# dB scale misunderstanding

1. Dec 7, 2018

### wirefree

G’day.

I am here to clear my doubt about the dB scale, in general, and it’s use in calculating received power in a microwave cavity.

Given that I have arrived at received power (Pr in Watts) for a given input power (Pt), I understand that

10*log (Pt/Pr)

yields Pr in dB.

Now, say, the cavity’s a Directonal Coupler, which has an Insertion Loss (Il) of 0.5 dB. If I substract Il from Pr (in dB), and obtain a negaive number, that would violate basic principles as it implies Pr > Pt since log10 of a number less than 1 is negative.

Best Wishes for the Festive Season,
with kind regards,
wirefree

2. Dec 7, 2018

### Staff: Mentor

That doesn't sound right. 10*log (Pt/Pr) is the ratio of Pt to Pr in dB, If Pt>Pr, it is a positive number, and if Pt<Pr, it is a negative number.

If you want an absolute number, you can use the units of dBm, which is dB above 1mW in a 50 Ohm system. You can express Pt and Pr in dBm, and just subtract to see what your insertion loss is... Does that help?

3. Dec 7, 2018

### jim hardy

Is this a real life experiment?

I would be more inclined to question
1. How precise are my "arrived at" numbers for Pt and Pr ?
2. Is that coupler's insertion loss number "0.5db" of the flavor "guaranteed not greater than" , or "guaranteed 0.5±η db " ?
I'd find a way to measure its actual loss.

.................................... that said. .....................................................................................

Power expressed in db is always a ratio and it assumes a reference power.
We usually state what that reference power is The number of db tells how much larger or smaller (compared to that reference) is the power you are describing.
So your expression gives, as Berkeman said, the ratio of transmitted to received powers - not an absolute number for either.

I know nothing of microwaves
I assume there's convenient reference power that's so commonplace people forget to state it.

I do know that in telephony there's a widely used reference power, one milliwatt, that's so commonplace it's on multimeter faces.
When selected to AC volts the bottom scale indicates power referenced to one milliwatt in 600 ohms, about 0.77 volts
In my day we suffixed our db reading with "m", eg dbm to indicate that one milliwatt reference power.

(purple marks mine - jh)

old jim

4. Dec 8, 2018

### tech99

The ratio of Pt and Pr, which you have expressed in dB, is presumably the measured attenuation of the system under test, say L dB. If your system includes, a component, such as a directional coupler, which has 0.5dB attenuation, then the attenuation of the remainder of the system is L-0.5 dB. Yes, the remainder of the system has less attenuation.
On the other hand if you have calculated Pr, then you must have already included the directional coupler in the calculation, so subtracting it will not create negative attenuation.

5. Dec 9, 2018

### poor mystic

Attenuation is measured in dB; so that for example a level of -10dB is the outcome when a signal at a level of 0dB is attenuated by a factor of ten. :)

6. Dec 9, 2018

### sophiecentaur

Nah. People are just sloppy and assume they're talking the same terms as the other chap. dBW or dBmW should always be established when talking Power, dB only ever (even with microwaves) refers to a ratio of Powers. (Not Volts!!! - although that's another common example of sloppiness).
PS I know you know that, Jim.

7. Dec 9, 2018

### jim hardy

Yep.
$P = \frac{V^2}{R}$

so 1 milliwatt into 600 ohms requires
$V = \sqrt{ {P}X{R}} = \sqrt {{0.001}X{600}} = \sqrt{0.6} = ~0.775 volts$
and 0db on Decibel scale is right underneath that many volts on red 2.5 VAC scale
(note the 2.5VAC scale is ever so slightly nonlinear - that's to handle nonlinearity of the diode rectifiers inside the meter)

3 db doubles power (actually 3.01db)
so $P = Pref X 10^{\frac{dbm}{10}}$ and ${Pref} = 1 milliwatt$

so at 3dbm
10^0.3 = 1.995 , and
to make 1.995 mw in 600 ohms would require $V = \sqrt{ {P}X{R}} = \sqrt {{0.001995}X{600}} = \sqrt{1.197 } = ~1.094 volts$
which indeed is right above the 3db mark

at 6dbm
10^0.6 = 3.981 , and
to make 3.981 mw in 600 ohms would require $V = \sqrt{ {P}X{R}} = \sqrt {{0.003981}X{600}} = \sqrt{2.389 } = ~ 1.545 volts$

at 10 dbm
10^1 = 10 , and
to make 10 mw in 600 ohms would require $V = \sqrt{ {P}X{R}} = \sqrt {{.01}X{600}} = \sqrt{6 } = ~ 2.449 volts$

hmm.
at 6db I see 1.525 volts which would be 5.88 db not 6
either my purple line is just a smidge off, or the scale is.
I'd wager it's my hand drawn purple line.

old jim

8. Dec 9, 2018

### sophiecentaur

@jim hardy I wasn't thinking in terms of Analogue meters because people don't use them very much these days - but there's a different thing:
My problem is that a 240V 1kW electric heater and a 240V 20W lightbulb could be (if the sloppiness were followed through) 0dB different. . . . . and there's the problem with transformers that are sometimes awarded a 'Gain in dB'.

And you can hear the question "Is that dB Volts or dB Power?" Aaargggh.

9. Dec 9, 2018

### jim hardy

That's a "Big Ten-Four" on sloppy.....

I taught myself ,
upon encountering any mention of db
to immediately stop and ask "db referred to what ? "

It's a self defense reflex now, against proceeding on a mistaken premise and compounding errors -
"db means a ratio and before i go any further i want to know what's the denominator? "

I flogged this horse for benefit of OP because it seemed to me he wasn't thinking "ratio"
and the analogue meter is my anchor to reality.
After all, at the level of our sensory abilities, Mother Nature is analog

old jim

10. Dec 9, 2018

### jim hardy

HiFi gear response is always plotted in db - it looks so much better on the log scale.

11. Dec 13, 2018 at 8:58 AM

### wirefree

Many thanks for all the responses. A special thanks to @jim hardy for the historical/analog, and especially his personal, perspective.

A big thanks to @berkeman & @tech99 for taking the bull by its horns.

Thanks also to @sophiecentaur and @poor mystic for chipping in.

All help appreciated.

Before I proceed, I really must address @jim hardy about the source of this query:

This problem appears in a locally published and prescribed textbook for an undergraduate course in Microwave Engineering.

Next, a note on @tech99’s suggestion:

The second paragraph especially interests me. But I maintain my qualms about the subtraction.

Also, you prefer to term the ratio ‘Attenuation’ and not what I refer to it, i.e. ‘Pr’.

Perhaps that’s what’s contentious.

If we were to call it ‘Attenuation’, then I can clearly see now that the attenuation of the Circuit Under Test, with the Directional Coupler removed, would be L-0.5 dB. But, in terms of Pr, I cannot make peace with the given solution — see below the part of the solution to the question after all the mechanics of the DC are complete and all power values have been calculated.

In keen anticipation of your responses,
best regards,
wirefree

12. Dec 13, 2018 at 1:14 PM

### tech99

We only have a snip of the book, but it seems be mixing up decibel and absolute quantities. I am not at all happy about the calculation as shown or the units being used. For instance, power cannot be expressed in dB, only dBW, dBm etc. There must be a reference.

13. Dec 13, 2018 at 2:07 PM

### jim hardy

i lack the vocabulary to decipher this one.

What the heck are "C" and "D" ?
What the heck are "Pf" and "Pb" ?
Where did that 284.6 X10^-6 come from ? I notice it's 55db down from 90 (Pi ?) and 55 is the sum of "C" and "D". Is that coincidence?

Is "Effective Received Power" Pr' defined someplace?
If -0.445 = $10log\frac{P_r'}{90}$
then 10^-0.0445 = Pr' /90
and Pr' = 90 X 10^-.0445 = 90 X .90261 = 81.2 watts ?

14. Dec 13, 2018 at 4:06 PM

### DaveE

YES! dB is a ratio, otherwise write the reference in the unit i.e. dBm, dBuV, dBc etc.
Then everyone needs to be on the same page about the 10 or 20 factor. 10 for power, or power related things like dBc. 20 for voltages and currents.
Expressing a ratio of voltages in dB is different formula than expressing a ratio of power in dB and there is often no way of knowing without understanding the context.

15. Dec 13, 2018 at 5:03 PM

### davenn

I agree ... really odd
and particularly this part

@wirefree C and D are not even defined in the text you have given us

16. Dec 14, 2018 at 4:11 AM

### sophiecentaur

It's possible that the person who writes "dBV" knows what they mean by it but a Bell is a ratio of Powers and the Voltage ratio may or may not be the square root of the ratio of the Powers. Inside a circuit, the impedances will very likely be all over the place so dBV is an even more risky thing to use. There is never any actual need to quote dBV so why introduce it?

17. Dec 14, 2018 at 8:29 AM

### Averagesupernova

Do you feel the same way about dBmV and dBuV as well? dBmV is a very established standard in the cable TV industry.

18. Dec 14, 2018 at 10:09 AM

### sophiecentaur

Yes I do feel the same. They get away with it because they are, presumably using (or are assuming) the same impedance everywhere and the numbers they happen to be using happen to suit them. It is sloppy usage because it is a non-existent unit. A ratio of Powers is not a ratio of Voltages and neither are the logs10.
So called Standard Practice is often very dodgy and 'in house'. Talk to a plumber about the way pressure actually works and you will get a nonsense answer,. That's despite the fact that the guy can be trusted to fit a central heating system in your house very competently and with no mistakes.
I like a simple life and I resent having to double think such things as dBV before I can decide what they mean.

19. Dec 14, 2018 at 10:34 AM

### Averagesupernova

20. Dec 14, 2018 at 11:31 AM

### f95toli

It is a directional coupler so it is probably the coupling and the directivity