Question 7.15: Determine v(t) and vo (t) in the circuit of Fig. 7.57.

This problem can be solved in two ways, just like the previous example. However, we will apply only the second method. Since what we are looking for is the step response, we can apply Eq. (7.53) and write

v(t)=v(\infty)+[v(0)-v(\infty)] e^{-t / \tau}          t > 0       (7.15.1)
where we need only find the time constant τ, the initial value v(0), and the final value v(∞). Notice that this applies strictly to the capacitor voltage due a step input. Since no current enters the input terminals of the op amp, the elements on the feedback loop of the op amp constitute an RC circuit, with

\tau=R C=50 \times 10^{3} \times 10^{-6}=0.05      (7.15.2)
For t < 0, the switch is open and there is no voltage across the capacitor. Hence, v(0) = 0. For t > 0, we obtain the voltage at node 1 by voltage division as

v_{1}=\frac{20}{20+10} 3=2 V      (7.15.3)
Since there is no storage element in the input loop, v_{1} remains constant for all t. At steady state, the capacitor acts like an open circuit so that the op amp circuit is a noninverting amplifier. Thus,

v_{o}(\infty)=\left(1+\frac{50}{20}\right) v_{1}=3.5 \times 2=7 V      (7.15.4)
But

v_{1}-v_{o}=v      (7.15.5)

so that
v(∞) = 2 − 7 = −5 V
Substituting τ,v(0), and v(∞) into Eq. (7.15.1) gives

v(t)=-5+[0-(-5)] e^{-20 t}=5\left(e^{-20 t}-1\right) V,          t > 0        (7.15.6)
From Eqs. (7.15.3), (7.15.5), and (7.15.6), we obtain

v_{o}(t)=v_{1}(t)-v(t)=7-5 e^{-20 t} V,            t > 0          (7.15.7)