I have a confusion between the Laplace transform of a circuit and the frequency domain AC analysis. Suppose we have a series RLC resonant circuit connected to an ideal sinusoidal voltage source.

We can write $$V_m \sin(\omega t)=i R + \frac{1}{C} \int i dt + L \frac{d i}{dt} $$

now if we take the Laplace transform (assume initial inductor current to be zero),

$$\frac{V_m \omega }{s^2+\omega^2} =I(s) \big(R + \frac{1}{s C} + s L \big) \Rightarrow I(s) = \frac{V_m \omega }{(s^2+\omega^2) \big(R + \frac{1}{s C} + s L \big)} \tag{A}$$

On the other hand, using frequency domain analysis we can derive current as, $$ I= \frac{V_m }{R +j \omega L + \frac{1}{j\omega C}} \tag{B}$$

My question is: How can we deduce (B) directly from (A) without inverse Laplace transform? just putting \$s=j \omega\$ will not simply reproduce (A).

To further elaborate my question: If we have a current of a branch as the form of $$I_n=\frac{ V_n (s \sin\theta+\omega_1 \cos\theta)}{s^2+\omega_1^2} \frac{1}{Z_1(s)} $$

can we directly deduce (without taking the inverse) that particular branch is effectively acting as it is connected to a voltage source of \$ V_n \sin(\omega_1 t + \theta) \$ and impedance of the branch is \$ Z_1(j \omega) \$?

  • 1
    \$\begingroup\$ Substituting s=jω does not necessarily obtain the FT from LT. There are some conditions that must be true to yield the correct answer. See: dsp.stackexchange.com/a/37266 \$\endgroup\$
    – BB ON
    Nov 23, 2018 at 20:32
  • \$\begingroup\$ (A) contains a transient, which decays to zero - so it's not there in the sinusoidal steady state. \$\endgroup\$
    – Chu
    Nov 24, 2018 at 2:03

2 Answers 2


$$I(s) = \frac{V_m \omega_0 }{(s^2+\omega_0^2) \big(R + \frac{1}{s C} + s L \big)} \tag{A}$$

For a linear system with sinusoidal voltage input, the current also will be sinusoidal. In which case, \$I(s)\$ can be written as,

$$I(s) = \frac{I_m \omega }{(s^2+\omega_0^2)}$$

Substituting this into \$(A)\$ and replacing \$s\$ with \$j\omega\$ should will result in \$(B)\$.

  • \$\begingroup\$ Thanks, got the point. However, when we only have Laplace transform of current in this form, can we deduce it to be sinusoidal? for example, if we have \$I=\frac{s \sin\theta+\omega \cos\theta}{s^2+\omega^2} \frac{1}{Z_1(s)} \$, can we say it is effectively driven by a \$\sin(\omega t + \theta)\$ and impedance of \$Z_1\$? \$\endgroup\$
    – Pojj
    Nov 23, 2018 at 21:58
  • \$\begingroup\$ @Pojj If you have Laplace transform, you can find the inverse to see if it's sinusoidal or not. \$\endgroup\$
    – nidhin
    Nov 23, 2018 at 22:44
  • \$\begingroup\$ In fact, inverse transform will do. But my question here is that if it is possible guess without inverse. If the above \$ Z_1(s)\$ is a complicated polynomial of higher orders of s, it will be difficult to take inverse. Can we deduce it directly as in my previous comment? \$\endgroup\$
    – Pojj
    Nov 24, 2018 at 8:22

You need the \$ \frac{V(s)}{I(s)} = (R + \frac{1}{cs} + sL)\$ so the transfer function for \$ \omega \$ is :

\$ \frac{I( \omega)}{V(\omega)} = \frac{1}{(R + \frac{1}{cj\omega} + j \omega L)}\$

you insert the Laplace transform of the input source to transfer function.

Note: the Laplace can get any input function and you can solve the equation for time response of the system to any arbitrary input. But in frequency model, you know that the input is sinusoidal.

  • \$\begingroup\$ Thanks, But I am not interested in defining it as a transfer function. Just wondering what information we can get about frequency response directly from Laplace transform. \$\endgroup\$
    – Pojj
    Nov 24, 2018 at 8:24

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