From series to transform

Repeat a pulse 1 s wide every $T$ seconds, and let the period grow from 2 s to 16 s, until the pulses hardly ever repeat. Watch the lines of its series crowd together on one dashed curve that never moves.

A signal that does not repeat

A Fourier series needs a wave that repeats. Most signals do not: a single click, one struck bell, one flash of a lamp. This page asks what the lines of 7.2 become when the period is not 1 s or 2 s but infinite, so that nothing ever repeats.

The answer is a curve instead of a row of lines. It is called the Fourier transform, and it describes a signal that happens once. The sums of Fourier series coefficients (7.2) become integrals, and everything else carries over.

I use one signal for the first half of the page: the unit pulse, written $\mathrm{rect}(t)$. It is 1 for $\lvert t\rvert<\tfrac12$ and 0 elsewhere, so it is centred on $t=0$, 1 s wide and 1 high. A pulse of width $W$ is $\mathrm{rect}(t/W)$.

Stretch the period: the lines crowd onto one curve

Repeat the unit pulse every $T$ seconds. Its coefficients are 7.2’s pulse-train formula, $c_k=\tfrac1T\cdot\dfrac{\sin(k\omega_0/2)}{k\omega_0/2}$, where $\omega_0=2\pi/T$ is the fundamental, in rad/s. For $T=2$ s the first lines are $0.5,\ 0.318,\ 0,\ -0.106$. For $T=8$ s they are $0.125$, $0.122$, $0.113$, $0.098$, $0.080$.

The lines get closer together and shorter. If they keep shrinking, what is left when $T$ is infinite? Nothing would be left, which cannot be the answer. The fix is to multiply each line by $T$, so I will do that: the quantity I follow from here is $T\,c_k$.

One thing is new. These pulses are centred, so every $c_k$ is a real number. The strips of 7.2 showed only its size $\lvert c_k\rvert$, but here the sign stays, so a line can point down.

That is what the picture at the top of the page draws. With pulses every 2 s, each line is multiplied by 2. The lines sit at $0,\ 3.14,\ 6.28,\ 9.42$ rad/s, with heights $1,\ 0.637,\ 0,\ -0.212$.

As the period grows, the pulses move apart and the lines slide along the curve, closer together. With pulses every 4 s there are twice as many lines, on the same curve. They are $1.571$ rad/s apart, with heights $1$, $0.900$, $0.637$, $0.300$, $0$, $-0.180$. At 16 s the lines are $0.393$ rad/s apart. The curve never moved. Only how closely the lines sit on it changed.

When the clip has finished, drag along the pulses to set the period yourself, from 2 s to 16 s. Think of a camera flash that fires every 2 s, then every 16 s: the flash is the same, and only the gap changes.

Why does the curve stay still? For a pulse that fits inside one period, 7.2’s analysis equation, with the period taken from $-T/2$ to $T/2$, gives

Here $x(t)$ is the single pulse. It is zero outside the period, so the limits can be pushed out to $\pm\infty$ without changing anything. Now let the frequency $k\omega_0$ be any number $\omega$ and write what is left:

This is the Fourier transform of $x$. The lines of the instrument are the values $X(jk\omega_0)$, so $c_k=\tfrac1T X(jk\omega_0)$. A longer period only samples the same curve more often.

For a signal that never quite ends, such as a decaying exponential, the lines land on the curve only in the limit $T\to\infty$. For a pulse that fits inside one period they are on it for every such $T$.

Winding at any rate adds up to $X(j\omega)$

What does the integral mean? 7.2’s machine wound one period of the wave around the origin and took the balance point, which is an average over one period. A single pulse has no period to divide by. So I wind it at any rate $\omega$ and add up instead: every instant contributes the point $x(t)e^{-j\omega t}$ times a thin slice of time $dt$, as the strips of Continuous-time convolution (5.3) added up area.

The next instrument is 7.2’s winding panel with a rate that changes continuously, not only in whole laps per period. The unit pulse is wound into an arc, $e^{-j\omega t}$ for $\lvert t\rvert<\tfrac12$, covering the angles from $-\omega/2$ to $+\omega/2$. Start with the dot, the sum, and watch it as the rate rises.

With no winding the arc is a single point at 1, and adding up the pulse gives its area, 1. Winding faster, the arc wraps further round the circle, and the strip traces the sum. At $\pi$ rad/s the arc is half a circle and it adds up to 0.637.

At $4\pi$ rad/s the arc is two full circles. Whole circles add up to 0, and in between the sum swings negative. The strip shows the whole curve, with zeros at $2\pi$ and $4\pi$ rad/s and a dip to $-0.217$ near 9 rad/s. Afterwards, drag along the strip to choose your own winding rate. The dot stays on the real axis throughout.

The dot stays on the real axis because the pulse is symmetric about $t=0$: the arc is a mirror image of itself, so the imaginary parts cancel. A circle that closes adds up to nothing, as in 7.2. A part of a circle adds up to something, and between the zeros it points to the negative side.

Now the integral by hand. For the unit pulse, with $\int e^{at}\,dt=e^{at}/a$ and $a=-j\omega$,

The third step uses a line that is new here, the partner of the cosine rule on Complex exponentials & phasors (3.4): $\sin\theta=\dfrac{e^{j\theta}-e^{-j\theta}}{2j}$, with $\theta=\omega/2$. At $\omega=0$ the formula reads $0/0$, and the limit is $X(0)=1$, the area. At $\omega=\pi$ it is $2/\pi=0.637$, and at $3\pi$ it is $-2/(3\pi)=-0.212$.

Two more things to read from this curve. First, the units. A signal in volts, multiplied by seconds, gives volts times seconds, which is why the axes say “signal × s”. A height of $X$ is therefore an amount per unit of frequency, not an amount at a frequency.

Second, the other direction. Adding up arrows over all frequencies rebuilds the signal:

This is the inverse transform, and it is 7.2’s synthesis equation with the sum turned into an integral. To see why, write $c_k=\tfrac1TX(jk\omega_0)$ into the synthesis equation, and use $\tfrac1T=\tfrac{\omega_0}{2\pi}$: each term is $\tfrac1{2\pi}X(jk\omega_0)\,e^{jk\omega_0t}\cdot\omega_0$, a thin strip of width $\omega_0$. As $T$ grows $\omega_0$ shrinks, the strips become thin, and the sum is the integral, as in 5.3.

The thin strips give a way to read $X$: it is a density. The band from $\omega$ to $\omega+\Delta\omega$ carries about $X(j\omega)\,\Delta\omega/2\pi$, in the same units as the signal, and a single frequency carries nothing, because a band of zero width has zero area. For the unit pulse, the band from 3.1 to 3.2 rad/s carries about $0.637\times0.1/6.283=0.0101$. The mirror band, from $-3.2$ to $-3.1$, carries the same, and the two arrows turn in opposite directions to make a cosine of height about $0.0202$.

Narrower pulse, wider spectrum

Keep the pulse’s area at 1 and squeeze it to width $W$, so its height is $1/W$. Then the same integral gives

The height at $\omega=0$ stays 1, because it is the area. The curve stretches sideways. Its first zero, the first frequency at which it crosses 0, is at $\omega W/2=\pi$, so at $2\pi/W$ rad/s. Then width times first zero is $2\pi$, whatever the width.

In the next instrument the width shrinks from 2 s to 0.125 s. Watch the product of the width and the first zero, not only the picture.

A wide pulse has a narrow spectrum. At the start the width is 2 s, the height is 0.5, the first zero is $\pi=3.14$ rad/s and the product reads 6.283, which is $2\pi$. The pulse narrows and grows taller, and its area stays 1. At a width of 0.5 s the height is 2 and the first zero is $4\pi=12.57$ rad/s: four times narrower, four times wider in frequency, and still 1 at the centre, the area. At the end the width is $0.125$ s, the height is 8 and the first zero is $16\pi=50.27$ rad/s. Squeezed further, the pulse heads toward $\delta(t)$, whose spectrum is flat at 1: every frequency, equally.

The pulse has not yet become $\delta(t)$ at the end: there $X(j20)=0.759$, still below 1. The unit-area pulse that Impulse, step and ramp (3.1) shrank towards $\delta(t)$ is this one, and as $W\to0$ the spectrum becomes flat at 1.

A click is the example to remember. It is so short that it contains every pitch. A long steady note lasts a long time and contains one.

A decaying exponential: magnitude and phase

The pulse was symmetric, so its sum stayed on the real axis. A struck bell is not: its sound starts at once and then decays. I model it with $x(t)=e^{-t}u(t)$, which is 0 before $t=0$ and $e^{-t}$ after.

Wound at rate $\omega$, it becomes $e^{-t}e^{-j\omega t}=e^{-(1+j\omega)t}$, which is a shrinking spiral, as on Complex exponentials & phasors (3.4). With $a=-(1+j\omega)$ in $\int e^{at}\,dt=e^{at}/a$, and the spiral shrinking to 0,

This is a complex number, so I read it by its size $\lvert X\rvert$ and its angle $\angle X$. Dividing by a complex number divides the lengths and subtracts the angles, as on Fourier series and LTI systems (7.4). So $\lvert X\rvert=1/\sqrt{1+\omega^2}$ and $\angle X=-\arctan\omega$.

The winding instrument appears again, now with $e^{-t}$ for $t\ge0$, wound into the spiral $e^{-(1+j\omega)t}$. Watch the sum shrink and turn back as the rate rises from 0 to 10 rad/s.

With no winding, the spiral is a straight segment from 1 to 0 and the sum is 1, the area under $e^{-t}$. Winding faster, the spiral turns more tightly, and the sum shrinks and turns back. At 1 rad/s it is 0.707 long and 45° behind.

Those are the numbers $0.707\angle-45°$ of the RC circuit in 7.4, since $1/(1+j)$ has length $1/\sqrt2$ and angle $-45°$. Why a circuit’s response is the transform of its own decay is the subject of the page Frequency response and Bode plots (8.4). At 10 rad/s the sum is smaller and turned nearly a quarter turn back: $0.0995\angle-84.3°$, which the readouts show as 0.100 and −84.3°.

The strips show only $\omega\ge0$. For a real signal, the negative side is the mirror image: $X(-j\omega)=X^*(j\omega)$, so the size is even and the angle is odd. Here $X^*$ is the conjugate, the mirror image of $X$, from Complex numbers for signals (3.3).

Build it back from a band of frequencies

The inverse transform adds up arrows over all frequencies. Suppose a system keeps only the band $\lvert\omega\rvert<\omega_c$ and drops the rest, as a speaker that cannot reproduce high frequencies does to a sharp click. What comes back? Here $\omega_c$ is the edge of the kept band.

For the unit pulse, the rebuilt signal is $\tfrac1{2\pi}\int_{-\omega_c}^{\omega_c}X(j\omega)e^{j\omega t}\,d\omega$. It comes back rounded, with a bump beside each edge of the pulse. A wider band narrows the bump towards the edge, but the top of the bump heads for $1.09$ and not for 1. That is about 9 % of the jump from 0 to 1, the same bump as on Convergence and the Gibbs phenomenon (7.3).

The band widens from $\omega_c=4\pi$ to $32\pi$ rad/s in the next instrument. Watch the bump beside each edge of the rebuilt pulse, and the tallest point, in 7.3’s meaning: the highest point of the rebuilt curve.

With frequencies up to $4\pi$ rad/s, 12.566 rad/s, the rebuilt pulse is rounded and its top wobbles. The tallest point is $1.123$ at $\pm0.250$ s and the value at the edge is $0.475$. As the band widens, more frequencies are added back, and the bump moves toward the edge.

At $16\pi$ rad/s the tallest point is $1.096$ at $\pm0.438$ s and the edge value is $0.494$: the bump has moved to the edge and narrowed, and its top barely fell. At $32\pi$ rad/s the tallest point is $1.093$ at $\pm0.469$ s and the edge value is $0.497$.

The top heads for 1.09, about 9 % of the jump from 0 to 1, as in 7.3. At the edge itself, the curve passes through the middle, 0.5. The three held frames sit on high points. Between them the tallest point is not steady: it wobbles between about 1.06 and 1.12 as the band widens, while its position moves towards the edge, and it settles towards 1.09. It stays well above 1 however wide the band, and the edge value approaches $0.5$.

Worked example

1. Unit pulse. $X(j\omega)=\displaystyle\int_{-1/2}^{1/2}e^{-j\omega t}\,dt=\frac{e^{j\omega/2}-e^{-j\omega/2}}{j\omega}=\frac{2\sin(\omega/2)}{\omega}=\frac{\sin(\omega/2)}{\omega/2}$. Then $X(0)=1$, $X(j\pi)=0.6366$, $X(j2\pi)=0$ and $X(j3\pi)=-0.2122$.
2. Lines and curve. Unit pulses every $T$ s have $c_k=\tfrac1TX(jk\omega_0)$. For $T=2$ ($\omega_0=\pi$), $c_k=0.5,\ 0.3183,\ 0,\ -0.1061$ for $k=0$ to 3, and $T\,c_k=1,\ 0.6366,\ 0,\ -0.2122$. For $T=4$, $T\,c_k=1$, $0.9003$, $0.6366$, $0.3001$, $0$, $-0.1801$ for $k=0$ to 5. For $T=8$, $c_k=0.1250$, $0.1218$, $0.1125$, $0.0980$, $0.0796$ for $k=0$ to 4.
3. Density. The band from 3.1 to 3.2 rad/s carries about $X(j\pi)\,\Delta\omega/2\pi=0.637\times0.1/6.283=0.0101$. The mirror band carries the same. At exactly one frequency the band is zero wide and carries nothing.
4. Unit-area pulse of width $W$. $X(j\omega)=\dfrac{\sin(\omega W/2)}{\omega W/2}$, with first zero $2\pi/W$, so width times first zero is $2\pi$. For $W=0.125$ s the first zero is $50.27$ rad/s and $X(j20)=\sin(1.25)/1.25=0.7592$.
5. Exponential, $a=1$. $\displaystyle\int_0^\infty e^{-(1+j\omega)t}\,dt=\frac1{1+j\omega}$. At $\omega=1$: $\lvert1+j\rvert=\sqrt2$, so $0.7071\angle-45.00°$. At $\omega=2$: $0.4472\angle-63.43°$. At $\omega=5$: $0.1961\angle-78.69°$. At $\omega=10$: $0.0995\angle-84.29°$.
6. Band-limited pulse. Keeping only $\lvert\omega\rvert<\omega_c$, the tallest point of the rebuilt unit pulse is $1.1226$ at $\omega_c=4\pi$, $1.1039$ at $8\pi$, $1.0962$ at $16\pi$ and $1.0928$ at $32\pi$. The limit is $1.0895$, about 9 % above the top of the pulse.

Where you’ll meet this

A spectrum analyser on a single sound, a click or a drum hit, shows a curve and not a row of lines, because the sound does not repeat. The trade of step 4 is why a short click is spread over many frequencies, and why a high-quality loudspeaker needs a wide band to reproduce one cleanly. The transform exists, in the sense of this page, when the area under $\lvert x\rvert$ is finite, which a pulse and a decaying exponential satisfy. A signal that never dies out, such as a steady tone, needs spikes, and that is the subject of the page Transforms of common signals (8.3).

Each operation on a signal has a matching change in its spectrum. A delay turns the angles, a squeeze stretches the curve, and these rules are the page Properties of the Fourier transform (8.2). The same limit, applied to sampled data, gives The DTFT (12.2). Band-limiting and the rebuild of a signal from its samples return in The sampling theorem (10.2) and Reconstruction (10.3).

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