Phased Summations 

Given a function f(x) with period 2p, consider the function s_{n}(x) consisting of the sum of n copies of f(x) shifted in phase at uniform increments of 2p/n. In other words, we define the function s_{n}(x) as follows 

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If we expand f(x) into an exponential Fourier series 

_{} 

we can evaluate the previous summation on a termbyterm basis. This gives 

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Factoring e^{ikx} out of each term in the parentheses, this can be written as 

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Clearly if k is a multiple of n, the angular argument of each exponential in the parentheses is an integer multiple of 2p, so each of them equals 1, and the quantity in parentheses is simply n. On the other hand, if k is not a multiple of n, then let g denote the greatest common divisor of k and n, so that we have k = gk and n = gn where k and n are coprime integers. In this case the expression in parentheses can be written as 

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The angular arguments have factors of 2p times the ratios 0/n, 1/n, 2/n, ..., (gn1)/n, so we can reduce these modulo 2p to g copies of the sequence 0/n, 1/n, 2/n, ..., (n1)/n. Thus the above quantity can be written as 

_{} 

Since k is coprime to n, the set of integers 0k, 1k, 2k, ..., (n1)k comprises a complete set of residues modulo n, so the expression in parentheses is just the sum of n complex numbers uniformly located around the unit circle, so the sum vanishes identically. Therefore, all the components with k not a multiple of n in the expression for s_{n}(x) vanish, and we have the result 

_{} 

Essentially the same derivation can be performed for strictly realvalued functions. For a realvalued function f(x) the coefficients c_{j} and c_{j} in the exponential Fourier series must be complex conjugates of each other, and if all the coefficients are strictly real the function can be expanded into a sum of cosines, i.e., 



The kth term of the cosine series is c_{k} cos(kx), and we can consider the sum of n copies of this general term with x replaced by x, x+2p/n, x+2(2p/n), x+3(2p/n), ..., x+(n1)(2p/n). Using the trigonometric identity 

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each cosine in this
summation can be written as 
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On the other hand, if k is not a multiple of n, then the quantities kj(2p/n) give evenly spaced divisors of 2p, so the sums all vanish. (The proof is the same as in the exponential derivation above.) This applies to both sine and cosine functions, as can be seen for the case n = 5 in the figure below. 

The x coordinates of the five evenly spaced points on the circle are the cosines of the five values, and the y coordinates are the sines. By symmetry it's obvious that the average of those five points (or in general, the average of n points evenly spaced around a circle) is the center of the circle, which implies that the sum of the cosines and the sum of the sines are both zero for any n. Thus we have 

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If the only nonzero
coefficients of the Fourier series expansion of the original function f(x)
are for indices less than n, then obviously the resulting overall summation
is just the constant nc_{0}. On the other hand, if the coefficient c_{n}
is nonzero but all the higher order terms (with coefficient indices that are
multiples of n) are zero, than the overall sum is n[c_{0} + c_{n}
cos(nx)], so it's a pure cosine wave with a constant bias. In general, the
overall summation is an arbitrary function (not a pure sine wave) with
frequency n times the frequency of the original function. Of course, this
need not be the fundamental frequency, because it's possible for some of the leading
coefficients in the original function expansion to be zero. For example, if
the only nonzero coefficient of the original function is c_{3n},
then the fundamental frequency of the summation will be 3n times the
frequency of the original function. But this occurs only because the
spectrum of the original function has no lowerorder components that are
multiples of n. In general the frequency is the greatest common divisor of
the frequencies of the nonzero components. 
To illustrate, consider the function shown below. 



If we sum this function over 3 copies separated in phase by 2p/3, we get the result shown in the figure below. 



This sum is simply s(x) = 3[2 + cos(6x)], because only the constant term and the 6th order term from the original function contribute. On the other hand, if we add a term 2cos(9x) to the function f(x) we have 



In this case there are two nonconstant terms whose orders are multiples of 3. The sum of 3 copies of this function separated in phase by 2p/3 is therefore 3[2 + cos(6x) + 2cos(9x)], which is plotted in the figure below. 


This has components with frequencies proportional to 6x and 9x, and the overall frequency is the greatest common divisor of these, which is 3x. 
