ASN3 - Class Notes October 5

Back to my Home Page

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

$\underset{T\to \mathrm{\infty }}{\lim }$

$1/T⟶df$

$n/T⟶f$

$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{T}⟶\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}()df$

Using the relations above, can we make an unperiodic signal and make it periodic by taking the limit?

$x(t)=\underset{T\to \mathrm{\infty }}{\lim }\frac{1}{T}(\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}x(t^{\prime })e^{\frac{-j2\pi n{t}^{\prime }}{T}}d{t}^{\prime })e^{\frac{j2\pi nt}{T}}$

note that $X(F)={ℱ}[x(t)]\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}x(t^{\prime })e^{\frac{-j2\pi n{t}^{\prime }}{T}}d{t}^{\prime }$

becomes as the limit is taken n/t becomes f $x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}[\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\prime })e^{-j2\pi f{t}^{\prime }}d{t}^{\prime }]e^{j2\pi ft}df$

$x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\prime })[\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi f(t^{\prime }-t)}df]d{t}^{\prime }$

note that the defination of the delta function is $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{j2\pi f(t^{\prime }-t)}df$

$x(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\prime }){\delta }_{(}{t}^{\prime }-t)d{t}^{\prime }$

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

$x(t)⟶\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\prime }){\delta }_{(}{t}^{\prime }-t)d{t}^{\prime }$ Superposition

$X(f)⟶\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X(f^{\prime }){\delta }_{(}{f}^{\prime }-f)d{f}^{\prime }$ Superposition