0
$\begingroup$

How to do a step function based Fourier series? What I am confused about is how to calculate the time period since the step function doesn't end? And I don't really know the period since the exponential function doesn't repeat unlike the trigonometric functions.

$f(t) = 8e^{t} \cdot u(t-6)$

Thanks in advance.

$\endgroup$
2
  • $\begingroup$ Fourier series exist for periodic functions. Perhaps you're interested in the Fourier transform (or, rather, Laplace transform)? $\endgroup$ Sep 22, 2013 at 21:11
  • $\begingroup$ @Jonathan Y. Yes thats exactly what I am looking for. $\endgroup$ Sep 22, 2013 at 21:50

3 Answers 3

1
$\begingroup$

First, a quick reminder regarding Fourier series. It's well-known that in the Hilbert space of measurable functions $f:[0,T]\to\mathbb{C}$ such that $\int_0^T|f(t)|^2dt<\infty$ (with inner product $\langle f,g\rangle = \int_0^T\overline{f(t)}g(t)dt$), henceforth denoted $L^2([0,T])$, the trigonometric system forms a closed orthonormal system. Therefore, every such function can be represented as a series of trigonometric functions (which converges, I should clarify, in norm, and not necessarily pointwise). That's the Fourier series for such functions.

However, in our case $t\mapsto 8e^tu(t-6)$, defined on $\mathbb{R}$, isn't periodic. As such, it has no Fourier series. We therefore might turn to the notion of the Fourier transform: if a measurable function has the property $\int_\mathbb{R}|f(t)|dt<\infty$ (we say, $f\in L^1(\mathbb{R})$) we can define $F(\omega) = \int_\mathbb{R}f(t)e^{-i\omega t}dt$ (i.e., the integral converges, so $F(\omega)$ is well-defined). This is the Fourier transform. Another well-known fact is that in the Hilbert space of measurable functions $f:\mathbb{R}\to\mathbb{C}$ such that $\int_\mathbb{R}|f(t)|^2dt<\infty$ (with inner product $\langle f,g\rangle = \int_\mathbb{R}\overline{f(t)}g(t)dt$), henceforth denoted $L^2(\mathbb{R})$, functions which are members of $L^1(\mathbb{R})\cap L^2(\mathbb{R})$ are dense; this--along with the fact that the Fourier transform is an isometry for such functions (see Parseval's equality)--allows us to extend the transform to functions which are $L^2$ but not $L^1$.

However, once more, $f(t)=8e^tu(t-6)$ is neither $L^1$ nor $L^2$. In such cases, sometimes we can still define a Laplace transform, which helps 'dampen' growing rates which are exponential or sub-exponential. This is done by defining $F(z) = \int_\mathbb{R}f(t)e^{-zt}dt$, which usually converges on a band or half-plane of $\mathbb{C}$. The transform of the function you mentioned could probably be found in transformation tables on Wiki, but for $z\in\mathbb{C}$ the function $e^{-zt}8e^tu(t-6)\in L^1(\mathbb{R})$ iff $\operatorname{Re}(z)>1$, and there we have $$\int_{\mathbb{R}}e^{-zt}8e^tu(t-6)dt = 8\int_6^\infty e^{(1-z)t} = \frac{8}{z-1}e^{6(1-z)}.$$ This is therefore the Laplace transform (and its area of convergence) of $f$.

$\endgroup$
4
  • $\begingroup$ Thank you. However, I am confused with what did you do with the infinite? And I also have a question regarding the original equation you had used. In my notes, where it has e^(-zt) mine also has n and omega there as well. I am confused as to why you held the n as 1. $\endgroup$ Sep 22, 2013 at 22:25
  • $\begingroup$ @mk1, the primitive of the integrand vanishes at infinity, which is why we don't see its contribution there. I forgot a minus sing, though. With regards to the rest, you're looking at the definition of the Fourier transform. Note that the function is neither $L^1$ nor $L^2$, so it has no Fourier transform; what we did here is a Laplace transform (and indeed, the imaginary axis isn't part of its area of convergence). $\endgroup$ Sep 22, 2013 at 23:15
  • $\begingroup$ Thank you so much. One last thing. Can you please expand on L1 or L2 concept? That would be really appreciated. $\endgroup$ Sep 23, 2013 at 0:41
  • $\begingroup$ @mk1, how's that? $\endgroup$ Sep 23, 2013 at 11:45
1
$\begingroup$

\begin{align} \Theta\left(t\right) &\equiv \left\{% \begin{array}{ll} 0\,, &\quad\mbox{if}\quad x < 0 \\ 1\,, &\quad\mbox{if}\quad x > 0 \end{array}\right. \\[3mm] \Theta\left(t\right) &= -\lim_{\epsilon \to 0^{+}}\int_{-\infty}^{\infty}{{\rm d} k \over 2\pi{\rm i}}\, {{\rm e}^{-{\rm i}kx} \over k + {\rm i}\epsilon} \equiv -\int_{-\infty}^{\infty}{{\rm d} k \over 2\pi{\rm i}}\, {{\rm e}^{-{\rm i}kx} \over k + {\rm i}0^{+}} \end{align}

$\endgroup$
0
$\begingroup$

There is a Fourier series for the $\theta(x-1)$ function which takes a unit unit step at $x=1$. However, it's an infinite series of Fourier series versus a single Fourier series. Please see Illustration of Fourier Series for $\theta(x-1)$ Function

I believe the following answer I posted to one of my own questions provides a fair amount of insight into the theory of the Fourier series representation of $U(x)=-1+\theta(x+1)+\theta(x-1)$ and it's derivatives.

What is Relationship Between Distributional and Fourier Series Frameworks for Prime Counting Functions?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .