How can I expand function into a series of exponentially decaying (co)sines?

Question

A Fourier series is an expansion of a periodic function into a series of (co)sines.
Non-periodic functions can be expanded using a Fourier transform.
Using some simple steps the transition can be made from a Fourier series to a Fourier transform.
A Fourier transform can be expressed without using complex numbers.

I want to construct similar reasoning ending up with the Laplace transform.

I want to start with expanding a function $f(t)$ in $e^{a_kt} \sin(k\omega_0 t)$ and $ e^{a_kt} \sin(k\omega_0 t)$
$$ f(t)= \sum_{k=0}^\infty e^{c_kt}(a_k \sin(k\omega_0 t) \ + \ b_k \cos(k\omega_0 t)) $$ Slightly reformulated:
$$ f(t)= \sum_{k=0}^\infty (A_k \sin(k\omega_0 t) \ + \ B_k \cos(k\omega_0 t)) $$ where $$ A_k = a_k e^{c_kt} \ and \ B_k = b_k e^{c_kt} $$

I want to arrive at expressions for both the transform and its inverse without using complex numbers.
I want to do the transition to complex numbers as the final step to simplify the expressions.

Can someone help me to get started?

Answer 1

$$f{\left(t\right)}=\frac{1}{2\pi j}\lim_{T\to\infty}{\int_{\gamma-jT}^{\gamma+jT}F{\left(s\right)}e^{+st}ds}$$

$$f{\left(t\right)}=\frac{1}{2\pi j}\lim_{T\to\infty}{\int_{\gamma-jT}^{\gamma+jT}F{\left(\sigma+j\omega\right)}e^{\left(\sigma+j\omega\right)t}d\left(\sigma+j\omega\right)}$$

$$f{\left(t\right)}=\frac{1}{2\pi j}\lim_{T\to\infty}{\int_{\gamma-jT}^{\gamma+jT}F{\left(\sigma+j\omega\right)}e^{\sigma t}e^{\left(j\omega\right)t}d\left(\sigma+j\omega\right)}$$

$$f{\left(t\right)}=e^{\gamma t}\frac{1}{2\pi j}\lim_{T\to \infty}{\int_{\gamma-jT}^{\gamma+jT}F{\left(\sigma+j\omega\right)}e^{\left(j\omega\right)t}d\left(\sigma+j\omega\right)}$$

$$f{\left(t\right)}=e^{\gamma t}\frac{1}{2\pi j}\lim_{T\to\infty}{\int_{\gamma-jT}^{\gamma+jT}F{\left(\gamma+j\omega\right)}e^{\left(j\omega\right)t}d\left(j\omega\right)}$$

$$f{\left(t\right)}=e^{\gamma t}\frac{j}{2\pi j}\lim_{T\to\infty}{\int_{\gamma-jT}^{\gamma+jT}F{\left(\gamma+j\omega\right)}e^{\left(j\omega\right)t}d\left(\omega\right)}$$

$$f{\left(t\right)}=e^{\gamma t}\frac{1}{2\pi}\lim_{T\to\infty}{\left(\int_{\gamma-jT}^{\gamma+0j}F{\left(\gamma+j\omega\right)}e^{\left(j\omega\right)t}d\left(\omega\right)+\lim_{T\to\infty}{\int_{\gamma+0j}^{\gamma+jT}F{\left(\gamma+j\omega\right)}e^{\left(j\omega\right)t}d\left(\omega\right)}\right)}$$

$$f{\left(t\right)}=e^{\gamma t}\frac{2}{2\pi}\lim_{T\to\infty}{\left(\lim_{T\to\infty}{\int_{\gamma+0j}^{\gamma+jT}F{\left(\gamma+j\omega\right)}\frac{\left(e^{\left(j\omega\right)t}+e^{\left(-j\omega\right)t}\right)}{2}d\left(\omega\right)}\right)}$$

$$f{\left(t\right)}=e^{\gamma t}\frac{2}{2\pi}\lim_{T\to\infty}{\int_{\gamma+0j}^{\gamma+jT}F{\left(\gamma+j\omega\right)}\cos{\left(\omega t\right)}d\omega}$$

$$f{\left(t\right).e^{-\gamma t}}=\frac{1}{\pi}\int_{0}^{\infty}F{\left(\gamma+j\omega\right)}\cos{\left(\omega t\right)}d\left(\omega\right)$$

$$f{\left(t\right).e^{-\gamma t}}=\mathcal{F}^{-1}\left(F{\left(\gamma+j\omega\right)}\right)$$

$$\mathcal{F}\left(f{\left(t\right).e^{-\gamma t}}\right)=F{\left(\gamma+j\omega\right)}$$

$$\mathcal{F}\left(f{\left(t\right).e^{-\gamma t}}\right)=\mathcal{L}\left(f{\left(t\right)}\right)$$

When $\gamma=0$: $$\mathcal{F}\left(f{\left(t\right).e^{0}}\right)=\mathcal{L}\left(f{\left(t\right)}\right)\ when\ \gamma=0$$ $$\mathcal{F}\left(f{\left(t\right)}\right)=\mathcal{L}\left(f{\left(t\right)}\right)\ when\ \gamma=0$$

So the conclusion is 'the expansion in the Laplace transform' uses the Fourier transform, but it is the Fourier transform of the original function $f\left(t\right)$ multiplied with exponential decay $e^{-\gamma t}$.

Is there any literature developing a similar deduction?

Can anyone help by clarifying how to interpret calculating the Fourier transform of a function $f\left(t\right)$ multiplied with $e^{-\gamma t}$?

Answer 2

Some of your premise stated here is incorrect, so I'll address that first.

The Fourier-transform of the impulse-response, $F(\omega) := \mathscr{F}[f(t)](\omega) \in \mathbb{C}$, of a linear time-invariant system is the ratio of the system's output to its input when the input is $e^{j\omega t}$ with $\omega \in \mathbb{R}$. If you have instead a purely real input like $\sin(\omega t)$, the output will be $|F(\omega)|\sin\big{(}\omega t + \angle F(\omega)\big{)}$. Note that, $$ \frac{|F(\omega)|\sin\big{(}\omega t + \angle F(\omega)\big{)}}{\sin(\omega t)} \neq F(\omega)\ \ \text{or}\ \ |F(\omega)| $$

Similarly, the unilateral Laplace-transform of the impulse-response is the ratio of the output to the input when the input is $e^{s t}h_0(t)$ where $h_0(t)$ is the unit-step function and $s \in \mathbb{C}$. Complex numbers are unavoidable in these definitions, but I don't see why you would insist on avoiding them anyway. A real system's output will still be real if the input is real, don't worry. Considering complex inputs has great utility and unifying beauty.

It is true that $F(\omega)$ is equivalent to the bilateral Laplace-transform evaluated on the imaginary axis, provided both transforms converge. In that regard, the Fourier-transform is "contained within" the Laplace-transform, and like you seem to have deduced, since the Fourier-basis $\{e^{j\omega t}\ \forall \omega\}$ was already complete, the values obtained for $\mathcal{R}(s) \neq 0$ in the Laplace-transform are "extra", though be mindful of caveats related to convergence and possible definition mismatches.

Also notice that the inverse Laplace-transform is not as simple as the inverse Fourier-transform, because the Laplace-transform is not associated with a minimal and orthogonal basis. All in all, I think this makes the original pursuit of your question moot. The steps used in the derivation of the Fourier-coefficients rely on orthogonality. It then becomes somewhat unclear what you are even after - a proof of the inverse Laplace-transform that "feels" like that of the Fourier-transform? Hopefully that helps you see why people have been a little confused in the couple places you've posted this question.