0
\$\begingroup\$

I have a problem, I know how to analyze circuits in Laplace domain to get the transfer Laplace function. However, I don't know how to translate the answer to "real life"

Suppose that I have a transfer system with Vout=1/sC, where C is a capacitance (say of 10uF). And suppose now that I give the system Vin = 2*sin(pai/2 + 2pai). How do I determain the output signal from the laplace form of my transfer function?

Thanks!

\$\endgroup\$
  • 1
    \$\begingroup\$ Your question has unclear details such as "Vout=1/sC where C is a capacitance" and "Vin=2*sin(pai/2+2pai)". They simply look out nonsense. You should insert the schematic of your circuit and the calculations you have already done. I bet most of us wait to see time in the expression of Vin. But you still got an answer. \$\endgroup\$ – user287001 May 30 at 9:13
  • 1
    \$\begingroup\$ what do you mean by "real life"? \$\endgroup\$ – Unknown123 May 30 at 9:14
0
\$\begingroup\$

The Laplace domain transfer function, if properly derived, is Vout/Vin where Vout and Vin are the Laplace transforms of output and input voltages. You multiply the Laplace transform of the input voltage with the transfer function and find the inverse transform of the result. That's your Vout in time domain.

Pure sinusoidal inputs can be handled in frequency domain, you simply calculate the gain or attenuation and the phase shift. Your transfer function is a multiplying phasor when you calculate it as complex number in the operating frequency.

If you calculate the inverse Laplace transform of the Laplace domain transfer function, you get what the system outputs in case the input is the unit impulse function (=Dirac's delta)

Unfortunately often the inverse transform is too complex. Then you must calculate the output numerically. Circuit analyzing programs do it automatically.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.