2
\$\begingroup\$

enter image description here

Dear All, Greetings,

I am studying Response of the RC series circuit to a Pulse input. For calculating voltage across capacitor after pulse input ceases,author considers a time shift. I am not able to understand what for & why the time shift is considered here ( Highlighted in Red ).

\$\endgroup\$

2 Answers 2

3
\$\begingroup\$

The pulse starts at t = 0 and ends at t = T. And you're trying to find the response after the pulse ends.

First you solved for the response during the pulse, so you know the capacitor voltage at t = T. Now you want to find a response starting at time T. You're basically solving a whole new problem starting at time T, using the previous calculation to tell you the "initial" conditions at time T.

So the problem (and solution) is "shifted" in time compared to what they probably taught you previously about finding a response starting at t = 0, given initial conditions at t = 0.

\$\endgroup\$
1
\$\begingroup\$

It's not a phase shift, it's a time shift and equals the width of the pulse that charged up the capacitor initially.

\$\endgroup\$
2
  • \$\begingroup\$ Sir, not clear yet. \$\endgroup\$
    – Aditya
    Commented Dec 19, 2013 at 17:34
  • 1
    \$\begingroup\$ @aditya I'm assuming it's the "t-T" part in the formula that concerns you? Because t is time, to "reset" time at the point T, T is subtracted from t so that the result of (t-T) begins at zero and goes positive - this then properly defines the shape of the decay starting at t=T and ending at t=infinite. Does this make anymore sense? \$\endgroup\$
    – Andy aka
    Commented Dec 19, 2013 at 18:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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