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In EMC there is an equation for charactristic impedance (If I am not wrong) that is defined as: $$Z_w = \frac{E}{H}$$

As you know \$E\$ is expressed in Volts and \$H\$ as Ampers.

I had the exam today and for sake of confusion of students, the teacher brough this values in dB. But I actually needed this value in pure ohm (no dB) to solve the next part of question. The values were:

\$E = 100\frac{mVdB}{m}\$ and \$H=80 \frac{mA dB}{m}\$

I am not sure if I did correct, but I went ahead and did this: $$Z_w = 20\log\frac{100}{80} = 1.93 \Omega$$

Is this the correct? If not, how to handle this?

This is really confusing, in our study book the same question is presented by just E and H in normal way! for example E is 50mV/meter and H is 40 mA/meter...either I am so stupid or teacher is a troll!

UPDATE

First, Thank to all for the feedback. I asked my teacher the correct soloution and it appeard to be solved in this way:

$$Z_w=\frac{E}{H}=\frac{10^{\frac{100}{20}}}{10^{\frac{80}{20}}}=10\Omega$$

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Is this the correct?

No.

If not, how to handle this?

Simply convert the E and H values from dB to \$V/m\$ and \$A/m\$ respectively. Although the notation you've used isn't conventional, it's pretty clear what it means. The E field is 100 dB referenced to \$1mV / m\$ and the H field is 80dB referenced to \$1mA / m\$

Now, when converting a voltage or current to dB, you take \$20 \ log\$ of the value divided by the reference unit.

So, in the case of the E field, to convert from dB, you divide 100 by 20 to get the exponent of 10 and then multiply that by the reference unit:

\$100 dB (\frac{mV}{m}) \rightarrow (10^{\frac{100}{20}}) \frac{mV}{m} = 100\frac{V}{m} \$

Similarly, for the H field:

\$80 dB (\frac{mA}{m}) \rightarrow (10^{\frac{80}{20}}) \frac{mA}{m} = 10\frac{A}{m} \$

Now that you have E & H in familiar units, you proceed as usual to find the characteristic impedance.


A more sophisticated approach would be to recall that division becomes subtraction of logs. You do have to be careful of reference units though. The difference of the logs is \$20 dB (\frac{mV}{mA})\$ which gives the same result as the procedure above: \$10 \Omega\$

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Decibels (dB) represent a dimensionless ratio. There is no way to express Ohms or any other non-dimensionless quantity in dB directly. You can express a ratio of any value in dB, since that is always dimensionless.

In some cases, specific numeric scales have been created based on dB with 0 dB defined as some particular value. That is the same as saying representing the ratio of the actual value to the reference value in dB. For example, dBm is a common unit used in RF, with 0 dB representing 1 mW of power. This is the same as saying that dBm expresses the ratio of some power to 1 mW.

In addition, dB is meant to express a power ratio. It's basic definition is 10 Log10 (power1/power2). Since power is proportional to voltage squared, dB is sometimes thought of as 20 Log10(volts1/volts2). Since resistance does not relate in a direct way to power, there is no easy way to scale the ratio of two resistances to be meaningful in dB. For example, if it were implied that the voltage is kept constant, then you could use -10 Log10(R1/R2), whereas if the current were kept constant then it would be 10 Log10(R1/R2).

So in short, there is no unambiguous way to express resistance in dB without some additional previously agreed upon convention. Basically this is a bad idea since it only invites confusion.

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  • \$\begingroup\$ +1 for emphasizing that decibels measure dimensionless ratios. The instructor of the course ought to be ashamed of himself. \$\endgroup\$ – Dilip Sarwate Jul 13 '12 at 19:56
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No, 1.93 Ω is not correct. It's not the 1.93 (I didn't even calculate it), it's the dimension. A dB value is dimensionless; you divide two number of the same quantity, so their units cancel, and that's a good thing, otherwise you couldn't calculate the logarithm of the ratio. A logarithm returns a dimensionless number, so times 20 (also dimensionless, and why 20??) it can never suddenly become a resistance.

Besides, sinds dBs are found by taking the logarithm you don't divide them, you subtract them. At least if they use the same reference. Apples and oranges. You can still convert dBm to dBW, because they both express power, but there's no relationship between a current in dB and a voltage in dB.

I think the answer is that there is no answer here.

edit
On the other hand(!), nobody can stop you from defining a dB scale for resistors, only to go there from voltage and current you have to define references for the three of them.

Let's assume the following

0 dB\$_{SV}\$ = 1 V ("SV" = Steven's Volts :-))
0 dB\$_{SA}\$ = 1 A
0 dB\$_{SR}\$ = 1 Ω

Then a voltage of 10 V = 20 dB\$_{SV}\$ across a 1 Ω = 0 dB\$_{SR}\$ resistance causes a 10 A = 20 dB\$_{SA}\$ current, and we can find dB\$_{SR}\$ = dB\$_{SV}\$ - dB\$_{SA}\$. (A factor 20 for current is because power is proportional to current squared, so it's the same as for voltage.)
Other example: 1000 V across a 10 Ω resistor = 100 A. Then 60 dB\$_{SV}\$ = 20 dB\$_{SR}\$ + 40 dB\$_{SA}\$. So this seems to work. Why? Because we started by choosing our zero dB for a correct relationship between the three. If we had chosen 0 dB\$_{SR}\$ to be 1 mΩ then we would have to add an offset of 60 dB each time: 10 V across a 1 Ω resistor = 10 A. Then 20 dB\$_{SV}\$ = 60 dB\$_{SR}\$ + 20 dB\$_{SA}\$ - 60 dB.

So you can define a dB scale for resistance, but its reference is tied to those for your voltage and current scales, on penalty of an offset. Due to Ohm there's a relationship between voltage, current and resistance, but note that you can't go from one quantity to the other without the third.

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Looking at your expressions, \$\frac{mVdB}{m}\$ is wrong.

It is common to express electric field intensity in \$dB(\mu V/m)\$. This means an electric filed intensity expressed in dB relative to \$1 \mu V/m\$ (one microvolt per meter).

If this is the case, then \$100 dB (\mu V/m)\$ means that

$$100 = 20 \log \frac{E}{1 \mu V/m}$$

From where your field E is \$10^5\$ microvolt per meter, or 0.1 V/m.

Same thing for current.

Edit: added parentheses around units affected by dB

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  • \$\begingroup\$ I'd go for dB(μV), dB(mV), \$dB_{μV}\$ or \$dB_{mV}\$ to make the unit more clearly readable. Maybe it is a matter of taste? At least I wouldn't argue about if m or μ is intended. I agree the unit mVdB is nonsense. \$\endgroup\$ – jippie Jul 10 '12 at 18:06
  • \$\begingroup\$ You're correct. I'll add parentheses around the units to make it clear. \$\endgroup\$ – Juancho Jul 10 '12 at 18:36
  • \$\begingroup\$ I would suggest to not put your comment Edit: added parentheses around units affected by dB into the answer itself. It just produces noise. A log of your edit history is saved and accessable anyway. \$\endgroup\$ – PetPaulsen Jul 10 '12 at 19:10
  • \$\begingroup\$ Most commonly used absolute terms are at a fixed impedance such as; 0 dBm = 1 mW at either 50Ω (RF) or 75Ω (TV) or 0 dBmV = 1 mV @ 75Ω cable TV (=−48.75 dBm) and a few others \$\endgroup\$ – Tony Stewart EE75 Jul 10 '12 at 20:56
  • \$\begingroup\$ @TonyStewart: No, dBm is just power referenced to 1 mW. No specific impedance is specified or implied. dBm can be used with 50 and 75 Ohm systems, but it doesn't need to be. 1 mW is still 1 mW regardless of the impedance. \$\endgroup\$ – Olin Lathrop Jul 10 '12 at 23:34
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Resistance shouldn't be calculated in dB. dB is almost always used for power. Since the power in a resitor depends on voltage (or current), you would use the voltage (or power) to calculate dB.

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