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Example 1
Compute the Fourier Transform of $ x(t)=e^{-t}u(t) $.
$ \chi(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt $
$ =\int_{-\infty}^{\infty}e^{-t}u(t)e^{-j\omega t}dt $
$ =\int_{0}^{\infty}e^{-t}e^{-j\omega t}dt $
$ =\int_{0}^{\infty}e^{-(1+j\omega )t}dt $
$ =[\frac {e^{-(1+j\omega )t}}{-(1+j\omega)}]|_0^\infty $
$ =\frac {e^{-(1+j\omega )\infty}}{-(1+j\omega)}-\frac {e^{-(1+j\omega )0}}{-(1+j\omega)} $
$ =0+\frac {1}{(1+j\omega)} $
$ =\frac {1}{1+j\omega} $
Example 2
The impulse response of an LTI system is $ h(t)=e^{-2t}u(t)+u(t+2)-u(t-2) $. What is the Frequency response $ H(j\omega) $ of the system?
$ H(j\omega)=H(\omega)=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}dt=\int_{-\infty}^{\infty}(e^{-2t}u(t)+u(t+2)-u(t-2))e^{-j\omega t}dt=\int_{-\infty}^{\infty}e^{-2t}u(t)e^{-j\omega t}dt+\int_{-\infty}^{\infty}u(t+2)e^{-j\omega t}dt-\int_{-\infty}^{\infty}u(t-2)e^{-j\omega t}dt $
Using the previous example and the time shifting property,
$ H(j\omega)=\frac {1}{2+j\omega}+\frac {2sin(2\omega)}{\omega} $
Example 3
What is the Fourier Transform of the signal $ x(t)=e^{j\omega _0t} $?
To solve this look at the the inverse Fourier transform, but the inverse transform of what?
Take $ \chi(\omega)=2\pi\delta(\omega-\omega _0) $
$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\chi (\omega)e^{j\omega t}d\omega $ $ =\frac{1}{2\pi}\int_{-\infty}^{\infty}2\pi\delta(\omega-\omega _0)e^{j\omega t}d\omega $ $ =\int_{-\infty}^{\infty}\delta(\omega-\omega _0)e^{j\omega t}d\omega $
by sifting property,
$ \int_{-\infty}^{\infty}\delta(\omega-\omega _0)e^{j\omega t}d\omega=e^{j\omega t}|_{\omega=\omega _0} $
$ x(t)=e^{j\omega _0 t} $
Thus, the fourier transform of $ x(t)=e^{j\omega _0t} $ is $ \chi(\omega)=2\pi\delta(\omega-\omega _0) $.
Example 4
Show that the Fourier transform of $ x(t)=cos(2\pi t) $ is $ \chi (\omega)=\pi\delta(\omega+2\pi)+\pi\delta(\omega-2\pi) $.
