Laplace Transforms and the imaginary input

Question

When we get an exponential on the RHS of our differential equation (or if we give an exponential input) it's quite easy to solve the equation, but if we provide a sinsusoidal or cosinusoidal input, the calculation becomes horrendous. So we go for a trick, we give an imaginary, complex exponential input and solve the differential equation as usual by assuming \$A \, e^{i \omega t}\$ is the solution. And finally take the real part for a cosinsusoidal input and imaginary part for sinusoidal input (superposition theorem). The only big thing here is finding the complex amplitude \$A\$, once we find \$A\$ we can just multiply it with \$ e^{i\omega t}\$ take real part and arrive at the solution. Here there's a neat trick, when we observe the response in a simple circuit, say an RC circuit we are measuring the voltage across the capacitor. The voltage across the cap is so very similar to an Voltage divider but with a slight change, instead of \$R_2\$ it has a \$\frac {1}{i\omega C}\$

So whenever we are given a circuit with capacitances and inductances, we convert it into an abstract circuit, considering only the complex amplitudes. Find the required amplitude and multiply \$ e^{i\omega t}\$

Now the Laplace transform somehow performs all these neat little tricks in a neat little integral. But I do not understand how, rather I can't relate these two processes eventhough they are the same.

Answer 1

I think you can easily see this from the \$i-v\$, characteristics of a capacitor, $$i_c=C\frac{dv_c}{dt}$$ If voltage is of the form, \$v_c=Ve^{j\omega t}\$, then for LTI system current is of the form \$i_c = I_ce^{j\omega t}\$, $$I_ce^{j\omega t} = C\frac{d}{dt}(Ve^{j\omega t})$$ $$I_ce^{j\omega t} = Cj\omega Ve^{j\omega t}$$ $$\frac{V_c}{I_c}=\frac{1}{j\omega C}$$ Thus, for sinusoidal inputs you can replace the capacitor by this equivalent impedance. Similar calculation can be performed for resistor or inductor to calculate their respective impedances.

In response to edit:
Intuitively, Fourier transform is decomposing a signal into its exponential components. Its amplitude at a given frequency implies how much a given frequency contributes to the total signal.
For a purely exponential signal, you just have a single frequency and so the amplitude of the exponential is itself the Fourier Transform.
Here is a more mathematical approach:
For a purely exponential input, \$v(t) = Ve^{j\omega_ot}\$, \$V\$ and \$\omega_o\$ are constants. Taking the Fourier Transform, $$V(j\omega) = V\delta(\omega-\omega_o)$$ Thus, $$V(j\omega_o) = V$$ In other words, the amplitude of the exponential is equivalent to the Fourier Transform.