's. It would be
improper to work with just V and I in this case. Now, if you want to go on
and convert this to dBA/m, that may be done as follows:
20 Log ( ((10^(X/20)*10^-6)/377)/1 ) =
20 ( Log(10^(X/20)) + Log(10^(-6)) - Log(377) ) =
20 ( (X/20) + (-6) - (2.576) ) =
X - 120 - 51.53 =
X - 171.53 (dBA/m)
I think this is correct as a mathematical proof, of sorts. I guess that
means I agree with you. Apologies to your colleague. As you pointed out,
his conversion would be correct if the initial units were dBV/m rather than
dBuV/m as the 20Log(10^(-6)) = -120 term would not be present.
Rob Hinz
Senior Electromagnetics Specialist
SiQual Corporation
rob@xxxxxxxxxx
phone (503)885-1231 x30
fax (503)885-0550
http://www.siqual.com
At 11:39 AM 8/13/2001 -0700, Chris Padilla wrote:
>Perhaps some of you can help me out here:
>
>In EMC/EMI, we often deal with units of dBuV/m (dB-microVolts per meter).
>
>What if I wish to convert the dBuV/m to A/m?
>
>For simplicity, use 120*pi Ohms or 377 Ohms as the impedance of free-space
>(assume a plane wave, in the far-field, etc.) so Ohm's law in logarithms is:
>
>20*log I (dB-A/m) = 20*log V (dB-V/m) - 20* log R (dB-Ohms) (1)
>
>since our meter reads in dB-uV/m, we need to convert the voltage units:
>
>20*log V (dB-V/m) = [20*log V (dB-uV/m) - 120 dB] (2)
>
>next, we need to convert the impedance:
>
>20*log R (dB-Ohms) = [20*log R (dB-uOhm) - 120 dB] (3)
>
>Put (2,3) into (1) and:
>
>20*log I (dB-A/m) = [20*log V (dB-uV/m) - 120 dB] - [20*log R (dB-uOhm) -
>120 dB] (4)
>
>
>Now here is where my colleague and I differ:
>
>He simply cancels out the "120s" and gets as a final answer:
>
>20*log I (dB-A/m) = 20*log V (dB-uV/m) - 20*log R (dB-uOhm) or
>
>20*log I (dB-A/m) = meter reading (dB-uV/m) - 51.53 (dB-uOhm) (A)
>
>(recall that R = 377)
>
>
>I don't think one can simply cancel the "120s" as that changes the
>equation. The 120 in (2) must stay or the fact that it is (dB-uV/m) no
>longer is true. Once the 120 is "cancelled" the units go back to
>(dB-V/m)! Same thing with (3), the units are now (dB-Ohm).
>
>Therefore, I say that the final answer is:
>
>20*log I (dB-A/m) = meter reading (dB-uV/m) - 171.53 (dB-uOhm)
>
>
>How can I mathematically validate either answer...I am just not seeing it
>to make it elegant nor rigorous. Anyone clear me up here? I believe I am
>right (I have done some simple circuits to prove it to myself).
>
>BTW, to be really rigorous, I should be doing this with E & H's but it
>doesn't matter as that isn't the argument here I am trying to
>understand--we cold drop the /m part on I and V to be more proper but that
>doesn't change the problem.
>
>
>Thanks----->Chris
>
>
>
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Rob Hinz
Senior Electromagnetics Specialist
SiQual Corporation
rob@xxxxxxxxxx
phone (503)885-1231 x30
fax (503)885-0550
http://www.siqual.com
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