time avg. amplitude of 2 equal sine waves at random phase angles

Discussion in 'Math' started by PeteHL, May 22, 2016.

  1. PeteHL

    Thread Starter Member

    Dec 17, 2014
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    If two voltages of equal amplitude and frequency that are sine waves add at random phase angles (constantly varying), then the time average amplitude of their summation equals 1.414 times the amplitude of one of them. So what I want to do is calculate the time average amplitude of the summation of two voltages V1 and V2 given that I am able to restrict the range of phase angles in which the two voltages add. What I have below is what I hope is a correct way to calculate the time avg. amplitude where the phase angle can be anywhere in the range of 0 degrees to 364 degrees. If it is correct, then I would use this type of analysis to calculate time avg. amplitude where some relative phase angles are omitted or not allowed.

    |V1| = |V2| = 1 Volt peak, both voltages of equal sine wave frequency
    Relative phase angle, degrees---(V1 + V2), peak volts---(V1 + V2)^2
    0---2.00---4.00
    45---1.85---3.42
    90---1.41---2.00
    135---0.76---.058
    180---0.00---0.00
    225---0.76---0.58
    270---1.41---2.00
    315---1.85---3.42 ​
    Sum of peak voltage squares = 16.00
    Average peak voltage square = 16.00/ 8 samples = 2.00
    Thus, time avg. peak voltage = SQRT(2.00) = 1.414

    Is this a valid way to arrive at 1.414 as the time average amplitude? Lord Rayleigh analyses this problem in Vol. 1 of his book "Theory of Sound", but his analysis with my limited knowledge of math is not understandable to me. So I realize this is probably a rather crude way to arrive at an answer, but is the only method that I could think of given my non-engineer range of comprehension.

    Primarily I would like to know if this way of analyzing the problem is correct or not. Please do keep in my mind that I am seeking a solution comprehensible to non-engineers, if that is possible.

    Best Regards,
    Pete
     
    Last edited: May 22, 2016
  2. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    :confused: If by phase angle you mean the phase of one relative to the other and their frequencies are identical, why is the phase angle varying?
    Do you really mean 'average', or should that be 'rms'?
     
    Last edited: May 22, 2016
  3. WBahn

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    Mar 31, 2012
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    I think the phase angle is varying only because that is part of the statement of the problem he is trying to solve.

    If I understand what he is asking, the problem could be phrased to ask about either the average or the expected amplitude.

    For instance: Two sinusoids of equal amplitude and frequency but arbitrary phase are added together. What is the expected amplitude of the summed signal?

    This would be looking for an average value.

    Closer to what the TS is doing: Two sinusoids of equal amplitude and frequency but with randomly varying phase are summed together. What is the amplitude of the summed signal?

    This could reasonably be answered either with the average amplitude or with the effective amplitude (i.e., rms) of the resulting waveform. It depends on how the answer is to be used, which is not given. So the TS really needs to specify things a bit tighter.
     
  4. WBahn

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    Mar 31, 2012
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    The peak voltage is not going to be the square of the sum of the individual peak voltages. If nothing else, that yields voltage squared.

    What kind of math background is your target audience assumed to have?
     
  5. PeteHL

    Thread Starter Member

    Dec 17, 2014
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    Think of the voltages V1 and V2 as being the output of first and second microphones that are picking up a sound source that is located at different distances from the mikes. If the difference of distance is significant relative to the wavelength of the sound, then relative phase of V1 and V2 can be considered to be random. The problem is to solve what on average would be the output level of a mixer equally mixing V1 and V2.

    The problem as is posed is idealized to make solving it easier. But for example if in a first case you have two loudspeakers driven by separate and uncorrelated signal generators to equal levels, the combined level is 3 dB greater than that of just one. In a second case If the same signal generator drives both loudspeakers, then the combined level is 6 dB greater than that of just one loudspeaker. In other words in the first case of the generators being uncorrelated (or random), the acoustic intensity adds in quadrature. In the second case there is linear summing of the outputs of the two speakers.

    My intention was to say average, not root mean square which is different.
     
  6. PeteHL

    Thread Starter Member

    Dec 17, 2014
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    I agree. What I'm hoping is correct is that the average peak voltage is the square root of the average of the addition of each of the various summation peak voltages squared. My "table" could be better, but right now my computer is lacking some software that I've made use of in the past.

    My reason for posting was to get some feedback as to the correctness/ incorrectness of my approach to the problem, without thinking about who might be able to see a mistake in what I did.

    Thanks,
    Pete
     
  7. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hello,

    If you have two sine waves where one has one constant phase shift and the other has another constant phase shift then you can reference all the calculations to the sine wave with the first phase shift and call that phase shift zero and the second phase shift (the only one left now) as phb-pha, assuming for now that phb>pha.
    This simplifies everything to:
    A*sin(w*t)+B*sin(w*t+phb)

    Unfortunately, the time average of this is always zero unless you want to consider the full wave rectified average that we sometimes look for. The RMS value however is not hard to calculate:
    Vrms=sqrt(A^2+2*cos(phb)*A*B+B^2)/sqrt(2)

    You can try that and see if you get the results you want.
    Once you have that, you can calculate the peak with Vrms*sqrt(2), and once you have that you can calculate what we sometimes call the sinusoidal 'power line' average by multiplying by 2/pi.
     
    Last edited: May 22, 2016
  8. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Can you understand Rayleigh's result, even if you don't understand his method? What is the result? On what page in his book is his analysis to be found?
     
  9. MrAl

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    Jun 17, 2014
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    Hello again,

    If both sine waves are of the same amplitude then we have a much simpler solution:
    Vrms=A*sqrt(cos(phb)+1)

    A plot of this is shown in the attachment, with Vrms normalized. This shows the variation in Vrms (and hence Vpk) as the phase shift of the second wave changes.
    Note when phb (phase shift in wave b) is pi the amplitude is zero because then the two waves are 180 degrees out of phase and so cancel each other.
     
    Last edited: May 23, 2016
  10. PeteHL

    Thread Starter Member

    Dec 17, 2014
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    Unfortunately no, because it isn't clear to me which part of his analysis is the result. There is very little explanation, and it is very math intensive. It is on pp. 35- 42.
     
  11. PeteHL

    Thread Starter Member

    Dec 17, 2014
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    The time average RMSV would then be the one-half times the area underneath the curve of your plot, is that correct? That should be equal to 0.707 RMSV. Taking the integral (calculus) of Vrms=A*sqrt(cost(phb)+1) would yield the time average rmsv?

    Even if that is true, it won't help, as I need to calculate time average amplitude where using signal processing the amplitude corresponding to phase difference is other than the summation of the two sine waves.

    It would be helpful if you could comment on whether or not the method of my first post in arriving at time average amplitude is correct. If that method is correct, then it is easy for me to calculate time average amplitude where the summing of the two sine waves is manipulated to be other than straight summation.

    Thanks,
    Pete
     
  12. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Well no, you dont have to integrate anything it is all already done for you in that equation and that graph. The graph shows for example that when the phase difference is exactly pi (3.14159...) then the rms voltage is zero, and that is from that one equation with cos() in it. You just plug in the phase difference. If you also have amplitude differences, then you must plug in the A and B in the previous equation too.
    For example, with both amplitudes equal to 1v peak and zero phase difference, the RMS value is 1.41, so the peak is 2, and the average is 2*2/pi=4/pi=1.27, keeping in mind this is the average of the full wave rectified wave which is sometimes a useful measurement, as the actual time average is always zero (sine wave). Keep in mind that the result of the addition is always a sine wave too.

    Another example, the phase difference is 1 rad. Looking at the graph, that corresponds to Vrms=1.2 volts (approximate reading from the graph). The peak is then 1.2*1.41=1.7, and the average is 1.08 (assuming both sines have amplitude 1v peak again).
    If the two sines have amplitudes of 3v peak, then the average is 3.24 because 3*1.08=3.24 volts.
    Note again here the time average is zero, but the 'power line' sine average is 3.24 which has some usefulness in some calculations and some measurements. That's the same as if the wave was full wave rectified before finding the average.

    Just to note, the 'power line' sine average can be found from:
    (1/T)*integrate(|sin(w*t)+sin(w*t+phb)|)
    although it's a little tricky because we have to solve for time values first, so calculating the RMS value first then calculating the peak and average is a little easier. Both methods will yield the same result.

    I went by what you stated in post 1:
    You said that the two waves were ADDED, and so the math i developed went according to that statement in your first post. If they are not added but there is some other arrangement, then you should state what that is.

     
    Last edited: May 23, 2016
  13. PeteHL

    Thread Starter Member

    Dec 17, 2014
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    Mr. Al, thanks for your response, and others too.

    Maybe it will help if I try to restate the problem. There are two voltages V1 and V2. The peak voltage of both is equal to 1V. The frequency of V1 and V2 is the same . These two voltages will add at any phase angle; at some point in time it could be 0 degrees, at another point in time 33 degrees, and so on. In other words relative phase between V1 and V2 over a period of time is throughout the possible range of phase angle. The question to answer is, taking into account that over some period of time V1 and V2 randomly are in a relative phase relationship throughout the full range of 0 degrees to 359 degrees, then what is the time-average voltage value of (V1 + V2)? Applied to my example, that time-average amplitude is known to be 1.414 times 1V peak.

    In my first post, I'm showing a method for arriving at the time-average amplitude based on first calculating (V1 + V2) at eight different relative phase angles, 0 deg., 45 deg., 90 deg., 135 deg., 180 deg., 225 deg., 270 deg. and 315 deg. Those summations are done using phasor algebra. The second step is to square each of the eight different summations of (V1 + V2) as power is proportional to the square of voltage. The third step is to add the squared summation voltages. The 4th and final step is take the square root of the addition of the third step. The result is a peak voltage of 1.414 volts, which is what it should be. That is, V1 and V2 summed in quadrature is equivalent to the time-average amplitude of V1 and V2 continuously summing at random relative phase angles.
     
  14. BR-549

    Well-Known Member

    Sep 22, 2013
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    The phase angle is never random.

    If you want only additive components, limit phase to 180 degrees in your cypher.

    Phase angle will be sequential at a rate, which itself may be varied.

    The phase angle between microphones will always be proportional to the distance between them, never random.

    Sound engineers can use microprocessors to set all speaker and microphones in a sound setup using this principle.

    Some systems can indicate a satisfactory location for these transducers automatically for you.

    If the physical location has limits, phase adjustments can help compensate.

    Don't know if my comments help, still not sure what your trying to figure out.
     
  15. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Rayleigh's result is that if you have n voltages like yours, the average amplitude in the situation you have described is simply SQRT(n).

    If you only omit single phase angles, there will be no change from the SQRT(n) result. You would have to omit a range of phase angles to get any change.
     
  16. PeteHL

    Thread Starter Member

    Dec 17, 2014
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    Random is probably not the best way to describe it. In each particular case there is definite relative phase relationship. Maybe it would be better to say that if for example the frequency of sound produced by a source is changing, then the relative phase of the output voltages of two microphones spaced unequal distances from the source varies over a wide range, over time.
     
  17. PeteHL

    Thread Starter Member

    Dec 17, 2014
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    Thanks! But what exactly does n stand for? Is that the quantity of voltages to be summed or the equal amplitude of each of the voltages to be summed? In my example. what is the value of n?

    Regards,
    Pete
     
  18. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    It's the number of voltages to be summed.

    In your example, it's 2.
     
    Last edited: May 24, 2016
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