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General Input Functions - Fourier Transformation


General input functions

 
Let’s consider the absolutely integrable $f=f(t) input function (Figure 20.).

Image
Figure 20.

 
Let’s consider such a $g(t) ) function being periodic and equivalent with $f(t) in the $ \left [ -\frac{T}{2} ; \frac{T}{2}\right ] interval:

$ g(t)=\begin{cases} f(t) \quad \text{and\quad periodic, \quad for} \quad t\in\left [ -\frac{T}{2} ; \frac{T}{2}\right ] \\ \text{otherwise} \quad g(t)\neq f(t) \end{cases}

Consequently, both $g(t) and $f(t) can be expanded into Fourier series in the $ \left [ -\frac{T}{2} ; \frac{T}{2}\right ] interval:

$ g(t)=f(t)=\sum_{k=-\infty}^{\infty}c_ke^{j\omega kt};\quad \text{where\quad}c_k=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-j\omega kt}dt,\quad \text{and } t\in \left [ -\frac{T}{2} ; \frac{T}{2}\right ]

$ g(t)=f(t)=\sum_{k=-\infty}^{\infty}\left ( \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-j\omega kt}dt \right )e^{j\omega kt}

Apply the following denoting: $ \omega_k=k\omega and attempt to determine the following limit: $ T \to \infty

$ \lim_{T \to \infty}g(t)=\lim_{T \to \infty}f(t)=\lim_{T \to \infty}\sum_{k=-\infty}^{\infty} \frac{1}{T}\left (\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-j\omega_kt}dt \right )e^{j\omega_kt}

Sufficient condition of existing of $ \int_{-\infty}^{\infty}f(t)e^{-j\omega_kt}dt < \infty , as well as $ f(t) < \infty is the
$ \int_{-\infty}^{\infty} \left | f(t) \right | dt < \infty , i.e. $f(t) should be absolutely integrable.

Substitute the following expression: $ T=\frac{2\pi }{\omega_k} .

$ \lim_{T \to \infty}f(t)=\lim_{T \to \infty}g(t)=\sum_{k=-\infty}^{\infty} \frac{1}{T}\left(\int_{-\infty}^{\infty}f(t)e^{-j\omega_kt}dt \right )e^{j\omega_kt}=\sum_{k=-\infty}^{\infty} \frac{\omega_k}{2\pi}\left(\int_{-\infty}^{\infty}f(t)e^{-j\omega_kt}dt \right )e^{j\omega_kt} .

Apply the following denoting: $ F(j\omega_k)=\int_{-\infty}^{\infty}f(t)e^{-j\omega_kt}dt.

$ f(t)=\frac{1}{2\pi}\sum_{k=-\infty}^{\infty}F(j\omega_k)e^{j\omega_k t}\omega_k=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(j\omega)e^{j\omega t}d\omega .

$ f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left [ \int_{-\infty}^{\infty}f(t)e^{-j\omega t}dt \right ]e^{j\omega t}d\omega

The internal improper integral $ F(j\omega)=\mathfrak{F} \left \{ f(t) \right \}=\int_{-\infty}^{\infty}f(t)e^{-j\omega t}dt within the double integral expression is the Fourier transformation of $f=f(t) function. The sufficient condition of Fourier transformation is the $ \int_{-\infty}^{\infty} \left | f(t) \right | dt < \infty , i.e. $f(t) should be absolutely integrable.

The external improper integral of the double integral expression is
$ f(t)=\frac{1}{2\pi} \int_{-\infty}^{\infty}F(j\omega)e^{j\omega t}d \omega=\mathfrak{F}^{-1} \left \{ F(j\omega) \right \} the inverse Fourier transformation.
It is possible to see from the Fourier transformation deriving, that Fourier transformation is an operator. Fourier transformation maps from the set of absolutely integrable functions $f=f(t) to the $ F(j\omega)=\mathfrak{F} \left \{ f(t) \right \} function set. Other words: Fourier transformation maps $f=f(t) functions having domain in the real parameter space to the frequency space of the $f(t) functions (we can call it abstract frequency space due to “t” may represent any type of variable /time, position, temperature,….etc./).

Fourier transformation is a linear operator:

$ \left.\begin{matrix} \mathfrak{F} \left \{ \sum_{i=1}^{n}\lambda_if_i(t) \right \}=\sum_{i=1}^{n}\lambda_i\mathfrak{F} \left \{ f_i(t) \right \}\\ \\ \mathfrak{F}\left \{f(\lambda t) \right \} =\frac{1}{| \lambda |}F(j\frac{\omega}{\lambda}) \end{matrix}\right\} \quad \text{where}\quad \lambda,\lambda_1,\lambda_2,...\lambda_n\in \Gamma

Physical interpretation of Fourier transformation

Definition: Fourier transformation of the absolutely integrable $f(t) functions $ F(j\omega)=\mathfrak{F}\left\{f(t)\right\} is called complex spectrum of $f(t).

$ F(j\omega)=\mathfrak{F}\left\{f(t)\right\} =| F(j\omega) | e^{ j arc(F(j\omega))} , where is possible to present the phase and amplitude response depending on frequency.

$ \left.\begin{matrix} |F(-j\omega)|=|F(j\omega)| \\ \\ arc F(j\omega)=-arcF(j\omega) \end{matrix}\right\} , where it is possible to see, the amplitude response is even function, while the phase function is odd function.

Consequence of the mapping properties of Fourier transformation is: $ F(-j\omega)= F^*(j\omega) , where * is the operation of complex conjugate. Figure 21. shows a possible way to present the complex spectrum. It is quite frequently used at the synthesis and analysis of linear systems depending on the frequency.

Image
Figure 21.

 
Definition: Energy content of an absolutely integrable $f=f(t) function is as follow:

$ E=\int_{-\infty}^{\infty}f^2(t)dt

Let’s describe the definition of E by another way:

$ E=\int_{-\infty}^{\infty}f^2(t)dt=\int_{-\infty}^{\infty}f(t)f(t)dt , here substitute one of the $f(t) factor by the inverse Fourier transformation formula:

$ f(t)=\mathfrak{F}^{-1} \left \{ F(j\omega) \right \} =\frac{1}{2\pi} \int_{-\infty}^{\infty}F(j\omega)e^{j\omega t}d \omega

Next expression can be written:

$ E=\int_{-\infty}^{\infty}f(t)\left (\frac{1}{2\pi} \int_{-\infty}^{\infty}F(j\omega)e^{j\omega t}d \omega \right )dt =\frac{1}{2\pi}\int_{-\infty}^{\infty}F(j\omega)\left [ \int_{-\infty}^{\infty} f(t)e^{j\omega t}dt \right ]d\omega

It is possible to see: $ \int_{-\infty}^{\infty} f(t)e^{j\omega t}dt =F(-j\omega)=\mathfrak{F}^{*}\left \{ f(t) \right \}=F^*(j\omega) .

Consequently $ E=\frac{1}{2\pi}\int_{-\infty}^{\infty} F(j\omega)F^*(j\omega)d \omega=\frac{1}{2\pi}\int_{-\infty}^{\infty} |F(j\omega)|^2d \omega .

Since the amplitude spectrum $ |F(j\omega)|, and $ |F(j\omega)|^2 are even, the energy expression is possible to write as follow:

$ E=\frac{1}{2\pi}\int_{-\infty}^{\infty} |F(j\omega)|^2d \omega=2\frac{1}{2\pi}\int_{0}^{\infty} |F(j\omega)|^2d \omega=\frac{1}{\pi}\int_{0}^{\infty} |F(j\omega)|^2d \omega , which is called Parseval theorem.

 
(Click here for evolving Parzeval theorem)

$ |F(j\omega)|^2 function in the expression of $'E' is called energy spectra of $f(t)

Main features of the Fourier transformation:

Some special features, properties of the Fourier transformation will be described in the followings. Fourier transformation of some important and special functions will be discussed also as follow. Let’s consider at first the general step function (see Figure 22.).

Definition: $f=f(t) function is called as a general step function, if the following condition is satisfied:

$ f(t)= \left\{\begin{matrix} & \varphi (t), & \text{ for } & 0\leq t \\ & 0,& \text{ for } & t<0 \end{matrix}\right. \; and $ \varphi (t) should be absolutely integrable

Image
Figure 22.

 
If $f(t) is a general step function and absolutely integrable, then the complex spectra may be obtained as follow: $ \mathfrak{F} \left \{ f(t) \right \} = \int_{-\infty}^{\infty}f(t)e^{-j\omega t}d t= \int_{0}^{\infty}f(t)e^{-j\omega t}d t .
$f(t) function can be described by the real expressions of $ A(\omega) and $ B(\omega) of complex spectra by using Parseval theorem.

 
(Click here for evolving Parseval theorem)

$ f(t)= \frac{1}{2\pi} \int_{0}^{\infty}\left [ A(\omega)\text{cos}(\omega t)+B(\omega)\text{sin}(\omega t) \right ]d \omega,\quad \text{where} \quad f(t)=\left\{\begin{matrix} \varphi(t), & \text{ for} & 0\leq t \\ 0, & \text{ for} & 0> t \end{matrix} \right.

Apply the following denote: $ t=-\tau .

Consequently, $ f(\tau)= \frac{1}{2\pi}\int_{0}^{\infty}\left [ A(\omega)\text{cos}(\omega \tau)-B(\omega)\text{sin}(\omega \tau) \right ]d \omega=0,\quad \text{if} \quad \tau>0

$ \int_{0}^{\infty} A(\omega)\text{cos}(\omega \tau)d \omega=\int_{0}^{\infty}B(\omega)\text{sin}(\omega \tau) d \omega \quad \text{ if } \tau>0 \text{ and } t=-\tau

Fourier transformation of periodic function

As it is known, the sufficient condition of Fourier Transformation is the $ \int_{-\infty}^{\infty} \left | f(t) \right | dt < \infty i.e. $f(t) should be absolutely integrable.
Neither periodic function can satisfy the sufficient condition. Consequently, another way has to attempt in order to be able to the Fourier transformation. At first, let’s try to derive the Fourier transformation for the simplest case, if $ f(t)=\text{sin}(\omega_0t) or $ f(t)=\text{cos}(\omega_0t). According to the Euler relation’s let’s execute the Fourier transformation by $ e^{j\omega_0t} = \text{cos}(\omega_0t)+j\text{sin}(\omega_0t). Let’s make assumption, Fourier transformation of $ f(t)= e^{j\omega_0t} function is a kind of shifted Dirac-delta type, i.e. $ F(j\omega)=k\delta(\omega-\omega_0). (see the Figure 23., 24 as follow).

Image
Figure 23.

 

Image
Figure 24.

 
Right of the assumption will be proofed by the inverse Fourier transformation. Consequently, $ F(j\omega)=k\delta(\omega-\omega_0) inverse Fourier transformation will be carried out.

$ \mathfrak{F}^{-1}\left \{ k\delta(\omega-\omega_0) \right \}=\frac{1}{2\pi}\int_{-\infty}^{\infty}k\delta(\omega-\omega_0)e^{j\omega_0 t}d\omega=\frac{1}{2\pi} ke^{j\omega_0 t}\int_{-\infty}^{\infty}\delta(\omega-\omega_0)d\omega

It is known about Dirac-delta: $ \int_{-\infty}^{\infty}\delta(\omega-\omega_0)d\omega=1

$ f(t)=\mathfrak{F}^{-1}\left \{ k\delta(\omega-\omega_0) \right \}=\frac{1}{2\pi}ke^{j\omega_0t}. It is possible to see, if $k=2\pi, then $ f(t)= e^{j\omega_0t}. Consequently, $ \mathfrak{F}\left \{ e^{j\omega_0t} \right \}=2\pi \delta(\omega-\omega_0) .

Next, let’s consider an $ f(t)=f(t-nT) periodic function. Attempt to make the Fourier transformation. It is known, any periodic $f(t) function can be expanded in Fourier series, i.e.

$ f(t)=\sum_{n=-\infty}^{\infty}e^{j\omega_0nt}c_n,\quad \text{where}\quad\omega_0=\frac{2\pi}{T}

$ f(t)=\sum_{n=-\infty}^{\infty}e^{j\omega_0nt}c_n=_{\cdots} c_{-m}e^{j\omega_0(-m)t}+_{\cdots} +c_{-1}e^{j\omega_0(-1)t}+_{\cdots} +c_{1}e^{j\omega_0t}+c_{2}e^{j\omega_0(2)t}+_{\cdots}+c_{m}e^{j\omega_0(m)t}+_{\cdots}

Since Fourier transformation is a linear transformation, then the principle of superposition is applicable:

If $ \mathfrak{F}\left \{e^{j\omega_0nt}c_n \right \}=\mathfrak{F}\left \{f_n(t) \right \}=c_n2\pi\delta(\omega-n\omega_0) ,

then
$ \mathfrak{F}\left \{f(t) \right \}=\mathfrak{F}\left \{ \sum_{n=-\infty}^{\infty}c_n{e^{j\omega_0nt} \right \}=2\pi \sum_{n=-\infty}^{\infty} c_n \delta(\omega-n\omega_0)

 

Periodic step function

Let’s define the periodic step function: $f(t)=1(t)f_{1}(t), where $f_{1}(t) = \sum_{k=-\infty}^{\infty}c_{k}e^{jk\omega_{0} t} (see Figure 25.)

Image
Figure 25.

 
According to the above mentioned method, Fourier transformation of the periodic step function can be written as follow:

$ F(j\omega) = \mathfrak{F}\left\{1(t)f_{1}(t)\right\} = \mathfrak{F}\left\{1(t)\sum_{k=-\infty}^{\infty}c_{k}e^{jk\omega_{0} t}\right\} = \sum_{k=-\infty}^{\infty}c_{k}\left[\pi \delta (\omega-k\omega_{0})-\frac{j}{\omega-k\omega_{0}}\right]

Let’s see the case, if $f(t) can be written as the superposition of $f_{1}(t) absolutely integrable and $f_{2}(t) periodic functions.

$ f(t) = \left[f_{1}(t) + f_{2}(t)\right]1(t), where $\int_{-\infty}^{\infty}\left|f_{1}(t)\right|dt < \infty, and $ f_{2}(t) = \sum_{k=-\infty}^{\infty}c_{k}e^{jk\omega_{0} t}
Consequently Fourier transformation of $f(t) is:

$ F(j\omega) = \mathfrak{F}\left\{f(t)\right\}=\mathfrak{F}\left\{\left[f_{1}(t) + \sum_{k=-\infty}^{\infty}c_{k}e^{jk\omega_{0} t}\right]1(t)\right\} = F_{1}(j\omega) + \sum_{k=-\infty}^{\infty}c_{k}\left[\pi \delta(\omega-k\omega_{0})-\frac{j}{\omega-k\omega_{0}}\right]

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