I failed to integrate the exponential kernel against the sinc function... Does anybody know what is the Laplace transform of an ideal low pass filter? thanks

# What is the Laplace transform of an ideal low pass filter?

Started by ●December 13, 2005

Reply by ●December 14, 20052005-12-14

"lucy" <losemind@yahoo.com> wrote in message news:1134502603.115693.72580@g43g2000cwa.googlegroups.com...> I failed to integrate the exponential kernel against the sinc > function... > > Does anybody know what is the Laplace transform of an ideal low pass > filter? > > thanks >It would be something like (1-exp(-sT))/s Ing

Reply by ●December 15, 20052005-12-15

lucy wrote:> I failed to integrate the exponential kernel against the sinc > function... > > Does anybody know what is the Laplace transform of an ideal low pass > filter?Yes, many people know. People like Oppenheim, Manolakis, Bracewell, et al. They put that information in these nice little containers called "books." You should try one some time. --RY

Reply by ●December 15, 20052005-12-15

"lucy" <losemind@yahoo.com> wrote in message news:1134502603.115693.72580@g43g2000cwa.googlegroups.com...>I failed to integrate the exponential kernel against the sinc > function... > > Does anybody know what is the Laplace transform of an ideal low pass > filter?If you will admit the Bilateral Laplace Transform (which is also known as the Complex Fourier Transform) then the answer is yes. But the particular function you ask about simplifies so that the transform becomes the ordinary Fourier Transform - for which there is no label using "Laplace" that I know of .... because the Laplace Transform requires the functions be causal and the impulse response of the ideal lowpass filter isn't causal. It revolves around the region of "t" being either from -infinity to +infinity or from 0 to infinity (causal). The pairs are: Transform: t s Complex Fourier and Bilateral Laplace -inf to +inf sigma + jw ordinary Fourier -inf to +inf jw ordinary Laplace 0 to +inf sigma + jw Ideal lowpass filter: -inf to +inf jw Therefore......... ordinary Fourier.... or Bilateral Laplace with sigma=0 if you will. Fred

Reply by ●January 13, 20062006-01-13

> >"lucy" <losemind@yahoo.com> wrote in message >news:1134502603.115693.72580@g43g2000cwa.googlegroups.com... >>I failed to integrate the exponential kernel against the sinc >> function... >> >> Does anybody know what is the Laplace transform of an ideal low pass >> filter? > >If you will admit the Bilateral Laplace Transform (which is also known as>the Complex Fourier Transform) then the answer is yes. > >But the particular function you ask about simplifies so that thetransform>becomes the ordinary Fourier Transform - for which there is no labelusing>"Laplace" that I know of .... because the Laplace Transform requires the>functions be causal and the impulse response of the ideal lowpass filter>isn't causal.> >It revolves around the region of "t" being either from -infinity to >+infinity or from 0 to infinity (causal). >The pairs are: > >Transform: t s >Complex Fourier and Bilateral Laplace -inf to +inf sigma + jw >ordinary Fourier -inf to +inf jw >ordinary Laplace 0 to +inf sigma + jw > >Ideal lowpass filter: > -inf to +inf jw > >Therefore......... ordinary Fourier.... or Bilateral Laplace with sigma=0if>you will. > >Fred > > >your problem is to find the integral Integral(0->inf) exp(-st)sin(t)/t dt. You may use the following property of the Laplace transforms: L(f(t)/t) = Integral(s -> inf) F(u)du where L{f(t)}= F(s). regarding that L{sin(t)} = 1/s^2+1, the integral would be sth like pi/2-arctan(s) (Do it yourself!)

Reply by ●January 13, 20062006-01-13

> >"lucy" <losemind@yahoo.com> wrote in message >news:1134502603.115693.72580@g43g2000cwa.googlegroups.com... >>I failed to integrate the exponential kernel against the sinc >> function... >> >> Does anybody know what is the Laplace transform of an ideal low pass >> filter? > >If you will admit the Bilateral Laplace Transform (which is also known as>the Complex Fourier Transform) then the answer is yes. > >But the particular function you ask about simplifies so that thetransform>becomes the ordinary Fourier Transform - for which there is no labelusing>"Laplace" that I know of .... because the Laplace Transform requires the>functions be causal and the impulse response of the ideal lowpass filter>isn't causal.> >It revolves around the region of "t" being either from -infinity to >+infinity or from 0 to infinity (causal). >The pairs are: > >Transform: t s >Complex Fourier and Bilateral Laplace -inf to +inf sigma + jw >ordinary Fourier -inf to +inf jw >ordinary Laplace 0 to +inf sigma + jw > >Ideal lowpass filter: > -inf to +inf jw > >Therefore......... ordinary Fourier.... or Bilateral Laplace with sigma=0if>you will. > >Fred > > >your problem is to find the integral Integral(0->inf) exp(-st)sin(t)/t dt. You may use the following property of the Laplace transforms: L(f(t)/t) = Integral(s -> inf) F(u)du where L{f(t)}= F(s). regarding that L{sin(t)} = 1/s^2+1, the integral would be sth like pi/2-arctan(s) (Do it yourself!)