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== Laplace Transforms == | == Laplace Transforms == | ||
Laplace Transforms provide a very convenient method of solving differential equations, | 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 <math> S = \sigma + j\omega | + | transforming from the time domain to the s domain, where s is a complex number of form <math> S = \sigma + j\omega </math> |
The bilateral Laplace transform of f(t) is defined as | The bilateral Laplace transform of f(t) is defined as | ||
| + | |||
| + | :<math>X(s) = \int^{\infty}_{-\infty}x(t)e^{-st}dt</math> | ||
| + | |||
| + | |||
| + | Now let us deal with the '''region of convergence''' ( those values for which the laplace transform converges ) | ||
| + | |||
| + | Let :<math> x(t) = e^{-2t} u (t) </math> | ||
| + | |||
| + | Thus, <math> X(s) = \int^{\infty}_{0}e^{-2t}e^{-st}dt</math> | ||
| + | |||
| + | :<math>X(s) = \int^{\infty}_{0}e^{-(2+s)t}dt</math> | ||
| + | |||
| + | = <math>\frac{1}{s+2}</math> if Re(s) > -2 | ||
| + | = 0 else | ||
| + | |||
| + | Hence '''ROC''' : Re (s) > -2 | ||
| + | |||
| + | Let :<math> x(t) = -e^{-2t} u (-t) </math> | ||
| + | |||
| + | Thus, <math> X(s) = \int^{0}_{-\infty}-e^{-2t}e^{-st}dt</math> | ||
| + | |||
| + | :<math>X(s) = \int^{0}_{-\infty}-e^{-(2+s)t}dt</math> | ||
| + | |||
| + | = <math>\frac{1}{s+2}</math> 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 od inverse laplace transform we should pay utmost importance to the region of convergence specified with the given laplace transform. | ||
Revision as of 12:25, 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 od inverse laplace transform we should pay utmost importance to the region of convergence specified with the given laplace transform.
