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 $\lim _{T\to \infty }n/t=f$ then

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

Reordering order of integration

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

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

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

Using such techniquies as we did above (refered to as The Game in Signals and Systems), similar equations can be manipulated to find its output of Linear Invarient System.

                     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