How to extand the spectral analysis to non periodic continuous time funtions ?
Fourier transform Let $f\in L^1(\mathbb{R})\cap L^2(\mathbb{RR})$. The Fourier transform of $f$, denoted by $\hat f$ is given by $$ \hat f(\nu) = \int_{-\infty}^{+\infty} f(t) e^{-i2\pi\nu t} \mathrm{d} t\ . $$
Other definitions of the Fourier transforms exists. By replacing the frequency $\nu$ in Herz by the pulsation $\omega = 2\pi\nu$ in radian per second (or “pulsation”) $$ \hat f(\omega) = \int_{-\infty}^{+\infty} f(t) e^{-i\omega t} \mathrm{d} t $$ and by weighting by $\frac{1}{\sqrt{2\pi}}$ $$ \hat f(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} f(t) e^{-i\omega t} \mathrm{d} t $$
If $f\in L^1(\mathbb{R})$, ie. $f$ is summable, the quantity $\hat f$ is well defined for all $\nu\in\mathbb{R}$. We then have the following decreasing property
If $f\in L^1(\mathbb{R})$, then $\hat f$ is continuous, bounded and $$ \lim_{\nu\rightarrow +\infty} \hat f(\nu) = 0\ . $$ The more regular the function $f$ is, the faster $\hat f$ goes to $0$.
The Fourier transform ensures the Energy preservation, thanks to the Plancherel-Parseval theorem
Plancherel-Parseval Let $f\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$ and $\hat f$ its Fourier transform. We have then the energy preservation $$ \Vert f\Vert_2^2 = \Vert\hat f\Vert_2^2 $$ ie. $$ \int_{\mathbb{R}} |f(t)|^2 \mathrm{d} t = \int_{\mathbb{R}} |\hat f(\nu)|^2 \mathrm{d} \nu $$ Let $g\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$ and $\hat g$ its Fourier transform. We have then the inner product preservation $$ \langle f,g\rangle = \langle \hat f,\hat g\rangle $$ ie. $$ \int_{\mathbb{R}} f(t) \overline{g(t)} \mathrm{d} t = \int_{\mathbb{R}} \hat{f}(\nu) \overline{\hat{g}(\nu)} \mathrm{d} \nu $$
As for the periodic functions, we can define the spectrum of a continuous function
Spectrum Let $f\in L^1(\mathbb{R})\cap L^2(\mathbb{RR})$ and $\hat f$ its Fourier transform. The spectrum of $f$ is given by $$ \text{spectrum(f)}=\lbrace |\hat f(\nu)|^2 \rbrace $$
In practice, we are interested by the bandwidth of a signal, and more particularly by band limited signals
Bandwidth Let $f\in L^2(\mathbb{R})$ an analogical signal and $\hat f$ its Fourier transform. $\text{supp}\lbrace\hat f\rbrace$ is the {\em bandwidth} of the signal. It corresponds to the frequency spreading of the signal.
Band limited signal Let $f\in L^2(\mathbb{R})$ an analogical signal and $\hat s$ its Fourier transform. $f$ is said to be band limited iff it exists $B>0$ such that $$ \text{supp}\lbrace\hat f\rbrace\subset [-B,B]\ . $$
An important result in signal processing is given by the Paley-Wiener’s theorem
Paley-Wiener Let $f\in L^2(\mathbb{R})$ a non null function with compact support. Then its fourier transform cannot vanish on an interval. Likewise, if $\hat f$ is with compact support, then $f$ cannot vanish on an interval.
Inversion of the Fourier transform Let $f\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$ such that $\hat f\in L^1(\mathbb{R})$. If $f$ is continuous in $t$, then $$ f(t) = \int_{\mathbb{R}} \hat f(\nu) e^{i2\pi\nu t} \mathrm{d} \nu\ . $$
With the two other definitions, reconstruction formula becomes
$$ f(t) = \frac{1}{2\pi}\int_{\mathbb{R}} \hat f(\nu) e^{i\omega t}\mathrm{d}\omega $$ $$ f(t) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} \hat f(\omega) e^{i\omega t} \mathrm{d}\omega $$
Let $f,g\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$ and $\hat f, \hat g$ their Fourier transforms.
- Convolution: $$\widehat{f* g}(\nu) = \hat f(\nu) \hat g(\nu)$$
- Multiplication: $$\widehat{f . g}(\nu) = (\hat f * \hat g ) (\nu)$$
- Translation: let $g_a(t) = f(t-a)$ $$\hat g_a(\nu) = e^{-i2\pi a\nu} \hat f(\nu)$$
- Modulation: let $g_{\theta}(t) = e^{i2\pi \theta t}f(t)$ $$\hat{g}_\theta(\nu) = \hat f(\nu - \theta)$$
- Scaling: let $g_{s}(t) = f(t/s)$ $$\hat g_s(\nu) = |s|\hat f(s \nu )$$
- Time derivation: $$\widehat{f^{(p)}}(\nu) = (i2\pi\nu)^p\hat f(\nu)$$
- Frequency derivation: let $g_p(t) = (-i2\pi t)^pf(t)$ $$\hat g_p(t) = \hat f^{(p)}(\nu)$$
- Complexe conjugation $$\widehat{\bar f}(\nu) = \overline{ \hat{ f}(-\nu)}$$
- Hermitian symetry $$f(t)\in\mathbb{R} \Rightarrow \hat f(-\nu) = \overline{\hat{ f}(\nu)}$$