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Old February 5th 05, 08:33 AM
Ian Jackson
 
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In message t, Old Ed
writes

SNIP

The original author observed that many practical devices
(e.g., mixers) exhibit distortion levels that rise as the "power"
of the product in question. For example, third-order distortion
rises 3 times as fast (dB scale) as the desired (linear) signal.

Snip

Ed, where the increasing intermodulation distortion is simply a result
of increasing the level of the signals at the input of the mixer (or
amplifier), third order distortion actually rises TWICE as fast as the
desired signal. Third order distortion DOES rise on a 'three dB per dB'
basis, but the wanted signal also rises - at 1dB per dB. The difference
is 2dB. So the relationship is 2dB per dB.

If you continued to increase the signal levels, you might expect that
the level of the intermodulation would eventually catch up with - and
overtake - the level of the wanted signal (it doesn't, of course).

The third order intercept point is simply the hypothetical level where
the level of the intermodulation would have risen so much (at 2dB per
dB) that it equals the level of the wanted signal.

Ian.
--

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Old February 5th 05, 11:38 AM
Jason
 
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But why we can add or minus gain and IP3 which are in different unit(db
and dbm)?
Anyone knows?
Thank you

rgds
Jason

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Old February 5th 05, 01:50 PM
Richard Fry
 
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"Jason" wrote
But why we can add or minus gain and IP3
which are in different unit (db and dbm)?

_______________

The algebraic summation of decibel values is a mathematically legitimate,
and convenient way to determine system performance. Decibels are based on
logarithms. Adding/subtracting logs or (decibels) is easier than
manipulating the real values they represent. The final dB value in an
analysis can be converted back to whatever units are desired.

For example, below is an analysis of a UHF radio link system over a
free-space path. The 5 watt power of the transmitter is first converted to
dBm so it can be used with other dB values present to analyze the system.
The same result is reached when multiplying tx power in watts by system
gains and losses expressed as decimal values, but that process is more
awkward -- at least when using a pencil & paper or a pocket calculator
(computers don't care).

TX PWR OUTPUT 36.99 dBm
TX ANT 19.20 dBi
RX ANT 19.20 dBi
TOTAL GAINS 75.39 dB

DISTANCE 18.00 Miles
FREQ 950.00 MHz
PATH LOSS 121.26 dB
LINE LOSS TX 1.80 dB
LINE LOSS RX 3.00 dB
CONN LOSS 1.00 dB
OTHER 0.00 dB
TOTAL LOSSES 127.06 dB

RX SIGNAL -51.67 dBm (584 uV)
RX SIGNAL REQ'D -90.00 dBm
RAW FADE MARGIN 38.33 dB

RF

Visit http://rfry.org for FM transmission system papers.

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Old February 5th 05, 02:56 PM
Ian Jackson
 
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In message .com,
Jason writes
But why we can add or minus gain and IP3 which are in different unit(db
and dbm)?
Anyone knows?
Thank you

rgds
Jason


Think of it this way:
dBm indicates an absolute value. db indicates a relative value.
For example:
0dBm = 1mW
0dBm + 3dB = 1mW x 2 = 2mW = 3dBm
0dBm + 10dB = 1mW x 10 = 10mW = 10dBm
3dBm + 10dB = 2mW x 10 = 20mW = 13dBm
20dBm - 30dB = 100mW/1000 = 0.1mW = -10dBm

What you can't do is to add dBm values directly.
If you have power combiner, and add 10dBm and 13dBm, you can't add 10dBm
and 13dBm and get 23dBm. 23dBm would be 200mW (because 20dB is x 100,
3dB is x 2, so 100 x 2 =200), and this is incorrect.

What you have to do is to convert the dBm values into mW, then add the
mW.
10dBm = 10mW
13dBm = 20mW
Total power = 30mW (and not 200mW)
30mW can then be converted back into dBm (= appx 14.5dBm)

Do you see the pattern?
Ian.
--

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Old February 5th 05, 08:05 PM
Caveat Lector
 
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Wow -- well written Ian

--
Caveat Lector (Reader Beware)
Help The New Hams
Someone Helped You
Or did You Forget That ?



"Ian Jackson" wrote in message
...
In message .com, Jason
writes
But why we can add or minus gain and IP3 which are in different unit(db
and dbm)?
Anyone knows?
Thank you

rgds
Jason


Think of it this way:
dBm indicates an absolute value. db indicates a relative value.
For example:
0dBm = 1mW
0dBm + 3dB = 1mW x 2 = 2mW = 3dBm
0dBm + 10dB = 1mW x 10 = 10mW = 10dBm
3dBm + 10dB = 2mW x 10 = 20mW = 13dBm
20dBm - 30dB = 100mW/1000 = 0.1mW = -10dBm

What you can't do is to add dBm values directly.
If you have power combiner, and add 10dBm and 13dBm, you can't add 10dBm
and 13dBm and get 23dBm. 23dBm would be 200mW (because 20dB is x 100, 3dB
is x 2, so 100 x 2 =200), and this is incorrect.

What you have to do is to convert the dBm values into mW, then add the mW.
10dBm = 10mW
13dBm = 20mW
Total power = 30mW (and not 200mW)
30mW can then be converted back into dBm (= appx 14.5dBm)

Do you see the pattern?
Ian.
--





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Old February 5th 05, 09:05 PM
Ian Jackson
 
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In message MF9Nd.29160$xt.24350@fed1read07, Caveat Lector
writes
Wow -- well written Ian

After over 40 years in Cable TV, I think I am beginning to get the hang
of it!
Ian ; ))
--

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Old February 5th 05, 04:34 PM
Airy R.Bean
 
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Let's talk loosely, and talk about money.

If I've got twice as much money than Ian has, then
I've got 3dB more.

How much do I have? Don't know.

If Ian has three times as much as Richard, then he
has 4.7 dB more than Richard, and I have 3 + 4.7 =7.7dB
more than Richard.

How much do I have? Don't know.
How much does Ian have? Don't know.
How much does Richard have? Don't know.

OK, assuming that we could deal in 1/10ths of a cent (1 milli-dollar!)
let's assume that Richard has $100 = 50dBm.

Ian therefore has 50 + 4.7 = 54.7 dBm.
And I have 54.7 + 3 = 57.7 dBm.

The answer to your question is that you can start off with an
actual reading in dBm, but everything else relative to that is
in dB only (although it does give a result in dBm).

If the above doesn't answer your question, then, sorry,
but I give up. (Which doesn't mean that my interest is 0dBm
but -173 dBm, ie, indiscernible below the noise)

"Jason" wrote in message
oups.com...
But why we can add or minus gain and IP3 which are in different unit(db
and dbm)?
Anyone knows?
Thank you

rgds
Jason



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Old February 5th 05, 04:25 PM
Airy R.Bean
 
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I think that you are confusing the _RATE_
or _SLOPE_ of each individually with
the differential increase per dB of input signal

"Ian Jackson" wrote in message
...
In message t, Old Ed
writes
The original author observed that many practical devices
(e.g., mixers) exhibit distortion levels that rise as the "power"
of the product in question. For example, third-order distortion
rises 3 times as fast (dB scale) as the desired (linear) signal.

Snip
Ed, where the increasing intermodulation distortion is simply a result
of increasing the level of the signals at the input of the mixer (or
amplifier), third order distortion actually rises TWICE as fast as the
desired signal. Third order distortion DOES rise on a 'three dB per dB'
basis, but the wanted signal also rises - at 1dB per dB. The difference
is 2dB. So the relationship is 2dB per dB.



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Old February 5th 05, 11:17 PM
Old Ed
 
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Hi Ian -

Thanks for trying to clarify, but I think you misread my post
somehow.

I said "...third-order distortion rises 3 times as fast (dB scale)
as the desired (linear) signal."

You said "Third order distortion DOES rise on a 'three dB
per dB' basis, but the wanted signal also rises - at 1dB per dB."

The content of our statements is the same. But you went on
to address the slope DIFFERENCE, which I did not discuss.

I believe Airy is making the same point I am making here
with his (2/5/05 8:25) post.

73, Ed, W6LOL


"Ian Jackson" wrote in message
...
In message t, Old Ed
writes

SNIP

The original author observed that many practical devices
(e.g., mixers) exhibit distortion levels that rise as the "power"
of the product in question. For example, third-order distortion
rises 3 times as fast (dB scale) as the desired (linear) signal.

Snip

Ed, where the increasing intermodulation distortion is simply a result
of increasing the level of the signals at the input of the mixer (or
amplifier), third order distortion actually rises TWICE as fast as the
desired signal. Third order distortion DOES rise on a 'three dB per dB'
basis, but the wanted signal also rises - at 1dB per dB. The difference
is 2dB. So the relationship is 2dB per dB.

If you continued to increase the signal levels, you might expect that
the level of the intermodulation would eventually catch up with - and
overtake - the level of the wanted signal (it doesn't, of course).

The third order intercept point is simply the hypothetical level where
the level of the intermodulation would have risen so much (at 2dB per
dB) that it equals the level of the wanted signal.

Ian.
--



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