Laplace transform vs phasor analysis in circuit analysis

[Wrichik Basu]

I recently acquainted myself with Laplace transform, and it appears that it has some relations with phasor analysis. This observation stems from the fact that while in Laplace transform, we have ##s = \sigma + j \omega## as the variable, in phasor analysis, we just use ##j\omega,## apparently just setting ##\sigma = 0.## Is there any other way the two methods are related?

I have seen phasor analysis being used only for sinusoidal steady state analysis, while Laplace transform gives the transient solution since the initial conditions are built into it, and we solve the circuit with that. Laplace transform also allows us to incorporate impulse-type voltages and currents in the circuit, wherefrom it seems to me that Laplace transform is a more powerful tool. Can Laplace transform be used to do sinusoidal steady state analysis as well? I don't see why not, but need a confirmation.

 

[DaveE]

Yes, absolutely related. When s=jω, you get the steady state sinusoidal solutions, otherwise there is an exponential growth (or decay) element in the solutions. So the Laplace transform is the more general case that can be used to obtain transient and/or steady state solutions.

 

[Wrichik Basu]

Yes, absolutely related. When s=jω, you get the steady state sinusoidal solutions, otherwise there is an exponential growth (or decay) element in the solutions. So the Laplace transform is the more general case that can be used to obtain transient and/or steady state solutions.

I will have to go through that video once again to get my brain straight, but I understood this much that if I use Laplace transform and get a decay term, then in the ##\lim_{t \rightarrow \infty}## that decay term basically ##\rightarrow 0,## so I am left with the steady state solution only. Thanks.

 

[jasonRF]

I recently acquainted myself with Laplace transform, and it appears that it has some relations with phasor analysis. This observation stems from the fact that while in Laplace transform, we have ##s = \sigma + j \omega## as the variable, in phasor analysis, we just use ##j\omega,## apparently just setting ##\sigma = 0.## Is there any other way the two methods are related?

I have seen phasor analysis being used only for sinusoidal steady state analysis, while Laplace transform gives the transient solution since the initial conditions are built into it, and we solve the circuit with that. Laplace transform also allows us to incorporate impulse-type voltages and currents in the circuit, wherefrom it seems to me that Laplace transform is a more powerful tool. Can Laplace transform be used to do sinusoidal steady state analysis as well? I don't see why not, but need a confirmation.

The two methods are related, but it depends on where the Laplace transform converges. I posted about this in a different thread awhile ago.

https://www.physicsforums.com/threa...-of-the-laplace-transform.971324/post-6174503

jason