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I want to truly understand the concept of Laplace transform. The Laplace transform of the cosine is

$$\mathcal L ( \cos (ω_0 t) ) = \frac{s}{s^2 + ω_0^2}$$

but the Fourier transform is two Dirac deltas in two symmetric positions of the $\omega$-axis, meaning it consists of two exponential component of complex frequency. In the $s$-domain, we perceive that cosine is not a pure tone but it's made of lots of tone. We know laplace transform and fourier transform difference is on just real value component added to $w$.

The question is: The Cosine mentioned $x(s) \to s=jw$ must be equal to Fourier$\{x\}$ doesn't ? I want a proof if this is really equal. Also I have seen this question.

It means ($j=\sqrt{-1}$)($s=\sigma+jω$) $$\mathcal B \mathcal L[ cos(ω_0t)] \to \frac{s}{s^2+ω_0^2} (s \to jw) \to \mathcal F[cos(ω_0t)] \to π [ δ (ω − ω_0) + δ (ω + ω_0)]$$

And I expect to $s/s^2+ω_0^2$ becoming equal to $π [ δ (ω − ω_0) + δ (ω + ω_0)]$ but this is not.

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    $\begingroup$ Please write your question clearly. Also use 'latex' to write mathematical expressions $\endgroup$
    – MAS
    Mar 24, 2022 at 8:08
  • $\begingroup$ The Laplace transform of $\cos(x)$ is $\mathcal{L}_x[\cos(x)](s)=\int\limits_0^{\infty} \cos(x)\,e^{-s x}\,dx=\frac{s}{s^2+1}$. The Fourier transform $\mathcal{F}_x[f(x)](\omega)=\int\limits_{-\infty}^{\infty } f(x)\,e^{-i \omega x}\,dx$ is equivalent to the two-sided Laplace transform $\int\limits_{-\infty}^{\infty} f(x)\,e^{-s x}\,dx$ evaluated at $s=i \omega$. $\endgroup$ Mar 24, 2022 at 16:57
  • $\begingroup$ @StevenClark Question updated I hope it clarify more. $\endgroup$ Mar 24, 2022 at 18:00
  • $\begingroup$ I assume what you mean by $j$ is $j=i=\sqrt{-1}$. You seem to be ignoring the lower limit of $-\infty$ in the two-sided Laplace transform, so what you have written is incorrect. The Wikipedia article at en.wikipedia.org/wiki/Laplace_transform#Fourier_transform seems to specifically address your question. $\endgroup$ Mar 24, 2022 at 18:29
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    $\begingroup$ And are you sure ω will be supported forever? I am sure \omega will. Also, think of making life easy for search engines. $\endgroup$ Mar 24, 2022 at 20:01

1 Answer 1

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This answer assumes the following definitions of the Fourier, Laplace, and two-sided Laplace transforms.

$$\mathcal{F}_t[f(t)](\omega)=\int\limits_{-\infty}^{\infty } f(t)\,e^{-i \omega t}\,dt\tag{1}$$

$$\mathcal{L}_t[f(t)](s)=\int\limits_0^{\infty} f(t)\,e^{-s t}\,dt\tag{2}$$

$$\mathcal{BL}_t[f(t)](s)=\int\limits_{-\infty}^{\infty} f(t)\,e^{-s t}\,\tag{3}dt$$


Note the Fourier transform defined in formula (1) above is equivalent to the bilateral Laplace transform defined in formula (3) above evaluated at $s=i \omega$.


The Fourier and Laplace transforms of $\cos(\omega_0\,t)$ are as follows:

$$\mathcal{F}_t[\cos(\omega_0\,t)](\omega)=\int\limits_{-\infty}^{\infty}\cos (\omega_0\,t)\,e^{-i \omega t}\,dt=\pi\,\delta(\omega+\omega_0)+\pi\,\delta(\omega-\omega_0),\quad\omega\in\mathbb{R}\tag{4}$$

$$\mathcal{L}_t[\cos(\omega_0\,t)](s)=\int\limits_0^{\infty} \cos(\omega_0\,t)\,e^{-s t}\,dt=\frac{s}{s^2+\omega_0^2},\quad\Re(s)>0\tag{5}$$


The Fourier and bilateral Laplace transforms of $\cos(\omega_0\,t)\,\theta(t)$ are

$$\mathcal{F}_t[\cos(\omega_0\,t)\,\theta(t)](\omega)=\int\limits_{-\infty}^{\infty}\cos(\omega_0\,t)\,\theta(t)\,e^{-i \omega t}\,dt$$ $$=\int_0^{\infty}\cos(\omega_0\,t)\,e^{-i \omega t}\,dt=\frac{\pi}{2} \delta(\omega+\omega_0)+\frac{\pi}{2} \delta(\omega-\omega_0)+\frac{i \omega}{\omega_0^2-\omega^2},\quad\omega\in\mathbb{R}\tag{6}$$

$$\mathcal{BL}_t[\cos(\omega_0\,t)\,\theta(t)](s)=\int\limits_{-\infty}^{\infty}\cos(\omega_0\,t)\,\theta(t)\,e^{-s t}\,dt=\int\limits_0^{\infty} \cos(\omega_0\,t)\,e^{-s t}\,dt=\frac{s}{s^2+\omega_0^2},\ \Re(s)>0\tag{7}$$

where

$$\theta(t)=\left\{ \begin{array}{cc} 0 & t<0 \\ 1 & t>0 \\ \end{array}\right.\tag{8}$$

is the Heaviside step function.


Consider the following limit representation of the bilateral Laplace transform of $\cos(\omega_0\,t)$.

$$\mathcal{BL}_t[\cos(\omega_0\,t)](s)=\underset{T\to\infty}{\text{lim}}\left(\int\limits_{-T}^{T}\cos(\omega_0\,t)\,e^{-s t}\,dt\right)=\underset{T\to\infty}{\text{lim}}\left(\frac{2 (s \sinh(s T) \cos(T \omega_0)+\omega_0\cosh (s T) \sin (T \omega_0))}{s^2+\omega_0^2}\right)\tag{9}$$


Substituting $T=2 \pi f$ and $s=i \omega$ in formula (9) above and accounting for the resulting equivalence to formula (4) above leads to the limit representation

$$\delta(\omega+\omega_0)+\delta(\omega-\omega_0)=\underset{f\to\infty}{\text{lim}}\left(\frac{2}{\pi} \frac{\omega \cos(2 \pi f \omega_0) \sin (2 \pi f \omega )-\omega_0 \sin(2 \pi f \omega_0) \cos(2 \pi f \omega )}{\omega ^2-\omega_0^2}\right),\quad\omega\in\mathbb{R}\tag{10}$$

which is equivalent to the limit representation

$$\delta(\omega+\omega_0)+\delta(\omega-\omega_0)=\underset{f\to\infty}{\text{lim}}\left(2 f\,\text{sinc}(2 \pi f (\omega+\omega_0))+2 f\,\text{sinc}(2 \pi f (\omega -\omega_0))\right),\quad\omega\in\mathbb{R}.\tag{11}$$


The relevant question seems to be whether $\mathcal{F}_t[\cos(\omega_0\,t)\,\theta(t)](\omega)$ can be approximated by $\underset{\epsilon\to 0^+}{\text{lim}}\left(\mathcal{BL}_t[\cos(\omega_0\,t)\,\theta(t)](\epsilon+i \omega)\right)=\underset{\epsilon\to 0^+}{\text{lim}}\left(\mathcal{L}_t[\cos(\omega_0\,t)](\epsilon+i \omega)\right)$ for $\omega\in\mathbb{R}$ which, based on formulas (5) to (7) above, is equivalent to the following.

$$\frac{\pi}{2} \delta(\omega+\omega_0)+\frac{\pi}{2} \delta(\omega-\omega_0)+\frac{i \omega}{\omega_0^2-\omega^2}\approx\underset{\epsilon\to 0^+}{\text{lim}}\left(\frac{\epsilon+i \omega}{(\epsilon+i \omega)^2+\omega_0^2}\right),\quad\omega\in\mathbb{R}\tag{12}$$


Figure (1) below illustrates the imaginary part of the right side of formula (12) in orange overlaid on the imaginary part of the left-side of formula (12) in blue where both sides are evaluated at $\omega_0=1$ and the right side is evaluated at $\epsilon=10^{-6}$. Note for $\epsilon=0$ the right-side of formula (12) above becomes equivalent to the last term on the left-side of formula (12) above (which is the imaginary part of the left-side of formula (12) above), so clearly $\mathcal{F}_t[\cos(\omega_0\,t)\,\theta(t)](\omega)\ne\mathcal{L}_t[\cos(\omega_0\,t)](i \omega)$.


Illustration of imaginary parts of right and left sides of formula (12)

Figure (1): Illustration of imaginary parts of right and left sides of formula (12) (orange overlaid on blue) evaluated at $\omega_0=1$ and $\epsilon=10^{-6}$


Figures (2) below illustrates the real part of the right-side of formula (12) above evaluated at $\omega_0=1$ and $\epsilon=10^{-6}$. Note the evaluation seems to suggest the real part of the right-side of formula (12) above may perhaps approximate a limit representation for the real-part of the left side of formula (12) above (i.e. the two Dirac delta terms).


Illustration of real part of right-side of formula (12)

Figure (2): Illustration of real part of right-side of formula (12) evaluated at $\omega_0=1$ and $\epsilon=10^{-6}$


The analysis and results illustrated above seem to suggest the left-side of formula (12) above can be approximated by the limit representation on the right-side of formula (12) above, but the following integral which assumes $\omega_0=1$ provides further evidence.

$$\int\limits_{-2}^2 \left(\frac{\pi}{2} \delta(\omega+1)+\frac{\pi}{2} \delta(\omega-1)+\frac{i \omega}{1-\omega^2}\right)\,d\omega =\pi\approx \underset{\epsilon\to 0^+}{\text{lim}}\left(\frac{i}{2} (\log(\epsilon (\epsilon-4 i)-3)-\log(\epsilon (\epsilon+4 i)-3))\right)=\underset{\epsilon\to 0^+}{\text{lim}}\left(\int\limits_{-2}^2 \frac{\epsilon+i \omega}{(\epsilon+i \omega)^2+1}\,d\omega\right)\tag{13}$$


Note for the integral of the imaginary part of the left-side of formula (12) above the Cauchy principal value $\mathcal{P}\int\limits_{-2}^2 \frac{i \omega}{1-\omega^2}\,d\omega=0$ is consistent with the fact that $\frac{i \omega}{1-\omega^2}$ is an odd function of $\omega$ as illustrated in Figure (1) above.


For $\epsilon=10^{-6}$ the right side of formula (13) above evaluates to $3.14159\,+0.i$ which is very close to the value of $\pi$.

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  • $\begingroup$ Why in equation (9) you had to including T in transform?(also could you please reference it?) Totally you are telling the Fourier transform and bilateral transform are the same but to see the sameness of these to we need to do some tricky math, doesn't it? $\endgroup$ Mar 25, 2022 at 21:22
  • $\begingroup$ @mohammadsdtmnd I included the limit $T$ because $\int\limits_{-\infty}^{\infty}\cos(\omega_0\,t)\,e^{-s t}\,dt$ doesn't converge but can be evaluated as the limit $\underset{T\to\infty}{\text{lim}}\left(\int\limits_{-T}^{T}\cos(\omega_0\,t)\,e^{-s t}\,dt\right)$. You don't really need any tricky math to see the Fourier transform defined in formula (1) above is equivalent to the bilateral Laplace transform defined in formula (3) above evaluated at s=iω. $\endgroup$ Mar 25, 2022 at 21:31
  • $\begingroup$ @mohammadsdtmnd But the function $f(t)=\cos(\omega_0\,t)$ is a bit tricky because the transform integrals don't converge in a normal sense (since they're related to distributions). $\endgroup$ Mar 25, 2022 at 21:41
  • $\begingroup$ What is the distribution? Transform converges in lateral but not in bilateral, why? In bilateral case: converge on imaginary S axis but not in whole complex S axis. In general by multiplying and summing two cosine with same frequency and phase shift, must cause infinity! $\endgroup$ Mar 25, 2022 at 22:17
  • $\begingroup$ This not converging means cosine must always being calculating by using limit. But still what is distribustion, and sorry for double asking since I have lots of interest to knowing it. :) $\endgroup$ Mar 25, 2022 at 22:30

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