Catto / Topic Map · Signals and Systems Level 2
Signals and Systems · 144 h · Topic 15 of 34

Even and odd signals

Every signal splits into an even part and an odd part, in exactly one way. That split decides which harmonics it can have and explains why a push-pull stage cancels the even ones.

Even and odd Decomposition Hermitian symmetry Harmonics Push-pull

01Two symmetries

A signal is even if mirroring it in time does not change it, and odd if mirroring it is equivalent to changing its sign:

\[ \text{even: } x(-t) = x(t) \qquad\qquad \text{odd: } x(-t) = -x(t) \] In discrete time it is the same, with \( n \) in place of \( t \).

The cosine is even, the sine is odd, and an arbitrary signal is neither one nor the other. From this comes a minor observation: every odd signal is zero at the origin, because \( x(0) = -x(0) \) forces \( x(0) = 0 \).

The interesting thing is not classifying signals, but that these two symmetries split any signal in two, and that this split says concrete things about the spectrum and about the distortion of an amplifier.

02Every signal is the sum of an even one and an odd one

And in only one way. The recipe is so simple it looks like a trick, but it is verified in one line:

\[ \mathcal{E}v\{x(t)\} = \frac{x(t) + x(-t)}{2}, \qquad \mathcal{O}d\{x(t)\} = \frac{x(t) - x(-t)}{2} \]

The first is even by construction —replacing \( t \) with \( -t \) leaves the same two terms— and the second is odd for the same reason. And added together they give \( x(t) \), because the \( x(-t) \) terms cancel.

The example to keep at hand is the step. Its even part equals \( 1/2 \) everywhere, and its odd part is half the sign function:

\[ u(t) = \underbrace{\frac{1}{2}}_{\text{even}} + \underbrace{\frac{1}{2}\operatorname{sgn}(t)}_{\text{odd}} \] A step is a DC level plus an antisymmetric jump. This decomposition shows up when computing its Fourier transform.
Lab · splitting a signal in two

Above, the signal and its mirror image; below, the even part and the odd part. Check that added together they give back the original, and see how the energy is divided.

03The rules for products and integrals

OperationResultLooks like
even × eveneven\( (+)(+) = + \)
odd × oddeven\( (-)(-) = + \)
even × oddodd\( (+)(-) = - \)

And over a symmetric interval, the integral of an odd signal is zero, while that of an even one is twice the integral over half:

\[ \int_{-a}^{a} x_{\text{odd}}(t)\,dt = 0, \qquad \int_{-a}^{a} x_{\text{even}}(t)\,dt = 2\int_{0}^{a} x_{\text{even}}(t)\,dt \]

Those two lines are what make half of the coefficients of a Fourier series vanish before computing anything, and what let you save half of the work when there is symmetry.

The energy is divided too

Since the product of the even part and the odd part is odd, its integral over the whole axis vanishes. The cross term disappears and what remains is \( E = E_{\text{even}} + E_{\text{odd}} \): the energy of the signal is the sum of the energies of its two parts, with no interference between them. The lab verifies it number by number.

04What symmetry does to the spectrum

This is where it stops being a curiosity. The symmetry of a periodic signal determines which harmonics it can have:

If the signal is…Its Fourier series…Example
EvenHas only cosines: \( b_n = 0 \)Centered triangle wave
OddHas only sines: \( a_n = 0 \)Centered sawtooth
Half-wave: \( x(t \pm T/2) = -x(t) \)Only odd harmonicsSymmetric square wave
Real, in general\( X(-j\omega) = X^{*}(j\omega) \)Any measurable signal

The last row is called Hermitian symmetry and it is the reason for something you see every day: on a spectrum analyzer only the positive frequencies are plotted. The negative part is not left out for convenience, but because it adds no information: for a real signal, the magnitude of the spectrum is even and the phase is odd.

Why a push-pull sounds different

The nonlinearity of an amplifier is expanded as \( y = a_1x + a_2x^2 + a_3x^3 + \cdots \). The even terms generate even harmonics —the second above all— and the odd ones, odd harmonics. A push-pull stage has odd symmetry by construction: the two halves process opposite half-cycles, and when subtracted they cancel all the even terms. That is why a push-pull has almost no second harmonic and its residual distortion is third-order. A single-stage class A amplifier, on the other hand, lets the second harmonic through, which is the one associated with a “warm” sound.

05In the lab

Exercise 1 · Decomposing by hand

For \( x(t) = e^{-t}u(t) \), obtain \( \mathcal{E}v\{x\} \) and \( \mathcal{O}d\{x\} \) and check that they equal \( \frac{1}{2}e^{-|t|} \) and \( \frac{1}{2}\operatorname{sgn}(t)e^{-|t|} \). Check that they add up to the original.

Exercise 2 · The harmonics of a push-pull

Measure with a spectrum analyzer the distortion of a class A stage and of a push-pull stage, with the same input signal and the same output power. Compare the amplitude of the second harmonic in each case and explain the difference through symmetry.

Exercise 3 · The square wave has no even harmonics

Generate a symmetric square wave and measure its spectrum. Check that only the odd harmonics appear and that they fall off as \( 1/n \). Then unbalance the duty cycle and watch how the even ones appear.

06Common mistakes

  • Believing every signal is even or odd. Almost none is: most have both parts.
  • Forgetting that odd × odd gives even. It is the rule of signs, and it is used all the time in Fourier analysis.
  • Confusing half-wave symmetry with odd symmetry. They are different conditions: the first eliminates even harmonics, the second eliminates cosines.
  • Thinking the negative spectrum is an artifact. It exists; what happens is that for real signals it is determined by the positive one.
  • Assuming \( x(0) \) can be anything in an odd signal. It has to be zero.

07Self-assessment

How do you get the even part of a signal?

\( \mathcal{E}v\{x\} = \frac{1}{2}\big(x(t) + x(-t)\big) \). The odd one comes out with a minus.

What is the even part of the step?

The constant \( 1/2 \). And the odd part is \( \frac{1}{2}\operatorname{sgn}(t) \).

What is the integral of an odd signal from −a to a?

Zero: what is on one side cancels what is on the other.

Which harmonics does a signal with half-wave symmetry have?

Only the odd ones. It is the case of the symmetric square wave.

Why does the spectrum analyzer show only positive frequencies?

Because of Hermitian symmetry: for a real signal, the magnitude is even and the phase odd, so the negative half adds nothing.

Why does a push-pull cancel the even harmonics?

Because its transfer characteristic is odd by construction, and the even terms of the expansion are subtracted between the two halves.

08Further reading

  • Alan V. Oppenheim and Alan S. Willsky. Signals and Systems. 2nd ed. (Spanish edition, “Señales y sistemas”), Prentice Hall, 1998. The even-odd decomposition and its role in the symmetry properties of Fourier analysis.
  • Adel S. Sedra and Kenneth C. Smith. Microelectronic Circuits. 7th ed. (Spanish edition, “Circuitos microelectrónicos”), Oxford University Press, 2016. The harmonic distortion of output stages, with the push-pull cancellation.
  • Simon Haykin and Barry Van Veen. Signals and Systems. (Spanish edition, “Señales y sistemas”), Limusa Wiley, 2001. The symmetries of Fourier series, tabulated with their consequences for the coefficients.
Development of the topic “Even and odd signals” of Signals and Systems (Level 2), based on the curriculum of the UTN Electronic Engineering program, 2023 curriculum — Ordinance No. 1849 of the UTN Higher Council. Back to the Topic Map · catto.ar