- Signals and Systems Tutorial
- Signals & Systems Home
- Signals & Systems Overview
- Signals Basic Types
- Signals Classification
- Signals Basic Operations
- Systems Classification
- Signals Analysis
- Fourier Series
- Fourier Series Properties
- Fourier Series Types
- Fourier Transforms
- Fourier Transforms Properties
- Distortion Less Transmission
- Hilbert Transform
- Convolution and Correlation
- Signals Sampling Theorem
- Signals Sampling Techniques
- Laplace Transforms
- Laplace Transforms Properties
- Region of Convergence
- Z-Transforms (ZT)
- Z-Transforms Properties
- Signals and Systems Resources
- Signals and Systems - Resources
- Signals and Systems - Discussion
Laplace Transforms (LT)
Complex Fourier transform is also called as Bilateral Laplace Transform. This is used to solve differential equations. Consider an LTI system exited by a complex exponential signal of the form x(t) = Gest.
Where s = any complex number = $\sigma + j\omega$,
σ = real of s, and
ω = imaginary of s
The response of LTI can be obtained by the convolution of input with its impulse response i.e.
$ y(t) = x(t) \times h(t) = \int_{-\infty}^{\infty}\, h (\tau)\, x (t-\tau)d\tau $
$= \int_{-\infty}^{\infty}\, h (\tau)\, Ge^{s(t-\tau)}d\tau $
$= Ge^{st}. \int_{-\infty}^{\infty}\, h (\tau)\, e^{(-s \tau)}d\tau $
$ y(t) = Ge^{st}.H(S) = x(t).H(S)$
Where H(S) = Laplace transform of $h(\tau) = \int_{-\infty}^{\infty} h (\tau) e^{-s\tau} d\tau $
Similarly, Laplace transform of $x(t) = X(S) = \int_{-\infty}^{\infty} x(t) e^{-st} dt\,...\,...(1)$
Relation between Laplace and Fourier transforms
Laplace transform of $x(t) = X(S) =\int_{-\infty}^{\infty} x(t) e^{-st} dt$
Substitute s= σ + jω in above equation.
$→ X(\sigma+j\omega) =\int_{-\infty}^{\infty}\,x (t) e^{-(\sigma+j\omega)t} dt$
$ = \int_{-\infty}^{\infty} [ x (t) e^{-\sigma t}] e^{-j\omega t} dt $
$\therefore X(S) = F.T [x (t) e^{-\sigma t}]\,...\,...(2)$
$X(S) = X(\omega) \quad\quad for\,\, s= j\omega$
Inverse Laplace Transform
You know that $X(S) = F.T [x (t) e^{-\sigma t}]$
$\to x (t) e^{-\sigma t} = F.T^{-1} [X(S)] = F.T^{-1} [X(\sigma+j\omega)]$
$= {1\over 2}\pi \int_{-\infty}^{\infty} X(\sigma+j\omega) e^{j\omega t} d\omega$
$ x (t) = e^{\sigma t} {1 \over 2\pi} \int_{-\infty}^{\infty} X(\sigma+j\omega) e^{j\omega t} d\omega $
$= {1 \over 2\pi} \int_{-\infty}^{\infty} X(\sigma+j\omega) e^{(\sigma+j\omega)t} d\omega \,...\,...(3)$
Here, $\sigma+j\omega = s$
$jdω = ds → dω = ds/j$
$ \therefore x (t) = {1 \over 2\pi j} \int_{-\infty}^{\infty} X(s) e^{st} ds\,...\,...(4) $
Equations 1 and 4 represent Laplace and Inverse Laplace Transform of a signal x(t).
Conditions for Existence of Laplace Transform
Dirichlet's conditions are used to define the existence of Laplace transform. i.e.
The function f(t) has finite number of maxima and minima.
There must be finite number of discontinuities in the signal f(t),in the given interval of time.
It must be absolutely integrable in the given interval of time. i.e.
$ \int_{-\infty}^{\infty} |\,f(t)|\, dt \lt \infty $
Initial and Final Value Theorems
If the Laplace transform of an unknown function x(t) is known, then it is possible to determine the initial and the final values of that unknown signal i.e. x(t) at t=0+ and t=∞.
Initial Value Theorem
Statement: if x(t) and its 1st derivative is Laplace transformable, then the initial value of x(t) is given by
$$ x(0^+) = \lim_{s \to \infty} SX(S) $$
Final Value Theorem
Statement: if x(t) and its 1st derivative is Laplace transformable, then the final value of x(t) is given by
$$ x(\infty) = \lim_{s \to \infty} SX(S) $$