Signals and Systems · 144 h · Topic 23 of 34

Fourier series and integrals

Every reasonable periodic signal is a sum of sinusoids whose frequencies are multiples of the fundamental. And since each one goes through an LTI system without changing shape, it is enough to know what happens to each harmonic.

Fourier series Orthogonality Dirichlet Gibbs Spectrum

01Building any periodic signal out of sinusoids

The result that opens the second half of the subject is this one: every reasonable periodic signal can be written as a sum of harmonically related complex exponentials, that is, of sinusoids whose frequencies are integer multiples of the fundamental.

\[ x(t) = \sum_{k=-\infty}^{\infty} a_k\,e^{jk\omega_0 t}, \qquad \omega_0 = \frac{2\pi}{T} \] The exponential form. It is the most convenient to work with, although the trigonometric one is easier to visualize.

And since those exponentials are eigenfunctions of any LTI system, the benefit is immediate: to know what a circuit does to a periodic signal, it is enough to know what it does to each harmonic separately and add them back up.

\[ x(t) = a_0 + \sum_{n=1}^{\infty}\Big(a_n\cos n\omega_0 t + b_n\operatorname{sin} n\omega_0 t\Big) \] The trigonometric form, equivalent. For real signals the two contain exactly the same information.

02Orthogonality: why they can be solved for

That the sum exists is one thing; being able to compute the coefficients is another. What makes it possible is a property of the harmonic exponentials:

\[ \int_{T} e^{jk\omega_0 t}\,e^{-jm\omega_0 t}\,dt = \begin{cases} T & k = m \\ 0 & k \ne m \end{cases} \] Two different harmonics are orthogonal: their product averaged over one period gives zero.

The idea is the same as for perpendicular vectors. To find out how much of the component \( m \) there is in a signal, multiply it by the conjugate of harmonic \( m \) and average: everything that is not that component cancels, and what is left is the coefficient being sought.

\[ a_k = \frac{1}{T}\int_{T} x(t)\,e^{-jk\omega_0 t}\,dt \]

It is exactly what a lock-in amplifier does in a lab: it multiplies the measured signal by a reference at the frequency of interest and averages. Everything that is not at that frequency —including noise, which is at all of them— averages to zero. A lock-in is, literally, a calculator of Fourier coefficients.

03Adding harmonics and watching the signal appear

Lab · harmonic synthesis

Above, the original signal in gray and the sum of the first harmonics in color. Below, the spectrum. Increase the number of terms and see what happens near the jumps.

Harmonics = 5

04Convergence, and the overshoot that does not go away

The Dirichlet conditions say when the series converges, and any engineering signal satisfies them: that it be absolutely integrable over one period, that it have a finite number of maxima and minima, and a finite number of discontinuities, all with finite jumps.

At points of discontinuity the series converges to the average of the two one-sided values. In a square wave that jumps from −1 to +1, the series equals exactly zero at the instant of the jump, no matter how many terms are added.

The Gibbs phenomenon

Near a jump, the partial sum always overshoots it, and the excess is about 8.95 % of the height of the jump. The remarkable thing is that this percentage does not go down when adding terms: the oscillation gets narrower and squeezes up against the jump, but its height stays the same. The lab measures it: with 5 harmonics and with 60 the overshoot is practically the same.

The practical consequence appears when filtering. An abrupt low-pass filter applied to a pulse produces that same overshoot —the ringing seen on the oscilloscope— and it cannot be eliminated by sharpening the cutoff: the transition of the filter has to be smoothed, which is what design windows do.

05From the series to the integral

The series serves for periodic signals. An isolated signal —a pulse, a transient— is thought of as periodic with period \( T \to \infty \). As \( T \) grows, the spacing between harmonics, \( \omega_0 = 2\pi/T \), shrinks, and the lines of the spectrum crowd together until they become a continuum:

\[ X(j\omega) = \int_{-\infty}^{\infty} x(t)\,e^{-j\omega t}\,dt, \qquad x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(j\omega)\,e^{j\omega t}\,d\omega \] The Fourier transform pair. The sum became an integral and the discrete coefficients, a continuous function of frequency.

That explains a difference worth having clear from the start: a periodic signal has a line spectrum, with lines spaced \( \omega_0 \) apart; an isolated signal has a continuous spectrum. And as a pulse train gets more widely spaced, its lines crowd together and its envelope is precisely the transform of the single pulse.

06In the lab

Exercise 1 · The coefficients of the square wave

Compute by hand the \( b_n \) of an odd square wave of amplitude 1 and check that they equal \( 4/(n\pi) \) for odd \( n \) and zero for even \( n \). Compare with the spectrum shown by the lab.

Exercise 2 · Measuring real harmonics

Generate a 1 kHz square wave and measure with the spectrum analyzer the amplitude of the third, fifth and seventh harmonics. Check that they fall off as \( 1/n \) and that the even ones do not appear.

Exercise 3 · Seeing the ringing

Pass a pulse train through a high-order low-pass filter with an abrupt cutoff and observe the overshoot on the edge. Measure its height relative to the jump and compare with the 8.95 % that Gibbs predicts.

07Common mistakes

Believing the Gibbs overshoot disappears with more terms. It narrows, but its height does not go down.
Forgetting the \( 1/T \) in the coefficient formula.
Expecting even harmonics in a signal with half-wave symmetry. It does not have them.
Confusing the line spectrum with the continuous one. Lines for periodic signals, continuous for isolated ones.
Assuming the series equals the value of the signal at a jump. It equals the average of the two one-sided values.

08Self-assessment

What property allows the coefficients to be solved for?

The orthogonality of the harmonic exponentials: the average of the product of two different ones is zero.

What is the value of the series at a point of discontinuity?

The average of the two one-sided values.

How much does the partial sum overshoot at a jump?

About 8.95 % of the height of the jump, and that value does not go down no matter how many terms are added.

How do the harmonics of a square wave fall off?

As \( 1/n \), and only the odd ones are present.

What happens to the spectrum when \( T \to \infty \)?

The lines crowd together into a continuum: the series becomes the Fourier integral.

What instrument computes a Fourier coefficient?

A lock-in amplifier: it multiplies by a reference at that frequency and averages.

09Further reading

Alan V. Oppenheim and Alan S. Willsky. Signals and Systems. 2nd ed. (Spanish edition, “Señales y sistemas”), Prentice Hall, 1998. Chapters 3 and 4: the series, the transform and the passage from one to the other.
Ronald N. Bracewell. The Fourier Transform and Its Applications. 3rd ed., McGraw-Hill, 1999. The graphical treatment of the Gibbs phenomenon and of convergence.
Simon Haykin and Barry Van Veen. Signals and Systems. (Spanish edition, “Señales y sistemas”), Limusa Wiley, 2001. Tables of coefficients for the classic waveforms, with their symmetries.

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