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

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

|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 = 16.000---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

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

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