Difference between revisions of "HW10 Monil Goklani ECE301Fall2008mboutin" - Rhea

Latest revision as of 11:26, 24 November 2008

Laplace Transforms

Laplace Transforms provide a very convenient method of solving differential equations, transforming from the time domain to the s domain, where s is a complex number of form $ S = \sigma + j\omega $

The bilateral Laplace transform of f(t) is defined as

$ X(s) = \int^{\infty}_{-\infty}x(t)e^{-st}dt $

Now let us deal with the region of convergence ( those values for which the laplace transform converges )

Let :$ x(t) = e^{-2t} u (t) $

Thus, $ X(s) = \int^{\infty}_{0}e^{-2t}e^{-st}dt $

$ X(s) = \int^{\infty}_{0}e^{-(2+s)t}dt $

           = $ \frac{1}{s+2} $  if Re(s) > -2
           = 0     else

Hence ROC : Re (s) > -2

Let :$ x(t) = -e^{-2t} u (-t) $

Thus, $ X(s) = \int^{0}_{-\infty}-e^{-2t}e^{-st}dt $

$ X(s) = \int^{0}_{-\infty}-e^{-(2+s)t}dt $

           = $ \frac{1}{s+2} $  if Re(s) < -2
           = 0     else

Hence ROC : Re (s) < -2

Here we observe that the two different signals have the same laplace transform

In such cases we should be careful while taking the inverse laplace transform.Thus while looking at the table of inverse laplace transform we should pay utmost importance to the region of convergence specified with the given laplace transform.