9.2 The Inverse Transform and Convolution

In Section 9.1 we went from a function $f(t)$ to its transform $F(s)$. When we solve a differential equation, we travel the other way: the algebra hands us $Y(s)$, and we need the function $y(t)$ it came from. This return trip is the inverse transform.

In practice $Y(s)$ is almost always a ratio of two polynomials in $s$. So the skill we need is concrete: break a rational function into simple pieces that appear in the table. That technique is partial fractions, and most of this section is about doing it reliably. At the end we meet the convolution, which answers the question "which function has transform $F(s)G(s)$?"

The inverse transform

Can two different functions have the same transform? Strictly speaking, yes, but only in a harmless way. Changing a function at a single point (for instance at a jump) does not change any integral, so it does not change the transform. The next theorem says this is the only freedom, and that for continuous functions there is none at all.

So when we find one continuous function with transform $F(s)$, we have found the inverse transform. Since $\mathcal{L}$ is linear, so is ${\mathcal{L}}^{-1}$:

The working method is therefore: split $F(s)$ into a sum of pieces that appear in the right-hand column of Table 9.1.1, then read the table from right to left.

Completing the square

A quadratic denominator $s^2+ps+q$ with no real roots ($p^2<4q$) cannot be factored over the real numbers. Instead we complete the square,

with $a=-p/2$ and $b^2=q-p^2/4>0$. Then we rewrite the numerator in terms of $s-a$ and use rows 9 and 10 of the table (the shifted sine and cosine).

These two examples already show something useful. The roots of the denominator (called the poles of $F$) decide what kind of function comes back. A real pole $s=a$ produces $e^{at}$; a pair of complex poles $a\pm ib$ produces $e^{at}\cos bt$ and $e^{at}\sin bt$. A pole in the left half-plane ($a<0$) means decay; a pole in the right half-plane means growth; poles on the imaginary axis mean steady oscillation. Engineers read stability straight off the pole locations.

Partial fractions

Suppose $F(s)=P(s)/Q(s)$, where $P$ and $Q$ are polynomials with no common factor, and $\mathrm{deg}P<\mathrm{deg}Q$ (a proper fraction). Every transform we meet in this chapter has this form; recall that a transform must tend to $0$ as $s\to \infty$, which a rational function does exactly when it is proper.

Algebra guarantees that $Q$ factors into real linear factors $(s-a)$ and real irreducible quadratic factors $(s-\alpha {)}^2+{\beta }^2$, possibly repeated. Partial fractions rewrites $P/Q$ as a sum with one block of terms for each factor:

To find the unknown constants, multiply the decomposition by $Q(s)$ and compare both sides. Two tools do almost all of the work.

• Substituting good values of $s$. If $s=a$ is a root of $Q$, substituting it kills every term that contains the factor $(s-a)$.
• Comparing coefficients of equal powers of $s$ on both sides. This always works and is the fallback for whatever is left over.

For a simple factor $s-a$ there is a shortcut, known as Heaviside's cover-up rule: $A$ equals $P(s)/Q(s)$ with the factor $(s-a)$ covered up (deleted), evaluated at $s=a$. It is just the substitution trick done in your head.

Practise with the solver below. Enter a numerator and a denominator (you may give the denominator already factored). It shows the decomposition, every constant, the completed squares and the final inverse transform. The preset is Example 9.2.5; try changing the numerator to $s^2+8s+12$ and predict which constants change.

Convolution

We know that the transform of a product is not the product of the transforms. But the opposite question is very natural: if $Y(s)=F(s)G(s)$ and we know $f$ and $g$, what is $y$? The answer is a special "product" of functions defined by an integral.

How should you picture this integral? Fix a time $t$. Flip $g$ around and slide it so that it ends at $\tau =t$; this is $g(t-\tau )$ as a function of $\tau$. Multiply by $f(\tau )$ and take the area from $0$ to $t$. As $t$ increases, the flipped copy of $g$ slides to the right, and the area changes: that changing area is $(f\ast g)(t)$ (Figure 9.2.2). In physical terms, $f(\tau )$ is an input that arrived at time $\tau$, and $g(t-\tau )$ says how much of its effect is still felt $t-\tau$ seconds later; the convolution adds up the effects of the whole past input.

Convolution behaves like a multiplication in several ways. For piecewise continuous $f,g,h$ and constants $c$:

• $f\ast g=g\ast f$ (substitute $u=t-\tau$ in the integral);
• $f\ast (g+h)=f\ast g+f\ast h$ and $(cf)\ast g=c(f\ast g)$;
• $(f\ast g)\ast h=f\ast (g\ast h)$;
• $f\ast 0=0$.

One familiar rule fails: $f\ast 1\ne f$ in general. In fact $(f\ast 1)(t)={\int }_0^{t}f(\tau )d\tau$, the running integral of $f$.

A useful special case is $g=1$, $G=1/s$: dividing a transform by $s$ integrates the function,

It mirrors the derivative rule, where multiplying by $s$ differentiates.

Equations of this kind, where the unknown appears under an integral with a kernel $k(t-\tau )$, describe systems with memory (viscoelastic materials, population models, circuits with distributed elements). The convolution theorem turns them into algebra exactly as it does for differential equations.

Use the convolution calculator to compute $f\ast g$ for your own pairs. It forms $F(s)G(s)$, inverts the product and shows the result. The preset is Example 9.2.6.