ขนาดวิดีโอ: 1280 X 720853 X 480640 X 360
แสดงแผงควบคุมโปรแกรมเล่น
เล่นอัตโนมัติ
เล่นใหม่
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
Thanks a lot ❤
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/
there will be one zero 1/RC and pole -1/RC ,please rectify it
How?
Okayy got it, only one pole.
For 2nd ckt, there is a zero at zero
Sir you are referring 5V As V ? In all the solutions ?
Yup
@@HimanshuAgarwal_ thankyou ☺️
❤❤❤❤❤
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
Slope at t=0+ will be Zero
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
Thanks a lot ❤
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/
there will be one zero 1/RC and pole -1/RC ,please rectify it
How?
Okayy got it, only one pole.
For 2nd ckt, there is a zero at zero
Sir you are referring 5V As V ? In all the solutions ?
Yup
@@HimanshuAgarwal_ thankyou ☺️
❤❤❤❤❤
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
Slope at t=0+ will be Zero