[ENGN2211 Home]

s-domain Method

We re-draw the circuit of Figure 25 using s-domain quantities, as shown in Figure 27.

  

Figure 27: s-domain RC circuit.

The capacitor current and voltage are related by

$$V(s) = \frac{1}{sC} I(s) + \frac{v(0^-)}{s},$$

and so

I(s) = sC V(s) -C v(0^-) .

Now applying KCL to equate this with the resistor current we get

$$\frac{5/s-V(s)}{R} = sC V(s) -C v(0^-) .$$

Solving for voltage gives

$$V(s) = \frac{5}{sRC(s+\frac{1}{RC})} + \frac{v(0^-)}{(s+\frac{1}{RC})}$$ (27)

Plugging in our numbers gives

$$V(s) = \frac{5}{sRC(s+\frac{1}{RC})} + \frac{2}{(s+\frac{1}{RC})}$$ (28)

with RC=1 $\mu$s. Notice that the voltage is a superposition of two terms, one due to the source voltage, and the other due to the initial condition.

Equation (28) gives the solution voltage V(s) in the s-domain. We now work out the time domain waveform v(t).

The second term in (28) transforms to

$$2e^{-\frac{t}{RC}} \ {\rm V},$$ (29)

using the LT pairs table in §3.2.2. The first term does not appear in the LT pairs table, so we need to re-express it as a sum of terms in the table. Using partial fractions, we have

$$\frac{5}{sRC(s+\frac{1}{RC})} = \frac{5}{s} - \frac{5}{s+\frac{1}{RC}} ,$$

which transforms back to

$$5u(t)-5e^{-\frac{t}{RC}}u(t) \ {\rm V} .$$ (30)

Combining (29) and (30) we get

$$v(t) = 5u(t)-3e^{-\frac{t}{RC}}u(t) \ {\rm V} ,$$ (31)

in agreement with (26).

[ENGN2211 Home]