ppat2006
Member level 2
does anybody know how to obtain impulse sensitivity function for single ended ring oscillators. the formula is
isf = f'(x)/[f'(x)2 + f''(x)2] where f'(x) is derivative of output voltage function
of ring osc and f''(x) is double derivative of output voltage function of ring osc.
for eg for colpitt osc, f(x) = sin(x) so we can calculate derivative and obtain isf plot.
so i dont know how to obtain f(x) for single ended 5 stage ring osc?
isf = f'(x)/[f'(x)2 + f''(x)2] where f'(x) is derivative of output voltage function
of ring osc and f''(x) is double derivative of output voltage function of ring osc.
for eg for colpitt osc, f(x) = sin(x) so we can calculate derivative and obtain isf plot.
so i dont know how to obtain f(x) for single ended 5 stage ring osc?