Part 3 · Chapter 11

Properties and Convergence of Fourier Series

Chapter 10 left two unfinished pieces of business: a set of structural rules that make the analysis integral almost never necessary, and an honest answer to the question of whether an infinite sum of smooth sinusoids can really equal a signal that jumps.

Signals and Systems Prof. Mithun Mondal Reading time ≈ 50 min
i What you'll learn
  • How linearity, shifting, reversal and scaling move a known set of coefficients onto a new signal without touching an integral.
  • Why a time shift leaves the magnitude spectrum untouched and tilts the phase spectrum by a straight line — the single most reused fact in the whole of Part 3.
  • The pair multiplication ↔ periodic convolution, and the duality between the two domains that it exposes.
  • The differentiation property \(jk\omega_0 a_k\), and the trick of differentiating a signal until it becomes impulses in order to read its spectrum off by inspection.
  • Parseval's relation: average power is the sum of the powers of the harmonics, so \(|a_k|^2\) is a genuine power spectrum.
  • The Dirichlet conditions — what a signal must satisfy to have a Fourier series at all, and the three classic signals that fail each one.
  • Why the partial sums are the best least-squares approximation, why the error energy goes to zero, and why the Gibbs overshoot nevertheless refuses to.
Section 11-1

What the Properties Are For

Chapter 10 supplied a complete recipe: given \(x(t)\), integrate to get \(a_k\); given \(a_k\), sum to get \(x(t)\). Followed literally, that recipe means a fresh integration by parts for every new waveform, which is both tedious and uninformative. Chapter 8 faced the same situation with convolution and answered it the same way — by establishing structural properties of the operation, so that new results follow from old ones by algebra rather than by calculus.

The properties below do two distinct jobs, and it is worth separating them. Some are labour-saving: a shifted, scaled or differentiated version of a signal whose coefficients you already know needs no new integral. Others are structural: they tell you what the spectrum of a signal must look like before you compute anything, and they are what allows an engineer to glance at a waveform and say "odd harmonics only, falling like \(1/k^2\)" without writing a line.

The second half of the chapter changes register. Everything so far has assumed that the infinite series in the synthesis equation converges to \(x(t)\). For a signal with jumps in it — a square wave, a sampled-and-held voltage, a switching waveform — that assumption deserves scrutiny, because each partial sum is a finite combination of smooth sinusoids and is therefore perfectly continuous. Something must happen in the limit, and what happens is more interesting than either "it works" or "it doesn't".

A convention for this chapter. Throughout, \(x(t)\) and \(y(t)\) are periodic with the same period \(T\), with \(\omega_0 = 2\pi/T\), and \(x(t) \stackrel{\mathcal{FS}}{\longleftrightarrow} a_k\), \(y(t) \stackrel{\mathcal{FS}}{\longleftrightarrow} b_k\). Every property below has a discrete-time counterpart obtained by replacing \(T\) with \(N\), integrals with sums over one period, and \(\omega_0\) with \(2\pi/N\); the differences that matter are pointed out where they arise.
Section 11-2

Linearity, Shifting, Reversal and Scaling

Linearity is immediate, and it is immediate because the analysis equation is an integral, which is itself linear. For any constants \(A\) and \(B\), the coefficients of \(Ax(t)+By(t)\) are \(Aa_k + Bb_k\). It is the property used most often and noticed least.

Time shifting is the one that repays study. Let \(y(t)=x(t-t_0)\) and compute its coefficients directly, substituting \(\tau = t-t_0\). The interval of integration shifts, but it is still an interval of one full period, so the notation \(\int_T\) absorbs it:

Deriving the time-shift property
\[ b_k = \frac{1}{T}\int_T x(t-t_0)e^{-jk\omega_0 t}dt = \frac{1}{T}\int_T x(\tau)e^{-jk\omega_0(\tau+t_0)}d\tau = e^{-jk\omega_0 t_0}\cdot\frac{1}{T}\int_T x(\tau)e^{-jk\omega_0 \tau}d\tau \]
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Time shifting
\[ x(t-t_0) \;\stackrel{\mathcal{FS}}{\longleftrightarrow}\; a_k\,e^{-jk\omega_0 t_0} \]

The factor \(e^{-jk\omega_0 t_0}\) has unit magnitude, so \(|b_k| = |a_k|\): delaying a signal does not change its magnitude spectrum at all. What it changes is the phase, by \(-k\omega_0 t_0\) — an amount proportional to \(k\), which plots as a straight line through the origin whose slope is the delay.

Both halves of that statement matter. The first says that magnitude alone cannot tell you when anything happened; all timing information in a signal lives in its phase. The second says that a pure delay is exactly a linear phase characteristic. Chapter 15 will invert this observation: a filter whose phase response is linear in \(\omega\) delays every harmonic by the same amount of time and therefore does not distort the waveform's shape, while a filter with non-linear phase moves the harmonics relative to one another and mangles it, even if every magnitude is left alone.

|a k | k 0 unchanged by the shift ∠b k k 0 a straight line of slope −ω₀t₀
Delay is linear phase — the magnitude spectrum cannot see it

Time reversal follows the same substitution, this time \(\tau = -t\), which flips the limits and cancels the sign of \(d\tau\), leaving \(x(-t) \leftrightarrow a_{-k}\). Combined with the conjugation results of the next section, this is what turns statements about even and odd signals into statements about their spectra.

Time scaling is the property most often misremembered, because in the Fourier transform of Chapter 14 scaling stretches the spectrum, while here it does not. Let \(\alpha \gt 0\) and consider \(x(\alpha t)\). This is a different signal with a different period, \(T/\alpha\), and therefore a different fundamental frequency, \(\alpha\omega_0\). Substituting into the analysis equation with the new period returns the same numbers \(a_k\).

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Time scaling
\[ x(\alpha t) = \sum_{k=-\infty}^{\infty} a_k\,e^{jk(\alpha\omega_0)t}, \qquad \alpha \gt 0 \]

The coefficients are unchanged; the frequencies they sit at are multiplied by \(\alpha\). Speeding a record up does not alter the relative strengths of its harmonics — it moves every one of them up in pitch by the same factor, which is exactly what it sounds like.

Why discrete time has no scaling property. Chapter 4 showed that \(x[\alpha n]\) is not defined for non-integer \(\alpha\) — you cannot ask for the sample halfway between two samples. Decimation and interpolation exist, but they are different operations with different spectral consequences, and Chapter 22 treats them properly. The absence of a clean scaling rule in discrete time is not an oversight; it is the reason sampling-rate conversion is a subject in its own right.
Section 11-3

Conjugation and the Symmetries of Real Signals

Conjugating the analysis equation gives \(\overline{a_k}\) as the coefficient of \(x^*(t)\) at index \(-k\), which is the compact statement \(x^*(t) \leftrightarrow a_{-k}^*\). Chapter 10 extracted the most important consequence — for real \(x\), \(a_{-k}=a_k^*\) — and the rest of the family follows by combining that with time reversal.

The chain of reasoning is short enough to reconstruct rather than memorise. If \(x\) is real, \(a_{-k}=a_k^*\). If \(x\) is additionally even, time reversal gives \(a_{-k}=a_k\); putting the two together forces \(a_k = a_k^*\), so \(a_k\) is real. If instead \(x\) is odd, time reversal gives \(a_{-k}=-a_k\), so \(a_k = -a_k^*\) and \(a_k\) is purely imaginary. Two facts, combined two ways, produce four results.

If \(x(t)\) is…then the coefficients satisfy…and the spectrum shows…
real\(a_{-k}=a_k^{*}\)\(|a_k|\) even, \(\angle a_k\) odd
real and even\(a_k\) real and even in \(k\)phase is \(0\) or \(\pi\) only
real and odd\(a_k\) imaginary and odd in \(k\); \(a_0=0\)phase is \(\pm\pi/2\) only
imaginary\(a_{-k}=-a_k^{*}\)\(|a_k|\) even, \(\angle a_k\) even about \(\pi/2\)

One more consequence is worth stating because it is used in the next chapter. Any real signal splits into its even and odd parts, \(x = \mathrm{Ev}\{x\} + \mathrm{Od}\{x\}\), as Chapter 4 established. Applying the table to each part separately shows that the even part carries exactly the real part of the spectrum and the odd part carries exactly the imaginary part:

Even and odd parts of a real signal
\[ \mathrm{Ev}\{x(t)\} \;\stackrel{\mathcal{FS}}{\longleftrightarrow}\; \mathrm{Re}\{a_k\}, \qquad\qquad \mathrm{Od}\{x(t)\} \;\stackrel{\mathcal{FS}}{\longleftrightarrow}\; j\,\mathrm{Im}\{a_k\} \]
Section 11-4

Multiplication and Periodic Convolution

What happens to the coefficients when two periodic signals of the same period are multiplied? Write both as series and multiply them out. A term of the product carries \(e^{j(l+m)\omega_0 t}\), so to collect the coefficient of the \(k\)th harmonic we must gather every pair with \(l+m=k\):

Multiplying two Fourier series
\[ x(t)y(t) = \left(\sum_{l} a_l e^{jl\omega_0 t}\right)\left(\sum_{m} b_m e^{jm\omega_0 t}\right) = \sum_{k}\left(\sum_{l=-\infty}^{\infty} a_l\,b_{k-l}\right)e^{jk\omega_0 t} \]
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Multiplication in time is convolution of the coefficients
\[ x(t)\,y(t) \;\stackrel{\mathcal{FS}}{\longleftrightarrow}\; \sum_{l=-\infty}^{\infty} a_l\,b_{k-l} \;=\; a_k * b_k \]

The sum on the right is precisely the discrete convolution of Chapter 7, performed on the coefficient sequences instead of on signals. Multiplying waveforms mixes their spectra together; it is the mechanism behind every modulator and every mixer.

The companion property runs the other way. Ordinary convolution of two periodic signals diverges — the integral over an infinite range of a signal that never decays cannot converge — so the operation is redefined over a single period. Periodic convolution is

Periodic convolution
\[ z(t) = \int_{T} x(\tau)\,y(t-\tau)\,d\tau \]

and its coefficients follow from swapping the order of integration and applying the time-shift property to the inner integral. The inner integral is the analysis integral of \(y(t-\tau)\), which by Section 11-2 equals \(T\,b_k e^{-jk\omega_0\tau}\):

Deriving the periodic-convolution property
\[ c_k = \frac{1}{T}\int_T\!\left[\int_T x(\tau)y(t-\tau)d\tau\right]e^{-jk\omega_0 t}dt = \frac{1}{T}\int_T x(\tau)\Big[T\,b_k e^{-jk\omega_0\tau}\Big]d\tau = T\,a_k b_k \]
The duality is the point. Convolution in time ↔ multiplication of coefficients; multiplication in time ↔ convolution of coefficients. The same pairing reappears for the Fourier transform in Chapter 14, for the Laplace transform in Chapter 18, and for the \(z\)-transform in Chapter 24, and in every case it is the reason transforms are worth the trouble: the operation that is hard in one domain is trivial in the other.
Section 11-5

Differentiation and Integration

Differentiate the synthesis equation term by term. Each \(e^{jk\omega_0 t}\) contributes a factor \(jk\omega_0\) and nothing else, because the exponential is its own derivative up to that constant:

Differentiating the series
\[ \frac{dx}{dt} = \frac{d}{dt}\sum_{k}a_k e^{jk\omega_0 t} = \sum_{k} (jk\omega_0)\,a_k\,e^{jk\omega_0 t} \]
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Differentiation and integration
\[ \frac{dx(t)}{dt} \;\stackrel{\mathcal{FS}}{\longleftrightarrow}\; jk\omega_0\,a_k \qquad\qquad \int_{-\infty}^{t}\! x(\tau)d\tau \;\stackrel{\mathcal{FS}}{\longleftrightarrow}\; \frac{a_k}{jk\omega_0} \]

The integration result holds only when \(a_0=0\). If the signal has a non-zero average, integrating it produces a ramp that grows without bound, the result is not periodic, and no Fourier series exists for it — the \(k=0\) term of the formula, dividing by zero, says so plainly.

These two rules turn into a technique. Differentiating a piecewise-linear waveform repeatedly reduces it, by Chapter 3's chain, to a train of steps and then to a train of impulses — and the coefficients of an impulse train are trivial, since sifting evaluates the analysis integral instantly. Compute those, then divide by \((jk\omega_0)^m\) to climb back up. Worked Example 2 does exactly this for the triangular wave.

There is also a structural reading, and it is one of the most useful pieces of intuition in the subject. Differentiation multiplies \(a_k\) by \(k\); so if \(dx/dt\) is to have coefficients that stay bounded, \(a_k\) must fall off at least as fast as \(1/k\). Running the argument in reverse relates the smoothness of the waveform to the decay of its spectrum.

WaveformWorst featureDecay of \(a_k\)Example
Contains impulsesimpulsedoes not decayimpulse train
Discontinuousjump in \(x\)\(\propto 1/k\)square wave, sawtooth
Continuous, kinkedjump in \(dx/dt\)\(\propto 1/k^2\)triangular wave
Smooth to order \(m\)jump in \(d^{m}x/dt^{m}\)\(\propto 1/k^{m+1}\)raised-cosine pulse
Read a spectrum backwards. If a measured line spectrum falls off as \(1/k\), the waveform it came from has a discontinuity somewhere, whatever the oscilloscope's finite bandwidth may have made it look like. If it falls as \(1/k^2\), the waveform is continuous but has a corner. This inference — from decay rate to smoothness — is used constantly in Part 4, where it becomes a statement about the tails of the Fourier transform.
Section 11-6

Parseval's Relation and the Power Spectrum

Chapter 2 defined the average power of a periodic signal as \(P = \frac{1}{T}\int_T |x(t)|^2 dt\). The question Parseval answers is whether that power can be attributed to individual harmonics, and the answer — that it can, with no cross-terms whatsoever — is a direct consequence of orthogonality.

Write \(|x|^2 = x\,x^*\), substitute the series for each factor, and integrate:

Deriving Parseval's relation
\[ \frac{1}{T}\int_T |x(t)|^2 dt = \frac{1}{T}\int_T \sum_{k}a_k e^{jk\omega_0 t}\sum_{m}a_m^{*}e^{-jm\omega_0 t}\,dt = \sum_{k}\sum_{m} a_k a_m^{*}\left[\frac{1}{T}\int_T e^{j(k-m)\omega_0 t}dt\right] \]

The bracket is \(1\) when \(k=m\) and \(0\) otherwise, so the double sum collapses onto its diagonal.

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Parseval's relation for Fourier series
\[ \frac{1}{T}\int_{T}|x(t)|^2\,dt \;=\; \sum_{k=-\infty}^{\infty}|a_k|^2 \qquad\qquad \frac{1}{N}\sum_{n=\langle N\rangle}|x[n]|^2 = \sum_{k=\langle N\rangle}|a_k|^2 \]

Total average power equals the sum of the powers of the individual harmonics. Because the cross-terms vanish, \(|a_k|^2\) is genuinely the power carried by the \(k\)th harmonic, and the plot of \(|a_k|^2\) against \(k\) is called the power spectrum of the signal.

For a real signal the \(\pm k\) lines have equal magnitude, so they can be merged: \(P = a_0^2 + 2\sum_{k\ge1}|a_k|^2\), which in amplitude–phase notation reads \(P = C_0^2 + \tfrac12\sum_{k\ge1}C_k^2\). The factor of one half is the familiar statement that a sinusoid of peak amplitude \(C\) has mean-square value \(C^2/2\); Parseval simply says that these add.

The practical use is to decide how many harmonics matter. A specification such as "total harmonic distortion below 1%" is a statement about the tail of this sum, and truncating a series is only defensible once you know what fraction of the power the discarded terms carried. Worked Example 3 works that out for the square wave, with a result that surprises most people the first time.

Section 11-7

The Dirichlet Conditions

Everything up to this point assumed that the synthesis equation reproduces \(x(t)\). Fourier asserted it for all signals; that assertion is false, and identifying the correct hypotheses occupied mathematicians for most of the nineteenth century. Dirichlet's conditions are sufficient — not necessary — and they are satisfied by every signal an engineer is likely to meet.

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The Dirichlet conditions
\[ \text{(1)}\ \int_{T}|x(t)|\,dt \lt \infty \qquad \text{(2)}\ \text{finitely many maxima and minima in one period} \qquad \text{(3)}\ \text{finitely many finite discontinuities} \]

If all three hold, the Fourier series converges to \(x(t)\) at every point where \(x\) is continuous, and to the midpoint \(\tfrac12\big[x(t^{+})+x(t^{-})\big]\) at every discontinuity.

Condition (1) is what makes the analysis integral exist at all: if \(\int_T|x|\,dt\) is infinite then \(a_k\) itself is undefined and there is nothing to discuss. Conditions (2) and (3) exclude signals that oscillate or jump infinitely often within a finite time. Each condition has a standard signal that violates it and only it.

→ ∞ 1 0 x(t) = 1/t — fails (1)
Not absolutely integrable
1 0 x(t) = sin(2π/t) — fails (2)
Infinitely many maxima
1 0 endless staircase — fails (3)
Infinitely many jumps

None of these three is a signal any instrument could produce, and that is the honest summary of the conditions: they are a fence around pathologies, not a constraint on engineering. The clause that does bite in practice is the second half of the theorem — the behaviour at a jump. At a discontinuity the series does not converge to either one-sided value but to their average, and it does so for a reason that is easy to see: the series has no way of distinguishing the two sides, since every one of its terms is continuous. Confronted with a choice between \(x(t^-)\) and \(x(t^+)\), it splits the difference.

This is why Chapter 10's sawtooth series returned \(T/2\) at \(t=0\) while the signal was defined to be \(0\) there, and why the square-wave series returns \(0\) at every switching instant. In both cases the series is right and the point value is irrelevant — as Chapter 3 observed, changing a signal at isolated instants changes no integral, and integrals are all the coefficients ever saw.

Section 11-8

Mean-Square Convergence and the Gibbs Phenomenon

There is a second convergence theorem, with a weaker hypothesis and a weaker conclusion, and for engineering purposes it is the more important of the two. It asks only that the signal have finite energy in a period, \(\int_T |x|^2dt \lt \infty\), and it concludes not that the series matches \(x\) at each instant, but that the energy of the error goes to zero.

To state it, define the \(N\)th partial sum and its error. In fact it pays to be more ambitious and ask which coefficients \(c_k\) give the best \(N\)-term harmonic approximation, rather than assuming the Fourier ones are optimal. Let \(\hat{x}_N(t)=\sum_{|k|\le N} c_k e^{jk\omega_0 t}\) and expand the mean-square error, using orthogonality at every step:

Minimising the mean-square error over all choices of coefficients
\[ E_N = \frac{1}{T}\int_T \big|x-\hat{x}_N\big|^2 dt = \underbrace{\frac{1}{T}\int_T|x|^2dt}_{P} \;-\; \sum_{|k|\le N}\big(c_k^{*}a_k + c_k a_k^{*}\big) \;+\; \sum_{|k|\le N}|c_k|^2 \]

Now complete the square on the last two sums, adding and subtracting \(\sum|a_k|^2\):

The same expression, rearranged
\[ E_N = P - \sum_{|k|\le N}|a_k|^2 \;+\; \sum_{|k|\le N}\big|c_k - a_k\big|^2 \]

The last sum is a sum of squares and is the only part that depends on our choice. It is smallest — zero — precisely when \(c_k = a_k\) for every \(k\).

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The partial sum is the best approximation, and its error is what Parseval leaves out
\[ \min_{c_k} E_N = P - \sum_{|k|\le N}|a_k|^2 \;\ge\; 0 \quad\Longrightarrow\quad \sum_{|k|\le N}|a_k|^2 \le \frac{1}{T}\int_T |x|^2 dt \]

The Fourier coefficients are not merely a choice; they are the optimal choice in the least-squares sense, and adding a harmonic can never make a truncated series worse. The inequality on the right is Bessel's inequality, and Parseval's relation is exactly the statement that it becomes an equality as \(N\to\infty\) — which is to say that \(E_N \to 0\).

So the error energy vanishes. What does not vanish is the maximum of the error, and the discrepancy between those two statements is the Gibbs phenomenon.

Look again at the square-wave partial sums of Chapter 10. Near each jump the partial sum overshoots, and adding harmonics does not remove the overshoot; it only squeezes it into a narrower interval on either side of the discontinuity. The peak height converges to a fixed number, and that number is not 1.

The location of the peak can be found exactly. Differentiating the partial sum through harmonic \(2M-1\) and using the identity \(\sum_{m=1}^{M}\cos(2m-1)\theta = \sin(2M\theta)/(2\sin\theta)\) shows that the derivative first vanishes when \(2M\omega_0 t = \pi\), that is at \(t = T/(4M)\). Evaluating the sum there and letting \(M\to\infty\) turns it into a Riemann sum for an integral:

The limiting peak of the partial sums
\[ \lim_{M\to\infty} x_M\!\left(\frac{T}{4M}\right) = \frac{2}{\pi}\int_{0}^{\pi}\frac{\sin u}{u}\,du = \frac{2}{\pi}\,\mathrm{Si}(\pi) = 1.17898\ldots \]
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The Gibbs phenomenon
\[ \text{overshoot} \;\to\; 8.95\% \ \text{of the height of the jump, for every } M \]

The square wave jumps by 2, from \(-1\) to \(+1\), and the partial sums peak at \(1.179\) — an excess of \(0.179\), which is \(8.95\%\) of \(2\). Increasing \(M\) narrows the ripple but never lowers it. Because the ripple's width shrinks in proportion, its energy does go to zero, which is exactly how mean-square convergence and a permanent overshoot coexist without contradiction.

1.179 +1 jump at t = 0 — through 5th harmonic — through 29th harmonic
The overshoot narrows but does not shrink — Gibbs at a jump
Why this is not a curiosity. Any real processing chain truncates a spectrum: a filter has finite bandwidth, a digital system a finite sample rate. Truncating a spectrum is exactly forming a partial sum, so ringing at edges is not a defect of a particular design but a mathematical consequence of band-limiting a signal with sharp transitions. Window functions, which reappear in Chapter 26, exist to trade this overshoot against a wider transition — you can suppress the ringing, but never for free.
Section 11-9

Worked Examples

1 A shifted square wave, without integrating

Problem. Chapter 10 found that the odd square wave \(s(t)\), equal to \(+1\) on \((0,T/2)\) and \(-1\) on \((-T/2,0)\), has coefficients \(s_k = -2j/(\pi k)\) for odd \(k\) and \(0\) otherwise. Find the Fourier series of \(y(t) = s(t-T/4)\).

Solution. Time shifting multiplies each coefficient by \(e^{-jk\omega_0 t_0}\) with \(t_0 = T/4\). Since \(\omega_0 T/4 = \pi/2\), the factor is \(e^{-jk\pi/2}\), which for odd \(k\) is \(-j\) when \(k \equiv 1 \pmod 4\) and \(+j\) when \(k \equiv 3 \pmod 4\).

Working — the first three coefficients
\[ b_1 = \left(\frac{-2j}{\pi}\right)(-j) = -\frac{2}{\pi}, \qquad b_3 = \left(\frac{-2j}{3\pi}\right)(+j) = \frac{2}{3\pi}, \qquad b_5 = \left(\frac{-2j}{5\pi}\right)(-j) = -\frac{2}{5\pi} \]

Every coefficient has become real, and they alternate in sign. Real coefficients mean an even signal — which is right, because shifting the square wave by a quarter period places it symmetrically about \(t=0\). Pairing \(\pm k\) and using \(2b_k\cos k\omega_0 t\):

Working — result
\[ y(t) = -\frac{4}{\pi}\left[\cos\omega_0 t - \frac{1}{3}\cos 3\omega_0 t + \frac{1}{5}\cos 5\omega_0 t - \cdots\right] \]

Check at \(t=0\), where \(y(0)=s(-T/4)=-1\). The bracket is \(1-\tfrac13+\tfrac15-\cdots = \pi/4\), so the series gives \(-\tfrac{4}{\pi}\cdot\tfrac{\pi}{4} = -1\). Correct — and obtained without a single integral, from coefficients computed once in the previous chapter.

2 The triangular wave by differentiation

Problem. Let \(y(t) = 1 - 4|t|/T\) for \(|t|\le T/2\), repeated with period \(T\) — a triangular wave running between \(+1\) and \(-1\). Find its coefficients using the differentiation property.

Solution. Differentiate. On \((0,T/2)\) the slope is \(-4/T\); on \((-T/2,0)\) it is \(+4/T\). That is a square wave of amplitude \(4/T\), inverted relative to \(s(t)\):

Working — the derivative
\[ \frac{dy}{dt} = -\frac{4}{T}\,s(t) \quad\Longrightarrow\quad \text{its coefficients are} \quad -\frac{4}{T}\,s_k = -\frac{4}{T}\cdot\frac{-2j}{\pi k} = \frac{8j}{T\pi k} \ \ (k \ \text{odd}) \]

By the differentiation property those coefficients also equal \(jk\omega_0 b_k\). Equating and solving, with \(\omega_0 = 2\pi/T\):

Working — solving for \(b_k\)
\[ b_k = \frac{1}{jk\omega_0}\cdot\frac{8j}{T\pi k} = \frac{8}{T\pi k^2\omega_0} = \frac{8}{T\pi k^{2}}\cdot\frac{T}{2\pi} = \frac{4}{\pi^{2}k^{2}} \quad (k \ \text{odd}), \qquad b_k = 0 \ (k \ \text{even}) \]

The \(k=0\) coefficient is not determined by this method and must be supplied separately; here the triangle is symmetric about zero, so \(b_0 = 0\). Assembling the series and pairing \(\pm k\):

Working — result
\[ y(t) = \frac{8}{\pi^{2}}\sum_{k \ \text{odd}}\frac{\cos k\omega_0 t}{k^{2}} \]

Two checks. At \(t=0\) the series gives \(\frac{8}{\pi^2}\big(1+\tfrac19+\tfrac1{25}+\cdots\big) = \frac{8}{\pi^2}\cdot\frac{\pi^2}{8}=1 = y(0)\). And the coefficients fall as \(1/k^2\) rather than the square wave's \(1/k\), exactly as Section 11-5 predicts for a signal that is continuous but has corners.

3 How much power lives in the first few harmonics

Problem. For the unit square wave, find the fraction of the total average power carried by the harmonics up to and including the 1st, 3rd, 5th and 7th.

Solution. The total power is easy without any series: \(|x(t)|=1\) everywhere, so \(P = \frac{1}{T}\int_T 1\,dt = 1\). Each harmonic pair \(\pm k\) contributes \(2|a_k|^2 = 2\big|{-2j}/{\pi k}\big|^2 = 8/(\pi^2 k^2)\).

Harmonic \(k\)Power \(8/\pi^2k^2\)CumulativeFraction of \(P\)
10.810570.8105781.06%
30.090060.9006390.06%
50.032420.9330693.31%
70.016540.9496094.96%

The fundamental alone carries 81% of the power of a square wave, and the first four non-zero harmonics carry 95%. This is why a square wave passed through a narrow filter still looks recognisably periodic at the right frequency, and it is also a warning: a representation that captures 95% of the power can still be visibly wrong in shape, because the missing 5% is concentrated entirely at the edges. Parseval measures energy, not resemblance.

As a consistency check, the full sum must give 1: \(\sum_{k \ \text{odd}} 8/(\pi^2k^2) = \frac{8}{\pi^2}\cdot\frac{\pi^2}{8} = 1\), using \(\sum_{k \ \text{odd}}1/k^2 = \pi^2/8\). It does.

4 Multiplication as spectral shifting

Problem. A periodic signal \(x(t)\) with coefficients \(a_k\) is multiplied by \(\cos\omega_0 t\). Find the coefficients of the product, and verify the result for \(x(t)=1+\cos\omega_0 t\).

Solution. The cosine has only two non-zero coefficients, \(b_1 = b_{-1} = \tfrac12\). The multiplication property then gives a convolution with just two terms:

Working — the general result
\[ c_k = \sum_{l} a_l\,b_{k-l} = \tfrac12 a_{k-1} + \tfrac12 a_{k+1} \]

Every line of the original spectrum is halved and copied to the two neighbouring harmonics. That is amplitude modulation in miniature, and Chapter 28 shows the same formula moving an entire baseband spectrum up to a carrier frequency.

Verify with \(x(t)=1+\cos\omega_0 t\), whose coefficients are \(a_0=1\) and \(a_{\pm1}=\tfrac12\):

Working — applying the formula
\[ c_0 = \tfrac12(a_{-1}+a_1) = \tfrac12, \qquad c_{\pm1} = \tfrac12(a_0 + a_{\pm2}) = \tfrac12, \qquad c_{\pm2} = \tfrac12(a_{\pm1}+a_{\pm3}) = \tfrac14 \]

Now multiply out directly: \((1+\cos\omega_0t)\cos\omega_0 t = \cos\omega_0 t + \cos^2\omega_0 t = \tfrac12 + \cos\omega_0 t + \tfrac12\cos 2\omega_0 t\). Reading off its coefficients gives \(c_0 = \tfrac12\), \(c_{\pm1}=\tfrac12\), \(c_{\pm2}=\tfrac14\). The two routes agree, and the product contains a second harmonic that neither factor possessed — multiplication, unlike any LTI operation, creates new frequencies.

5 What the series does at a jump

Problem. Evaluate the square-wave series at \(t=0\) and at \(t=T/4\), and say what each result means.

Solution. The series is \(x(t) = \frac{4}{\pi}\sum_{k \ \text{odd}}\sin(k\omega_0 t)/k\). At \(t=0\) every sine is zero, so the series returns \(0\). The signal itself jumps from \(-1\) to \(+1\) there, and the midpoint of that jump is \(\tfrac12(-1+1)=0\). The series has landed exactly where Dirichlet's theorem says it must.

At \(t=T/4\) we have \(\omega_0 t = \pi/2\), so \(\sin(k\pi/2)\) cycles through \(+1, -1, +1, \ldots\) for \(k = 1,3,5,\ldots\). The signal is continuous there with value \(+1\), so the series must converge to \(1\):

Working — a by-product
\[ 1 = \frac{4}{\pi}\left[1-\frac13+\frac15-\frac17+\cdots\right] \quad\Longrightarrow\quad \frac{\pi}{4} = 1-\frac13+\frac15-\frac17+\cdots \]

which is Leibniz's series for \(\pi\), obtained here as a free consequence of a square wave. A convergence theorem about signals has produced a fact about numbers, and the same trick applied to the triangular wave of Example 2 at \(t=0\) yields \(\sum_{k \ \text{odd}}1/k^2 = \pi^2/8\).

6 Two kinds of error, measured

Problem. For the unit square wave approximated by harmonics up to the 9th, compute (a) the mean-square error \(E_N\), and (b) the largest pointwise error. Comment on the comparison.

Solution (a). By the result of Section 11-8, \(E_N = P - \sum_{|k|\le N}|a_k|^2\) with \(P=1\). Summing the harmonic powers from Example 3 and adding the ninth, \(8/(81\pi^2)=0.01001\):

Working — error energy
\[ \sum_{|k|\le 9}|a_k|^2 = 0.81057+0.09006+0.03242+0.01654+0.01001 = 0.95960, \qquad E_N = 1 - 0.95960 = 0.0404 \]

So just over 4% of the power is unaccounted for.

Solution (b). The largest pointwise error occurs at the Gibbs peak. With harmonics through the 9th we have \(2M-1=9\), so \(M=5\) and the peak sits at \(t = T/20\). Evaluating the partial sum there,

Working — the peak
\[ x_5\!\left(\frac{T}{20}\right) = \frac{4}{\pi}\sum_{m=1}^{5}\frac{\sin\!\big((2m-1)\pi/10\big)}{2m-1} = \frac{4}{\pi}(0.9286) = 1.1823 \]

The pointwise error there is \(0.1823\), or 18.2% of the signal's amplitude — more than four times the fractional power error, and it barely improves with more terms: the same calculation with harmonics through the 29th gives \(1.1793\), and the limit is \(1.17898\).

The comparison is the lesson of the chapter. Averaged over a period, the approximation is excellent and improving steadily. At the worst instant it is poor and hardly improving at all. Which measure you should care about depends entirely on the application: a power calculation is served well by the first, a timing or threshold-crossing calculation is misled badly by it.

Review

Chapter Summary

Shifting

\(x(t-t_0)\leftrightarrow a_k e^{-jk\omega_0 t_0}\): magnitude untouched, phase tilted by a straight line. All timing lives in the phase.

Symmetry

Real gives \(a_{-k}=a_k^*\); real and even gives real \(a_k\); real and odd gives imaginary \(a_k\). Two facts, four results.

The duality

Periodic convolution in time is \(T a_k b_k\); multiplication in time is convolution of the coefficient sequences.

Differentiation

\(jk\omega_0 a_k\). Hence the decay of \(a_k\) measures smoothness: \(1/k\) for a jump, \(1/k^2\) for a corner.

Parseval

\(\frac1T\int_T|x|^2dt = \sum_k|a_k|^2\). No cross-terms, so \(|a_k|^2\) is a true power spectrum.

Convergence

Dirichlet gives pointwise convergence, to the midpoint at a jump. Finite energy gives \(E_N\to0\) — but the Gibbs overshoot stays at 8.95%.

Practice

Problems

Problems 1 to 4 should be solved with the properties alone — if you find yourself setting up an analysis integral, look again. Problems 5 to 8 concern power and convergence.

  1. A periodic signal \(x(t)\) with period \(T\) has coefficients \(a_k\). Write down, without integrating, the coefficients of \(x(t-T/2)\), \(x(-t)\), \(x(t)-x(t-T/2)\), and \(\mathrm{Ev}\{x(t)\}\) assuming \(x\) is real.
  2. Show that if \(x(t) = x(t+T/2)\) — that is, the signal actually repeats twice as fast as claimed — then \(a_k = 0\) for all odd \(k\). Compare this with the half-wave symmetry result of Chapter 10 and explain why the conclusions are opposite.
  3. Starting from the sawtooth coefficients \(a_k = jT/(2\pi k)\) of Chapter 10, use the properties to find the coefficients of the reversed sawtooth \(x(-t)\) and of the shifted sawtooth \(x(t - T/2)\). Sketch both waveforms and confirm your answers make sense.
  4. A signal has coefficients \(a_k\) that fall off as \(1/k^3\). What can you say about the continuity of \(x(t)\) and of its first derivative? What would \(a_k\) look like for a waveform containing a train of impulses?
  5. The unit triangular wave of Worked Example 2 has \(b_k = 4/\pi^2k^2\) for odd \(k\). Compute its average power directly from the waveform, then verify Parseval's relation to three decimal places using the first three non-zero harmonics and comment on the residue.
  6. Two signals of the same period \(T\) have coefficients \(a_k\) and \(b_k\). Prove the generalised Parseval relation \(\frac1T\int_T x(t)y^*(t)\,dt = \sum_k a_k b_k^{*}\), and deduce the ordinary form by setting \(y=x\).
  7. Exactly one of the following satisfies all three Dirichlet conditions. Say which, and for each of the other two name the specific condition it violates: (a) \(x(t)=\tan t\) on \((-\pi/2,\pi/2)\), repeated with period \(\pi\); (b) a square wave of amplitude \(10^6\); (c) \(x(t)=t^{-1/2}\) on \((0,1]\), repeated with period 1. For (c), check condition (1) carefully before deciding.
  8. A square wave is reconstructed from its harmonics up to the 49th. Estimate the peak overshoot, the fraction of power captured, and the width of the overshoot relative to the period. Which of the three numbers improves fastest as more harmonics are added?
Tip: when a problem hands you a waveform that resembles one you have already analysed, resist the analysis integral. Ask instead which sequence of operations — shift, reverse, scale, add a constant, differentiate — turns the known waveform into the new one, then apply the matching property to the known coefficients. The chain is almost always shorter than the integral, and it leaves you understanding why the new spectrum looks as it does.