Weekly Blog 3: More About Integral Transforms

Welcome to the 3rd installment of the weekly controls blog! I was going to do a blog on first and second-order systems, but wow, those Laplace transforms last time just got my blood pumped to extend the discussion on integral transforms a bit! I've also recently become interested in applications of integral equations, Fredholm theory, and fractional calculus to control and I figure recounting a bit of common knowledge on integral transforms is a good way to dive into this.

Of course I won't do hardly any more on integral transforms in the large scope, there are 500 page books for that. Instead I'll work a little bit more with the Laplace Transform
$$(\mathcal{L}f)(s) = \int_{0}^{\infty}e^{-st}f(t)\ dt,$$
and the Fourier transform
$$(\mathcal{F}f)(\omega) =  \int_{-\infty}^{\infty}e^{-j\omega t}f(t)\ dt,$$
where, since I'm an engineer, I prefer to annoy mathematicians by defining \(j^2 = -1\).

1. Laplace Transform of a Delay

We've seen the widely used result \(\mathcal{L}\dot{y}(t) = sY(s)\) in the previous blog. This can also be seen as the Laplace transform of the operator \(d/dt\) which gives us the operator equation

$$ \mathcal{L}\frac{d}{dt} = s,$$

or, recursively

$$ \mathcal{L}\frac{d^n}{dt^n} = s^n.$$

We proved this by taking an arbitrary sample function, in this case \(y\), and using derivatives of this function (with the known Laplace transform \(Y(s)\)) we were able to compute the Laplace transform for the derivative itself. 

Now let \(D_\tau\) be the delay operator which is defined as the operator for which

$$D_\tau f(t) = f(t-\tau).$$

That is, \(D_\tau\) simply delays an arbitrary function \(f\) by some amount of time \(\tau\). The question of this section will be to compute \(\mathcal{L}D_\tau\). We do this again by using a sample function, say \(f\) so that we have

$$ \mathcal{L}D_\tau f = \int_{0}^{\infty}e^{-st}D_\tau f(t)\ dt = \int_{0}^{\infty}e^{-st}f(t-\tau)\ dt.$$

Letting \(\sigma(t) = t-\tau\) we find \(d\sigma = dt\) and \(\sigma(0) = -\tau,\sigma(\infty)=\infty\). Then

$$\int_{0}^{\infty}e^{-st}f(t-\tau)\ dt = \int_{-\tau}^{\infty}e^{-s(\sigma+\tau)}f(\sigma)\ d\sigma = e^{-s\tau}\left(\int_{-\tau}^{0}e^{-s\sigma}f(\sigma)\ d\sigma + \int_{0}^{\infty}e^{-s\sigma}f(\sigma)\ d\sigma \right).$$

As before we define \(F(s) = \mathcal{L}f\). Now we need only take care of that bizarre term with the delay in the integral bounds. We have a couple of options: most people define \(f\) as zero on \(0\leq t\leq\tau\) which you will find makes the integral above work out to have only \(F(s)\) and not the second term, but this is ad hoc for me. We can equivalently simply define \(f = 0\) for all \(t < 0\), but again why should we? We obviously need to do something. For now let's just say

$$\int_{-\tau}^{0}e^{-s\sigma}f(\sigma)\ d\sigma + F(s) \approx F(s)$$

but keep that issue in the back of our minds. We thus have

$$\mathcal{L}(D_\tau) = e^{-s\tau}.$$

2. Fourier Transform of the Derivative and Neper Frequency

I'm finished with the Laplace transform for now, though I make no promises of never returning to it again...Let's go ahead and figure out what the Fourier transform of the derivative is. Say \(y\) is our sample function whose Fourier transform is \(Y(\omega)\). Then

$$(\mathcal{F}\dot{y})(s) = \int_{-\infty}^{\infty}e^{-j\omega t}\dot{y}(t)\ dt = -j\omega e^{-j\omega t}y(t)|_{-\infty}^{\infty} - \left(-j\omega\int_{-\infty}^{\infty}e^{-j\omega t}y(t)\ dt\right)$$

so, assuming again the surface term vanishes, 

$$(\mathcal{F}\dot{y})(s)  = j\omega Y(\omega),$$

or

$$\mathcal{F}\frac{d}{dt} =  j\omega.$$

This in particular allows us to easily map linear systems from the \(s\)-domain to the frequency domain by sending \(s\mapsto j\omega\). This is the so-called Joukowski substitution, though it is unclear whether or not Joukowski was the first one to notice this fact. 

Although the Joukowski substitution is the most typically used way of mapping into the frequency domain, it should be noted for completeness that it assumes the forcing frequency is perfect and non-attenuating. Consider for instance the signal 

$$y(t) = \sin(\omega t)$$

The Joukowski substitution works (in the complexified case) and we have no problem, but what about 

$$y(t) = A(t)\sin(\omega t),$$

where \(A\rightarrow 0\) as \(t\rightarrow \infty\)? In this case the Joukowski substitution is not the correct map to the frequency domain. Instead we must introduce the Neper Frequency, \(\sigma\), which keeps track of the signal attenuation, and send \(s\mapsto \sigma+j\omega\) to get into the frequency domain. In practice however, the frequency domain representation of a function is regarded as being consistent with a series of ideal simple harmonic oscillators being used to represent a function and the Neper frequency rarely comes into play when analyzing systems at a purely mathematical level.