Symmetries and Fourier series

Before making a quantitative analysis of the Fourier series of the classical waveforms, we pause to make two useful observations about symmetries in waveforms and the corresponding symmetries in the Fourier series. First, a Fourier series may consist only of even or odd-numbered harmonics; this will be reflected in symmetries comparing a waveform to its displacement by half a cycle. Second, the Fourier series may contain only real- or pure imaginary-valued coefficients (corresponding to the cosine or sine functions). This is reflected in symmetries comparing the waveform to its reversal in time.

In this section we will assume that our waveform has an integer period $N$, and furthermore, for simplicity, that $N$ is even (if it isn't we can just up-sample by a factor of two). We know from chapter 9 that any (real or complex valued) waveform $X[n]$ is equal to its Fourier series (whose coefficients we'll denote by $Y[k]$):

$$X[n] = Y[0] + Y[1]{U^n} + \cdots + Y[N-1]{U^{(N-1)n}}$$

or, equivalently,

$$X[n] = Y[0] + Y[1](\cos(\omega n) + i \sin(\omega n)) + \cdots$$

$$+ Y[N-1](\cos(\omega (N-1) n) + i \sin(\omega (N-1) n))$$

where $\omega=2\pi/N$ is the fundamental frequency of the waveform, and $U = \cos(\omega) + i \sin(\omega)$ is the unit-magnitude complex number whose argument is $\omega$.

To analyze the first symmetry we delay the signal $X[n]$ by a half-cycle. Since ${U^{N/2}} = -1$ we get:

$$X[n+N/2] = Y[0] - Y[1]{U^n} + Y[2]{U^{2n}} \pm \cdots$$

$$+ Y[N-2]{U^{(N-2)n}} - Y[N-1]{U^{(N-1)n}}$$

We combine this with the original series in two different ways. Letting $X'$ denote half the sum of the two:

$$X'[n] = {{X[n] + x[n+N/2]}\over 2} = Y[0] + Y[2]{U^{2n}} + \cdots + Y[N-2]{U^{(N-2)n}}$$

and $X''$ the difference:

$$X''[n] = {{X[n] - x[n+N/2]}\over 2} = Y[1]{U^n} + Y[3]{U^{3n}} + \cdots + Y[N-1]{U^{(N-1)n}}$$

we see that $X'$ consists only of even-numbered harmonics (including DC) and $X''$ only of odd ones.

Furthermore, if $X$ happens to equal itself shifted a half cycle, that is, if $X[n] = X[n+N/2]$, then looking at the definition of $X'$ we get $X'[n] = X[n]$ and $X''[n] = 0$. This implies that, in this case, $X[n]$ has only even numbered harmonics. Indeed, this should be no surprise, since in this case $X[n]$ would have to repeat every $N/2$ samples, so its fundamental frequency is twice as high than in general for period $N$.

In the same way, if $X[n] = -X[n+N/2$, then $X$ can have only odd-numbered harmonics. This allows us easily to split any desired waveform into its even- and odd-numbered harmonics. (This is equivalent to using a comb filter to extract even or odd harmonics; see Chapter 7.)

For the second symmetry we compare $X[n]$ with its time reversal, $X[-n]$ (or. equivalently, since $X$ repeats every $N$ samples, with $X[N-n]$). The Fourier series becomes:

$$X[-n] = Y[0] + Y[1](\cos(\omega n) - i \sin(\omega n)) + \cdots$$

$$+ Y[N-1](\cos(\omega (N-1) n) - i \sin(\omega (N-1) n))$$

(since the cosine function is even and the sine function is odd). In the same way as before we can extract the cosines by forming $X'[n]$ as half the sum:

$$X'[n] = {{X[n] + X[-n]}\over 2} = Y[0] + Y[1]\cos(\omega n)+ \cdots + Y[N-1]\cos(\omega (N-1) n)$$

and $X''[n]$ as half the difference over $i$:

$$X'[n] = N {{X[n] + X[-n]}\over {2i}} = Y[1]\sin(\omega n)+ \cdots + Y[N-1]\sin(\omega (N-1) n)$$

So if $X[n]$ satisfies $X[-n] = X[n]$ the sinusoidal components are all cosines; and if $X[-n] = -X[n]$ they are all sines; and as before we can decompose any $X[n]$ (that repeats every $N$ samples) into the two.