(New page: This page shows an example of LT transform computation let <math>x(t) = -e^{-2t}u(-t)</math> then :<math>X(s) = \int^{\infty}_{-\infty}x(t)e^{-st}dt</math> :<math>X(s) = \int^{\infty}_...) |
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:<math>X(s) = \int^{0}_{-\infty}-e^{-(2+a)t}e^{-jwt}dt</math> | :<math>X(s) = \int^{0}_{-\infty}-e^{-(2+a)t}e^{-jwt}dt</math> | ||
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| + | :<math>X(s) = \left[-\frac{e^{-(2+a)t}e^{-jwt}}{-(2+a+jw)}\right]^{0}_{-\infty}</math> | ||
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| + | From that, we can conclude that if <math>2 + a</math> is greater or equal to 0 then the integral diverges. Else: | ||
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| + | :<math>X(s) = -\frac{1}{-(2+a+jw)} - 0</math> | ||
| + | |||
| + | :<math>X(s) = \frac{1}{2+s}</math> | ||
Latest revision as of 13:35, 23 November 2008
This page shows an example of LT transform computation
let $ x(t) = -e^{-2t}u(-t) $
then
- $ X(s) = \int^{\infty}_{-\infty}x(t)e^{-st}dt $
- $ X(s) = \int^{\infty}_{-\infty}-e^{-2t}u(-t)e^{-st}dt $
- $ X(s) = \int^{0}_{-\infty}-e^{-2t}e^{-st}dt $
Now let $ s = a + jw $
- $ X(s) = \int^{0}_{-\infty}-e^{-2t}e^{-(a+jw)t}dt $
- $ X(s) = \int^{0}_{-\infty}-e^{-(2+a)t}e^{-jwt}dt $
- $ X(s) = \left[-\frac{e^{-(2+a)t}e^{-jwt}}{-(2+a+jw)}\right]^{0}_{-\infty} $
From that, we can conclude that if $ 2 + a $ is greater or equal to 0 then the integral diverges. Else:
- $ X(s) = -\frac{1}{-(2+a+jw)} - 0 $
- $ X(s) = \frac{1}{2+s} $
