Fourier Series Representation of Periodic Signals

Chapter 3 ended with a decomposition that is exact, universal and free of conditions: every signal is a sum of scaled, shifted impulses. Chapter 7 pushed that through an LTI system and obtained convolution. Nothing about that chain is wrong. The difficulty is that convolution is expensive — every output value requires an entire integral over the input's history — and that the result is opaque. Look at \(y(t)=\int h(\tau)x(t-\tau)\,d\tau\) and you cannot see at a glance which features of \(x\) the system keeps and which it discards.

Chapter 6 hinted at the way out. Feed \(e^{st}\) into an LTI system and the convolution integral factors:

For this one family of inputs, the integral is done once and for all and produces a single complex number. No convolution survives. The system's entire action on \(e^{st}\) is a multiplication by \(H(s)\) — a stretch of amplitude and a shift of phase, nothing more.

That suggests an obvious strategy, and it is the strategy of the next eight chapters. If we can write an arbitrary input as a combination of exponentials, then linearity finishes the job: multiply each exponential by its own \(H\), add the results, and the output is known without ever performing a convolution. The impulse decomposition gave us building blocks that are easy to write down; the exponential decomposition gives us building blocks that are easy to push through a system.

The remaining question is which exponentials to use, and the answer depends on the signal. For a periodic signal the answer is startlingly economical: a countable set of exponentials, all at integer multiples of one frequency, suffices exactly. That is the Fourier series, and it is this chapter. For an aperiodic signal the set must become a continuum, which is the Fourier transform of Chapter 13.

Take a signal \(x(t)\) that is periodic with fundamental period \(T\), so \(x(t+T)=x(t)\), and define the fundamental frequency \(\omega_0 = 2\pi/T\) as in Chapter 2. If we intend to build \(x\) out of complex exponentials \(e^{j\omega t}\), the very first constraint is that the pieces must themselves be periodic with period \(T\) — a sum of signals none of which repeats every \(T\) seconds cannot repeat every \(T\) seconds either.

So ask which frequencies survive that test. The exponential \(e^{j\omega t}\) is periodic with period \(T\) exactly when \(e^{j\omega(t+T)} = e^{j\omega t}\), that is when \(e^{j\omega T}=1\), that is when \(\omega T\) is an integer multiple of \(2\pi\). Writing that integer as \(k\) gives \(\omega = k\,(2\pi/T) = k\omega_0\), and nothing else is admissible.

The word harmonic is not decoration. The \(k\)th member completes exactly \(k\) cycles in the time the fundamental completes one, which is why a violin string vibrating at 220 Hz also carries energy at 440, 660 and 880 Hz and not at 300 Hz. A periodic waveform of period \(T\) is physically constrained to the frequencies \(k\omega_0\); the shape of the waveform decides only how much sits at each.

Two details deserve care. First, \(\phi_k\) is periodic with period \(T\) but its fundamental period is \(T/|k|\); it repeats \(k\) times within one period of \(x\), which is precisely why it fits. Second, negative \(k\) is not redundant. In the real-signal case \(\phi_{-k}\) is the conjugate of \(\phi_k\), and it is the pairing of the two that produces a real cosine — Section 10-5 makes that explicit. Discarding negative \(k\) would leave us unable to represent a real signal at all.

Fourier's claim — and it was a claim, disbelieved by Lagrange and Laplace for years before it was proved — is that the harmonic set is not merely admissible but complete: every reasonable periodic signal is a weighted sum of its members, and the weights are unique. Chapter 11 states precisely what "reasonable" means. For now, take the representation as given.

Before working out how to find the \(a_k\), it is worth seeing what the representation is for. Suppose this periodic \(x(t)\) is applied to an LTI system with impulse response \(h(t)\). Each term \(a_k e^{jk\omega_0 t}\) is an exponential, so by the eigenfunction property it emerges as \(a_k H(jk\omega_0) e^{jk\omega_0 t}\), where \(H(j\omega)=\int h(\tau)e^{-j\omega\tau}d\tau\) is the system's frequency response. Linearity then adds the pieces back up.

Read the two sides carefully, because this single line is the reason the subject exists. The input has coefficients \(a_k\); the output has coefficients \(a_k H(jk\omega_0)\). The convolution integral has become an ordinary multiplication, performed separately on each harmonic. And the output is still periodic with the same period \(T\), because no new frequencies were created — an LTI system can amplify, attenuate and delay the harmonics it is given, but it cannot invent one that was not there.

We now need the coefficients. The trick is the same one used to extract a component of a vector: find something that is "perpendicular" to every block except the one you want, and take an inner product. For signals, the inner product is an integral over one period, and the perpendicularity statement is the following.

Both cases are one line. If \(k=n\) the integrand is \(e^{0}=1\) and the integral over an interval of length \(T\) is \(T\). If \(k \ne n\), write \(m = k-n\), a non-zero integer, and integrate directly over \(0\) to \(T\):

because \(\omega_0 T = 2\pi\) and \(e^{j2\pi m}=1\) for every integer \(m\). Geometrically the reason is even simpler: \(e^{jm\omega_0 t}\) traces \(|m|\) complete circles about the origin as \(t\) crosses one period, and the average of a point going round a full circle is the centre.

Now apply this. Multiply the synthesis equation by \(e^{-jn\omega_0 t}\) and integrate over one period. Interchanging sum and integral — legitimate under the conditions of Chapter 11 — every term dies except \(k=n\):

Setting \(k=0\) collapses the exponential to 1 and leaves \(a_0 = \frac{1}{T}\int_T x(t)\,dt\): the coefficient of the zeroth harmonic is simply the average value of the signal over a period, its DC component. This is worth checking first in any problem, because it is free and it catches sign errors immediately.

The pair of boxed equations is the whole apparatus. Analysis takes a waveform and returns a list of numbers; synthesis takes the list and rebuilds the waveform. They are inverses, and the standard shorthand for the pairing is

The exponential form is the one to compute with, but it has an uncomfortable feature: a real signal like a voltage is written as a sum of complex terms. The discomfort is illusory, and seeing why produces two alternative forms that are often more useful for interpretation.

Suppose \(x(t)\) is real. Then \(x^*(t)=x(t)\), and conjugating the synthesis equation gives \(x(t)=\sum_k a_k^* e^{-jk\omega_0 t}\). Replacing the summation index \(k\) by \(-k\) turns this into \(x(t)=\sum_k a_{-k}^* e^{jk\omega_0 t}\). Two Fourier series for the same signal must have the same coefficients, so:

Now group the terms in \(\pm k\) pairs. Writing \(a_k = |a_k|e^{j\theta_k}\), the pair contributes

which is manifestly real. Summing over \(k \ge 1\) and adding the \(k=0\) term — real by the symmetry rule, since \(a_0=a_0^*\) — gives the amplitude–phase form. Expanding each cosine with the compound-angle identity gives the trigonometric form.

The dictionary between them follows from \(\cos(k\omega_0t+\theta_k) = \cos\theta_k\cos k\omega_0 t - \sin\theta_k \sin k\omega_0 t\), which identifies \(A_k = C_k\cos\theta_k\) and \(B_k = -C_k\sin\theta_k\), so that \(a_k = \tfrac{1}{2}C_k e^{j\theta_k} = \tfrac{1}{2}(A_k - jB_k)\). Collecting the results:

Chapter 2 classified signals as even, odd or neither, and Chapter 4 showed how to split any signal into those two parts. The classification earns its keep here: each symmetry of \(x(t)\) forces a whole family of coefficients to vanish, and recognising one before integrating can halve or quarter the work.

Start with an even real signal, \(x(-t)=x(t)\). Use the symmetric period \(-T/2\) to \(T/2\) and split the exponential into its cosine and sine parts. The product \(x(t)\sin k\omega_0 t\) is even times odd, hence odd, and an odd function integrated over a symmetric interval gives zero. Only the cosine part survives, so \(B_k = 0\) and \(a_k\) is real. The mirror argument for an odd signal kills the cosine part instead.

The third symmetry is less familiar and more powerful. A signal has half-wave symmetry when shifting it by half a period inverts it: \(x(t \pm T/2) = -x(t)\). A square wave has it; a sawtooth does not. The claim is that such a signal contains no even harmonics whatsoever, and the proof is a substitution.

Split the analysis integral at \(T/2\), and in the second half put \(t = \tau + T/2\). Then \(x(\tau+T/2) = -x(\tau)\) by hypothesis, while the exponential picks up a factor \(e^{-jk\omega_0 T/2} = e^{-jk\pi} = (-1)^k\):

The bracket \(1-(-1)^k\) is \(0\) for even \(k\) and \(2\) for odd \(k\). Every even coefficient vanishes, including \(a_0\) — a half-wave symmetric signal necessarily has zero average — and every odd coefficient reduces to an integral over half a period. Recognising this symmetry in an examination question typically saves more time than any other single observation.

A Fourier series is a list of complex numbers indexed by an integer, and the natural way to display it is a pair of stem plots: \(|a_k|\) against \(k\), and \(\angle a_k\) against \(k\). Because the frequencies present are isolated multiples of \(\omega_0\), the picture consists of discrete lines rather than a continuous curve, and it is called the line spectrum. The gaps between the lines are real: a periodic signal genuinely contains no energy at any frequency other than \(k\omega_0\).

Take the odd square wave of amplitude 1 — equal to \(+1\) on \((0,T/2)\) and \(-1\) on \((-T/2,0)\). It is real and odd, so \(A_k=0\) and only sine terms appear; it has half-wave symmetry, so only odd \(k\) survives. Both facts drop out of the one integral we still have to do:

which is \(4/\pi k\) when \(k\) is odd and zero when \(k\) is even, exactly as the symmetry arguments promised.

The second canonical example is the periodic rectangular pulse train: a pulse of unit height and width \(2T_1\) centred on the origin, repeating with period \(T\). It is real and even, so the coefficients will be real. Integrating only over the interval where the signal is non-zero,

The last simplification uses \(\omega_0 T = 2\pi\). The \(k=0\) value cannot be read from the formula — it is \(0/0\) — so it is obtained separately as the average, which is the duty cycle \(2T_1/T\), and it is also the limit of the general expression as \(k \to 0\).

Two features of this result are worth fixing in memory. First, the coefficients are samples of a smooth envelope: writing \(a_k = \frac{2T_1}{T}\cdot\frac{\sin(k\omega_0T_1)}{k\omega_0 T_1}\) exposes the function \(\sin\xi/\xi\), the sinc shape promised at the end of Chapter 3. Second, the envelope has zeros wherever \(k\omega_0T_1\) is a multiple of \(\pi\), so the spectrum has evenly spaced nulls whose position depends on the pulse width alone. Narrow the pulse and the nulls move outward: a short pulse needs a wide band of harmonics, a long one does not. That reciprocity between duration and bandwidth is a theme of the rest of the book.

Everything above has a discrete-time counterpart, and as usual in this subject the discrete version is the simpler one. Let \(x[n]\) be periodic with fundamental period \(N\), so \(x[n+N]=x[n]\), and set \(\Omega_0 = 2\pi/N\). The candidate building blocks are \(\phi_k[n]=e^{jk\Omega_0 n}\), and the argument of Section 10-2 applies unchanged: these are periodic in \(n\) with period \(N\) exactly for integer \(k\).

But now Chapter 3's warning about discrete-time frequency intervenes, and it changes the shape of the theory. Because \(n\) is an integer, \(e^{j2\pi n}=1\), so

The \((k+N)\)th harmonic is the \(k\)th harmonic — not similar to it, identical to it, sample for sample. There are therefore only \(N\) distinct harmonics, not infinitely many, and the synthesis sum runs over any \(N\) consecutive values of \(k\), written \(\sum_{k=\langle N\rangle}\). A periodic discrete-time signal has \(N\) independent values per period, and the series has exactly \(N\) coefficients: no information is gained or lost, and there is nothing to converge.

The analysis equation follows from the same orthogonality argument, with the integral replaced by a finite geometric sum. For \(m\) not a multiple of \(N\), summing the series \(\sum_{n=0}^{N-1}(e^{jm\Omega_0})^n\) gives \(\big(1-e^{jm2\pi}\big)/\big(1-e^{jm\Omega_0}\big) = 0\), the numerator vanishing while the denominator does not; for \(m\) a multiple of \(N\), every term is 1 and the sum is \(N\).

The absence of convergence questions is not a technicality; it is the reason digital signal processing is possible. The continuous-time series is an infinite sum that may or may not reproduce the signal at every instant, and Chapter 11 is largely devoted to the fine print. The discrete-time series is a finite sum of \(N\) terms that reproduces \(N\) numbers exactly, always, for every periodic sequence without exception. When Chapter 26 arrives at the discrete Fourier transform and the FFT, it will be this equation — rescaled — that is being computed.

Problems 1 to 3 are inspection exercises — no integral should be needed. Problems 4 to 6 require the analysis equation, and in each case the symmetry should be identified first. Problems 7 and 8 push the ideas towards the systems view of Chapter 12.