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From Continuous Fourier Transform to Laplace Transform

Forward Laplace Transform

The Fourier transform of a continuous signal $x(t)$ is defined as:

\begin{displaymath}X(j\omega)={\cal F}[x(t)]=\int_{-\infty}^\infty x(t) e^{-j\om...
...(f)={\cal F}[x(t)]=\int_{-\infty}^\infty x(t) e^{-j2\pi ft} dt \end{displaymath}

provided $x(t)$ is absolutely integrable, i.e.,

\begin{displaymath}\int_{-\infty}^\infty \vert x(t)\vert dt<\infty \end{displaymath}

Obviously many functions do not satisfy this condition and their Fourier transform do not exist, such as $x(t)=t$, $x(t)=t^2$, and $x(t)=e^{t}$. In fact signals such as $x(t)=1$, $x(t)=u(t)$ and $x(t)=cos(2\pi f_0t)$ are not strictly integrable and their Fourier transforms all contain some non-conventional function such as $\delta(f)$.

To overcome this difficulty, we can multiply the given $x(t)$ by an exponential decaying factor $e^{-\sigma t}$ so that $x(t)$ may be forced to be integrable for certain values of the real parameter $\sigma$. Now the Fourier transform becomes:

\begin{displaymath}\int_{-\infty}^\infty x(t)e^{-\sigma t} e^{-j\omega t} dt
=\int_{-\infty}^\infty x(t)e^{-st} dt \end{displaymath}

The result of this integral is a function of a complex variable $s=\sigma+j\omega$, and is defined as the Laplace transform of the given signal $x(t)$, denoted as:

\begin{displaymath}X(s)={\cal L}[x(t)]\stackrel{\triangle}{=}
\int_{-\infty}^\infty x(t) e^{-st} dt \end{displaymath}

provided the value of $\sigma$ is such that the integral converges, i.e., the function $X(s)$ exists. Note that $X(s)$ is a function defined in a 2-D complex plane, called the s-plane, spanned by $\sigma$ for the real axis and $\omega$ for the imaginary axis.

(Pierre-Simon Laplace 1749-1827)

Inverse Laplace Transform

Given the Laplace transform $X(s)$, the original time signal can be obtained by the inverse Laplace transform, which can be derived from the corresponding Fourier transform. We first express the Laplace transform as a Fourier transform:

\begin{displaymath}{\cal L}[x(t)]=X(s)=X(\sigma+j\omega)=\int_{-\infty}^\infty x...
...{-\sigma t}] e^{-j\omega t} dt
={\cal F}[x(t) e^{-\sigma t}]

then $x(t)e^{-\sigma t}$ can be obtained by the inverse Fourier transform:

\begin{displaymath}x(t) e^{-\sigma t}={\cal F}^{-1}[X(\sigma+j\omega)]
...}\int_{-\infty}^\infty X(\sigma+j\omega) e^{j\omega t} d\omega

Multiplying both sides by $e^{\sigma t}$, we get:

\begin{displaymath}x(t)=\frac{1}{2\pi}\int_{-\infty}^\infty X(\sigma+j\omega) e^{(\sigma+j\omega) t} d\omega

To represent the inverse transform in terms of $s$ (instead of $\omega$), we note

\begin{displaymath}ds=d(\sigma+j\omega)=j\;d\omega, \;\;\;\;i.e., \;\;\;\;d\omega=ds/j \end{displaymath}

and the inverse Laplace transform can be obtained as:

\begin{displaymath}x(t)={\cal L}^{-1}[X(s)]
=\frac{1}{j2\pi}\int_{\sigma-j\infty}^{\sigma+j\infty}X(s) e^{st} ds

Note that the integral with respect to $\omega$ from $-\infty$ to $\infty$ becomes an integral in the complex s-plane along a vertical line from $z=\sigma-j\infty$ to $z=\sigma+j\infty$ with $\sigma$ fixed.

Now we have the Laplace transform pair:

\begin{displaymath}X(s)={\cal L}[x(t)]=\int_{-\infty}^\infty x(t)e^{-st} dt \end{displaymath}

\begin{displaymath}x(t)={\cal L}^{-1}[X(s)]
=\frac{1}{j2\pi}\int_{\sigma-j\infty}^{\sigma+j\infty}X(s) e^{st} ds

The forward and inverse Laplace transform pair can also be represented as

\begin{displaymath}x(t) \stackrel{ {\cal L} }{\longleftrightarrow} X(s) \end{displaymath}

In particular, if we let $\sigma=0$, i.e., $s=j\omega$, then the Laplace transform becomes the Fourier transform:

\begin{displaymath}X(s)\bigg\vert _{s=j\omega}=\int_{-\infty}^\infty x(t) e^{-j\omega t} dt=X(j\omega) \end{displaymath}

This is the reason why sometimes the Fourier spectrum is expressed as a function of $j\omega$.

Different from the Fourier transform which converts a 1-D signal $x(t)$ in time domain to a 1-D complex spectrum $X(j\omega)$ in frequency domain, the Laplace transform $X(s)$ converts the 1D signal $x(t)$ to a complex function defined over a 2-D complex plane, called the s-plane, spanned by the two variables $\sigma$ (for the horizontal real axis) and $\omega$ (for the vertical imaginary axis).

In particular, if this 2D function $X(s)=X(\sigma+j\omega)$ is evaluated along the imaginary axis $Re[s]=\sigma=0$, it becomes a 1D function $X(j\omega)$, the Fourier transform of $x(t)$. Graphically, the spectrum of the signal, can be found as the cross section of the 2D function $X(s)=X(\sigma+j\omega)$ along the line $Re[s]=\sigma=0$.


Transfer Function of LTI system

Recall that the output $y(t)$ of a continuous LTI system with input $x(t)$ can be found by convolution:

\begin{displaymath}y(t)={\cal O}[x(t)]=h(t)*x(t)=\int_{-\infty}^\infty h(\tau)x(t-\tau) d\tau \end{displaymath}

where $h(t)$ is the impulse response function of the system. In particular if the input is a complex exponential $x(t)=e^{st}=e^{(\sigma+j\omega)t}$, then the output of the system can be found to be:

\begin{displaymath}y(t)={\cal O}[e^{st}]=\int_{-\infty}^\infty h(\tau)e^{s(t-\ta...
...^{st} \int_{-\infty}^\infty h(\tau)e^{-s\tau} d\tau=H(s)e^{st} \end{displaymath}

This is an eigenequation with the complex exponential $x(t)=e^{st}$ being the eigenfunction of any LTI system, corresponding to its eigenvalue defined as:

\begin{displaymath}H(s)\stackrel{\triangle}{=}\int_{-\infty}^\infty h(t) e^{-st} dt \end{displaymath}

which is the Laplace transform of its impulse response $h(t)$, called the transfer function of the LTI system. In particular, when $\sigma=0$, i.e., $s=j\omega$, the transfer function $H(s)$ becomes the frequency response function, the Fourier transform of the impulse response:

\begin{displaymath}H(j\omega)\stackrel{\triangle}{=}\int_{-\infty}^\infty h(t) e^{-j\omega t} dt \end{displaymath}

next up previous
Next: Region of Convergence (ROC) Up: Laplace_Transform Previous: Laplace_Transform
Ruye Wang 2012-01-28