[SOLVED] Power Levels At Different Frequencies

Does a system which has a 10V peak to peak signal generate the same power level at 1 MHz than the same as a 10 GHz 10 V peak to peak system?

Does a system which has a 10V peak to peak signal generate the same power level at 1 MHz than the same as a 10 GHz 10 V peak to peak system?

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Hi chiques
I think your question is so general . what kind of system ? an amplifier ? if yes it depends on BW of your amplifier .
without auxiliary information it is really hard to answer you .
Best Wishes
Goldsmith

Are you generating the signal? If you're generating the signal, it probably will cost more power to make a 10 GHz 10Vpp signal. If you're putting this signal into a system, it depends on the system, as goldsmith said.

In CMOS technology, power is split into two categories: dynamic and static. Static power is easy: P = IV, no problem. Dynamic power is P = CV^2 * f, where C is the capacitance of your load and f is the frequency. For any given system in CMOS, you can see that power scales linearly with frequency.

However, in a load without any frequency dependence, this isn't the case. In this case, the average power is the same for all frequencies (I think) (https://en.wikipedia.org/wiki/Alternating_current#Mathematics_of_AC_voltages).

Hope this helps!

as you 2 guys are responding ... I too am not sure about how to answer the OP cuz of the lack of info

I initially assumed that he may be referring to the output of the transmitter ... say into a dummy load
that measured across the load he has 10V p-p.
We would still need to know what the RF current flow was at the 2 different freq's ??

so ... say assuming a dummy load of 50 Ohms, then with 10V p-p across it, the current would be the same
therefore the power out is the same

the difference then would be what it took to get that final RF voltage of 10V p-p

Experience would tell me ... more difficult at 10 GHz than at 1MHz because device efficiency is likely to be less at 10GHz.

Dave

10V peak to peak, 50 Ohm impedance, Generated by a Power Amplifier (not necessarily the same one), CW signal,

Sorry for being to broad.

So it looks like the energy it would take to get the PA to generate the 10 v pk pk would be the issue.

Is it safe to say my "average power" would be the same if the voltages measured the same (peak to peak)?

Although AC is at an average level of zero, it carries energy in its fluctuations.

The greater the frequency of fluctuations, the greater the energy (in certain cases anyway).

For instance, notice when designing a switched-coil power supply, that you can use smaller components when it is operated at a higher frequency.

Although AC is at an average level of zero, it carries energy in its fluctuations.

The greater the frequency of fluctuations, the greater the energy (in certain cases anyway).

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I fear, this doesn't answer the present question, or even further confuses it.

Please consider that average voltage of zero doesn't mean average power of zero. You don't need to refer to "fluctuations" (whatever it is), just apply simple math
The square operation eliminates the voltage sign, that's all.

So it looks like the energy it would take to get the PA to generate the 10 v pk pk would be the issue.

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This makes sense if you refer to PA efficiency, respectively DC input to RF output power ratio. You can hardly compare energy to power, because the quantities differ by the unit time...

Fvm:

Given what you mentioned about signs, you seem to be referring to instantaneous power, and might be more appropriately written as P=V(t)²/R. In which case, it might be helpful consider to Vrms for a standard period signal for the time-average power. I feel P_time-average = V_rms²/R seems more appropriate in this case. At least, it's helpful for me to think of it that way! It hides the nastiness of integrals in a nifty "rms"

Chiques:

Is it safe to say my "average power" would be the same if the voltages measured the same (peak to peak)?

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Given a load with no frequency dependence, I believe that would be an accurate statement

There is no such Power Amplifier that provides a same amount power from 1MHz to 10GHz in whole band.Especially 33dBm !!
In signal generators-so that those generators don't supply such high RF Power levels- operating frequency band has been divided into few sub-bands and they generate the signal but the amplification is done by dedicated independent amplifiers.
For low level RF signal, it's possible but the amplifier can not guarantee flatness except some digitally calibrated ones.

OK, so getting to a system in the real world which generated 10 V peak to peak is obviously a debate in itself.

Let's say we are talking to a PhD. trying to sound smart but can't really build anything:
A standing wave with voltage rating of 10V peak to peak has the same average power at 1 Mhz as it does at 10 GHz? Hypothetically, the average power is independent of frequency and dependent on peak to peak voltage (and impedance).

The phrasing of the question seems to address a larger scope than the classical engineering definition of power. More than the friction of electrons in the wires.

As in the practical and/or philosophical side of things.

Compared with DC, AC has fluctuations (or changes of state).

These fluctuations convey greater activity, or instigation, or information.

As read by a wattmeter it may not be greater power... but we do find that AC can accomplish things which DC cannot.

Furthermore we find that slower frequencies cannot do these things as efficiently as higher frequencies can. This is a benefit of the more numerous changes of state, or change of polarity, etc.

So if we confine ourselves to basic theory only, a higher frequency does not convey greater power per se. However from the practical standpoint, we find it does, when we examine various applications involving transfer of power.

I see that you are trying to elaborate your comments in post #6.

In my view, we don't need to refer to philosophy where elementary natural science explains things satisfyingly. In addition, the "fluctuations" point can be discussed more clearly in physical terms, I think.

If you refer to electromagnetic theory, change in time like dV/dt or dI/dt plays a role. Only AC fields can be radiated.

But the original question is about power delivered to a real (frequency independent) load. And in this case, power is only a function of voltage and load resistance, as stated in post #10.

Great discussion for me. Thank you everyone for chiming in.