Question B.12: Pulse response of a parallel RC circuit A constant-current s...

For a pulse signal, the input current in Laplace’s domain is I_{ S }(s)=I_{ S } / s .
(a) Using the current divider rule, we can find the capacitor current I_{ C } in Laplace’s domain:

I_{ C }(s)=\frac{R}{R+1 / C s} I_{ S }(s)=\frac{s}{s+1 / R C} I_{ S }(s)=\frac{s}{s+1 / R C} \times \frac{I_{ S }}{s}=\frac{I_{ S }}{s+1 / R C}

=I_{ S } \frac{I}{s+1 / R C}                                                 (B.43)

The inverse transform of I_{ C }(s) in Eq. (B.43) gives

i_{ C }(t)=I_{ S } e^{-t / \tau}                                                 (B.44)

where τ = RC.

(b) The instantaneous current i_{ R }(t) through resistance R is

i_{ R }(t)=I_{ S }-i_{ C }(t)=I_{ S }\left(1-e^{-t / \tau}\right)                                                   (B.45)

(c) \tau=R C=100 \times 10^{3} \times 0.1 \times 10^{-6}=10 ms \text {, and } T=0.5 ms \text {. Therefore, } \tau>>T , and Eq. (B.42) gives

S=\frac{\Delta V}{V_{ S }}=\frac{V_{ S } T / \tau}{V_{ S }}=\frac{T}{\tau}=\frac{T}{R C}                  (B.42)

S=T / \tau=0.5 / 10=5 \%