Properties and the inverse Laplace transform

Here is the signal $e^{-t}u(t)$: 0 before $t=0$, then a jump to 1 and a slow decay. A point rides along it with its tangent, and its slope is traced below. Before you watch, make a guess: what is the slope at the jump?

Differentiating becomes multiplying

The Laplace transform (9.1) turned a signal $x(t)$ into a function $X(s)$, with the integral counted from $0^-$, just before the start. This page is about what that buys. The plan is short: differentiating a signal will turn into multiplying $X(s)$ by $s$, so an equation with slopes in it will turn into algebra.

The signal in the picture at the top is $x(t)=e^{-t}u(t)$, which is 0 before $t=0$ and then dies away. Its transform is $X(s)=\dfrac1{s+1}$. I want the transform of its slope, $\dot x$, the slope of the curve as in Differential equations and analog systems (6.2).

Before the jump the slope is 0. At the jump the tangent goes vertical, and the slope is an impulse of area 1. This is the fact from Impulse, step and ramp (3.1): the slope of $u(t)$ is $\delta(t)$, an arrow read by its area. So after the jump the slope is $\delta(t)-e^{-t}u(t)$: the impulse, then the downhill slope of the curve, $-e^{-t}$, flattening out.

Now read the row under the slope plot. It builds the transform of that slope piece by piece. It reads 0 before the jump, then 1 for the impulse, then $1-\dfrac1{s+1}$, since the impulse has transform 1 and the decaying piece has $-\dfrac1{s+1}$. At the end it shows the whole line,

so the slope’s transform is $s$ times the signal’s. The slope of $x$ has the transform $sX(s)$.

What if the signal does not start at 0

A capacitor may already hold a charge before the switch closes. Write $x(0^-)$ for the value just before $0$. If the signal already sits at $x(0^-)$, the jump at $t=0$ is smaller by that much, so the impulse is smaller too. The general rule, for the transform counted from $0^-$, is

Here is the reason, in one line. First the rule for the slope of a product, in words: the slope of $f$ times $g$ is $f$‘s slope times $g$, plus $f$ times $g$‘s slope. Apply it to $x(t)$ times $e^{-st}$:

Add up both sides from $0^-$ to $\infty$. The left side adds up to the end value minus the start value. The end value is 0, because $x\,e^{-st}\to0$ for every $s$ in the ROC, which is what being in the ROC means (The Laplace transform, 9.1). The start value is $x(0^-)$. So

and the first integral is the transform of $\dot x$, which gives the rule above.

Integration goes the other way, and divides by $s$. The running total $w(t)=\int_{0^-}^{t}x$ starts at 0 and has slope $x$. By the rule just shown, $s\,W(s)-0=X(s)$, so $W(s)=X(s)/s$.

Two more rules take one sentence each. Delaying a signal by $t_0$ multiplies its transform by $e^{-st_0}$, as $e^{-j\omega t_0}$ did in Properties of the Fourier transform (8.2). Convolution turns into a product, $X(s)H(s)$, as in 8.2.

The final value, read from sX(s)

Take the charging curve of 6.2, $x(t)=1-e^{-t}$ for $t\ge0$. It is $u(t)-e^{-t}u(t)$, so by linearity and the table in 9.1,

I will not solve for the curve this time. I want to read where it ends up from $X(s)$ alone. The guess to make: if $x(t)$ settles to a final value, which quantity made from $X(s)$ would give it?

Watch the value of $sX(s)$ as $s$ slides from 10 toward 0, beside the charging curve.

At $s=10$, $sX(s)=0.091$. At $s=1$ it is $0.5$, and at $s=0.01$ it is $0.990$. As $s$ goes to 0, $sX(s)$ goes to 1: the value $x$ settles to.

That is the final-value theorem:

It lets me read the end of a curve without finding the curve. The RC circuit’s final voltage comes out as 1 V from $X(s)$ alone.

The theorem needs $x$ to settle. Every pole of $sX$ must lie left of the spin-rate axis, the axis from First- and second-order systems (6.3). Take $\sin t$, with $X=\dfrac1{s^2+1}$. The limit of $sX$ as $s\to0$ exists and is 0, but 0 is not a value that $\sin t$ settles to, because $\sin t$ never settles. Its poles, $\pm j$, sit on the spin-rate axis.

The same idea works at the other end. The initial-value theorem is

For our $X$, $sX=\dfrac1{s+1}\to0$, and $x(0^+)=1-1=0$. This one has a condition too: $X$ must be strictly proper, meaning the top polynomial has a lower degree than the bottom one. The slope’s transform $\dfrac{s}{s+1}$ above is not strictly proper, and indeed that signal holds an impulse at 0 and has no single starting value.

One piece per pole: the cover-up

Getting back to time is the harder direction. The idea is to split $X(s)$ into one simple piece for each pole, because each simple piece is one exponential from the table in 9.1.

Take

The poles are $p_1=-1$ and $p_2=-2$, and I write the weights $r_1$ and $r_2$. To find $r_1$, multiply both sides by $(s+1)$ and then set $s=-1$. The second piece is multiplied by $(s+1)$ and then by $0$, so it vanishes, and what remains is $r_1$.

In practice this is the cover-up rule: cover the factor of the pole you want and put that pole into the rest. For the pole at $-1$,

and for the pole at $-2$,

Each piece is one exponential, so $x(t)=2e^{-t}-e^{-2t}$ for $t\ge0$. Software does the same for any number of poles: SciPy’s residue returns the weights and the poles.

Watch the cover-up done for each pole in turn, and then the two exponentials add.

Each weight times its pole’s exponential gives $2e^{-t}$ and $-e^{-2t}$, and their sum is $x(t)$: one exponential per pole. At $t=0$, 1 and 2 it is $1.000$, $0.600$ and $0.252$.

I can check the start with the initial-value theorem: $sX=\dfrac{s(s+3)}{(s+1)(s+2)}\to1$ as $s\to\infty$, and $x(0)=2-1=1$. ✓

Two poles pushed together

What if two poles land on the same spot? The cover-up needs two different poles, so I start with two close ones and push them together. Put the poles at $-1+\varepsilon$ and $-1-\varepsilon$. Here $\varepsilon$ is the half-gap between them, a symbol I use only on this page.

The cover-up gives the weights $\dfrac1{2\varepsilon}$ and $-\dfrac1{2\varepsilon}$. As the poles come together, each weight grows large, with opposite signs. Together the two pieces make

What happens as $\varepsilon\to0$? The top goes to $0$ and so does the bottom, so I need one more fact. For small $\varepsilon$, $e^{\pm\varepsilon t}\approx1\pm\varepsilon t$, the tangent at 0 from 6.2. Then the top is about $e^{-t}\cdot2\varepsilon t$, and dividing by $2\varepsilon$ leaves $t\,e^{-t}$.

So a double pole at $-1$ gives $\dfrac1{(s+1)^2}\leftrightarrow t\,e^{-t}u(t)$. In general, a double pole at $p$ gives $t\,e^{pt}$. The extra factor $t$ is what a double pole costs.

In the next instrument the two poles start $0.5$ either side of $-1$ and slide together. Watch the curve as they merge.

With the poles 1 apart, $x(1)=0.383$ and the peak is at $1.099$ s. Closer, at $\varepsilon=0.25$, the curve changes only a little: $x(1)=0.372$ and the peak is at $1.022$ s. With one double pole the curve is $t\,e^{-t}$, with $x(1)=0.368$ and its peak at $1.000$ s.

The curve changes smoothly into $t\,e^{-t}$. Nothing jumps when the poles meet, so the double pole is not a new kind of thing, only two poles pushed together.

A mirror pair is a damped sine

Poles can also be complex. The signal with

has the poles $-1\pm2j$, a mirror pair as in 6.3. The cover-up works the same way. For the upper pole,

Dividing by a complex number was Fourier series and LTI systems (7.4): $4j$ has length 4 and angle $90°$, so its inverse has length $\tfrac14$ and angle $-90°$, which is $-\pi/2$ rad. The lower pole has the mirror weight $r^*=+0.25j$, because the conjugate of $-0.25j$ is $0.25j$ (Complex numbers for signals, 3.3).

The two exponentials add up, and Euler’s formula (Complex exponentials and phasors, 3.4), $\sin\theta=\dfrac{e^{j\theta}-e^{-j\theta}}{2j}$ turns the sum into a sine:

Watch the two arrows, one for each pole, and where their sum stays.

Each pole gives a spinning, shrinking arrow. At the start they cancel. They turn opposite ways, and their sum stays on the real line: at $t=0.5$ s it is $0.255$, and by $t=4$ s it has decayed to $0.009$.

This is the point of the page. The lower pole’s weight is the mirror of the upper one’s, $-0.25j$ and $+0.25j$: work out one, write down the other. So a mirror pair needs one cover-up, not two.

The round trip

Now I use all of it on one problem. In 6.2 a capacitor started with $0.5$ V and a 1 V battery was connected, with $RC=1$ s. The equation is $\dot y+y=1$ for $t>0$ with the starting value $y(0^-)=0.5$, and the answer was $1-0.5e^{-t}$. That needed a trial solution and a natural part and a forced part. Here I solve it in four stations: time, $s$, algebra, time.

Before you watch, make a guess. The battery is $1$ V, a constant switched on at $t=0$, so its transform is $\dfrac1s$ by 9.1’s table. What will happen to the starting value $0.5$?

Watch the starting value as it travels through the four stations.

The capacitor starts at 0.5 V and heads for 1 V. At the second station each term goes to $s$, and the slope rule brings in the starting value. The slope $\dot y$ becomes $sY-0.5$, where $0.5$ is the starting value, $y$ becomes $Y$, and the battery’s 1 becomes $\tfrac1s$:

At the third station the calculus has gone: only algebra is left. Collect the $Y$ terms, $(s+1)Y=\tfrac1s+0.5$, and divide:

Cover-up gives the weights. At $s=0$ the rest is $\dfrac{0+1}{0+1}=1$, and at $s=-1$ it is $\dfrac{-0.5+1}{-1}=-0.5$, so

The calculus became algebra. The starting value went into the equation as the $-0.5$ in the second line, and it comes out as the weight $-0.5$ on the pole at $-1$.

At the last station each piece goes back to time, one exponential per pole: $1$ for the pole at 0, and $-0.5e^{-t}$ for the pole at $-1$. Their sum lands on 6.2’s curve, with $y(1)=0.816$: the same answer as 6.2, with no calculus in the middle.

Two checks, using the theorems above. The initial value is $\lim_{s\to\infty}sY=0.5$, the starting charge, since $Y$ is strictly proper. The final value is $\lim_{s\to0}sY=1$, the battery’s voltage.

Worked example

1. Round trip. $\dot y+y=1$ for $t>0$, $y(0^-)=0.5$. Transform: $sY-0.5+Y=\tfrac1s$. Solve: $Y=\dfrac{0.5s+1}{s(s+1)}$. Cover-up: the weight at $s=0$ is $1$ and at $s=-1$ it is $-0.5$. Back to time: $y=1-0.5e^{-t}$, so $y(1)=0.81606$ and $y(2)=0.93233$, the values of 6.2.
2. A second-order equation from rest. Take $\ddot y+3\dot y+2y=u(t)$ with every starting value 0. Applying the slope rule twice gives $s^2Y+3sY+2Y=\tfrac1s$, because the starting terms are 0. So $Y=\dfrac1{s(s+1)(s+2)}$, and cover-up gives the weights $\dfrac1{(1)(2)}=0.5$ at $0$, $\dfrac1{(-1)(1)}=-1$ at $-1$ and $\dfrac1{(-2)(-1)}=0.5$ at $-2$. Then $y=0.5-e^{-t}+0.5e^{-2t}$. Checks: $y(0)=0$, $\dot y(0)=1-1=0$, and $y(\infty)=0.5=\lim_{s\to0}sY$. At $t=1$, $y(1)=0.19979$.
3. Reading the ends of a curve. For $X=\dfrac1{s(s+1)}$, $sX$ at $s=10,\ 1,\ 0.1,\ 0.01$ is $0.0909,\ 0.5,\ 0.909,\ 0.990$, closing in on $x(\infty)=1$. For $X=\dfrac{s+3}{(s+1)(s+2)}$, $sX\to1$ as $s\to\infty$, so $x(0^+)=1$, and the curve $2e^{-t}-e^{-2t}$ gives $x(0)=1$ too.
4. Two poles pushed together. $x_\varepsilon(1)=\dfrac{e^{-1+\varepsilon}-e^{-1-\varepsilon}}{2\varepsilon}$ is $0.383$ for $\varepsilon=0.5$ and $0.372$ for $\varepsilon=0.25$. The limit is $e^{-1}=0.368$, the value of $t\,e^{-t}$ at $t=1$.
5. Mirror pair. For $\dfrac1{s^2+2s+5}$ the weight at $-1+2j$ is $-0.25j$, and $x(t)=\tfrac12e^{-t}\sin2t$. At $t=0.5$ it is $0.5e^{-0.5}\sin1=0.255$, and at $t=4$ it is $0.5e^{-4}\sin8=0.00906$.

Where you’ll meet this

Every time a circuit or a mechanical system is described by a differential equation with a known starting state, the round trip of this page applies. Transforming each term turns the slopes into powers of $s$, and the starting state enters as the extra terms. The poles you find in the algebra are the same poles as the free motion of 6.3.

Software does the splitting. scipy.signal.residue(b, a) returns the weights and poles of $b(s)/a(s)$ for you, the same cover-up numbers as above: for $\dfrac{0.5s+1}{s^2+s}$ it gives the weights $1$ and $-0.5$ at the poles $0$ and $-1$.

The next page, Poles, zeros and the s-plane (9.3), reads a system’s behaviour straight from where its poles sit. The same steps for sampled signals come in Properties and the inverse z-transform (16.2), and analog filter designs are built from poles in Analog prototype filters (20.1).

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