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

$${\cal L}[x(t)]=X(s),\;\;\;\;ROC=R_x,\;\;\;\;\;\mbox{and}\;\;\;\;\;\; {\cal L}[y(t)]=Y(s),\;\;\;\;ROC=R_y$$

• Linearity
  
  $${\cal L}[a x(t)+b y(t)]=aX(s)+bY(s), \;\;\;\;ROC \supseteq (R_x \cap R_y)$$
  
  
  ( $A \supseteq B$ 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
  

  $$X(s)={\cal L}[x(t)]=\frac{1}{s+1},\;\;\;\;Re[s]>-1,\;\;\;\;\;\;\;\; Y(s)={\cal L}[y(t)]=\frac{1}{(s+1)(s+2)},\;\;\;\;Re[s]>-1$$
  
  
  then
  
  $${\cal L}[x(t)-y(t)]=\frac{1}{s+1}-\frac{1}{(s+1)(s+2)} =\frac{s+1}{(s+1)(s+2)}=\frac{1}{s+2}, \;\;\;\;Re[s]>-2$$
  
  
  We see that the ROC of the combination is larger than the intersection of the ROCs of the two individual terms.

• Time Shifting
  
  $${\cal L}[x(t-t_0)]=e^{-t_0s} X(s),\;\;\;\;ROC=R_x$$
  
  

• Shifting in s-Domain
  
  $${\cal L}[e^{s_0t}x(t)]=X(s-s_0),\;\;\;\;ROC=R_x+Re[s_0]$$
  
  
  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
  
  $${\cal L}[x(at)]=\frac{1}{\vert a\vert} X(\frac{s}{a}),\;\;\;\;ROC=\frac{R_x}{a}$$
  
  
  Note that the ROC is horizontally scaled by $1/a$, which could be either positive ($a>0$) or negative ($a<0$) in which case both the signal $x(t)$ and the ROC of its Laplace transform are horizontally flipped.

• Conjugation
  
  $${\cal L}[x^*(t)]=X^*(s^*), \;\;\;\;\;ROC=R_x$$
  
  
  Proof:
  
  $$X^*(s^*)=[\int_{-\infty}^\infty x(t)e^{-s^*t} dt ]^* =\int_{-\infty}^\infty x^*(t)e^{-st} dt={\cal L}[x^*(t)]$$
  
  

• Convolution
  
  $${\cal L}[x(t)*y(t)]=X(s) Y(s), \;\;\;\;ROC \supseteq (R_x \cap R_y)$$
  
  
  Note that the ROC of the convolution could be larger than the intersection of $R_x$ and $R_y$, due to the possible pole-zero cancellation caused by the convolution, similar to the linearity property.

  Example Assume
  

  $$X(s)={\cal L}[x(t)]=\frac{s+1}{s+2},\;\;\;Re[s]>-2\;\;\;\mbox{and}\;\;\;\; Y(s)={\cal L}[y(t)]=\frac{s+2}{s+1},\;\;\;Re[s]>-1$$
  
  
  then
  
  $${\cal L}[x(t)*y(t)]=X(s) Y(s)=1,\;\;\;\;\mbox{ROC is the entire s-plane}$$
  
  

• Differentiation in Time Domain
  
  $${\cal L}[\frac{d}{dt}x(t)]=sX(s), \;\;\;\;\;ROC \supseteq R_x$$
  
  
  This can be proven by differentiating the inverse Laplace transform:
  
  $$\frac{d}{dt}x(t)=\frac{1}{j2\pi}\int_{\sigma-j\infty}^{\sigm... ...{1}{j2\pi}\int_{\sigma-j\infty}^{\sigma+j\infty}sX(s)e^{st}ds$$
  
  
  In general, we have
  
  $${\cal L}[ \frac{d^n}{dt^n}x(t)]=s^n X(s)$$
  
  
  Again, multiplying $X(s)$ by $s$ may cause pole-zero cancellation and therefore the resulting ROC may be larger than $R_x$.

  Example: Given
  

  $${\cal L}[u(t)]=1/s,\;\;\;\; Re[s]>0$$
  
  
  we have:
  
  $${\cal L}[\frac{d}{dt} u(t)]={\cal L}[\delta(t)]=1,\;\;\;\;\;\... ...elta(t)]=s^n, \;\;\;\;\;\;\; \mbox{ROC is the entire s-plane}$$
  
  

• Differentiation in s-Domain
  
  $${\cal L}[t x(t)]=-\frac{d}{ds}X(s),\;\;\;\; ROC=R_x$$
  
  
  This can be proven by differentiating the Laplace transform:
  
  $$\frac{d}{ds}X(s)=\int_{-\infty}^\infty x(t) \frac{d}{ds} e^{-st} dt =\int_{-\infty}^\infty (-t) x(t) e^{-st} dt$$
  
  
  Repeat this process we get
  
  $${\cal L}[t^n x(t)]=(-1)^n\frac{d^n}{ds^n}X(s),\;\;\;\; ROC=R_x$$
  
  

• Integration in Time Domain
  
  $${\cal L}[\int_{-\infty}^t x(\tau) d\tau ]=\frac{X(s)}{s}, \;\;\;\;ROC \supseteq (R_x\cap \{Re[s]>0\})$$
  
  
  This can be proven by realizing that
  
  $$x(t)*u(t)=\int_{-\infty}^\infty x(\tau) u(t-\tau) d\tau =\int_{-\infty}^t x(\tau) d\tau$$
  
  
  and therefore by convolution property we have
  
  $${\cal L}[x(t)*u(t)]=X(s)\frac{1}{s}$$
  
  
  Also note that as the ROC of ${\cal L}[u(t)]=1/s$ is the right half plane $Re[s]>0$, the ROC of $X(s)/s$ is the intersection of the two individual ROCs $R_x \cap \{Re[s]>0\}$, except if pole-zero cancellation occurs (when $x(t)=d\delta(t)/dt$ with $X(s)=s$) in which case the ROC is the entire s-pane.