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.

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    \$\begingroup\$ "considering only the complex amplitudes" are you sure that is how you solve and RC circuit? ignoring the R. I suggest you study the math behind both transforms, the integral is not what makes Laplace work. In both cases you are changing from time dependency to frequency dependency that is what allows you to easily deal with derivatives and integrals. \$\endgroup\$ – Juan Aug 5 '19 at 3:17
  • \$\begingroup\$ @Juan The complex amplitude of R is just R right ? So after performing some complex algebra we multiply \$ e^{i\omega \$ What's wrong here ? \$\endgroup\$ – Aravindh Vasu Aug 5 '19 at 6:19
  • \$\begingroup\$ careful, it is correct to say the amplitude of R is just R, but that does not make it complex. for example for RC in series you do the square root of R squared plus the Xc squared where Xc is 1/wC \$\endgroup\$ – Juan Aug 5 '19 at 7:13
  • \$\begingroup\$ @Juan "complex" means considering "real" and "imaginary" terms so I think "complex" is the correct term. \$\endgroup\$ – Andy aka Aug 5 '19 at 10:50
  • \$\begingroup\$ webhome.phy.duke.edu/~rgb/Class/phy51/phy51/node15.html \$\endgroup\$ – Scott Seidman Aug 5 '19 at 13:08

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.

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    \$\begingroup\$ nicely shown,sir. \$\endgroup\$ – analogsystemsrf Aug 5 '19 at 10:22
  • \$\begingroup\$ Sorry, but then I don't understand how you showed how Laplace transform does this. I mean I understand the impedance model but I don't get how Laplace does the same \$\endgroup\$ – Aravindh Vasu Aug 5 '19 at 11:23
  • \$\begingroup\$ @AravindhVasu Just take the laplace transform of the differential equation to get the same result. I have updated my answer, hopefully it is clear now. \$\endgroup\$ – sarthak Aug 5 '19 at 11:37
  • \$\begingroup\$ Yeah, that I noted, but how does the integral do this, that's my question. How is taking a integral equal to giving an complex exponential input \$\endgroup\$ – Aravindh Vasu Aug 5 '19 at 11:40
  • \$\begingroup\$ Excuse me, if you had already explained the same, but I don't understand how the integral encodes this process. \$\endgroup\$ – Aravindh Vasu Aug 5 '19 at 11:53

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