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The description of linear system will be extended into the complex frequency domain {EQUATION(size="80")}$s=\sigma + j\omega{EQUATION} in the following chapters, in order to investigate what kind of association may be created between the various system responses.
__Definition:__ Let {EQUATION(size="80")}$f=f(t){EQUATION} real variable and real as well as complex value general step function:
{EQUATION(size="80")}$
f(t)=\left\{\begin{matrix}
\varphi (t), &\text{ for } t \geq 0\\
\\
0, & \text{ otherwise }
\end{matrix}\right.
{EQUATION}, as well as {EQUATION(size="80")}$\int_{-\infty}^{\infty}\left|f(t)\right| dt < \infty{EQUATION}
{EQUATION(size="80")}$F(s)=\mathcal{L}\left\{f(t)\right\} = \int_{0}^{\infty}f(t)e^{-st}dt{EQUATION} operation is the Laplace transformation of {EQUATION(size="80")}$f(t){EQUATION}. Sufficient condition of Laplace transformation is the {EQUATION(size="80")}$f(t){EQUATION} function should be absolutely integrable (the condition is not necessary, but sufficient).
Laplace transformation is linear, i.e.:
{EQUATION(size="80")}$
\left.\begin{matrix}
\mathcal{L}\left\{\sum_{(i)}\lambda_i f_i (t) \right\} = \sum_{(i)}\lambda_i \mathcal{L} \left\{f_i (t)\right\} \\
\\
\mathcal{L} \left\{ f(\lambda t)\right\} = \frac{1}{\lambda} F(\frac{s}{\lambda})
\end{matrix}\right\}
{EQUATION}, where {EQUATION(size="80")}$\lambda \in \Gamma{EQUATION}
Letâ€™s see, the relation between the Laplace and Fourier transformation. It is known from the above mentioned, the sufficient condition of both Laplace and Fourier transformation is {EQUATION(size="80")}$ \int_{-\infty}^{\infty}|f(t)| dt < \infty{EQUATION}, i.e. {EQUATION(size="80")}$f=f(t){EQUATION} should be absolutely integrable.
Fourier transformation: {EQUATION(size="80")}$\mathfrak{F}\left\{f(t)\right\} = \int_{-\infty}^{\infty}f(t)e^{-j\omega t}dt{EQUATION}
Laplace transformation: {EQUATION(size="80")}$\mathcal{L}\left\{f(t)\right\} = \int_{0}^{\infty}f(t)e^{-st}dt{EQUATION}
Apply the following substitution in the Laplace transformation expression: {EQUATION(size="80")}$s=\sigma +j\omega{EQUATION}.
{EQUATION(size="80")}$
\mathcal{L}\left\{f(t)\right\} = \int_{0}^{\infty}f(t)e^{-st}dt = \int_{0}^{\infty}e^{-(\sigma +j\omega)t}dt = \int_{0}^{\infty}f(t)e^{-\sigma t}e^{-j\omega t}dt
{EQUATION}
It is known from the definition of Laplace transformation, that {EQUATION(size="80")}$f(t){EQUATION} should be general step function. If {EQUATION(size="80")}$f=f(t){EQUATION} is general step function, then {EQUATION(size="80")}$f_1 (t) = f(t)e^{-\sigma t}{EQUATION} is general step function too.
Consequently
{EQUATION(size="80")}$
\left.\begin{matrix}
\mathcal{L}\left\{f(t)\right\} = \int_{0}^{\infty}f_1 (t)e^{-j\omega t}dt = \mathfrak{F}\left\{f_1 (t)\right\} = \mathfrak{F}\left\{f(t)e^{-\sigma t}\right\}\\
\\
\mathfrak{F}\left\{f(t)\right\} = \mathcal{L}\left\{f(t)\right\} \big{|}_{s=j\omega}
\end{matrix}\right\} \Rightarrow \begin{matrix}
F(s)=\mathfrak{F}\left\{f(t)e^{-\sigma t}\right\} \\
\\
F(j\omega) = F(s) \big{|}_{s=j\omega}
\end{matrix}
{EQUATION}
The conclusions are the followings:
Under the conditions, that {EQUATION(size="80")}$f=f(t){EQUATION} is general step function and absolutely integrable, the Laplace transformation of {EQUATION(size="80")}$f(t){EQUATION} among {EQUATION(size="80")}$s=j\omega{EQUATION} complex axis (i.e. among the imaginary axis) gives the Fourier transformation. Consequently, it is possible to see, the domain of Fourier transformation is the {EQUATION(size="80")}$j\omega{EQUATION} imaginary axis, while the domain of Laplace transformation is the complete complex plane. Another approach is existing also, that the Laplace transformation of {EQUATION(size="80")}$f=f(t){EQUATION} is possible to derive from the Fourier transformation of {EQUATION(size="80")}$f_1 (t) = f(t)e^{-\sigma t}{EQUATION} function. Consequently {EQUATION(size="80")}$f_1 (t){EQUATION} can be expressed by the inverse Fourier transformation as follow:
{EQUATION(size="80")}$
f_1 (t) = f(t)e^{-\sigma t} = \mathfrak{F}^{-1}\left\{F(\sigma+j\omega )\right\} = \frac{1}{2\pi} \int_{-\infty}^{\infty}F(\sigma+j\omega )e^{j\omega t}d\omega
{EQUATION}
{EQUATION(size="80")}$
f(t) = \frac{1}{2\pi j}\int_{-j\omega+\sigma}^{j\omega+\sigma}F(\sigma+j\omega)e^{(j\omega+\sigma)t}d(\sigma+j\omega)
{EQUATION}
{EQUATION(size="80")}$
f(t) = \frac{1}{2\pi j}\int_{\sigma-j\omega}^{\sigma+j\omega}F(s)e^{st}ds = \mathcal{L}^{-1}\left\{f(t)\right\}
{EQUATION} where the expression of inverse Laplace transformation is obtained,
and called Rieman-Melling integral.