ASN3 - Class Notes October 5

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When T is very large (approaching infinity) the quanity on the left transforms to be approximately the quanity on the right.

$\lim _{T\to \infty }$

$1/T\longrightarrow df$

$n/T\longrightarrow f$

$\sum _{n=-\infty }^{\infty }{\frac {1}{T}}\longrightarrow \int _{-\infty }^{\infty }()df\!$

Using the relations above, can we make an unperiodic signal such as the one given below and make it periodic by taking the limit?

$x(t)=\lim _{T\to \infty }{\frac {1}{T}}(\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t')e^{\frac {-j2\pi nt'}{T}}dt')e^{\frac {j2\pi nt}{T}}\!$

note that $X(F)={\mathcal {F}}[x(t)]\int _{-{\frac {T}{2}}}^{\frac {T}{2}}x(t')e^{\frac {-j2\pi nt'}{T}}dt'\!$

Using the Fourier Transform property along with $\lim _{T\to \infty }n/t$ is $f$ $x(t)=\int _{-\infty }^{\infty }[\int _{-\infty }^{\infty }x(t')e^{-j2\pi ft'}dt']e^{j2\pi ft}df\!$

$x(t)=\int _{-\infty }^{\infty }x(t')[\int _{-\infty }^{\infty }e^{j2\pi f(t'-t)}df]dt'\!$

note that the defination of the delta function is $\int _{-\infty }^{\infty }e^{j2\pi f(t'-t)}df\!$

$x(t)=\int _{-\infty }^{\infty }x(t')\delta _{(}t'-t)dt'\!$

                     THE GAME
            LTI (Linear Time Invariant System) 
Input     LTI                             Output                                  Reason

$x(t)\longrightarrow \int _{-\infty }^{\infty }x(t')\delta _{(}t'-t)dt'\!$ Superposition

$X(f)\longrightarrow \int _{-\infty }^{\infty }X(f')\delta _{(}f'-f)df'\!$ Superposition