# Properties of Laplace Transforms(Bi-lateral)

1.Linearity Property:-

$x_{1}(t)\leftrightarrow&space;X_{1}(S)&space;\&space;\&space;\&space;ROC&space;:&space;R_{1}$

$x_{2}(t)\leftrightarrow&space;X_{2}(S)&space;\&space;\&space;\&space;ROC&space;:&space;R_{2}$

$a\&space;x_{1}(t)+b\&space;x_{2}(t)\leftrightarrow&space;\&space;\&space;?$.

we know that  Laplace Transform of a signal  $x(t)$  is  $X(S)&space;=&space;\int_{-\infty&space;}^{\infty&space;}\&space;x(t)&space;\&space;e^{-St}&space;\&space;dt$ .

$L\left&space;\{a\&space;x_{1}(t)+b\&space;x_{2}(t)\right&space;\}&space;=&space;\int_{-\infty&space;}^{\infty&space;}\&space;\left&space;\{&space;a\&space;x_{1}(t)+b\&space;x_{2}(t)&space;\right&space;\}&space;\&space;e^{-St}\&space;dt$.

$L\left&space;\{a\&space;x_{1}(t)+b\&space;x_{2}(t)\right&space;\}&space;=&space;\int_{-\infty&space;}^{\infty&space;}\&space;\left&space;\{&space;a\&space;x_{1}(t)\&space;e^{-St}\&space;dt+b\&space;x_{2}(t)&space;\&space;e^{-St}\&space;dt\right&space;\}$

$L\left&space;\{a\&space;x_{1}(t)+b\&space;x_{2}(t)\right&space;\}&space;=&space;\&space;\&space;a&space;\int_{-\infty&space;}^{\infty&space;}\&space;x_{1}(t)\&space;e^{-St}\&space;dt+b&space;\int_{-\infty&space;}^{\infty&space;}\&space;x_{2}(t)&space;\&space;e^{-St}\&space;dt$.

$L\left&space;\{a\&space;x_{1}(t)+b\&space;x_{2}(t)\right&space;\}&space;=&space;\&space;\&space;a&space;X_{1}(S)&space;+b&space;X_{2}(S)$ .          $ROC:&space;R_{1}&space;\cap&space;R_{2}$.

2.Time-shifting Property:-

$x(t)\leftrightarrow&space;X(S)&space;\&space;\&space;\&space;ROC&space;:&space;R$

$x(t-t_{o})\leftrightarrow&space;\&space;?$.

we know that  $X(S)&space;=&space;\int_{-\infty&space;}^{\infty&space;}\&space;x(t)&space;\&space;e^{-St}&space;\&space;dt$ .

$L\left&space;\{&space;x(t-t_{o})&space;\right&space;\}&space;=&space;\int_{-\infty&space;}^{\infty&space;}\&space;x(t-t_{o})&space;\&space;e^{-St}\&space;dt$.    Let$t-t_{o}=\lambda&space;\Rightarrow&space;dt=&space;d\lambda$

$t&space;\&space;limits&space;\&space;:&space;\&space;-\infty&space;\&space;to&space;\&space;\infty,&space;\&space;\&space;\&space;\lambda&space;\&space;\&space;limits&space;\&space;:&space;\&space;\infty&space;\&space;to&space;\&space;-\infty$

$L\left&space;\{&space;x(t-t_{o})&space;\right&space;\}&space;=&space;\int_{\lambda&space;=-\infty&space;}^{\infty&space;}\&space;x(\lambda&space;)&space;\&space;e^{-S(\lambda&space;+t_{o})}\&space;d\lambda$ .

$L\left&space;\{&space;x(t-t_{o})&space;\right&space;\}&space;=e^{-St_{o}}&space;\int_{\lambda&space;=-\infty&space;}^{\infty&space;}\&space;x(\lambda&space;)&space;\&space;e^{-S\lambda&space;}\&space;d\lambda$.

$x(t-t_{o})\leftrightarrow&space;\&space;e^{-St_{o}}\&space;X(S)&space;\&space;,&space;\&space;\&space;ROC:R$.

from the above equation $x(t-t_{o})$  forms a Laplace Transform pair with $e^{-St_{o}}&space;\&space;X(S)$.

3.Frequency-shifting Property:-

$x(t)\leftrightarrow&space;X(S)&space;\&space;\&space;\&space;ROC&space;:&space;R$

$?\&space;\leftrightarrow&space;\&space;X(S-S_{o})$.

we know that  $X(S)&space;=&space;\int_{-\infty&space;}^{\infty&space;}\&space;x(t)&space;\&space;e^{-St}&space;\&space;dt$ .

$L\left&space;\{&space;e^{S_{o}t}\&space;x(t)&space;\right&space;\}&space;=&space;\int_{t=-\infty&space;}^{\infty&space;}\&space;e^{S_{o}t}\&space;x(t)&space;\&space;e^{-St}\&space;dt$.

$L\left&space;\{&space;e^{S_{o}t}\&space;x(t)&space;\right&space;\}&space;=&space;\int_{t=-\infty&space;}^{\infty&space;}\&space;\&space;x(t)&space;\&space;e^{-(S-S_{o})t}\&space;dt$ .

$e^{S_{o}t}x(t)\leftrightarrow&space;\&space;X(S-S_{o})&space;\&space;,&space;\&space;\&space;ROC:R$.

from the above equation $e^{S_{o}t}x(t)$  forms a Laplace Transform pair with $X(S-S_{o})$.

4. Differentiation in time-domain:-

$x(t)\leftrightarrow&space;X(S)&space;\&space;\&space;\&space;ROC&space;:&space;R$

$\frac{dx(t)}{dt}\leftrightarrow&space;\&space;?$.

we know that  inverse Laplace Transform  $x(t)&space;=\frac{1}{2\pi&space;\&space;j}&space;\int_{\sigma&space;-&space;j\infty&space;}^{\sigma&space;+j\infty&space;}\&space;X(S)&space;\&space;e^{St}&space;\&space;dS$ .

$\frac{dx(t)}{dt}&space;=\frac{1}{2\pi&space;\&space;j}&space;\int_{\sigma&space;-&space;j\infty&space;}^{\sigma&space;+j\infty&space;}\&space;X(S)&space;\&space;\frac{d(e^{St})}{dt}&space;\&space;dS$.

$\frac{dx(t)}{dt}&space;=\frac{1}{2\pi&space;\&space;j}&space;\int_{\sigma&space;-&space;j\infty&space;}^{\sigma&space;+j\infty&space;}\&space;X(S)&space;\&space;S&space;\&space;e^{St}&space;\&space;dS$ .

$\frac{dx(t)}{dt}&space;=\frac{1}{2\pi&space;\&space;j}&space;\int_{\sigma&space;-&space;j\infty&space;}^{\sigma&space;+j\infty&space;}&space;(\&space;S\&space;X(S))&space;\&space;e^{St}&space;\&space;dS$.

$\frac{dx(t)}{dt}\leftrightarrow&space;\&space;S\&space;X(S)$.

from the above equation $\frac{dx(t)}{dt}$  forms a Laplace Transform pair with $S\&space;X(S)$

Similarly  $\frac{d^{n}x(t)}{dt^{n}}\leftrightarrow&space;\&space;S^{n}\&space;X(S)$.

5.Differentiation in S-domain:-

$x(t)\leftrightarrow&space;X(S)&space;\&space;\&space;\&space;ROC&space;:&space;R$

$?\leftrightarrow&space;\frac{dX(S)}{dS}$.

we know that  $X(S)&space;=&space;\int_{-\infty&space;}^{\infty&space;}\&space;x(t)&space;\&space;e^{-St}&space;\&space;dt$ .

$\frac{dX(S)}{dS}&space;=&space;\int_{-\infty&space;}^{\infty&space;}\&space;x(t)&space;\&space;\frac{de^{-St}&space;}{dS}\&space;dt$.

$\frac{dX(S)}{dS}&space;=&space;\int_{-\infty&space;}^{\infty&space;}\&space;x(t)&space;\&space;e^{-St}&space;\&space;(-t)\&space;dt$ .

$\frac{dX(S)}{dS}&space;=&space;\int_{-\infty&space;}^{\infty&space;}\&space;(-t&space;\&space;x(t))&space;\&space;e^{-St}&space;\&space;dt$.

$\frac{dX(S)}{dS}\leftrightarrow&space;\&space;-t\&space;x(t)&space;\&space;\&space;\&space;ROC:R$.

from the above equation $\frac{dX(S)}{dS}$  forms a Laplace Transform pair with $-t\&space;x(t)$.

6. Time-reversal property:-

$x(t)\leftrightarrow&space;X(S)&space;\&space;\&space;\&space;ROC&space;:&space;R$

$x(-t)\leftrightarrow&space;\&space;\&space;?$.

we know that  $X(S)&space;=&space;\int_{-\infty&space;}^{\infty&space;}\&space;x(t)&space;\&space;e^{-St}&space;\&space;dt$ .

$L\left&space;\{&space;x(-t)&space;\right&space;\}&space;=&space;\int_{-\infty&space;}^{\infty&space;}\&space;x(-t)&space;\&space;e^{-St}&space;\&space;dt$.          Let  $-t=\&space;\lambda$ ,      $-dt=\&space;d\lambda$,  $t&space;\&space;limits&space;\&space;:&space;\&space;-\infty&space;\&space;to&space;\&space;\infty,&space;\&space;\&space;\&space;\lambda&space;\&space;\&space;limits&space;\&space;:&space;\&space;\infty&space;\&space;to&space;\&space;-\infty$.

$L\left&space;\{&space;x(-t)&space;\right&space;\}&space;=&space;\int_{\lambda&space;=\infty&space;}^{-\infty&space;}\&space;x(\lambda&space;)&space;\&space;e^{S\lambda&space;}&space;\&space;(-d\lambda&space;)$ .

$L\left&space;\{&space;x(-t)&space;\right&space;\}&space;=&space;\int_{\lambda&space;=-\infty&space;}^{\infty&space;}\&space;x(\lambda&space;)&space;\&space;e^{-(-S)&space;\lambda&space;}&space;\&space;d\lambda$.

$x(-t)\leftrightarrow&space;\&space;X(-S)$.

from the above equation $x(-t)$  forms a Laplace Transform pair with $X(-S)$.

7. Time-Scaling property:-

$x(t)\leftrightarrow&space;X(S)&space;\&space;\&space;\&space;ROC&space;:&space;R$

$x(at)\leftrightarrow&space;\&space;\&space;?$.

we know that  $X(S)&space;=&space;\int_{-\infty&space;}^{\infty&space;}\&space;x(t)&space;\&space;e^{-St}&space;\&space;dt$ .

$L\left&space;\{&space;x(at)&space;\right&space;\}&space;=&space;\int_{-\infty&space;}^{\infty&space;}\&space;x(at)&space;\&space;e^{-St}&space;\&space;dt$.          Let  $at=\&space;\lambda$ ,      $dt=\&space;\frac{d\lambda}{a}$,  $t&space;\&space;limits&space;\&space;:&space;\&space;-\infty&space;\&space;to&space;\&space;\infty,&space;\&space;\&space;\&space;\lambda&space;\&space;\&space;limits&space;\&space;:&space;\&space;-\infty&space;\&space;to&space;\&space;\infty$.

$L\left&space;\{&space;x(at)&space;\right&space;\}&space;=&space;\frac{1}{a}\int_{\lambda&space;=-\infty&space;}^{\infty&space;}\&space;x(\lambda&space;)&space;\&space;e^{(\frac{-S}{a})\lambda&space;}&space;\&space;d\lambda$ .

$x(at)\leftrightarrow&space;\frac{1}{a}&space;\&space;X(\frac{S}{a}),&space;\&space;\&space;\&space;if&space;\&space;a>0$.

$x(-at)\leftrightarrow&space;\frac{1}{a}&space;\&space;X(\frac{-S}{a}),&space;\&space;\&space;\&space;if&space;\&space;a<0&space;\&space;and&space;\&space;(a\neq&space;-1)$.

8. Convolution in Time-domain:-

$x_{1}(t)\leftrightarrow&space;X_{1}(S)&space;\&space;\&space;\&space;ROC&space;:&space;R_{1}$

$x_{2}(t)\leftrightarrow&space;X_{2}(S)&space;\&space;\&space;\&space;ROC&space;:&space;R_{2}$

$x_{1}(t)&space;*&space;x_{2}(t)\leftrightarrow&space;\&space;\&space;?$.

we know that  $X(S)&space;=&space;\int_{t=-\infty&space;}^{\infty&space;}\&space;x(t)&space;\&space;e^{-St}&space;\&space;dt$ .

$L\left&space;\{&space;x_{1}(t)&space;*&space;x_{2}(t)&space;\right&space;\}&space;=&space;\int_{t=-\infty&space;}^{\infty&space;}\&space;\left&space;\{&space;x_{1}(t)&space;*&space;x_{2}(t)&space;\right&space;\}\&space;e^{-St}&space;\&space;dt$.

$L\left&space;\{&space;x_{1}(t)&space;*&space;x_{2}(t)&space;\right&space;\}&space;=&space;\int_{t=-\infty&space;}^{\infty&space;}\&space;\int_{\tau&space;=-\infty&space;}^{\infty&space;}\&space;x_{1}(\tau&space;)\left&space;\{&space;x_{2}(t-\tau&space;)&space;\right&space;\}\&space;e^{-St}&space;\&space;dt&space;\&space;d\tau$.

$L\left&space;\{&space;x_{1}(t)&space;*&space;x_{2}(t)&space;\right&space;\}&space;=&space;\int_{\tau&space;=-\infty&space;}^{\infty&space;}\&space;x_{1}(\tau&space;)\left&space;\{&space;\int_{t=-\infty&space;}^{\infty&space;}&space;x_{2}(t-\tau&space;)\&space;e^{-St}&space;\&space;dt&space;\right&space;\}&space;\&space;d\tau$.

$L\left&space;\{&space;x_{1}(t)&space;*&space;x_{2}(t)&space;\right&space;\}&space;=&space;\int_{\tau&space;=-\infty&space;}^{\infty&space;}\&space;x_{1}(\tau&space;)\left&space;\{&space;e^{-S\tau&space;}&space;X_{2}(S)&space;\right&space;\}&space;\&space;d\tau$.

$L\left&space;\{&space;x_{1}(t)&space;*&space;x_{2}(t)&space;\right&space;\}&space;=&space;\left&space;\{&space;\int_{\tau&space;=-\infty&space;}^{\infty&space;}\&space;x_{1}(\tau&space;)&space;e^{-S\tau&space;}&space;\&space;d\tau&space;\right&space;\}&space;\&space;X_{2}(S)$.

$L\left&space;\{&space;x_{1}(t)&space;*&space;x_{2}(t)&space;\right&space;\}&space;=&space;\&space;X_{1}(S)&space;\&space;X_{2}(S)$

$x_{1}(t)&space;*&space;x_{2}(t)\leftrightarrow&space;\&space;X_{1}(S)&space;\&space;X_{2}(S),&space;ROC&space;:&space;R_{1}&space;\cap&space;R_{2}$.

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## Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems.

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