# Earth Cubed

## Laplace Transform of f(t) Related to smoothed f(t)?

When reading (Comment#18839) I started to wonder if there was a relationship between the Fourier Transform of a smoothed signal and the Laplace transform.  I assumed there was a relationship (Comment#18854). After further derivation, I recommenced that if the goal is to derive the Laplace transform from the Fourier transom of the filtered signal:

1) The signal be properly windowed.

2) The FFT of the windowed Fourier Transform, needs to be compensated for the frequency effects that resulted from the low pass filter.

Weather it is a good idea to compute the Laplace transform from a windowed FFT of a filtered signal is outside of the scope of this thread (but feel free to comment bellow) .

The Laplace transform is given by:

1)           $(\mathcal{L}f)(s) = \int_{0^-}^\infty e^{-st}f(t)\,dt$

The Fourier transform is given by:

2)           $\hat{f}(\omega) := \int_{-\infty}^{\infty} f(x)\ e^{- \omega i t }\,dt,$

The Two Sided Laplace Transform is given by:

3)           $\mathcal{B} \left\{f(t)\right\} = F(s) = \int_{-\infty}^{\infty} e^{-st} f(t) dt.$

Therefor the Fourier transform is the two sided Laplace transform evaluated at $s=i\omega$

Returning to the one sided Laplace transform:

4)            $(\mathcal{L}f)(s) = \int_{a}^b e^{-st}f(t)\,dt= \int_{0^-}^\infty e^{-(\alpha+iw)t}f(t)\,dt$

5)            $(\mathcal{L}f)(s) = \int_{a}^b e^{-st}f(t)\,dt= \int_{0^-}^\infty e^{-iw t}f(t)e^{- \alpha t}\,dt$

Let:

6)            $u=e^{- iw t}$
7)           $dv=f(t)e^{- \alpha t}\,dt$

8)            $V=\int_{a}^tf(t')e^{-\alpha t'}\,dt'=e^{-\alpha t}\int_{t'=a}^tf(t')e^{\alpha (t-t')}\,dt'$

where the low pass filtered version of f(t):

9)            $g(t)=f(t)*e^{\alpha t} =e^{-\alpha t} \int_{a}^tf(t')e^{\alpha (t-t')}\,dt'$

and is the convolution of f(t) and the impulse response of a filter (or atleast aproximatly so) with bandwidth $\alpha$.

Plugging this result into integration by parts gives:

10)         $(\mathcal{L}f)(s)=[e^{i \omega t}g(t)]_a^b-\int_a^b-i \omega e^{-i \omega t}e^{-\alpha t}g(t)dt$

or equivalently:

11)          $(\mathcal{L}f)(s)=g(b)e^{-i\omega b}-g(a)e^{-i \omega a}+i\int_a^b \omega e^{-i \omega t}e^{-\alpha t}g(t)dt$

The first two terms show how the endpoints chosen effect the transfrom.  These two terms will cancel for a given frequency if the distance between the endpoints is some multiple of the period. The last term is the Fouier transform of the smoothed function with the frequencies weighted by $\omega$ and using a windowing function $e^{-\alpha t}$
(note the multiple $i$ is there because the Fourier transform variable $\omega$ is the Laplace transform variable but rotated by 90 degrees.)

The effect of the windowing function is to smooth the frequency response. This is because multiplication in the time domain is equivalent to convolution in the frequency domain. The following Fourier transform relationship is useful (relationship 205):

12)           $\mathcal{F}(e^{- \alpha t} u(t)) \,=\frac{1}{\sqrt{2 \pi} (\alpha + i \omega)}=\frac{1}{\sqrt{2 \pi (\alpha^2 + \omega^2)}}e^{(-i \ tan^{-1}(\alpha,\omega))}$

Note, that if a non causal filter was used for the smoothing the relationship is much simpler.

13)           $\mathcal{F}({e}^{-a|t|}) \,=\sqrt{\frac{2}{\pi}} \cdot \frac{a}{a^2 + \omega^2}$

In both cases to properly deal with the end points the time shifting property of the Fourier transform is needed (relationship 102):

14)           $f(x - t)\, =e^{- i a \omega} \hat{f}(\omega)\,$

Applying this property to the last two relationships gives:

15)           $\mathcal{F}(e^{- \alpha (t-a)} (u(t-a)-u(t-b)) \,=\frac{e^{-ia \omega}-e^{\alpha a-ib\omega}}{\sqrt{2 \pi} (\alpha + i \omega)}$

16)           $\mathcal{F}({e}^{-a|(t-b)|}) \,=\sqrt{\frac{2}{\pi}} \cdot \frac{a}{a^2 + \omega^2}e^{-i\omega\frac{1}{2}(a+b)}$

Strictly dealing with the case where a causal filter is used and applying the rule for the Fourier transform of a convolution (Rule 109) we obtain:

17)           $(\mathcal{L}f)(s)=g(b)e^{-i\omega b}-g(a)e^{-i \omega a}+i\omega\frac{1}{\sqrt{2\pi}}\mathcal{F}(g(t))*\left( \frac{e^{-ia \omega}-e^{\alpha a-ib\omega}}{\sqrt{2 \pi} (\alpha + i \omega)} \right)$

of equivalently:

18)           $\mathcal{L}(f(y))=g(b)e^{-i\omega b}-g(a)e^{-i \omega a}+i\omega\frac{1}{\sqrt{2\pi}}\mathcal{F}(g(t))*\left( \frac{e^{-ia \omega}-e^{\alpha a-ib\omega}}{\sqrt{2 \pi(\alpha^2 +\omega^2)}}e^{ -i tan^{-1}(\alpha,\omega) } \right)$

1) If  $\alpha$ is negative the system is causal, and the filtered version $g(t)$ of the signal $f(t)$ will be causal.