Average Power

[ptc-1218903]

Hi

In the attached sheet, I am trying to find the average power of sum of two signals. The signals are separated by a delta frequency. On the power waveform, I am expecting a DC value with a signal of frequency delta riding, but I am not egtting this. Appreciate your comments.
Thanks
Leyo

[LouP]

The instantaneous power is the r(t)^2 in your graph, which exhibits exactly the behavior you expect. Increase the righthand x-axis limit to see the full period). The integral you define is the average power (which = 1) , and is independent of t. In the integral, t is the dummy variable of integration; the integral does not depend on t.

Lou

[TomGutman]

P is not a function of t at all. It is the average over the first 15 μs. Since there is no free x in the definition of P, the value of the argument to P is irrelevant to its value. In addition you are calculating an average over fifteen cycles, that is going to be essentially constant no matter where you start it.
__________________
� � � � Tom Gutman

[ptc-1218903]

Hi
Thanks for the reply.
Could you please let me know how to plot the envlope of r(t)^2 vs time.
Thanks

[ptc-1218903]

Hi
How to calculate the crest factor and Peak to average power ratio of r(t)?

[ptc-1368288]

Your attachment is not "Save as" readable for lower versions than 13, 14.

jmG

[ptc-1218903]

Saved as Mathcad11

[LouP]

Average power of orthogonal sinusoids is additive. In doing a numeric integral as in your sheet, you need to do the averaging over a full period, which is 1/delta, not 1/f.

The average power of a sinusoid A*sin(w*t) is just (A^2)/2. with A = 1, this gives Pavg = 1/2 for each such component, for a total Pavg = 1 in your example.

The instantaneous peak power is given by the square of the sum of the amplitudes. For two orthogonal sinusoids (differing freq's) of amplitudes A1 and A2, the avg power is (A1^2)/2 + (A2^2)/2, and the peak power is (A1+A2)^2. In your exampe, this is (1+1)^2 = 4.

Lou

[ptc-1218903]

I am still confused.... Each sinusoid has the peak power of 1W and average power of 0.5W. As per your analysis, peak power of sum of these sinusoids is 4W. Since each sinusoid has peak power of 1W, peak of the sum sinusoid will be 2W, noit 4W, right. Where is the extra 2W coming?

[ptc-1368288]

I am still confused.... Each sinusoid has the peak power of 1W and average power of 0.5W.

==> You decide that the integrand r(t)� is representative of "Power", but nothing from the work sheet explicit such an assumption. r(t)� is just a plot of anything in the time domain !

How to calculate the crest factor and Peak to average power ratio of r(t) ?

==> If you say that r(t)� is "Power" in the time domain, then there is no crest power unless you consider the peak power as the max of r(t)�. The cumulative integral delivers instantaneous cumulated delivered "Energy"


I assume you are looking for the Root Mean Square value as that is the DC equivalent of period wave(s)

==> I don't understand: RMS of what ? r(t)� is DC varying from 0...4 [maybe 4 W ?]

jmG

[LarryLederer-di]

>==> I don't understand: RMS of
>what ? r(t)� is DC varying
>from 0...4 [maybe 4 W ?]
>
>jmG

The average power of a sinusoidal wave (or any symmetrical wave that varies from +y to -y about the x axis is zero. However, if there is resistance in the circuit then you can calculate the DC equivalent of the wave using the Root Mean Square method.

Instead of an average value which is zero you get the effective value.

So Ieff*Veff=Peff or Ieff^2*R=Peff etc.

[ptc-1368288]

...OK, then what is Peff in the collab work sheet ?
Just add in the work sheet I have returned.


jmG

[TomGutman]

>>The average power of a sinusoidal wave (or any symmetrical wave that varies from +y to -y about the x axis is zero. <<

No, that is the amplitude. Amplitude is not power, don't ever get the two confused.
__________________
� � � � Tom Gutman

[LarryLederer-di]

On 3/7/2009 3:33:12 PM, Tom_Gutman wrote:
>>>The average power of a sinusoidal wave (or any symmetrical wave that varies from +y to -y about the x axis is zero. <<
>
>No, that is the amplitude.
>Amplitude is not power, don't
>ever get the two confused.
>__________________
>� � � � Tom Gutman

That is not correct. If there is no resistance (load or line losses) in the circuit there is conservation of energy and the power is zero. Even if you have a capacitor or an inductor in the circuit no net power is dissipated. I am an electrical engineer and this was made clear to us in our basic electrical classes. By using the RMS voltage or current(which is the DC equivalent) then you can multiply either times the equivalent resistance of the circuit to get power.
Veff=.707*Vmax for a sine wave.

[TomGutman]

>>If there is no resistance (load or line losses) in the circuit there is conservation of energy and the power is zero.<<

That is correct. But if the circuit has no impedance, then the voltage is zero. The only possible amplitude is then current. The instaneous power is therefore zero (the product of the voltage and the current).

While any symmetric waveform (same shape above and below zero) will have an average amplitude of zero, the power, both instanteous and average, depends on the relationship between the current and voltage waveforms. Either can be zero.

But still, the amplitude of either current or voltage is not power, and, in general, neither is the square of the amplitude. I repeat -- don't get amplitude and power confused.
__________________
� � � � Tom Gutman

[ptc-1368288]

>But still, the amplitude of either current or voltage is not power, and, in general, neither is the square of the amplitude. I repeat -- don't get amplitude and power confused <.
____________________________

Absolutely correct. If there is not infinitesimal change between open circuit and closed circuit it means the closed circuit has infinite impedance therefore 0 power. Vice versa: the max power dissipated occurs at minimum amplitude of the measured wave form.

jmG

[LarryLederer-di]

>But still, the amplitude of
>either current or voltage is
>not power, and, in general,
>neither is the square of the
>amplitude. I repeat -- don't
>get amplitude and power
>confused.

Getting back to the original question "How to determine the power of two sinusoidal waves"

The correct solution is

V1eff^2/R + V2eff^2/R = Pavg

V1eff and V2eff are the RMS values of the wave forms

[TomGutman]

>>V1eff^2/R + V2eff^2/R = Pavg<<

That is incorrect. To show this, consider the simple case where V1 and V2 have equal amplitudes but are 180° out of phase. Then V1=-V2. The sum is identically zero, the power is identically zero, so the average power, over any time interval, is zero.

The question asked is for a more complicated case, where the two amplitudes are of different frequencies so that the relative phase is constantly changing. Thus so is the instanteous average power. Note that "instaneous average" is an oxymoron, but yet that is what is wanted (and it actually makes sense).
__________________
� � � � Tom Gutman

[LarryLederer-di]

On 3/8/2009 2:31:56 PM, Tom_Gutman wrote:
>>>V1eff^2/R + V2eff^2/R = Pavg<<
>
>That is incorrect. To show
>this, consider the simple case
>where V1 and V2 have equal
>amplitudes but are 180�
>out of phase. Then V1=-V2.
>The sum is identically zero,
>the power is identically zero,
>so the average power, over any
>time interval, is zero.
>
>The question asked is for a
>more complicated case, where
>the two amplitudes are of
>different frequencies so that
>the relative phase is
>constantly changing. Thus so
>is the instanteous average
>power. Note that "instaneous
>average" is an oxymoron, but
>yet that is what is wanted
>(and it actually makes sense).

Correct Tom I used the super position theorem incorrectly in this instance my mistake. You must first sum the two sinusoidals vectorially then find the RMS value of the resultant wave form. Your comment about the "instantaneous average" is unclear to me. By using the RMS value of the resultant, you have an equivalent DC source. As DC does not vary with time it is the average value. Of course all of this assumes that the load is purely resistive. If there is inductance or capacitance involved then the currents will be needed to be calculated as even the out of phase portion of the resultant current will produce I^2 R losses.

[TomGutman]

>>Your comment about the "instantaneous average" is unclear to me.<<

The original problem is the sum of two sinusoids with different periods. But equal amplitude. Using various trigonometric identies one can rewrite that as the product of two sinusoids. These will also have different periods, both different from either of the periods of the two original sinusoids. Because of the frequencies involved, these two will have very different periods. One can consider the one with the high frequency to be the carrier, or base, and the one with the low frequency to be a modulation. The low frequency component is the envelop of the signal, and can be considered as the instaneous amplitude of the wave. Its square can then be thought of as the instaneous average power (assuming a pure resistive load).
__________________
� � � � Tom Gutman

[LarryLederer-di]

On 3/8/2009 5:39:48 PM, Tom_Gutman wrote:
>>>Your comment about the "instantaneous average" is unclear to me.<<
>
>The original problem is the
>sum of two sinusoids with
>different periods. But equal
>amplitude. Using various
>trigonometric identies one can
>rewrite that as the product of
>two sinusoids. These will
>also have different periods,
>both different from either of
>the periods of the two
>original sinusoids. Because
>of the frequencies involved,
>these two will have very
>different periods. One can
>consider the one with the high
>frequency to be the carrier,
>or base, and the one with the
>low frequency to be a
>modulation. The low frequency
>component is the envelop of
>the signal, and can be
>considered as the instaneous
>amplitude of the wave. Its
>square can then be thought of
>as the instaneous average
>power (assuming a pure
>resistive load).

I have no problem with your analogy of a modulated wave, but it does not address my issue with the fact that merely squaring the signal gives you the power. Even taking the integral of the signal squared does not give you the average power it does give you the . See the attached spreadsheet to see how to calculate average power of two sine waves.

[ptc-1368288]

>...but it does not address my issue ..<<br> __________________________

Take the original work sheet and address the problem directly, including pure fictive resistance only. So a result can be seen, qualified and quantified as function of the components, especially as function of the "wave(s) form(s)".

jmG

[TomGutman]

>>but it does not address my issue with the fact that merely squaring the signal gives you the power<<

In general, no. But if the signal is voltage or current across or through a pure resistance, then it is proportional to the power. In signal processing it is common to treat the square of the signal as a power (look at the definitions of decibels). If you look at the original question, and the sheets attached thereto, it is clear that this is the case here.
__________________
� � � � Tom Gutman

[ptc-1368288]

On 3/8/2009 11:04:27 AM, Derbigdog wrote:
...
>Getting back to the original
>question "How to determine the
>power of two sinusoidal waves"
>
>The correct solution is
>
>V1eff^2/R + V2eff^2/R = Pavg
>
>V1eff and V2eff are the RMS
>values of the wave forms
____________________________

If that would be true, no single motor would turn.
If not forced in one specific direction either by hand or by capacitive/inductive winding they want to turn equally in the two opposite directions. If you want to use the RMS in the original problem, you must evaluate the average energy over time [cumulative integral], relate energy to time, calculate P then relate to a 60 Hz source and deduce an "equivalent RMS".
In reality the original problem is not specified.

jmG

[ptc-1368288]

...no single motor = no single PHASE motor

jmG

[LarryLederer-di]

On 3/8/2009 3:29:27 PM, jmG wrote:
>...no single motor = no single
>PHASE motor
>
>jmG

A single phase motor works because the windings as made to create an out of phase component between the rotor and the stator. That provides a torque on the rotor that causes it to turn. Power is consumed by moving what ever is connect ot the rotor. The out of phase component is proportional to the resistance (load) to the rotor being able to turn.
So I do not see your point.

[ptc-1368288]

On 3/8/2009 3:54:44 PM, Derbigdog wrote:
...
>A single phase motor works because the
>windings are made to create an out of
>phase component between the rotor and
>the stator. That provides a torque on
>the rotor that causes it to turn....
>___________________________

Totally wrong,

Single phase small motors have either a starter winding or combination of capacitor + winding or no other starting facility than hand push in the desired rotation. Larger single phase motors [synchronous motors] will have like a car, a motor starter, which motor starter is designed for reversible rotation.
3 phases motors, hexaphases motors of huge size (vg: 10000 Hp) also have starter motors. Power consumption is not measured as you describe, rather by the Hydro Power Watt/Hour meter.
Once appropriately connected, 3 phases motors have a unique direction of rotation.

For the original problem in this thread, some lab equipment is required to heat a heat bath and deduce the power delivered by the "generating circuit".

jmG

[PhilipOakley]

All this discussion shows that the term "Power" is used in many subtly different ways, just as "Energy" is.

As long as the reader remembers it is not a case of Right or Wrong, but 'In context' vs. 'Out of context'. The had part being to determine which/whos context!

Philip Oakley

[ptc-1368288]

...>I am an electrical engineer...<
_______________________
Power is an instantaneous "quantified quality", whether resistive, inductive, capacitive or mixed or not consumed. Energy is power accumulated over time. Power measured, potential power... the question from the originator is unclear.

jmG

[rrandall]

On 3/7/2009 11:47:58 AM, Derbigdog wrote:
>>==> I don't understand: RMS of
>>what ? r(t)� is DC varying
>>from 0...4 [maybe 4 W ?]
>>
>>jmG
>
>The average power of a
>sinusoidal wave (or any
>symmetrical wave that varies
>from +y to -y about the x axis
>is zero. However, if there is
>resistance in the circuit then
>you can calculate the DC
>equivalent of the wave using
>the Root Mean Square method.
>
>Instead of an average value
>which is zero you get the
>effective value.
>
>So Ieff*Veff=Peff or
>Ieff^2*R=Peff etc.

I believe you have a mistake in that you are not calculating the average power flow over the full period of the resultant waveshape. That is the period is tau = 4 for one frequency being 1.25 times the other. If you do this you will find that the average power generated by the waveshape (v(t)+v1(t))^2 is 1 not 1.1192. This is the same solution as if you use superposition and treat each sinusoid separately.