(New page: ===a=== <math> f(t) = \delta\big(t+1) + \delta(t-1)</math> ::<math> F(j\omega) = \int_{-\infty}^{\infty}\delta(t+1)e^{-j\omega t}\,dt + \int_{-\infty}^{\infty}\delta(t-1)e^{-j\omega t}\,...) |
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<math> f(t) = \delta\big(t+1) + \delta(t-1)</math> | <math> f(t) = \delta\big(t+1) + \delta(t-1)</math> | ||
Latest revision as of 12:08, 12 December 2008
a
$ f(t) = \delta\big(t+1) + \delta(t-1) $
- $ F(j\omega) = \int_{-\infty}^{\infty}\delta(t+1)e^{-j\omega t}\,dt + \int_{-\infty}^{\infty}\delta(t-1)e^{-j\omega t}\,dt $
by the sifting property of the delta function:
- $ F\big(j\omega) = e^{j \omega} + e^{-j \omega} = 2cos\big(\omega ) $
b
$ f(t) = \frac{d\lbrace u(-2-t) + u(t-2)\rbrace }{dt} = \delta(-2-t) + \delta(t-2) $ because $ \frac{d\{u\big(t)\} }{dt} = \delta(t) $
So very similar to part a we can take the integral and use the sifting property of the delta function
- $ F(j\omega) = \int_{-\infty}^{\infty}\delta(-2-t)e^{-j\omega t}\,dt + \int_{-\infty}^{\infty}\delta(t-2)e^{-j\omega t}\,dt $
Paying special attention to the first integral, the resulting exponential is negative because the delta function is time reversed
- $ F\big(j\omega) = -e^{-2j \omega} + e^{2j \omega} = -2jsin\big(2\omega ) $
