Consider the circuit shown in Fig. 7.13a. The input function is the voltage pulse shown in Fig. 7.13b. Since the source is zero for all negative time, the initial conditions for the network are zero [ i.e., v_C(0-)=0 ]. The response v_o(t) for 0

Question

Consider the circuit shown in Fig. 7.13a. The input function is the voltage pulse shown in Fig. 7.13b. Since the source is zero for all negative time, the initial conditions for the network are zero [ i.e., [latex]v_C[/latex](0-)=0 ]. The response [latex]v_o[/latex](t) for 0 < t < 0.3 s is due to the application of the constant source at t = 0 and is not influenced by any source changes that will occur later. At t =0.3 s the forcing function becomes zero, and therefore [latex]v_{o}[/latex](t) for t > 0.3 s is the source-free or natural response of the network.
Let us determine the expression for the voltage [latex]v_{o}(t)[/latex].

Accepted Answer

Since the output voltage [latex]v_{o}(t)[/latex] is a voltage division of the capacitor voltage, and the initial voltage across the capacitor is zero, we know that [latex]v_{o}[/latex](0+) = 0, as shown in Fig. 7.13c.

If no changes were made in the source after t = 0, the steady-state value of [latex]v_{o}[/latex](t)[ i.e., [latex]v_{o}(\infty)[/latex]] due to the application of the unit step at t = 0 would be

[latex]v_{o}(\infty) =\frac{9}{6 \mathrm{k}+4 \mathrm{k}+8 \mathrm{k}}(8 \mathrm{k}) [/latex]

[latex]=4 \mathrm{~V}[/latex]

as shown in Fig. 7.13d.

The Thévenin equivalent resistance is

[latex]R_{\mathrm{Th}} =\frac{(6 \mathrm{k})(12 \mathrm{k})}{6 \mathrm{k}+12 \mathrm{k}} [/latex]

[latex]=4 \mathrm{k} \Omega[/latex]

as illustrated in Fig. 7.13e.

Therefore, the circuit time constant [latex] au[/latex] is

[latex] au =R_{\mathrm{Th}} C [/latex]

[latex]=(4)\left(10^{3} ight)(100)\left(10^{-6} ight) [/latex]

=0.4 s

Therefore, the response [latex]v_{o}(t)[/latex] for the period 0 < t < 0.3 s is

[latex]v_{o}(t)=4-4 e^{-t / 0.4} \mathrm{~V} \qquad 0

The capacitor voltage can be calculated by realizing that using voltage division, [latex]v_{o}(t)=2 / 3 v_{C}(t)[/latex]. Therefore,

[latex]v_{C}(t)=\frac{3}{2}\left(4-4 e^{-t / 0.4} ight) \mathrm{V}[/latex]

Since the capacitor voltage is continuous,

[latex]v_{C}(0.3-)=v_{C}(0.3+)[/latex]

and therefore,

[latex]v_{o}(0.3+) =\frac{2}{3} v_{C}(0.3-) [/latex]

[latex]=4\left(1-e^{-0.3 / 0.4} ight) [/latex]

=2.11 V

Since the source is zero for t > 0.3 s, the final value for [latex]v_{o}(t)[/latex] as [latex]t ightarrow \infty[/latex] is zero. Therefore, the expression for [latex]v_{o}(t)[/latex] for t > 0.3 s is

[latex]v_{o}(t)=2.11 e^{-(t-0.3) / 0.4} \mathrm{~V} \quad t>0.3 \mathrm{~s}[/latex]

The term [latex]e^{-(t-0.3) / 0.4}[/latex] indicates that the exponential decay starts at t = 0.3 s. The complete solution can be written by means of superposition as

[latex]v_{o}(t)=4\left(1-e^{-t / 0.4} ight) u(t)-4\left(1-e^{-(t-0.3) / 0.4} ight) u(t-0.3) \mathrm{V}[/latex]

or, equivalently, the complete solution is

[latex]v_{o}(t)=\left\{\begin{array}{ll}0 & t<0 \4\left(1-e^{-t / 0.4} ight) \mathrm{V} & 0

which in mathematical form is

[latex]v_{o}(t)=4\left(1-e^{-t / 0.4} ight)[u(t)-u(t-0.3)]+2.11 e^{-(t-0.3) / 0.4} u(t-0.3) \mathrm{V}[/latex]

Note that the term [latex][u(t)-u(t-0.3)][/latex] acts like a gating function that captures only the part of the step response that exists in the time interval 0 < t < 0.3 s. The output as a function of time is shown in Fig. 7.13f.

Question 7.6: Consider the circuit shown in Fig. 7.13a. The input function...

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