Having established some attractive properties of the Laplace Transform that make solving linear differential equations relatively easy, the next logical question to ask is, under what conditions does the Laplace transform actually exist? In answering this, I will follow closely the individual arguments found in Modern Control Engineering by Katsuhiko Ogata, but I’ll be modifying their context to aim at the larger question of the relation between the Laplace and Fourier Transforms.
The first step into understanding what functions are Laplace transformable is to look at the basic definition
\[ {\mathcal L}[f(t)] = \int_0^{\infty} dt \, e^{-st} f(t) \; .\]
The Laplace Transform will only exist when the integral on the right-hand side is convergent. Note that the integral restricts itself to the portion of $$f(t)$$ that falls in the range $$[0,\infty]$$. Since the Laplace Transform doesn’t ask or care about the behavior of $$f(t)$$ for values of $$t < 0$$, the common prescription is to define $$f(t) = 0$$ for $$t < 0$$.
It is a fair question to ask why this requirement is imposed if the the Laplace Transform doesn’t ‘see’ the values of $$f(t)$$ for $$t<0$$. Later on in this column, a relationship between the Fourier and Laplace Transforms presents a plausible justification. For now, let's just say that there are matters of causality that say a signal didn't exist in perpetuity and, thus, at some point in the past it had to turn on. That is what is meant by $$t=0$$.
Of course, there are deep philosophical problems with this point of view, but they don’t touch on the day-to-day workings of the physicist or controls engineer.
So, the operative question is, what type of function, so restricted, has a convergent integral? The class of functions that allow the integral to exist are said to be of exponential order. Operationally, this means that
\[ \lim_{t\rightarrow0} \, e^{-\gamma} |f(t)| = 0 \]
for some positive real number $$\gamma$$ ($$s = \gamma + i \omega$$). If the limit is such that it tends to $$0$$ for $$\gamma > \sigma_c$$ and tends to $$\infty$$ for $$\gamma < \sigma_c$$, the real, positive number $$\sigma_c$$ takes on the name the abscissa of convergence. What this is telling us is that the Laplace Transform only converges when the real part of $$\Re(s) = \gamma$$ is greater than $$\sigma_c$$.
As we will discuss in the example below, this means that $$\gamma$$ must be greater than the right-most pole of the transform $$F(s) = \int_0^{\infty} dt \, e^{-st} f(t)$$.
Some additional words are in order. Functions that increase faster than an exponential do not have Laplace Transforms. The most-oft cited counter-example is the function $$e^{t^2} \; \forall t > 0$$. Note that $$e^{t^2}$$ is a perfectly fine signal if it is defined with compact support, such that it is zero for times $$t>t_{max}$$.
There is an obvious concern that arises at this point. Since the purpose of the Laplace Transform is to turn differential equations into algebraic equations, there is no easy way to restrict the resulting solution to that algebraic equation to have one simple pole. For example, the differential equation $$\ddot x(t) + 3 \dot x(t) + 2 x(t) = f(t)$$ has a Laplace-Transformed left-hand-side of $$s^2 + 3s + 2 = (s+2)(s+1)$$ times $${\mathcal L}[x(t)] \equiv X(s)$$. Assuming the Laplace Transform of $$x(t)$$ and $$f(t)$$ exists, then the following algebraic equation corresponds to the original differential equation with homogeneous initial conditions:
\[ (s+2)(s+1)X(s) = F(s)\; . \]
Solving this for $$X(s)$$ sets before us the task of finding an inverse Laplace Transform for
\[ X(s) = \frac{F(s)}{(s+2)(s+1)} \; .\]
Even assuming that $$F(s)$$ has no poles of its own, the above analysis would suggest that the minimum value of $$\Re(s)$$ would be $$\Re(s)=-1+\epsilon$$, where $$\epsilon$$ is a small positive number. This condition restricts $$s$$ to a portion of the complex plane that doesn’t include the other pole. But the $$s$$ value at the other pole corresponds to physically allowed frequencies, so, how does one reconcile this?
To answer this question, I quote a passage from Ogata’s book:
So, since the theory is now well supported, let’s return to the question as to whether or not the restriction of $$f(t) = 0 \; \forall t < 0$$ is needed. To try to justify this, bring in the Fourier Transform (this argument is based on the ideas of P.C. Chau).
The basic form of the Fourier Transform is
\[ {\mathcal F}[f(t)] = \int_{-\infty}^{\infty} dt \, f(t) e^{-i \omega t} \equiv F(\omega) \]
with the inverse transform being
\[ f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} d\omega \, F(\omega) e^{i\omega t} \; .\]
Now consider the Fourier Transform of the special form of
\[ f(t) = e^{-\gamma t} g(t) \; .\]
The identity $$f(t) = {\mathcal F}^{-1}\left[{\mathcal F}[f(t)] \right]$$ becomes
\[ f(t) \frac{1}{2\pi} \int_{-\infty}^{\infty} d \omega \, \left[ \int_{-\infty}^{\infty} d \tau \, f(\tau) e^{-i \omega \tau} \right] e^{i\omega t} \]
and
\[ e^{-\gamma t} g(t) = \frac{1}{2\pi} \int_{\infty}^{\infty} d \omega \, \int_{-\infty}^{\infty} d \tau \, e^{-\gamma \tau} g(\tau) e^{-i \omega \tau} e^{i \omega t} \; .\]
Bringing the $$e^{-\gamma t}$$ over to the right-hand side and rearranging gives
\[ g(t) = \frac{1}{2\pi} \int_{\infty}^{\infty} d \omega \, e^{(i\omega + \gamma)t} \int_{-\infty}^{\infty} d \tau \, e^{-(i\omega + \gamma) \tau} g(\tau) .\]
Now define the Laplace variable $$s = \gamma + i \omega$$ from which $$d\omega = ds/i$$ and substitute back in to change the outer integral to get the final form
\[ g(t) = \frac{1}{2\pi} \int_{\gamma – i\infty}^{\gamma + i \infty} d s \, e^{s t} \int_{-\infty}^{\infty} d \tau \, e^{-s \tau} g(\tau) .\]
The inner integral will not converge unless $$\tau$$ is restricted so that it is always positive. So, here is a plausible, well-supported argument for always defining $$f(t)$$ in a piece-wise fashion so that it is zero in the range $$t<0$$.
A couple of final notes. The contour implied in the final integral is known as the Bromwich contour. It is rarely used in practice to compute an inverse Laplace Transform with the usual technique being a combination of a clever use of already tabulated transforms and the use of partial fraction decomposition. Another technique for obtaining the inverse Laplace Transform also exists that does not depend on either of these approaches and I’ll devote some future space to examining it.
Next week, I’ll be taking a look at some additional properties of the Laplace Transform that make convenient the separation of transient and steady-state effects.