# [SI-LIST] Re: Units Conversion Question

• From: Wang Xiao-yun <wangxiaoyun@xxxxxxxxx>
• To: cpad@xxxxxxxxx, <si-list@xxxxxxxxxxxxx>
• Date: Tue, 14 Aug 2001 09:12:38 +0800

```Hello, Chris:
I think your colleague made a mistake when converting units of the=20
variables.
To solve this problem, we could replace the variables when changing the=
=20
units and here it goes:

V(dB-V/m) =3D Vu (dB-uV/m) / 10^6
R(dB-Ohms) =3D Ru (dB-uOhms) / 10^6

Then units can be omitted and the equation could be changed to:

20*log I =3D 20*log V - 20* log R
20*log I =3D [20*log Vu - 120] - [20*log Ru - 120]

Now we may understand that Ru is not R anymore and 20*log Ru is 171.53=20
instead of 51.53. In this way the two 120 can be cancelled and the final=20
result is

20*log I =3D 20*log uV - 171.53

In fact you could directly use R without converting it to Ru / 10^6.
Sometimes I feel that including units into the equations may cause more=
=20
confusions and that's why I don't use the units in MathCAD, or I should say=
=20
I haven't been able to master the feature by far.
Regards.

At 11:39 01-8-13 -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) =3D 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) =3D [20*log V (dB-uV/m) - 120 dB] (2)
>
>next, we need to convert the impedance:
>
>20*log R (dB-Ohms) =3D [20*log R (dB-uOhm) - 120 dB] (3)
>
>Put (2,3) into (1) and:
>
>20*log I (dB-A/m) =3D [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) =3D 20*log V (dB-uV/m) - 20*log R (dB-uOhm) or
>
>20*log I (dB-A/m) =3D meter reading (dB-uV/m) - 51.53 (dB-uOhm) (A)
>
>(recall that R =3D 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) =3D 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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--
Wang Xiao-yun
Shanghai, China, http://go.163.com/philharmania

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