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I am an Electrical Engineering student and in my Circuits class, I learned that using Laplace transforms is helpful while analyzing circuits. I see how it makes our calculations easier but I still don't get the intuition behind it. If someone can explain me that/point me to some links that explain the big idea behind using Laplace transforms. Also I don't get why we change s to jw when looking at impedances.

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    \$\begingroup\$ You answered it already. " I see how it makes our calculations easier" <--- thats the big deal. \$\endgroup\$ – efox29 May 29 '13 at 5:48
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For the domain of circuit analysis the use of laplace transforms allows us to solve the differential equations that represent these circuits through the application of simple rules and algebraic processes instead of more complex mathematical techniques. It also gives insight into circuit behaviour.

There are many different transform possible and s domain analysis is useful and perhaps easier to to teach that other techniques often before the full use of complex analysis is embraced. It doesn't make it any less powerful or useful, just well suited to the problem at hand. Often laplace transforms are taught even before differential equations are fully embraced so it is a complementary approach.

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The intuition behind it is that \$s\$ is a "complex frequency". \$s\$ is not changed to \$j\omega\$. Rather, \$s\$ is a complex number which can be broken down into its real and imaginary parts that are called sigma and omega: \$s = \sigma + j\omega\$. At all times, wherever you see \$s\$ you can substitute \$\sigma + j\omega\$!

What happens when \$s\$ is apparently being replaced by \$j\omega\$ is that we are taking into consideration just a particular slice of the domain of \$s\$: the (positive) imaginary axis. We are not changing \$s\$, but only dropping the real part \$\sigma\$ (or rather setting it to zero) and retaining the component \$j\omega\$.

This is because the positive imaginary axis in the complex frequency space of \$s\$ is where ordinary frequencies lie.

So for instance if we have a transfer function in terms of \$s\$, then if we look at the slice of that function along the imaginary axis, which is generated by \$j\omega\$ for various values of parameter \$\omega\$, then we are looking at the frequency domain of that transfer function: the Fourier transform!

The Laplace transform is a generalization of the Fourier transform. The Fourier transform ends up embedded in the Laplace domain along the imaginary axis. It is complex valued, but its domain is one-dimensional. The Fourier transform handles time-invariant functions (periodic), but Laplace generalizes to functions that incorporate exponential growth or decay. Fourier deals with signals of the form \$e^{jwt}\$, whereas Laplace deals with \$e^{st}\$, where \$s = \sigma + j\omega\$. This covers exponential growths/decays as well as growing or decaying oscillations which are not periodic.

When we set \$\sigma\$ to zero, retaining \$j\omega\$, we are taking a trip down the Fourier transform, because, for instance, we are interested how a transfer function treats time-invariant signals that are nicely represented in that transform; i.e. what is the frequency/phase response.

Here are some lecture notes on qualitative interpretations of Laplace transforms.

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    \$\begingroup\$ I wish I had this explanation when I was in school. I would have understood the subject a lot better. Our prof would never explain what 's' was, really. \$\endgroup\$ – jwygralak67 May 29 '13 at 19:56
  • \$\begingroup\$ I never took this in school, luckily. \$\endgroup\$ – Kaz May 29 '13 at 19:56

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