```>
>"lucy" <losemind@yahoo.com> wrote in message
>>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.
>
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
>
>
>

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!)

```
```>
>"lucy" <losemind@yahoo.com> wrote in message
>>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.
>
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
>
>
>

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!)

```
```"lucy" <losemind@yahoo.com> wrote in message
>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

```
```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

```
```"lucy" <losemind@yahoo.com> wrote in message
> 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

```
```I failed to integrate the exponential kernel against the sinc