3 dB bandwidth

[Don Bowey]

Yes. so 0dBm is 1 mW.


Yes too.

and -30dBm is 1uW. And it would be 60dB less ( 1 million times ) than
+30dBm.
I.e. the dB difference between the numbers yields the ratio.

How is that *not* a ratiometric measurement ? What is your point ?

What's with the "yes's?" I didn't ask if I was correct.

My point is, dB expresses a ratio, for example, of two powers. A 10dB
increase in power my be from 0.1 Watt, to 10 Watts; or 10 Watts to 100
Watts. It conveys no absolute info without being given a reference.

0 dBm is a specific amplitude.

BIG difference.

 

[Guy Macon]

you're being a twit.

"I was judging everything and it seemed only natural that I should.
After all, who else could do it? Heaven became a courtroom where I
not only judged myself and the Mother, but later on I was even asked
to settle little differences among the spirits. Even though I realized
this wasn't going anywhere, I felt locked into the role of judge.

"It seemed the best I could do was to be consistent based on my
previous judgments. It wasn't until I questioned the whole purpose
of judging that I realized I was going about it in the wrong way."

Source: [ http://www.godchannel.com/judgment.html ].

0.5 x Volts _IS_ -6dB Power

Perhaps in your haste to judge you missed the fact that I have
repeatedly specified that 0.5 x *WATTS* is -3dB power. Go look it
up. That's the specific statement that Poop Bear disagreed with.
Do you really believe that I would have gotten as far as I have
without knowing that 0.5 x Volts is -6dB power?

Please go re-take high school logarithms and algebra.

"The punishment and obvious result of negative judgments is guilt.
Guilt erodes the sense of self worth and makes it very difficult to
feel feelings that have been infected with it. Depression is often
the consequence of guilt-bound feelings that cannot move and are
therefore denied.

"Since you are fundamentally innocent, guilt is not in its right
place in you. Guilt is held in place by judgments. Release the
judgments, and the guilt goes too."



"You can very quickly and dramatically change your life for the
better when you release the judgments that have held you captive
and left you cut off from love and from your true emotional strength
and power. This is much easier to do than it's seemed.

"Because judgments are mental decisions, they are easy to change.
The secret is simply to decide again. Take back your original
judgment, change your mind, undecide, unjudge."

Source: [ http://www.godchannel.com/judgment.html ].

 

[Guy Macon]

-6dB was and is the correct answer, you were, are and continue
to be plain wrong. Read it and weep, sorry.

Here are exact quotes from my previous posts. Please explain what
part of them in "plain wrong."

"half the voltage equals -6dB but half the power equals -3dB."

"Half the voltage equals -6dB. Half the power equals -3dB.
Double the voltage equals +6dB. Double the power equals +3dB."

I see nothing about what I actually wrote that is incorrect, but
I welcome you showing me what part of the above is wrong[1].


([1]: I *will* admit to rounding -3.01029995... to -3.)

 

[Kevin Aylward]

Decibels are, as you say, dimensionless; but they are specifically a
way of expressing the ratio of two power levels - only power levels,
not voltages or currents or anything else.

Complete nonsense.

I would suggest that you increase you electronic knowledge by about
20db.

Kevin Aylward
[email protected]
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Pooh Bear]

What's with the "yes's?" I didn't ask if I was correct.

My point is, dB expresses a ratio, for example, of two powers. A 10dB
increase in power my be from 0.1 Watt, to 10 Watts; or 10 Watts to 100
Watts. It conveys no absolute info without being given a reference.

0 dBm is a specific amplitude.

As is 0dBV

BIG difference.

So how come you think you can't use dBs to denote voltage ratios ? I do this
'everyday' in audio !

No hang on - that was Adrian who said you can't use dBs for volts. Why are you
replying to my comment about his nonsence ?

Graham



Graham

 

[Pooh Bear]

It's a ratio to 1 milliwatt.

Somoene else finally brings some sense to this thread !

Graham

 

[Pooh Bear]

Precisely.

But he's promised to change:
http://www.godchannel.com/ahriman.html

But we each need to let go of the unlovingness within ourselves, to
really heal.

But can I give the loons a good spanking first please ?

Graham

 

[Pooh Bear]

The formula for calculating power decibels is:

Po
G = 10 log10 ----
Pi

Where
G = Gain in dB
Po = Power output from the device
Pi = Power input to the device

That is the formula for calculating decibels, decibels are defined with
respect to power.

The formula for calculating voltage decibels is:

[...]

There is no such thing as a 'voltage decibel' so please avoid using the
expression, it can only lead to confusion and sloppy thinking.

I agree 100%. I was trying to accomodate the existing sloppy
thinking of certain ignorant induhviduals, but I see now that doing
so was a mistake; they just run in cyircles shouting "0.5 is -6dB!
0.5 is -6dB!" and flaming anyone who tries to educate them.

Thank you for the excellent advice.

Hahahahahahaha ! You got caught with your pants down and instead of blushing
and going away quietly you simply continued broadcasting your insanity all
over the place !

Aren't you the guy who took over *moderating* an ng recently ?

God help whoever posts there !

Graham

 

[Pooh Bear]

Guy Macon wrote:

Perhaps in your haste to judge you missed the fact that I have
repeatedly specified that 0.5 x *WATTS* is -3dB power. Go look it
up. That's the specific statement that Poop Bear disagreed with.

No it damn well *isn't* !

You said terminated ( 600 ohm ) lines 'lost' 3dB ( half power ).

They don't - they 'lose' 6dB - quarter power or *half the damn volts*.

Now troll off ! Go back to moderating you personal little cave of idiocy.

Graham

 

[Pooh Bear]

-6dB was and is the correct answer, you were, are and continue
to be plain wrong. Read it and weep, sorry.

Here are exact quotes from my previous posts. Please explain what
part of them in "plain wrong."

"half the voltage equals -6dB but half the power equals -3dB."

"Half the voltage equals -6dB. Half the power equals -3dB.
Double the voltage equals +6dB. Double the power equals +3dB."

I see nothing about what I actually wrote that is incorrect, but
I welcome you showing me what part of the above is wrong[1].

([1]: I *will* admit to rounding -3.01029995... to -3.)

That's an awfully big shovel you've got there !

Graham

 

[Adrian Tuddenham]

Somoene else finally brings some sense to this thread !

You can *measure* the voltage but the calculations are based on power.

You can't have different definitions of a decibel, depending on the
units you use to measure it. That's like saying you have different
definitions for Ohms Law, depending on whether you are measuring volts
and amps or deriving it from Watts. The formulae work all right, but
the definition of an Ohm does not include Watts.

The other formulae are useful derivatives (under certain conditions
which should be specifically stated), but they are not the definition.
(They can also be very confusing to a beginner who thought he had just
got a grasp on what a decibel was.)



If you add a suffix '$' to denote some reference level, then dB$ becomes
an absolute unit.


Hope that makes sense.

 

[Winfield Hill]

Pooh Bear wrote...

No it damn well *isn't* !

You said terminated ( 600 ohm ) lines 'lost' 3dB ( half power ).
They don't - they 'lose' 6dB - quarter power or *half the damn volts*.
Now troll off ! Go back to moderating you personal little cave of idiocy.

Perhaps we can help Guy. If we drive a 600-ohm destination directly
from an opamp with an output of 6 volts (rms), we'll see 10mA current
and the load will receive 10mA*6 = 60 milliwatts. But if we add 600
ohms to the opamp output so as to "impedance match" the system, the
600-ohm destination sees 3V, as expected, and it gets 1/4 the power,
or 15mW. Both of these amount to -6dB.

Hmm, if the load gets 1/4 the power, or 15mW, where'd the other 3/4 of
the power, 45mW, go? If the opamps's source resistor dissipates the
same power as the load (matched impedances, remember), namely 15mW,
where'd the rest of our 60mW go? Aha! Don't forgot that the opamp's
new load is 1200 ohms rather than 600, so it now delivers 1/2 of the
former power, or 30mW instead of 60mW. Guy argues that the load gets
1/2 the delivered power or -3dB, which is only part of the picture.
The other part is only 1/2 as much power is sourced by an impedance-
matched amplifier. The bottom line is that matched-impedance systems
deliver -6dB (not -3dB), whether we're considering voltage or power.

Guy, your statement a few days ago, which precipitated the argument,
"Back to the topic, 3DB is also the amount of drop you get when you
have the output impedence and the input impedence matched, which was
common in the days of transformer-balanced audio lines," was wrong.
BTW, that's impedance, not impedence, and dB, not DB. :>)

 

[Pooh Bear]

Pooh Bear wrote...

Perhaps we can help Guy. If we drive a 600-ohm destination directly
from an opamp with an output of 6 volts (rms), we'll see 10mA current
and the load will receive 10mA*6 = 60 milliwatts. But if we add 600
ohms to the opamp output so as to "impedance match" the system, the
600-ohm destination sees 3V, as expected, and it gets 1/4 the power,
or 15mW. Both of these amount to -6dB.

Hmm, if the load gets 1/4 the power, or 15mW, where'd the other 3/4 of
the power, 45mW, go? If the opamps's source resistor dissipates the
same power as the load (matched impedances, remember), namely 15mW,
where'd the rest of our 60mW go? Aha! Don't forgot that the opamp's
new load is 1200 ohms rather than 600, so it now delivers 1/2 of the
former power, or 30mW instead of 60mW. Guy argues that the load gets
1/2 the delivered power or -3dB, which is only part of the picture.
The other part is only 1/2 as much power is sourced by an impedance-
matched amplifier. The bottom line is that matched-impedance systems
deliver -6dB (not -3dB), whether we're considering voltage or power.

Guy, your statement a few days ago, which precipitated the argument,
"Back to the topic, 3DB is also the amount of drop you get when you
have the output impedence and the input impedence matched, which was
common in the days of transformer-balanced audio lines," was wrong.
BTW, that's impedance, not impedence, and dB, not DB. :>)

What's staggering Win, is that we have to explain it to him, the guy who
thought that it was me who was the ineducable one earlier in this thread ! I
did mention he might like to calculate the power delivered by and dissipated in
the source but he missed the hint.

A lesson learnt. I'll know not to pay too much attention in future to any
'advice' he gives.

Cheers, Graham

 

[Guy Macon]

Pooh Bear wrote...

First indication that the above is what Pooh Bear was disagreeing with,
and it comes long after I killfiled the little pest.

Guy, your statement a few days ago, which precipitated the argument,
"Back to the topic, 3dB is also the amount of drop you get when you
have the output impedance and the input impedance matched, which was
common in the days of transformer-balanced audio lines," was wrong.

You are correct. Too bad it took so long for anyone to reference
the actual incorrect statement. Instead I kept posting things like

"Half the voltage equals -6dB. Half the power equals -3dB.
Double the voltage equals +6dB. Double the power equals +3dB."

and I kept getting strong disagreement and abuse when I did so.
A simple "yes, that is correct, but what you said *before* was
in error" would have avoided a lot of trouble, but instead I got
a load of "You are wrong" replies to the correct statement above.

(Typo flame deleted)

I should also point out that, when talking about power as opposed
to voltage, the incorrect statement...

"Back to the topic, 3dB is also the amount of drop you get when you
have the output impedance and the input impedance matched, which was
common in the days of transformer-balanced audio lines,"

....is under some conditions, *still* incorrect if you just change the
3dB to 6dB. The obvious question is "drop from what starting condition
to what ending condition?" Assume the ending condition is matched Z.
If the starting condition is a near-infinite-Z load, close to zero
power is being transfered as the starting condition, so changing to
a matched Z system is a large gain in power transfer, not a drop.

What I was thinking of when I wrote the above was the case of the
starting condition being a near-zero-Z source. I was thinking of
the way Joule misapplied Jacobi's theorem and concluded that an
electric motor driven by a battery or dynamo could not be more than
50% efficient. Edison & Upton showed this to be a bad assumption
when they made a 90% efficient system by making the source Z very low.

 

[Don Bowey]

As is 0dBV


So how come you think you can't use dBs to denote voltage ratios ? I do this
'everyday' in audio !

Use voltage ratios with my blessing. I use the frequently to good
advantage. Obviously, it wasn't me who said you couldn't.

No hang on - that was Adrian who said you can't use dBs for volts. Why are you
replying to my comment about his nonsence ?

After you carve up a quote, how the hell can anyone know who posted it?

 

[Pooh Bear]

Use voltage ratios with my blessing. I use the frequently to good
advantage. Obviously, it wasn't me who said you couldn't.


After you carve up a quote, how the hell can anyone know who posted it?

That's why trimming and snipping is good but needs to be done carefully.

Cheers, Graham

 

[Larry Brasfield]

Incorrect. Please see below about this.

True, it is not. See below.

First indication that the above is what Pooh Bear was disagreeing with,
and it comes long after I killfiled the little pest.

That is not the first indication. See below.

The first indication was before you could possibly have killfiled him,
seeing as how you responded to his innocuous correction in your post of
June 22, 1:33 AM Message-ID: <[email protected]>,
which I quote (with '|' for clarity) following:
| Pooh Bear wrote:
| >
| >Guy Macon wrote:
| >>
| >> Back to the topic, 3dB is also the amount of drop you get when you
| >> have the output impedance and the input impedance matched, which was
| >> common in the days of transformer-balanced audio lines.
| >
| >Nooooo ! That's a 6dB drop.
|
| Yeeeeees! That's a 3dB drop.

You are correct. Too bad it took so long for anyone to reference
the actual incorrect statement.

Graham's above quoted post ("... Nooooo! ..."), "referenced" (quoted,
actually) your mistake in post .
If you had been able to recognize your mistake then, instead of what you
did, (see above), this thread would have been somewhat shorter.

Instead I kept posting things like

"Half the voltage equals -6dB. Half the power equals -3dB.
Double the voltage equals +6dB. Double the power equals +3dB."

and I kept getting strong disagreement and abuse when I did so.
A simple "yes, that is correct, but what you said *before* was
in error" would have avoided a lot of trouble, but instead I got
a load of "You are wrong" replies to the correct statement above.

That's not how I read the thread. As you can see, either from my
above excerpts or by reviewing the thread yourself, at first you
affirmed your mistake. It was only later that you began stating
the common knowledge of what "dB" means, and nobody has
been disagreeing with that.

(Typo flame deleted)

Mr. Hill's remark concerning spelling was, verbatum:
"BTW, that's impedance, not impedence, and dB, not DB. :>)"
Nobody in their right mind can honestly characterize that as a flame.
I urge you to be more careful when characterizing unquoted material.

 

[keith]

Decibels are, as you say, dimensionless; but they are specifically a way
of expressing the ratio of two power levels - only power levels, not
voltages or currents or anything else.

You're as pig-ignorant as Guy! Decibels are a log ratio of *anything*.
Crap, what are they teaching engineers these days?

 

[keith]

-6dB was and is the correct answer, you were, are and continue
to be plain wrong. Read it and weep, sorry.

Here are exact quotes from my previous posts. Please explain what
part of them in "plain wrong."

"half the voltage equals -6dB but half the power equals -3dB."

"Half the voltage equals -6dB. Half the power equals -3dB.
Double the voltage equals +6dB. Double the power equals +3dB."

I see nothing about what I actually wrote that is incorrect, but
I welcome you showing me what part of the above is wrong[1].

([1]: I *will* admit to rounding -3.01029995... to -3.)

That's an awfully big shovel you've got there !

....and it's full of dung. I guess it's too dark there for him to admit we
is wrong now.

What a maroon!

 

[Tony Williams]

You're as pig-ignorant as Guy! Decibels are a log ratio of
*anything*. Crap, what are they teaching engineers these days?

The Decibel is a Unit of Attenuation, defined in
terms of the ratio of power levels only.

Attenuation = 10 * log10 (Pin/Pout) in Decibels.

If (and only if) the input/output impedances are identical,
then the Attenuation (in Decibels) can be calculated as the
log10 ratio of the *square* of the currents or voltages.

2 2
Attenuation = 10 * log10 (Vin/Vout) in Decibels.

Or, Attenuation = 20 * log10 (Vin/Vout) in Decibels.
/|\
|____ Notice where the squares went to?

That calculation using (Vin/Vout) produces units of dB
only when the input/output impedances are identical.

For example. An audio power amplifier requires an input
of 1mW into 600 ohms, for an output of 1W into 15 ohms.
The voltage/current ratios are 0.77/3.87 and 1.29/258.

10 * log10 (1mW/1W) is -30, Decibels of Attenuation.

20 * log10 (0.77/3.87) is -14, Units of Nothing.

20 * log10 (1.29/258) is -46, Units of Nothing.

Only the power ratio produces legitimate units of dB.