Question 9.20: Find υ_{C}(t) in the RC circuit in Figure 9–14 when the inpu......

The transform of the input is V_S(s) = V_A/(s + α). For the exponential input the response transform in Eq. (9–37) becomes

V_{C}(s) = \frac{V_{S}(s)/RC}{s  +  1/RC} + \frac{V_{0}}{s  +  1/RC} V-s           (9–37)

V_{C}(s) = \frac{V_{A}/RC}{(s  +  α)(s  +  1/RC)} + \frac{V_{0}}{s  +  1/RC} V-s           (9–38)

If α ≠ 1/RC, then the first term on the right is a proper rational function with two simple poles. The pole at s = −α came from the input and the pole at s = −1/RC from the circuit. A partial fraction expansion of the first term has the form

\frac{V_{A}/RC}{(s  +  α)(s  +  1/RC)} = \frac{k_{1}}{s  +  α} + \frac{k_{2}}{s  +  1/RC}

The residues in this expansion are

k_{1} = \frac{V_{A}/RC}{s  +  1/RC}|_{s  =  −α} = \frac{V_{A}}{1  −  αRC}

k_{2} = \frac{V_{A}/RC}{s  +  α}|_{s  =  −1/RC} = \frac{V_{A}}{αRC  −  1}

The expansion of the response transform V_C(s) is

V_{C}(s) = \frac{V_{A}/(1  −  αRC)}{s  +  α} + \frac{V_{A}/(αRC  −  1)}{s  +  1/RC} + \frac{V_{0}}{s  +  1/RC}V-s

The inverse transform of V_C(s) is

υ_{C}(t) = \left[\frac{V_{A}}{1  −  αRC}e^{−αt} + \frac{V_{A}}{αRC  −  1}e^{−t/RC} + V_{0}e^{−t/RC}\right] u(t)  V

The first term is the forced response, and the last two terms are the natural response. The forced response is an exponential because the input introduced a pole at s = −α. The natural response is also an exponential, but its time constant depends on the circuit’s pole at s = −1/RC. In this case the forced and natural responses are both exponential signals with poles on the real axis. However, the force dresponse comes from the pole introduced by the input, while the natural response depends on the circuit’s pole. If α = 1/RC, then the response just given is no longer valid (k₁ and k₂ become infinite). To find the response for this condition, we return to Eq. (9–38) and replace α by 1/RC:

V_{C}(s) = \frac{V_{A}/RC}{(s  +  1/RC)^{2}} + \frac{V_{0}}{s  +  1/RC} V-s

We now have a double pole at s = −1/RC = −α. The double pole term is the transform of a damped ramp, so the inverse transform is

υ_{C}(t) =\left[V_{A}\frac{t}{RC}e^{−t/RC} + V_{0}e^{−t/RC}\right] u(t)  V

When α = 1/RC, the s-domain poles of the input and the circuit coincide and the zero-state (V₀ = 0) response has the form αte^-αt . We cannot separate this response into forced and natural components since the input and circuit poles coincide.