= Solution
A <Fourier series> separates a periodic physical <signal> into harmonics. For example, a periodic <voltage> applied to a linear circuit can be treated one <frequency> at a time; the harmonic responses then superpose to give the output waveform.
Write $w=\Delta T$ for the pulse width and $\delta=w/T$ for its <duty cycle>, with $0<w<T$. The waveform is a train of positive <rectangular pulses> centred on $t=jT$, each with area one. It is even about $t=0$, so every <sine> coefficient vanishes. <Cosine> <orthogonality> gives
$$
a_0=\frac2T\int_{-T/2}^{T/2}f(t)\,dt=\frac2T,
$$
and, for $n\geq1$,
$$
a_n=\frac2{Tw}\int_{-w/2}^{w/2}\cos\frac{2\pi nt}{T}\,dt
=\frac{2}{\pi nw}\sin\frac{\pi nw}{T}.
$$
This is the <Fourier series of a periodic rectangular pulse>:
$$
\boxed{a_n=\frac2T\operatorname{sinc}(\pi n\delta)\quad(n\geq0),\qquad
\operatorname{sinc}z=\frac{\sin z}{z},\quad\operatorname{sinc}0=1.}
$$
The actual constant term in the series is $a_0/2=1/T$, not $a_0$. As usual for a <jump discontinuity>, the series converges to the average of the one-sided values at each pulse edge, namely $1/(2w)$, irrespective of the value assigned at that isolated endpoint. Elsewhere it converges to the rectangular waveform.
Counting the constant as the first term, the first three terms are
$$
\frac1T,\qquad \frac2T\operatorname{sinc}(\pi\delta)\cos\frac{2\pi t}{T},\qquad
\frac2T\operatorname{sinc}(2\pi\delta)\cos\frac{4\pi t}{T},
$$
so their sum is
$$
\boxed{S_2(t)=\frac1T+\frac2T\left[\operatorname{sinc}(\pi\delta)\cos\frac{2\pi t}{T}+\operatorname{sinc}(2\pi\delta)\cos\frac{4\pi t}{T}\right].}
$$
If “first three” instead means the first three nonconstant harmonics, add $a_3\cos(6\pi t/T)$; the original plot shows that alternative partial sum as well. The three-term approximation rounds the narrow peaks and ripples between them; many harmonics are needed to represent a sharp pulse.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2010/paper-2a-pulse-fourier.png]
{title=Rectangular pulse harmonics}
The original sketches use the illustrative <duty cycle> $\delta=0.15$ and cover $-T\leq t\leq T$. The coefficient panel also displays the first spectral zero.
For $n\delta\ll1$, expand the <sinc function>:
$$
\boxed{a_n=\frac2T\left[1-\frac{\pi^2n^2\delta^2}{6}+O(n^4\delta^4)\right]\simeq\frac2T.}
$$
Thus the low-index coefficients, including $a_0$, have nearly equal values, although their \i[constant-series contribution] still has the factor $1/2$. The first sinc zero occurs at $n\delta=1$, or $n=T/w$. If this is not an integer, the closest integer coefficient is near zero rather than exactly zero: writing $n\delta=1+\varepsilon$ gives $a_n\simeq-(2/T)\varepsilon$. This <pulse-width scale of Fourier coefficients> means that a narrower pulse requires significant harmonics extending to proportionally larger indices, the reciprocal time-width/frequency-width relation.
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