Consider the first order RC circuit shown in Figure 1. With v.(1) being a symmetrical square- voltage having period T and amplitude V, as shown in Figure 2(a), derive an expression for the capacitor voltage vo(1) in the first half-cycle, as shown in Figure 2(a), by following the procedure below: (a) Derive the differential equation (DE) for the capacitor voltage v (t) valid in the interval 0
Consider the first order RC circuit shown in Figure 1. With v.(1) being a symmetrical square- voltage having period T and amplitude V, as shown in Figure 2(a), derive an expression for the capacitor voltage vo(1) in the first half-cycle, as shown in Figure 2(a), by following the procedure below: (a) Derive the differential equation (DE) for the capacitor voltage v (t) valid in the interval 0
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can you solve from a to e with steps explanation
![Consider the first order RC circuit shown in Figure 1. With v.(1) being a symmetrical square-wave
voltage having period T and amplitude V, as shown in Figure 2(a), derive an expression for the
capacitor voltage vo(t) in the first half-cycle, as shown in Figure 2(a), by following the procedure
below:
(a) Derive the differential equation (DE) for the capacitor voltage v, (t) valid in the interval
0<t<T/2, where v, (t) = V. Note that the initial condition for this DE is v. (0) = V, which is
still unknown.
(b) Assume that the solution has the form vo(t) = A + Bea. Then substitute this into the DE and
determine the unknown parameters A, B and a.
(c) Find the unknown initial condition V (in terms of the parameters R, C, T and V) by imposing
the symmetry condition on the solution: v(T/2) = -v(0) = V.
(d) Find the time instant t₁ (in terms of R, C, T and V) at which the capacitor voltage becomes zero
in the first half-cycle.
(e) Calculate the numerical values of V, and t₁ for T=10 ms, Vs = 10 V, R = 10 k2 and C = 0.1 μµF.
Figure 1
vs(t)
risques bris 20
R = 10 kΩ
ww
C=0.1 µF
+
vo(t)
-1.0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b5d4c98-afbd-424f-9855-c5eac84955f5%2F1e017663-8a62-419f-90c4-a9cd1b9b0f5d%2F3ee5o39_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider the first order RC circuit shown in Figure 1. With v.(1) being a symmetrical square-wave
voltage having period T and amplitude V, as shown in Figure 2(a), derive an expression for the
capacitor voltage vo(t) in the first half-cycle, as shown in Figure 2(a), by following the procedure
below:
(a) Derive the differential equation (DE) for the capacitor voltage v, (t) valid in the interval
0<t<T/2, where v, (t) = V. Note that the initial condition for this DE is v. (0) = V, which is
still unknown.
(b) Assume that the solution has the form vo(t) = A + Bea. Then substitute this into the DE and
determine the unknown parameters A, B and a.
(c) Find the unknown initial condition V (in terms of the parameters R, C, T and V) by imposing
the symmetry condition on the solution: v(T/2) = -v(0) = V.
(d) Find the time instant t₁ (in terms of R, C, T and V) at which the capacitor voltage becomes zero
in the first half-cycle.
(e) Calculate the numerical values of V, and t₁ for T=10 ms, Vs = 10 V, R = 10 k2 and C = 0.1 μµF.
Figure 1
vs(t)
risques bris 20
R = 10 kΩ
ww
C=0.1 µF
+
vo(t)
-1.0
![V.
S
V
>V,
-V.
S
Figure 2
t₁ T/2
Vi
Vo
T
t](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b5d4c98-afbd-424f-9855-c5eac84955f5%2F1e017663-8a62-419f-90c4-a9cd1b9b0f5d%2Fwxqy8gy_processed.jpeg&w=3840&q=75)
Transcribed Image Text:V.
S
V
>V,
-V.
S
Figure 2
t₁ T/2
Vi
Vo
T
t
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