Question

Write the differentiation in time property of laplace transform

Answer 1

Answer:

Properties of Laplace Transform

The Laplace transform has a set of properties in parallel with that of the Fourier transform. The difference is that we need to pay special attention to the ROCs. In the following, we always assume

\begin{displaymath}{\cal L}[x(t)]=X(s),\;\;\;\;ROC=R_x,\;\;\;\;\;\mbox{and}\;\;\;\;\;\;

{\cal L}[y(t)]=Y(s),\;\;\;\;ROC=R_y \end{displaymath}

Linearity

\begin{displaymath}{\cal L}[a x(t)+b y(t)]=aX(s)+bY(s), \;\;\;\;ROC \supseteq (R_x \cap R_y) \end{displaymath}

( $ A \supseteq B$ means set $A$ contains or equals to set $B$, i.e,. $A$ is a subset of $B$, or $B$ is a superset of $A$.)

It is obvious that the ROC of the linear combination of $x(t)$ and $y(t)$ should be the intersection of the their individual ROCs $R_x \cap R_y$ in which both $X(s)$ and $Y(s)$ exist. But also note that in some cases when zero-pole cancellation occurs, the ROC of the linear combination could be larger than $R_x \cap R_y$, as shown in the example below.

Example: Let

\begin{displaymath}X(s)={\cal L}[x(t)]=\frac{1}{s+1},\;\;\;\;Re[s]>-1,\;\;\;\;\;\;\;\;

Y(s)={\cal L}[y(t)]=\frac{1}{(s+1)(s+2)},\;\;\;\;Re[s]>-1 \end{displaymath}

then

\begin{displaymath}{\cal L}[x(t)-y(t)]=\frac{1}{s+1}-\frac{1}{(s+1)(s+2)}

=\frac{s+1}{(s+1)(s+2)}=\frac{1}{s+2}, \;\;\;\;Re[s]>-2 \end{displaymath}

We see that the ROC of the combination is larger than the intersection of the ROCs of the two individual terms.

Time Shifting

\begin{displaymath}{\cal L}[x(t-t_0)]=e^{-t_0s} X(s),\;\;\;\;ROC=R_x \end{displaymath}

Shifting in s-Domain

\begin{displaymath}{\cal L}[e^{s_0t}x(t)]=X(s-s_0),\;\;\;\;ROC=R_x+Re[s_0] \end{displaymath}

Note that the ROC is shifted by $s_0$, i.e., it is shifted vertically by $Im[s_0]$ (with no effect to ROC) and horizontally by $Re[s_0]$.

Time Scaling