Electric Circuits and Networks, Ist Edition by Pearson

Solving for V

o

(s) and simplifying the expression, we get,

Substituting the numerical values for various parameters, we get,

The poles are at s  –10

9

Np/s and s  –11.07  10

6

Np/s. The zero is at s  1.6

 10

10

Np/s.

Note that compared to the pole at –11.07  10

6

the other pole and the zero are

located two orders away from it. The natural response term contributed by the pole at

s  –11.07  10

6

Np/s will have a time constant of 90.3 ns whereas the natural response

term contributed by the pole at s  –10

9

Np/s will have a time constant of 1 ns. Thus, the

natural response term contributed by the pole at s  –10

9

Np/s will disappear in about

5% of the time ...