The Laplace transform

Here a growing signal, $e^{2t}u(t)$, is divided by a test exponential that grows faster and faster, and the area under the ratio is shaded. Before you press play, decide: how fast must the test grow for the area to stop growing? Watch the readout “area”.

Divide by a faster exponential

Some signals grow, such as $e^{2t}u(t)$. The Fourier integral of From series to transform (8.1) cannot be added up for them, because the area under the winding never settles. The spikes of Transforms of common signals (8.3) rescue signals that hold steady, but not signals that grow.

The Laplace transform repairs this in one step. Divide the signal by a test exponential first, and keep track of which tests make the area exist. The border between the tests that work and the tests that fail turns out to be the signal’s own growth rate.

I need one meaning for the letter $\sigma$, and I will keep it for the whole page. It is the one it had in Complex exponentials & phasors (3.4) and First- and second-order systems (6.3): the rate of an exponential $e^{\sigma t}$. A positive $\sigma$ grows, a negative $\sigma$ dies, and the signal $e^{2t}$ grows at rate $+2$.

In the picture at the top of the page I divided $e^{2t}u(t)$ by a test exponential $e^{\sigma t}$, with $\sigma$ rising from 0 to 4. Divided by a flat test, $\sigma=0$, the area never stops growing. A test that grows slower than the signal leaves a ratio that still grows, and so does the area. At $\sigma=2$ the ratio is flat at 1, and the area grows by 1 every second, with no limit.

A test that grows faster makes the ratio die out, and then the area is a number: 1.000 at $\sigma=3$ and 0.500 at $\sigma=4$. The test must grow faster than the signal. The border is the signal’s own rate, +2, marked on the axis “decay rate σ” of 6.3 at the end of the clip. After the clip, drag σ along that axis and watch the readout.

The numbers agree with a one-line integral. The ratio is $e^{(2-\sigma)t}$, and for $\sigma>2$ the exponent is negative, so

At $\sigma=3$ this is $1$, and at $\sigma=4$ it is $0.5$; at $\sigma=2.5$ it would be $2$. For $\sigma\le2$ the exponent is zero or positive and the integral has no limit.

The definition

Now I let the test spin as well. A test exponential $e^{st}$ has the two numbers of 3.4: $s=\sigma+j\omega$, with $\sigma$ the rate at which it grows or dies and $\omega$ the rate at which it spins. Dividing by $e^{st}$ is multiplying by $e^{-st}$. The Laplace transform adds up the result:

For each test exponential $e^{st}$, $X(s)$ says how much of $x$ matches it. At a fixed $\sigma$ the integral is 8.1’s Fourier transform of the signal $x(t)/e^{\sigma t}$, evaluated at $\omega$, because $e^{-st}=e^{-\sigma t}e^{-j\omega t}$. So the new transform is the old one, applied after dividing by a test exponential.

The set of values of $\sigma$ where the integral has a value is the region of convergence, written ROC. For our signal it is $\mathrm{Re}\,s>2$, and there $X(s)=\dfrac{1}{s-2}$, the number we computed above with $\sigma=\mathrm{Re}\,s$. Here $\mathrm{Re}\,s$ is the real part of $s$.

Slicing the plane gives the Fourier transform

Take a signal that does not grow: $x(t)=e^{-t}u(t)$, the decaying exponential of 8.1. Its transform is

because the exponent is $-(s+1)t$ and it dies out only when $\mathrm{Re}\,(s+1)>0$. The formula blows up at $s=-1$. A value of $s$ where $X(s)$ blows up is a pole of the signal’s $X(s)$. It sits at the rate of one of the signal’s own exponentials, as the poles of First- and second-order systems (6.3) sat at the rates of a system’s free motion.

I draw $X$ on the same plane of rates as 6.3, decay rate $\sigma$ across and spin rate $\omega$ up. At every point I shade the length $\lvert X(s)\rvert$, as in Complex numbers for signals (3.3): more colour means bigger. I shade only where the integral has a value, which is the ROC, and leave the rest grey. The edge of the ROC is a line at $-1$, and the colour is strongest toward the pole on it.

The next instrument adds a vertical line at one $\sigma$. The values of $\lvert X\rvert$ along that line, plotted against $\omega$, are the profile. They are the Fourier transform of $x(t)e^{-\sigma t}$, since a fixed $\sigma$ is the situation of the last section. Watch the profile as the line moves from $\sigma=3$ to $\sigma=0$.

Far from the bright spot, at $\sigma=3$, the profile is low: its peak is $0.250$ at $\omega=0$, and it is $0.243$ at $\omega=1$. Closer to the pole the profile rises, to a peak of $0.400$ at $\sigma=1.5$, with $0.371$ at $\omega=1$. At the end the line sits on the spin-rate axis, $\sigma=0$. The peak is $1.000$, and the value at $\omega=1$ is $0.7071$.

On the spin-rate axis the slice is 8.1’s curve, with $\lvert X(j1)\rvert=0.7071$ as in 8.1. So the Fourier transform is the slice $\sigma=0$ of the Laplace transform, which is $X(j\omega)=X(s)\big\rvert_{s=j\omega}$. This holds only when the line $\sigma=0$ is inside the ROC. Here it is, because $0>-1$. When the clip ends, drag the line yourself.

The numbers come from the formula. On a vertical line at $\sigma$,

At $\sigma=3$ the peak is $1/4=0.25$, and at $\omega=1$ it is $1/\sqrt{17}=0.2425$. At $\sigma=1.5$ the peak is $1/2.5=0.4$, and at $\omega=1$ it is $1/\sqrt{7.25}=0.3714$. At $\sigma=0$ the value at $\omega=1$ is $1/\sqrt2=0.7071$.

One formula, two signals: the region decides

Is the formula enough to name the signal? Take the left-sided signal $-e^{-t}u(-t)$, which is $-e^{-t}$ for $t<0$ and zero afterwards. It lives before time 0, and going back in time it grows without limit: at $t=-3$ it equals $-e^{3}=-20.09$. To tame it, the test exponential $e^{\sigma t}$ must grow even faster going back, which needs $\sigma<-1$.

Its transform is

The integral has a value only if $e^{-(s+1)t}$ dies out as $t\to-\infty$, which needs $\mathrm{Re}\,(s+1)<0$. The formula is the same $1/(s+1)$ as for $e^{-t}u(t)$. The region is on the other side of the pole.

The next instrument puts the two signals side by side. For each, a line sweeps $\sigma$ from $-3$ to $3$, and a light beside it says whether the integral adds up. Watch which side of $-1$ each region falls on.

The first signal, $e^{-t}u(t)$, starts at 0 and dies out, and it adds up for decay rates to the right of $-1$. The second, $-e^{-t}u(-t)$, lives before 0 and grows going back, so the test must grow even faster going back: it adds up only to the left of $-1$. Same formula, opposite regions. The formula alone does not tell you the signal; the region does.

Only the first region contains the spin-rate axis, so only the first signal has a Fourier transform. In both pictures the region stops at the pole, because the integral cannot have a value where the formula blows up.

A signal that extends in both directions needs both conditions at once. For $e^{-\lvert t\rvert}$ the region is a strip, $-1<\mathrm{Re}\,s<1$; the worked example below computes it.

There is one rule for the general case. A right-sided signal is zero before some time, and a left-sided signal is zero after some time. For a right-sided signal that is a sum of exponentials, the test must outgrow every exponential in it, so its ROC is everything to the right of its rightmost pole. A left-sided signal has the mirror rule. The ROC never contains a pole.

I call $\sigma_1$ the rate of the pole that forms the border: the rightmost pole for a right-sided signal, the leftmost for a left-sided one. In the first example $\sigma_1=2$, and in the last two it is $-1$.

Where the clock starts

For a signal that starts at time zero, as in Kinds of signals (1.2), the whole integral can start at zero. But what if the signal has an impulse exactly at $t=0$? Write the lower limit as $0^-$, a time just before zero, and the impulse is counted. This is the unilateral transform:

An impulse is an arrow read by its area, as in 8.3, and the arrow at zero has area 1. A hammer blow at the instant the stopwatch starts must count, and the lower limit $0^-$ makes it count. The next lesson, Properties and the inverse Laplace transform (9.2), uses this to put starting values into the algebra.

The next instrument takes the signal $\delta(t)+e^{-t}u(t)$ at $s=1$, so the integrand is $x(t)e^{-t}$, and adds it up from a lower edge that slides from $-1$ to $+1$. Press play and watch the readout as the edge passes 0.

Starting before 0, the impulse (1) plus the rest (0.5) is 1.5. As the edge passes 0 the integral drops from 1.5 to 0.5: the drop is the impulse’s area, 1. Start later, at $+1$, and you lose what came before: 0.068 is left. When the clip holds, drag the edge yourself. The three numbers agree with the integrals. The impulse gives $1$, and the exponential part gives $\int_0^\infty e^{-2t}\,dt=0.5$. From $t=1$ the exponential part gives $\int_1^\infty e^{-2t}\,dt=\tfrac12e^{-2}=0.0677$.

Worked example

1. Growing exponential. $e^{2t}u(t)\to\dfrac{1}{s-2}$, $\mathrm{Re}\,s>2$. At $s=2.5$ it is $2$, at $s=3$ it is $1$, and at $s=4$ it is $0.5$.
2. Step. $u(t)\to\displaystyle\int_0^\infty e^{-st}\,dt=\dfrac1s$, $\mathrm{Re}\,s>0$. At $s=2$ it is $0.5$.
3. Decaying exponential. $e^{-t}u(t)\to\dfrac{1}{s+1}$, $\mathrm{Re}\,s>-1$. On $s=j\omega$ it is $1/(1+j\omega)$, so $\lvert X(j1)\rvert=0.7071$. The slices are those of the second instrument: $0.25$ and $0.2425$ at $\sigma=3$, $0.4$ and $0.3714$ at $\sigma=1.5$.
4. Two-sided signal. For $e^{-\lvert t\rvert}$ split the integral at 0. The part $t>0$ is $e^{-t}u(t)$, giving $\dfrac{1}{1+s}$ for $\mathrm{Re}\,s>-1$. The part $t<0$ is $e^{t}$ for $t<0$, giving $\dfrac{1}{1-s}$ for $\mathrm{Re}\,s<1$. Both must hold at once, so $$X(s)=\frac{1}{1-s}+\frac{1}{1+s}=\frac{2}{1-s^2},\qquad -1<\mathrm{Re}\,s<1 .$$ At $s=0$ this is $2$, the area under $e^{-\lvert t\rvert}$. At $s=0.5$ it is $2.667$.
5. Left-sided signal. $-e^{-t}u(-t)\to\dfrac{1}{s+1}$, $\mathrm{Re}\,s<-1$. At $s=-2$ it is $-1$, and at $t=-3$ the signal is $-e^{3}=-20.09$.
6. Unilateral and impulses. $\delta(t)+e^{-t}u(t)\to1+\dfrac{1}{s+1}$. At $s=1$ this is $1.5$. From $0^+$ it is $0.5$, and from $t=1$ it is $\tfrac12e^{-2}=0.0677$.
7. Cosine and sine from two arrows. Write $\cos\omega_1t$ as half the sum of two arrows, $\tfrac12\left(e^{j\omega_1t}+e^{-j\omega_1t}\right)$. Each has a transform of the form $e^{-at}u(t)\to\dfrac{1}{s+a}$, with $a=\mp j\omega_1$, valid for $\mathrm{Re}\,s>0$. Here $\omega_1$ is the frequency of one tone, as in 8.3. Then $$\cos\omega_1t\,u(t)\to\tfrac12\left(\frac{1}{s-j\omega_1}+\frac{1}{s+j\omega_1}\right)=\frac{s}{s^2+\omega_1^2}.$$ For the sine, subtract and divide by $2j$: $$\sin\omega_1t\,u(t)\to\frac{1}{2j}\left(\frac{1}{s-j\omega_1}-\frac{1}{s+j\omega_1}\right)=\frac{\omega_1}{s^2+\omega_1^2}.$$ At $\omega_1=1$ and $s=1$ both are $0.5$.

Where you’ll meet this

The unilateral transform is the one engineers use to solve circuits that start from a charged state, and 9.2 does exactly that with the capacitor of Differential equations and analog systems (6.2). The poles you saw as marks on a plane are the system’s own rates of motion, and Poles, zeros and the s-plane (9.3) reads a system’s behaviour off their positions.

The same idea, with sums in place of integrals, becomes the z-transform for sampled signals in a later chapter. There the ROC will matter for exactly the reason it did here: the formula alone does not name the signal.

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