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.
Best Answer
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.