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# Direct adding - a better reasoning needed

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#### newbie_rf

##### Junior Member level 3
dB and dB can be directly added up together; so is dB and dBm.

However, dBm and dBm can not be directly added up. They need to be converted to linear unit first, added up, then converted back to dBm. This is the correct way to do, but anyone can provide some insight on the reasoning for this?

When you add dB you are really multiplying. Think of 2 gain stages each with 20dB gain (x10 in voltage terms). If the signal goes through one then the other it is x10 x10 = x100.

If you have a 10dBm signal and apply gain of say 20dB, again you are multiplying by x10. However, in this case the 10dBm is a signal level - 10mW - not a 10dB gain.

When you have two signals in dBm say 10dBm and 20dBm you do not get 30dBm. You are not multiplying. You are most likely adding them together.

I hope that helps!

Keith.

newbie_rf

### newbie_rf

Points: 2
There are two kinds of measurement, absolute and relative.

You can say a bag weighs 1 kilo gram. Thats an absolute measurement. Alternatively, you can say this bag is heavier than that bag. Thats relative measurement.

Now, dBm is an absolute measurement. You can say a signal strength is -30dBm. You can say a signal is weaker by 10dB with respect to signal 1. So, dB is a relative sort of thing while dBm, dBW etc are with respect to a defined reference.

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### sequel

Points: 2
Thank you Keith for the insight.

I guess dBm should be always considered as absolute value, rather than a ratio, despite of the fact dBm is really referencing to 1mW. dBm=10log(signal_power/1mW). It sounds like a ratio similar to dB, but it is really not.

I guess dBm should be always considered as absolute value, rather than a ratio, despite of the fact dBm is really referencing to 1mW. dBm=10log(signal_power/1mW). It sounds like a ratio similar to dB, but it is really not.

Yes, that is absolutely correct. As ninju says - dBm is an absolute value. It is just a convenient way of describing power rather than saying 1pW and 1W, for example. The fact is has "dB" in the name doesn't alter the fact it is really an absolute.

Keith.

newbie_rf

### newbie_rf

Points: 2
An additional comment about "adding" dBm. The nature of problem and signal decides if you have to add power or voltage. Uncorrelated signals will always add as power, coherent signals may add as voltage in some situations, e.g. in an electronical adder.

Points: 2

### newbie_rf

Points: 2
Yes, that is absolutely correct. As ninju says - dBm is an absolute value. It is just a convenient way of describing power rather than saying 1pW and 1W, for example. The fact is has "dB" in the name doesn't alter the fact it is really an absolute.

Keith.

Thank you Keith and every one for the input and verification.

I have a few more comments for discussion:

1.) To have absolute power expressed in dBm does provide conveinence. The conveinence lies in the fact that it can be directly added with dB (for gain budget etc.) While using mW or Watts does not have this advantage. So absolute power expressed in the sort of decibel format does have its purpose, although it does not change its nature that it is still an absolute quantity, as seen in 2.) below.

2.) When two dBm numbers need to be added up (for the non-coherent case that FvM mentioned), one cannot avoid the pain of converting back to linear first. This is where even using decibel format could not help.

3.) When two signals are slightly off in frequency (for example, 10MHz apart), can the power be added up to get a total power? My guess is no, since the power of these two tones are at different frequencies. However, on the other hand, if the two tone freq separation is small enough, for example, a few Hz apart, it can be added, since one can be treated as the sideband of the other. Is this reasonig correct? (yeah, I know, this question is slightly off the subject this post was originally started..)

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Signals of different frequency are "orthogonal" and add by their power, also harmonics, or quadrature components.

newbie_rf

### newbie_rf

Points: 2
Signals of different frequency are "orthogonal" and add by their power, also harmonics, or quadrature components.

Thank you FvM. Are you saying the total power can be added up regardless of frequency difference?

As an after thought,you might be interested in Power spectral density and Energy spectral density.

newbie_rf

### newbie_rf

Points: 2
As an after thought,you might be interested in Power spectral density and Energy spectral density.

Then integrate over freq will yield total power. So total power is regardless with the difference of frequencies I guess?

Yeah, kind of. I like to view it this way: You are transmitting some signal from a transmitter. The power consumed by it is V into I. All this power goes into any frequency it transmits, even if that frequency is not intended to be transmitted.

We take PSD to measure electromagnetic emissions from an IC.

2.) When two dBm numbers need to be added up (for the non-coherent case that FvM mentioned), one cannot avoid the pain of converting back to linear first.

This is where even using decibel format could not help.
It's right in order to have exact values, but I don't agree with the second part:
In that case dB are useful for quick approximations. Concretely, if you add two powers expressed in dB:

= If they are equal, the result is 3dB higher than them
= if they differ by 6 dB, the result is the strongest + 1 dB (very good approx.)
= if they differ by 12 dB, the result is the strongest + 0.5 dB (very good approx.)
= If they are very different, the result is practically the strongest

Regards

Z

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newbie_rf

### newbie_rf

Points: 2
Zorro, thank you for the nice comments! These rules of thumb are very handy.
I did some quick calc, for 6dB, needs to add .97dB; for 12dB, needs to add 0.27dB.

Hi,

You are right. I made a mistake mixing numbers. Right is:
...
= if they differ by 9 dB, the result is the strongest + 0.5 dB
= if they differ by 12 dB, the result is the strongest + 0.25 dB
...

Thank you for pointing it out.
Regards

Z

Points: 2

### keith1200rs

Points: 2
Hi,

You are right. I made a mistake mixing numbers. Right is:
...
= if they differ by 9 dB, the result is the strongest + 0.5 dB
= if they differ by 12 dB, the result is the strongest + 0.25 dB
...

Thank you for pointing it out.
Regards

Z

Thank you very much again for your contribution!

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