How idt is behaving like this?

[Cornel]

Hi,

Why?

[KSKelvin]

idt is time integral and ddt is time derivite. mixed this up?

[Cornel]

A, I see, yes…now its better, they match

[Cornel]

But in the case of idt seems to be still a mistmatch:

[KSKelvin]

integration needs to consider initial condition. V=idt(V(sin),-1/2/pi,0)*(-2*pi), add -1/2/pi as initial condition. [you can just use V=idt(V(sin),-1/2/pi)*(-2*pi), if you don’t need to reset integrator during simulation.

[Cornel]

Ok, understood, thanks