Hi Airy -

I generally agree with your description of IP3; but I would add
a few points.

The IP3 model was first published in a now-classic article
back in the 60s. (I could probably dig up the specific
reference, if someone really wants to know.)

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.

If the subject distortion is plotted against input/output levels,
and approximated by a best-fit straight line, that line will
intersect a similar linear extrapolation of the desired signal
at a point dubbed the "Intercept Point."

The utility of all this is that you can use a single specification--
intercept point--to make quite good predictions of distortion
levels over a wide range of input conditions.

But it is important to remember that IP is only a MODEL,
and an empirical one at that. Real devices will never follow
the model exactly and completely--as you note in your
discussion of the saturation region.

73, Ed, W6LOL

"Airy R.Bean" wrote in message
...
From off the top of my head, without any revision.....

IP3, or "Third Order Intercept Point" is an indication
of how good a mixer is, but it is not a physical point!

If you were to plot the wanted output of a mixer stage against
the input signal (ignoring the local oscillator input), you would
get a graph that is a nearly-straight line from the origin
which then starts to flatten off.

At the point of the line where it starts to curve over to
flatness, and therefore starts to be non-linear, other
mixer products, mainly those based upon the third
harmonic of the input signals start to appear in the
output. if you plot these other products on your graph in
addition to the wanted output signal, they grow at a rate (the slope)
which is 3 times greater than was the initial straight line
of the wanted output.

If you take the original straight line of the wanted output, and
extrapolate it so that it meets the other line growing at 3 times
the slope, you get what is known as the "Third Order Intercept
Point". The reason that this is a theoretical point is because the
wanted output has long since flattened off!

The better a mixer is, the higher is IP3 for the outputs of the mixer.

IP3 will be given in terms of the power of the wanted output signal,
say, 50 dBm - other respondents have informed you that this is
50dB (or 10^5) times greater than 1mW, or 100W (Perhaps not
a good figure for an example - a mixer with an output of those
levels could be a PA stage!). In this case dBm gives us the power relative
to the mW.

If we now go back to the flattening off of the curve, at some point,
the curve will be 1dB less than what it would have been had the curve
not been a curve but had carried on as a straight line. This point is
known
as the "1dB Compression Point" - In this case we use dB and not dBm
because
we are talking relative to some other point on the line.

There is a mathematical derivation (which I don't know off-hand) which
shows that the 1dB Compression Point is 10.4dB below IP3.

So, I hope that I have gone some way to explaining (or increasing your
confusion) on the points that you raised!


"jason" wrote in message
ups.com...
May I know what actually the unit of dbm and db is different from one
another?
If they are different how can we minus the gain in unit of db from a
IP3 in unit of dbm?
Kindly enlighthen
Thank you all

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.
--

But why we can add or minus gain and IP3 which are in different unit(db
and dbm)?
Anyone knows?
Thank you

rgds
Jason

"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.

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.
--

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.
--

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 ; ))
--

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

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.

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.
--