# Please convert "dB/m" to "/m"

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1. Apr 13, 2015

### taoz

I want to get a input impedance from microwave.
if the attanuation constant is 0.13, the value is 0.13dB/m.
for get the Zin, the dB have to disappear.

0.13dB=10log(X)
0.013dB=log(X)
10^(0.013)dB=X
???????

2. Apr 13, 2015

### Staff: Mentor

Welcome to the PF.

dB per meter? I'm not familiar with that, except for attenuation throgh a medium...

3. Apr 13, 2015

### taoz

in Wikipedia,
https://www.physicsforums.com/wiki/Attenuation [Broken] constants, in fields such as https://www.physicsforums.com/wiki/Optical_fiber [Broken] communication and https://www.physicsforums.com/wiki/Radio_propagation [Broken] https://www.physicsforums.com/wiki/Path_loss [Broken], are often expressed as a https://www.physicsforums.com/wiki/Fraction_(mathematics) [Broken] or ratio to distance of transmission. dB/m means decibels per meter, dB/mi is decibels per mile, for example. These quantities are to be manipulated obeying the rules of https://www.physicsforums.com/wiki/Dimensional_analysis [Broken], e.g., a 100-meter run with a 3.5 dB/km fiber yields a loss of 0.35 dB = 3.5 dB/km × 0.1 km.

i thought the dB/m means microwave how much propagate while per meter. it also appears nepers/m. And I want to disappear the unit, dB.

Last edited by a moderator: May 7, 2017
4. Apr 14, 2015

### jim hardy

I think db is a dimensionless unit, since it's just a ratio .

X is power/power so the unit is already gone.

5. Apr 14, 2015

### meBigGuy

Not sure how you arrived at the conclusion that you can somehow determine Zin (input impedance?) from the attenuation factor.

That like trying to determine the input impedance of a cable from the losses. It can be lossless, or very lossy totally independent of the impedance.

Maybe you need to rethink or rephrase your question.

6. Apr 14, 2015

### Baluncore

Maybe it is a reference to dBm. That is dB relative to a signal of 1 milliwatt.

7. Apr 21, 2015

### taoz

I found the unit of attenuation constant is nepers/meter and nepers also dont have any unit, just neper=10ln(P2/P1)
and dBm also dont have any unit.
the problem was solved! thanks!!

8. Apr 22, 2015

### Averagesupernova

To say dBm has no unit is incorrect.

9. Apr 22, 2015

### Staff: Mentor

Interesting. Are you thinking [W]/[W]?

10. Apr 22, 2015

### Averagesupernova

I mentioned this in another thread today. 0 dBm is .001 watts into a specified impedance. Often it is 50 ohms. We go from there. It is a specific unit like an ampere or volt. I used to work on RF signal level and field strength meters that are used in the cable TV industry so I am quite familiar with these terms. dBmV is also a reference. It is .001 volts into a specified impedance. However, it is still treated as power. The volts part is just a reference. If the power goes from 0 dBmV to 10 dBmV we say it went up 10 dB and the actual power in watts increased by 10. The volts would increase only by a factor of about 3.

11. Apr 22, 2015

### Baluncore

When it comes to dimensional analysis, a power series has an interesting property.
The Logarithm of anything can have no unit. It must be a non-dimensional input such as a ratio.
dBm is short for dB relative to 1 mW.
dBm = 10 * Log10(power/1mW)

12. Apr 22, 2015

### Staff: Mentor

I agree with Baluncore on this one. dBm is dB above a milliwatt, with a 50 Ohm system. The units are log(W/W), which is dimensionless.

13. Apr 22, 2015

### Baluncore

I disagree with berkeman slightly on this one. One dBm is one dBm no matter what the impedance of the system. It is simply a power ratio.

The specification of impedance is only needed when converting between dBm and voltage. Most of the dBm to voltage equivalence lookup tables used by RF engineers are standardised for the now common 50 ohms specification.

The input and output of the transcendental functions must be non-dimensional as the functions are defined as infinite power series. To remove dimensions requires division by a parameter with identical dimensions, that is where the term “pure ratio” meaning “non-dimensional” comes from.

dB are so convenient that RF engineers, being human, often happily use and trust dB without understanding the fundamentals or the pitfalls.

There is a confusion that can arise where the output of an RF voltage buffer, (distribution amplifier), is very low. It may then be specified as 1 dBm into 50 ohms even though it has an output impedance of less than one ohm. It could also drive 1mW into each of three 50 ohm cables in parallel.

With transfer functions the output is usually implicitly reference to the input. On the other hand a spectrum analyser with a dB vertical axis must specify what zero on the dB scale really means. There must be a reference parameter somewhere. It will be either simply dBm, or dBV in the specified input impedance.

The most common confusion I find is in exam questions where a double negative appears. “A circuit has an attenuation of –20dB. If the input is one volt, what is the output voltage? ”. An attenuation of –20dB is a gain of +20dB so the output will be 10 volts. But that assumes the unspecified input and output impedances are identical.

This confusion also shows up where the graph of a passive filter transfer function is labelled “attenuation”, then numbered with zero at the top and negative dB below. The vertical axis should have been labelled “Gain”, (even though it has none). This is very common where the teacher does not understand the subject.

14. Apr 22, 2015

### Staff: Mentor

Yes, thanks for the correction.

15. Apr 22, 2015

### jim hardy

i posted this image in the other db thread
db scale on a voltmeter,
Simpson 260 is set up for telephone company standard 600 ohm, see lower left
it wouldn't be much use at rf but is fine for audio

db is bottom scale , just below red 2.5VAC scale
0dbm (edit-jh) = 1 milliwatt into 600 ohms observe they defined their reference there on the meter face

P = V2/R
0.001 = V2/600
V0db = √0.6 = 0.77 volts

+10db = .01watt into 600 ohms
V10db = √6 = 2.45 volts

Last edited: Apr 23, 2015
16. Apr 23, 2015

### meBigGuy

dBm has no units because it IS a unit. It like saying 5 inches and then asking what the units are for inches.

dB is like saying 3 inches is 2 times as long as 1.5 inch.

BTW. 0 dBm (not 0 dB) is 1mw into 600 ohms. 0 dB has no fixed reference level.

dBm is an absolute expression of power (as is dBu and dBW). dB is a ratio between two power quantities.

Now, I understand that dBm is just a ratio between two power quantities when 1 has a fixed reference value.

So you could say that 0 dBm is is 0 dB greater than 1mW into 600 ohms. (the 600 ohms pins down the voltage).

The Simpson meter is doing that. Saying that the measurement is 0 dB relative to the reference level. (they define the 0 dB reference level at the bottom left)

Anyway, that's the way I see it.

God, I love those old Simpson meters. You know you are measuring something real when it swings that needle.

.

17. Apr 23, 2015

### Baluncore

No. The impedance does not need to be specified for dBm. Only when you convert dBm to voltage or current must the impedance be specified.

It is probable that telephone and audio engineers will use 600 ohm systems while RF engineers will use 50 ohm systems.

The Simpson meter measures low frequency AC volts. It then calibrates the 2.5VAC voltage scale in dBm. It must therefore specify a reference impedance. Since it is a low frequency meter it uses the audio 600 ohm standard. Notice the table of additives in the lower right of the panel for use on different AC voltage ranges.

The 5000 ohms per volt spec on the 2.5VAC range offers an impedance of 12.5k ohm to the circuit.
For 600 ohm measurements you must sort out your own termination.

The thing that makes the Simpson meter an asset is not the dBm scale. It is the "Taut Band Suspension".

18. Apr 23, 2015

### meBigGuy

I agree with all that. I misspoke. The reference is 1 mW, pure and simple. I understand that the 600 ohms (or 50 ohms) only allows a voltage reference.

I did say "So you could say that 0 dBm is is 0 dB greater than 1mW into 600 ohms. (the 600 ohms pins down the voltage)."