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Average Power

ptc-1218903
1-Newbie

Average Power

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
52 REPLIES 52

The square of the amplitude is not power. Power is voltage times current.

In the special case when you are looking at a pure resistance voltage and current are proportional to each other, and so the power is proportional to the square of either.

To calculate an energy budget (which will balance) you have to consider the entire circuit. How are you adding the two sinusoids? Indeed, what are you measuring to get them? Once you specify the circuit that is generating this sum of sinusoids the energy budget can be calculated. But not before.
__________________
Tom Gutman

On 3/6/2009 11:25:51 PM, leyo123 wrote:
>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?

First, a word about "power." In the physical world, instantaneous power is voltage times current, which, depending on conditions, may have a positve, zero or negative average value. In signal processing, as Tom has pointed out, power is simply defined as the squared magnitude of the signal (equivalent to a voltage in a one ohm R). Often, there are no amplitude dimensions used, although time and frequency dimensions may be used when discussing averagong times or power spectral densities. I treat the case here as a signal processing question, since neither circuit nor units are given.

Also, AVERAGE power of orthogonal sinusoids (different freq's) is additive. Instantaneous peak power is not (you have assumed this should be true). In my earlier example, the instantaneuos peak power is the amplitude squared = (A1+A2)^2, which is not A1^2 +A2^2. When the two signals are orthogonal, then the cross term 2*A1*A2 will have a zero average, contributing no average power , but will still have a nonzero peak power component(as long as both terms are nonzero). If the signals are not orthogonal, then this cross term will have a nonzero average, thus contributing to the total average power also.


The following may help explain the original result. In example(43), the full period of the sum r(t) is 1us. The instantaneous power is shown in the plot. The peak of the signal r(t) is 2, and the peak power, the peak of r(t)^2, is = 4. This is the condition at t= 0, 1us, ... Near t=0, the local average (visualize over a single cycle) of the power is 2 (variation is from 0to 4). This decreases to zero (both instantaneous pk and local average) at the center of the graph at t = 0.5us, and increases again to a local average of 2 at t = 1us. Over the full 1us period of the power waveform, the average is 1. the calculation gives the average over the full period of the waveform; the true DC component.

The peak /average power ratio depends on the waveform. for a single sinusoid, pk/avg = 2. For other combinations, the result will vary; it must be calculated case by case. FOr example(43), with two sinusoidal components, the single sinusoid value of 2 just doesn't apply. The calc gives pk/avg = 4.

Lou
LouP
11-Garnet
(To:LouP)

I should add that the "average" calculated in example(43) is over the interval (0, 0.1us), which is just the first local cycle with pk=4 and avg=2. This is NOT the full period of r(t), which is 1us, and so will not give the true DC average value. The same local "average" over the interval (0.5us, 0.6us) will give a value close to zero.

An average over any interval which is an integral number of periods will give a result which is independent of the intial time of the interval. This is not the case in the example.

Lou

>In the physical world, instantaneous power is voltage times current, which, depending on conditions, may have a positve, zero or negative average value <<br> ___________________________

Lou,

I have difficulty to see "negative power" looking at my "Hydro power meter". Whichever direction my motors rotate, that bloody meter always totalizes. Power is instantaneous potential energy equivalent, Hydro Corporations have understood that and made their gadget work for them, not for me or you. The best way to measure the power delivered by the original project is to connect to a "Power meter". The mid point of the cumulative integral in relation to time should help.

jmG

THis is all about sign conventions, not the physical situation. For a given system with v(t), i(t) signs defined normally, the power "into" that port is as I described, and its average may be +, 0 or -. Whatever it is, if the "port' is defined in the opposite direction, looking 180 degrees the other way, the sign of the current is reversed, and the average power flow "into" that port is the negative of that in the previous case. Both cases define the physical power flow in the same direction, only the reference direction is changed. If the average power is negative, then the network you are looking a is a source of power, not a sink.

Lou

On 3/5/2009 2:49:28 PM, leyo123 wrote:
>Saved as Mathcad 11
_____________________

If what you want to do is to integrate r(t)�, the solution is the cumulative integral. It gives the integral value at any time (t) and the last value is the equivalent of the numerical integral you are trying to do. Make sure you doctor the cumulative integral as mentioned by Lou. I understand nothing more than integrate r(t)�. If something else applies, you must express it in term of function or relationship.

jmG



... add points for a better view.

jmG

Hi
The cumulative integral works good.
Thanks

replace limit of integration with t, and divide integral by t, instead of what you used.

You'll then get a sinusoidally varying value superimposed on another sinusoidally varying value.

TTFN,
Eden

On 3/5/2009 1:12:20 PM, leyo123 wrote:
>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.

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

>I am expecting a DC
>value with a signal of
>frequency delta riding, but I
>am not getting this.<<br> ______________________________

I have returned the DC c/w delta riding,
but you didn't acknowledge.

jmG

Turns out the answer is simple. I'm surprised no EE has popped up with it. Just use complex amplitudes.
__________________
� � � � Tom Gutman
LouP
11-Garnet
(To:TomGutman)

Not quite that simple. In the sinusoidal steady state, with v(t) and i(t) at a port having complex amplitudes V and I, the complex power is defined as P = (1/2) VI* = A+jB (assuming nonzero freq). the real part A is the average power into the port, and the imaginary part B gives the average reactive stored energy. The factor of 1/2 is important, since creating the complex form adds another term (the imaginary part) to the signal with as much real power as the real part. In your example, you have created a complex signal as a function of time, but you are not really using complex amplitudes.

The use of the complex amplitudes assumes an implicit time dependence on a single frequency, and the complex amplitudes are not fcts of time. As a result, when dealing with amplitudes representing different freqs, it is the restored time dependent signals that add, not the complex amplitudes. the freq components still need to be kept separate, and powers calc freq by freq, then added. In this case, we have two amplitudes R1=1 at f1, and R2=1 at f2. the total avg power is given by (1/2)Re(|R1|^2)+(1/2)Re(|R2|^2) = 1.

Lou

On 3/10/2009 8:37:52 AM, lpoulo wrote:

>Not quite that simple. In the

>sinusoidal steady state, with

>v(t) and i(t) at a port having

>complex amplitudes V and I,

>the complex power is defined

>as P = (1/2) VI* = A+jB



Actually complex power is S=VI* or S=P+jQ

where P and Q are the average values



P = 1/2(Vmax)(Imax)cos(theta) or

P = (Vrms)(Irms)cos(theta)



Q = 1/2 (Vmax)(Imax)sin(theta)

Q = (Vrms)(Irms)sin(theta)



In the original question I did not see a complex load mentioned. Can I ask where the P=1/2 VI* came from?

On 3/10/2009 10:43:32 AM, Derbigdog wrote:
In the original
>question I did not see a complex load
>mentioned. Can I ask where the P=1/2 VI*
>came from?

Usually these 1/2 factors occur when there is positive and negative frequencies to consider in the maths analysis (ffts) in the complex domain. Endless fun and confusion is available...

Philip Oakley

On 3/10/2009 12:13:33 PM, philipoakley wrote:
>On 3/10/2009 10:43:32 AM, Derbigdog
>wrote:
>In the original
>>question I did not see a complex load
>>mentioned. Can I ask where the P=1/2 VI*
>>came from?
>
>Usually these 1/2 factors occur when
>there is positive and negative
>frequencies to consider in the maths
>analysis (ffts) in the complex domain.
>Endless fun and confusion is
>available...
>
>Philip Oakley
________________________________

I have yet to see negative frequencies from fft's that take the Y values. It does not care time neither frequencies.

jmG



>In the original question I did not see a complex load mentioned. Can I ask where the P=1/2 VI* came from?

The complex quantities come from the complex generalization of the signals as well as the circuit elements. We can still analyze with complex amplitudes whether the circuit elements are real or complex.

If a real, single frequency signal a*cos(w*t+p) is taken as the real part of a complex signal A*exp(j*w*t), then the time dependence is all in the complex exponential exp(j*w*t), with the constant complex amplitude A = a*exp(j*p). The complex amplitude encodes both the magnitude and phase of the real signal. With these definitions of signals in terms of complex amplitudes V and I for voltage and current, the real average power P = Re{(1/2)VI*}.

The complex amplitudes are the scale factors of the complex signals corresponding to the real waveforms. They have no direct relation to whether any circuit elements are complex or not, but the whole point of using them is that amplitude and phase are automatically accounted for without dealing with factors such as "cos(theta)." Vrms, Vmax, theta, etc. are parameters of a real signal, not the complex equivalent. Using the complex forms.

A simple heuristic example: Let complex current amplitude I = I0*exp(jp) correspond to a real current i(t)= I0*cos(wt+p). The average (real) power dissipated in a resistance R is Pavg = avg[R*i(t)^2] = (1/2)R*I0^2, independent of phase p, as it should be. In terms of the complex amplitude I, this is Pavg = (1/2)*R*|I|^2 = (1/2)*R*I*Ic(conj) = (1/2)*V*Ic (the formula that gives the correct real power answer).

Lou
TomGutman
1-Newbie
(To:LouP)

I believe you are talking about a different use of complex numbers to represent voltages and currents, what I have previously seen referred to as phasors. As you say, with this representation the phasor for a simple sinusoid is a constant, representing the amplitude and phase of the sinusoid (or, if you prefer, the in phase and quadrature components relative to some arbitrary phase reference). Phasors are used only when analyzing systems with a single frequency, and I think they work only if the variations in the phasor are slow relative to the frequency.

The factor of ½ in the power equation is from the integral of the square of the sign, the same thing that gives the factor of half the square root of two for the RMS value. You can see this if you use a pure resistive system and the voltage as the phase reference. Then both I and E are real. But I and E, as phasors, are the peak current and voltage, not the RMS values, hence the power is ½IE.

As you say, what I have done is quite different. I've used a complex time varying signal, where the real part is the observed signal (voltage, typically), and the inaginary part is, well, imaginary -- or something, I'm not an EE and don't fully understand it. Here we are dealing with signal processing concepts (as discussed earlier), there is no specific circuit in mind and no actual current. Power is apparently just nominal, the power that would be dissipated if the signal were measured across a one ohm resistor. I am not quite sure of the terminology that goes with this procedure, ISTR the term "complex envelop", and that may be what applies.

IAC, the request, as clarified in one of the requestors early post, was for the envelop of the instaneous power (again, taken as simply the square of the instaneous amplitude), and this method produces that. The instaneous average power would be half of that. So this answers the original question. I believe, without evidence or proof, that it is reasonably general and would work for a more complicated system (such as having different amplitudes for the two sine waves, where the sum to product transformation does not work).
__________________
� � � � Tom Gutman
PhilipOakley
5-Regular Member
(To:TomGutman)

On 3/10/2009 3:38:45 PM, Tom_Gutman wrote:
..
>
>The factor of � in the
>power equation is from the
>integral of the square of the
>sign, the same thing that
>gives the factor of half the
>square root of two for the RMS
>value. You can see this if
>you use a pure resistive
>system and the voltage as the
>phase reference. Then both I
>and E are real. But I and E,
>as phasors, are the peak
>current and voltage, not the
>RMS values, hence the power is
>�IE.
>
..
>__________________
>� � � � Tom Gutman

Absolutely right. I'd forgotten that particular 'mistake' of peak voltage vs. rms voltage.

The "negative frequency" problem still happens as well, but in other places in an analysis 😉

Philip Oakley

On 3/10/2009 4:46:28 PM, philipoakley wrote:
...
>The "negative frequency" problem still
>happens as well, but in other places in
>an analysis 😉
>
>Philip Oakley
__________________________

No fft's produce negative frequencies as the transform(s) starts essentially at 0 index. That you came across some such negative frequencies would surely indicate a wrong implementation of the algorithm. Here attached is the TA DFT, so you can check if your other software is correct or not.

For your "other places in an analysis"

4. The signal and its Hilbert Transform have identical energy because phase shift do not change the energy of the signal only amplitude changes can do that.

Read more in the attached.

jmG





The range of freq's (temporal or spatial) of a Fourier transform covers the full real number line. Complex signals contain both pos and neg frequencies. That an FFT - an implementation of a DFT - starts with zero index is an artifice of the implementation. A DFT givers one period of a periodic spectrum with period fs, the sampling freq. The particular period over which it is defined is arbitrary, but some chices are more convenient than others. Matchad returns a DFT defined on the interval (0,fs), where the sampling freq fs= 2*fN(Nyquist freq). The spectrum on the upper half interval from fs/2(=fN) to fs is the same as the spectrum one period lower, which is (-fs/2,0). It's up to the users (us) to keep track of what freq labels to use. For convenience in implementing filter fcts, I often use DFT versions which have zero freq in the center of the array, rather than at the beginning.

Lou
LouP
11-Garnet
(To:TomGutman)

On 3/10/2009 3:38:45 PM, Tom_Gutman wrote:
>I believe you are talking
>about a different use of
>complex numbers to represent
>voltages and currents, what I
>have previously seen referred
>to as phasors. As you say,
>with this representation the
>phasor for a simple sinusoid
>is a constant, representing
>the amplitude and phase of the
>sinusoid (or, if you prefer,
>the in phase and quadrature
>components relative to some
>arbitrary phase reference).
>Phasors are used only when
>analyzing systems with a
>single frequency, and I think
>they work only if the
>variations in the phasor are
>slow relative to the
>frequency.

By definition, the complex amplitudes (phasors) are constants, and have no "slow variations."


>The factor of � in the
>power equation is from the
>integral of the square of the
>sign, the same thing that
>gives the factor of half the
>square root of two for the RMS
>value. You can see this if
>you use a pure resistive
>system and the voltage as the
>phase reference. Then both I
>and E are real. But I and E,
>as phasors, are the peak
>current and voltage, not the
>RMS values, hence the power is
>�IE.

In order to give the correct real power (and independent of phase), the formula needs to be (1/2)Iconj*E. The alternate choice (1/2)I*Econj gives the same real power, but changes the sign of the imaginary part. The choice is by convention, and makes inductive stored enrgy positive).

>As you say, what I have done
>is quite different. I've used
>a complex time varying signal,
>where the real part is the
>observed signal (voltage,
>typically), and the inaginary
>part is, well, imaginary -- or
>something, I'm not an EE and
>don't fully understand it.
>Here we are dealing with
>signal processing concepts (as
>discussed earlier), there is
>no specific circuit in mind
>and no actual current. Power
>is apparently just nominal,
>the power that would be
>dissipated if the signal were
>measured across a one ohm
>resistor. I am not quite sure
>of the terminology that goes
>with this procedure, ISTR the
>term "complex envelop", and
>that may be what applies.

You created the analytic signal corresponding to r(t). In general, the complex analytic signal of x(t) is xa(t) = x(t) +jHilbert[x(t)]. It's defining characteristic is that its FT(xa) is zero for f<0, and FT(xa) = 2*FT(x) for f>0. (The scale factor of 2 gives x=Re(xa), assuming x is real). This complex xa(t) is the generalization of the single freq complex signal x1(t)=X0*exp(2*pi*f*t). |X0| is the (constant) complex envelope of x1(t). |xa(t)| is the time varying complex envelope of x(t). As a vector in the complex plane, xa(t) is the generalization of the fixed amplitude single freq rotating "phasor."


>IAC, the request, as clarified
>in one of the requestors early
>post, was for the envelop of
>the instaneous power (again,
>taken as simply the square of
>the instaneous amplitude), and
>this method produces that.
>The instaneous average power
>would be half of that. So
>this answers the original
>question. I believe, without
>evidence or proof, that it is
>reasonably general and would
>work for a more complicated
>system (such as having
>different amplitudes for the
>two sine waves, where the sum
>to product transformation does
>not work).

To be pedantic about the use of words, the envelope (see above comments) of instantaneous power can be defined, but is not the function expected, or of use. It has no physical (nor a signal prcoessing equivalent), and is similar to "RMS power." Average power is the quantity of interest, and RMS voltage (by def) is the signal domain quantity whose square gives this average. RMS power squared will give avg(Pwr^2) - by def, not usually of interest. The power envelope is in the same category, and its avg does not give the avg power of the complex signal. However, as you noted, the square of the complex envelope of the signal does have the desired properties. If EnvX(t) is the envelope of the analytic signal xa(t) derived from real x(t), then avg power of x = avg{x(t)^2] = (1/2)avg[EnvX(t)^2], and EnvX(t)^2 may be interpreted as the "instantaneous," time varying envelope power of the complex signal.

Keeping track of the domains, the facotr of 1/2, etc. can be a full time job in signal analysis:-).

Lou
TomGutman
1-Newbie
(To:LouP)

>>By definition, the complex amplitudes (phasors) are constants, and have no "slow variations."<<

I beg to disagree. They are parameters of the system, and, like any parameter, may vary with time. Even in cases where they are nominally constant the reality generally includes some variation. Measure the voltage in your power lines -- while the power companies spend a lot of effort to keep the voltage and phase constant, there are small variations over time. Even the frequency is not absolutely constant, but varies a bit over time. Perhaps a bit more relevant is modulation. Amplitude and phase modulation is specifically about having A (the complex amplitude) vary with time. In general the modulating signal has a much lower frequency (variation rate) than the carrier frequency.

>>the formula needs to be (1/2)Iconj*E<<

My example was specifically for the case where both I and E were real, hence conjugation is an identity operation and can be ignored. The purpose was simply to demonstrate that the factor of ½ did not come from the imaginary part but rather from the averaging of a sinusoid (RMS value).

>>To be pedantic about the use of words, the envelope (see above comments) of instantaneous power can be defined, but is not the function expected, or of use. It has no physical (nor a signal prcoessing equivalent), and is similar to "RMS power." <<

But the envelop of r² is what was explicitly asked for (see message # 5 in this thread). I agree that it has limited, if any, meaning. In this simple case the peak instaneous power (hence the peak of the envelop) is 4. The instaneous average power is half of that, 2, from the sinusoidal variation in the carrier. The average average power is half of that again, 1, from the sinusoidal variation of the envelop (modulation).

>>You created the analytic signal corresponding to r(t). In general, the complex analytic signal of x(t) is xa(t) = x(t) +jHilbert[x(t)]. It's defining characteristic is that its FT(xa) is zero for f<0, and FT(xa) = 2*FT(x) for f>0. (The scale factor of 2 gives x=Re(xa), assuming x is real). This complex xa(t) is the generalization of the single freq complex signal x1(t)=X0*exp(2*pi*f*t). |X0| is the (constant) complex envelope of x1(t). |xa(t)| is the time varying complex envelope of x(t). As a vector in the complex plane, xa(t) is the generalization of the fixed amplitude single freq rotating "phasor."<<

Oh, good. So despite not having a full understanding nor the proper terminology, I did get the procedure correct.
__________________
� � � � Tom Gutman
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