Properties of Laplace Transforms(Bi-lateral)

1.Linearity Property:-

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

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

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

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

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

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

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

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

2.Time-shifting Property:-

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

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

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

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

$t \ limits \ : \ -\infty \ to \ \infty, \ \ \ \lambda \ \ limits \ : \ \infty \ to \ -\infty$

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

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

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

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

3.Frequency-shifting Property:-

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

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

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

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

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

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

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:-

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

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

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

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

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

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

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

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

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

5.Differentiation in S-domain:-

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

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

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

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

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

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

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

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

6. Time-reversal property:-

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

$x(-t)\leftrightarrow \ \ ?$ .

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

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

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

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

$x(-t)\leftrightarrow \ X(-S)$ .

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

7. Time-Scaling property:-

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

$x(at)\leftrightarrow \ \ ?$ .

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

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

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

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

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

8. Convolution in Time-domain:-

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

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

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

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

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

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

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