Lect. 1.2 RC Circuit with Step Input || Complete Analysis || Crack Interview with Me

[@karanjot5921]

In both the circuits, pole should be at -1/RC...because only when we keep s=(-1/RC) will we get transfer function as infinity, not at s=1/RC

[@harishmacharla1863]

Thanks a lot ❤

[@amanruke3961]

❤❤❤❤❤

[@HimanshuAgarwal_]

Correction: it should be t=0+, and not t=0 Visit my Website to see all my placement courses. You can simply follow my placement courses for the preparation of placements in Analog VLSI Domain. himanshu-agarwal.netlify.app/

[@samanwaypal4552]

Sir you are referring 5V As V ? In all the solutions ?

[@HimanshuAgarwal_]

Yup

[@samanwaypal4552]

@@HimanshuAgarwal_ thankyou ☺️

[@VLSI_Project]

there will be one zero 1/RC and pole -1/RC ,please rectify it

[@HimanshuAgarwal_]

How?

[@VLSI_Project]

Okayy got it, only one pole.

[@HimanshuAgarwal_]

For 2nd ckt, there is a zero at zero

[@arghya.7098]

Isn't V=5.u(t) ??? Then shouldn't dV/dt = 5.δ(t) ??? I'm confused, at 9:33 you have put dV/dt = 0

[@HimanshuAgarwal_]

Slope at t=0+ will be Zero