Transforms of common signals

Take a pulse of height 1 and widen it, from 2 s to 40 s, until it looks like a level that never ends. Watch its spectrum grow taller and narrower while the area under the whole curve stays put.

Signals that never end

The pairs on From series to transform (8.1) all belonged to signals that die out: a pulse, a decaying exponential. Most signals you meet in practice do not die out. A steady voltage, a hum, a switch left on: none of them ends. By How big is a signal (1.3) they have infinite energy, so the transform integral cannot be computed for them in the usual way.

Their spectra still exist, and they contain spikes. This page shows where a spike comes from and fixes how you read one. Together with the rules of Properties of the Fourier transform (8.2), a short list of pairs gives you every transform an exam asks for.

Here is the convention, stated once. A spike is drawn as an arrow. The number beside it is its area, and its height means nothing. It is the same convention as for $\delta(t)$ on Impulse, step and ramp (3.1), now drawn against frequency.

To get the first area, put $t=0$ into the inverse transform of 8.1. Every $e^{j\omega\cdot0}$ is 1, so

The area under a spectrum is therefore $2\pi\,x(0)$. This line decides the spikes of the next two sections.

A constant is a spike at zero

I will take a unit-height pulse, 1 for $\lvert t\rvert<W/2$ and 0 elsewhere, and widen it. By 8.1 and the scaling rule of 8.2, its spectrum is

It is $W$ at $\omega=0$ and first reaches zero at $2\pi/W$ rad/s. So a wider pulse gives a taller, narrower spectrum. The line above says its area is $2\pi$ for every $W$, because $x(0)=1$.

That is the picture at the top of the page. A pulse 2 s wide has height 2 and first zero $\pi$ rad/s, and the whole curve, dips and tails included, has area $2\pi=6.283$. Inside the plot the fill holds $6.297$; the tails beyond take back the difference. The central hump alone would hold more, and the dips take that back too.

Wider in time is narrower in frequency. At $8.94$ s wide, the height is $8.94$ and the first zero is $0.70$ rad/s, and the area still reads $6.283$. At 40 s the height is 40 and the first zero is $0.157$ rad/s. Then the fill shrinks into an arrow at 0 labelled $2\pi$: a level that never ends is one spike at zero frequency, of area $2\pi$.

That is the first pair, which you can read off the end frame:

Check it with the line from the start of the page: $\tfrac1{2\pi}\int2\pi\,\delta(\omega)\,d\omega=1=x(0)$. A steady level on a meter has all of its content at zero frequency, and that spike is the picture of it.

A tone is a pair of spikes

A cosine is two half-size arrows, one spinning each way, as on Complex exponentials and phasors (3.4). So I expect two spikes, each half of the $2\pi$ above. To see it, I cut a burst of $\cos5t$ and make it longer.

By the modulation rule of 8.2, a burst has two half copies of its window’s spectrum, centred at $+5$ and $-5$ rad/s. The same instrument shows them, while the burst lengthens from four to forty cycles of $\cos5t$. Watch the area of each copy.

Four cycles of the tone give two lobes, each of area $\pi$. The burst is $5.03$ s long, the lobes are $2.51$ high, and their first zeros are $1.25$ rad/s from each centre. Each whole-curve area reads $3.142$. A longer tone gives thinner, taller lobes with the same area: at about $12.65$ cycles the heights are $7.95$, the first zeros are $0.40$ rad/s from each centre, and the areas read $3.142$. A tone that never ends gives two spikes at $\pm5$ rad/s, each of area $\pi$.

Writing $\omega_1$ for the frequency of a tone, the pairs are

One spinning arrow alone gives one spike of area $2\pi$. The cosine is two half arrows, so it gives two spikes of area $\pi$. The sine is $\bigl(e^{j\omega_1t}-e^{-j\omega_1t}\bigr)/2j$, which gives the factor $\pi/j$ and the opposite signs. A pure tone on a spectrum analyser is a single line, and now you know what the line stands for.

The same step covers any periodic signal. If $x(t)=\sum_kc_ke^{jk\omega_0t}$ with $\omega_0=2\pi/T$, as on Fourier series coefficients (7.2), then each term gives a spike, and $X(j\omega)=\sum_k2\pi c_k\,\delta(\omega-k\omega_0)$.

The step: a spike plus 1/(jω)

The step $u(t)$ is also a signal that never ends. I get it as a limit: $e^{-at}u(t)$ with $a$ shrinking toward 0. By 8.1 its transform is $\dfrac1{a+j\omega}$.

This is complex, so I will split it. Multiply top and bottom by the conjugate $a-j\omega$, as on Complex numbers for signals (3.3). The product $zz^*=\lvert z\rvert^2$ turns the bottom into $a^2+\omega^2$:

The first term is the real part and the second is the imaginary part. From here I draw them as two curves against $\omega$.

In the next clip $e^{-at}u(t)$ lasts longer and longer, as $a$ falls from 2 to 0.02. Watch the real part narrow while its whole area stays, and the imaginary part settle onto $-1/\omega$.

A fast decay, $a=2$, has a low, wide real part. It peaks at $0.5$, and the imaginary part at $\omega=1$ is $-0.2$. Its whole area, tails included, is $\pi$, read as $3.142$, though only $2.652$ of it lies inside the plot. A slower decay, $a=0.2$, has a narrower real part peaking at 5, and the imaginary part at $\omega=1$ is $-0.962$, moving onto $-1/\omega$. At $a=0.02$ the real part peaks at 50 and the imaginary part at $\omega=1$ is $-0.9996$. A step is a spike of area $\pi$ at zero, plus $1/(j\omega)$.

The area of the real part is $\pi$ for every $a>0$, which you can check from $\int\frac{a}{a^2+\omega^2}\,d\omega=\pi$. As $a$ shrinks, the area stays and the curve collapses to a spike. The imaginary part settles onto $-1/\omega=1/(j\omega)$ away from zero. So

Two consequences. First, $\mathrm{sgn}(t)$ is $+1$ for $t>0$ and $-1$ for $t<0$. It equals $2u(t)-1$, so its transform is $2\bigl[\pi\delta(\omega)+\tfrac1{j\omega}\bigr]-2\pi\delta(\omega)=\dfrac2{j\omega}$. The spikes cancel, because $\mathrm{sgn}$ has no level.

Second, the integration rule that 8.2 left open. The step is the integral of $\delta$, and $\delta(t)\leftrightarrow1$. Integrating divides by $j\omega$, the reverse of the differentiation rule, and the spike repairs the division at $\omega=0$:

Put $x=\delta$, so $X(0)=1$, and you get the step’s pair again. The spike is there whenever the signal has a nonzero total area, $X(0)\neq0$.

A Gaussian keeps its shape

The bell curve with half-width $a$ is $x(t)=e^{-t^2/2a^2}$. Its transform is $a\sqrt{2\pi}\,e^{-a^2\omega^2/2}$. Working the integral out needs a calculus step this page does not teach, so take this pair on trust; the clip checks it numerically, frame by frame.

The next instrument squeezes the bell from half-width 2 s to $0.5$ s. Each width is the half-width where the curve has fallen to $0.61$ of its peak, which is $e^{-1/2}$. Watch the shape of the spectrum, and the product of the two half-widths.

A wide bell in time is a narrow bell in frequency: half-widths of 2 s and $0.5$ rad/s, with the spectrum peaking at $5.013$, and a product of 1.00. Squeezed to 1 s, it is still a bell, wider in frequency, 1 rad/s, with peak $2.507$. The product is still 1. At $0.5$ s and 2 rad/s the peak is $1.253$. It is the same shape every time: no ripples, no negative parts.

The half-widths are $a$ in time and $1/a$ in frequency, so their product is 1 for every Gaussian. This is the time–bandwidth trade of 8.2 again. Notice what differs from the pulse of 8.1: the spectrum is a bell too, and it never dips below zero. A short laser pulse has a broad spread of colours for the same reason.

With spread measured another way, as on the card of 8.2, the product of the two spreads can never be below $\tfrac12$. Only the Gaussian reaches that bound.

A comb in time is a comb in frequency

The impulse train $\sum_m\delta(t-mT)$, with $m$ counting the spikes, is a spike every $T$ seconds. It is periodic, so it has a Fourier series, and by sifting every coefficient is the same:

Here $\omega_0=2\pi/T$. So the series is equal arrows at every harmonic. Each term $e^{jk\omega_0t}$ is a spike of area $2\pi c_k=2\pi/T$ in frequency, by the rule for periodic signals above.

The next instrument adds the harmonics in pairs, for $T=1$ s. Its time panel shows the partial sum $S_K(t)=1+2\sum_{k=1}^{K}\cos2\pi kt$, as on Convergence and the Gibbs phenomenon (7.3). Watch the peaks in time as more arrows join.

With three equal arrows, $K=1$, the time signal already peaks once a second: the peak is 3 at $t=0,\pm1$ and the value is $-1$ at $t=\tfrac12$. With twenty-one arrows, $K=10$, the peaks are thin and tall, 21, and still 1 s apart. With 41 arrows, $K=20$, the peak is 41. A comb in time is a comb in frequency: teeth 1 s apart become teeth $2\pi$ rad/s apart.

The partial sum has a closed form, $S_K(t)=\dfrac{\sin\bigl((2K+1)\pi t\bigr)}{\sin\pi t}$. Its peak is $2K+1$ at every whole $t$, and it is $(-1)^K$ at $t=\tfrac12$. So the more equal lines you add, the thinner and taller the time spikes become, always one period apart. A strobe light is such a train. Sampling in Sampling and aliasing (10.1) multiplies a signal by one, and this pair is the reason it copies the spectrum.

Halve $T$ and the frequency teeth move twice as far apart, each with twice the area.

Worked example

A new pair, step by step. Find the transform of $e^{-at}\cos(\omega_1t)\,u(t)$. Look up: $e^{-at}u(t)\leftrightarrow\dfrac1{a+j\omega}$ (8.1). Apply modulation (8.2), which makes two half copies at $\pm\omega_1$:

Substitute $a=3$, $\omega_1=2$, $\omega=2$:

Check: the size is $0.24037$ and the angle is $-19.44°$. Reading a spike. The signal $3\cos(5t)$ has two spikes at $\pm5$ rad/s, each of area $3\pi=9.425$. Their height means nothing; the area carries the amplitude. Spacing. A train with $T=0.1$ s has spikes $2\pi/0.1=62.83$ rad/s apart, which is 10 Hz. Each has area $62.83$.

Where you’ll meet this

A spike in a spectrum is how a tone, a hum or a DC level shows up in every analyser. When you read one, read its area: it holds the amplitude, as the clips showed. The integration rule is why a step input to a system has a spike at zero frequency, and why a plain division by $j\omega$ cannot be the whole answer.

The impulse train is the heart of sampling, which comes back in Sampling and aliasing (10.1) and The sampling theorem (10.2). The ideal low-pass on the card below is the subject of Frequency response and Bode plots (8.4). The pairs for $u$ and $\mathrm{sgn}$ return in The Hilbert transform and the analytic signal (27.2).

Reference card