7.5 Step Response of an RC Circuit

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7.5 Step Response of an RC Circuit

When the dc source of an RC circuit is suddenly applied, the voltage or current source can be modeled as a step function, and the response is known as a step response.

The step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source.

Practice Problem 7.9

The step response is the response of the circuit due to a sudden applica tion of a dc voltage or current source.

Consider the RC circuit in Fig. 7.40(a) which can be replaced by the circuit in Fig. 7.40(b), where Vs is a constant dc v oltage source. Again, we select the capacitor voltage as the circuit response to be determined. We assume an initial v oltage V0 on the capacitor , although this is not necessary for the step response. Since the v oltage of a capacitor cannot change instantaneously,

$$v(0^{-}) = v(0^{+}) = V_0 \tag{7.40}$$

where v(0⁻) is the voltage across the capacitor just before switching and v( 0 ⁺) is its voltage immediately after switching. Applying KCL, we have

$$C\frac{dv}{dt} + \frac{v - V_s u(t)}{R} = 0$$ $$\frac{dv}{dt} + \frac{v}{RC} = \frac{V_s}{RC}u(t)$$

\n(7.41)

where v is the voltage across the capacitor. For t > 0, Eq. (7.41) becomes

$$\frac{dv}{dt} + \frac{v}{RC} = \frac{V_s}{RC}$$

(7.42)

Rearranging terms gives

$$\frac{dv}{dt} = -\frac{v - V_s}{RC}$$

______ dv v− Vs = − ___ dt RC (7.43)

Integrating both sides and introducing the initial conditions,

$$\ln(v - V_s) \Big|_{V_0}^{v(t)} = -\frac{t}{RC} \Big|_{0}^{t}$$ $$\ln(v(t) - V_s) - \ln(V_0 - V_s) = -\frac{t}{RC} + 0$$

or

or

or

$$\ln \frac{v - V_s}{V_0 - V_s} = -\frac{t}{RC}$$

\n(7.44)

Taking the exponential of both sides

$$\frac{v - V_s}{V_0 - V_s} = e^{-t/\tau}, \qquad \tau = RC$$ $$v - V_s = (V_0 - V_s)e^{-t/\tau}$$

or

$$v(t) = V_s + (V_0 - V_s)e^{-t/\tau}, \qquad t > 0 \tag{7.45}$$

Thus,

$$v(t) = \begin{cases} V_0, & t < 0 \\ V_s + (V_0 - V_s)e^{-t/\tau}, & t > 0 \end{cases}$$

(7.46)

This is known as the complete response (or total response) of the RC circuit to a sudden application of a dc voltage source, assuming the capacitor is initially charged. The reason for the term “complete” will become evident a little later . Assuming that Vs > V0, a plot of v(t) is sho wn in Fig. 7.41.

If we assume that the capacitor is uncharged initially, we set V0 = 0 in Eq. (7.46) so that

$$v(t) = \begin{cases} 0, & t < 0 \\ V_s(1 - e^{-t/\tau}), & t > 0 \end{cases}$$

(7.47)

which can be written alternatively as

$$v(t) = V_s(1 - e^{-t/\tau})u(t)$$

\n(7.48)

This is the complete step response of the RC circuit when the capacitor is initially uncharged. The current through the capacitor is obtained from Eq. (7.47) using i(t) = C dv ∕ dt. We get

$$i(t) = C\frac{dv}{dt} = \frac{C}{\tau}V_s e^{-t/\tau}, \qquad \tau = RC, \qquad t > 0$$ $$i(t) = \frac{V_s}{R} e^{-t/\tau} u(t) \tag{7.49}$$

Figure 7.42 shows the plots of capacitor v oltage v(t) and capacitor current i(t).

R

Rather than going through the derivations above, there is a systematic approach—or rather, a shortcut method—for finding the step response of an RC or RL circuit. Let us reexamine Eq. (7.45), which is more general than Eq. (7.48). It is evident that v(t) has two components. Classically there are two ways of decomposing this into tw o components. The first is to break it into a “natural response and a forced response” and the second is to break it into a “transient response and a steady-state response.” Starting with the natural response and forced response, we write the total or complete response as

Complete response = natural response + forced response stored energy independent source

v = vn + v^f (7.50)

or

or

where

and

$$v_f = V_s(1 - e^{-t/\tau})$$

vn = Vo e⁻t∕^τ

We are familiar with the natural response vn of the circuit, as discussed in Section 7.2. vf is known as the forced response because it is produced by the circuit when an external “force” (a voltage source in this case) is applied. It represents what the circuit is forced to do by the input excitation. The natural response e ventually dies out along with the transient component of the forced response, leaving only the steady-state component of the forced response.

Step response of an RC circuit with initially uncharged capacitor: (a) voltage response, (b) current response.

Response of an RC circuit with initially charged capacitor.

274 Chapter 7 First-Order Circuits

Another way of looking at the complete response is to break into two components—one temporary and the other permanent, that is

Complete response = transient response + steady-state response temporary part permanent part

or

$$v = v_t + v_{ss} \tag{7.51}$$

where

$$v_t = (V_o - V_s)e^{-t/\tau}$$

(7.52a)

and

$$v_{ss} = V_s \tag{7.52b}$$

The transient response vt is temporary; it is the portion of the complete response that decays to zero as time approaches infinity. Thus,

The transient response is the circuit’s temporary response that will die out with time.

The steady-state response vss is the portion of the complete response that remains after the transient reponse has died out. Thus,

The steady-state response is the behavior of the circuit a long time after an external excitation is applied.

The first decomposition of the complete response is in terms of the source of the responses, while the second decomposition is in terms of the permanency of the responses. Under certain conditions, the natural response and transient response are the same. The same can be said about the forced response and steady-state response.

Whichever way we look at it, the complete response in Eq. (7.45) may be written as

$$v(t) = v(\infty) + [v(0) - v(\infty)]e^{-t/\tau}$$

(7.53)

where v(0) is the initial voltage at t = 0 ⁺ and v(∞) is the final or steadystate value. Thus, to find the step response of an RC circuit requires three things:

• 1. The initial capacitor voltage v(0).
• 2. The final capacitor voltage v(∞).
• 3. The time constant τ.

Once we know x (0), x (∞), and τ, almost all the circuit problems in this chapter can be solved using the formula

x(t) = x(∞) + [x(0) − x(∞)] e⁻t∕^τ

We obtain item 1 from the given circuit for t < 0 and items 2 and 3 from the circuit for t > 0. Once these items are determined, we obtain the

This is the same as saying that the complete response is the sum of the transient response and the steady-state response.

response using Eq. (7.53). This technique equally applies to RL circuits, as we shall see in the next section.

Note that if the switch changes position at time t = t0 instead of at t = 0, there is a time delay in the response so that Eq. (7.53) becomes

$$v(t) = v(\infty) + [v(t_0) - v(\infty)]e^{-(t - t_0)/\tau}$$

(7.54)

where v(t0) is the initial value at t = t^{0 +} . Keep in mind that Eq. (7.53) or (7.54) applies only to step responses, that is, when the input e xcitation is constant.

The switch in Fig. 7.43 has been in position A for a long time. At t = 0, the switch moves to B. Determine v(t) for t > 0 and calculate its value at t = 1 and 4 s.

Solution:

For t < 0, the switch is at position A. The capacitor acts like an open circuit to dc, but v is the same as the voltage across the 5-k Ω resistor. Hence, the voltage across the capacitor just before t = 0 is obtained by voltage division as

$$v(0^{-}) = \frac{5}{5+3}(24) = 15 \text{ V}$$

Using the fact that the capacitor voltage cannot change instantaneously,

$$v(0) = v(0^-) = v(0^+) = 15$$

V

For t > 0, the switch is in position B. The Thevenin resistance connected to the capacitor is RTh = 4 kΩ, and the time constant is

$$\tau = R_{\text{Th}} C = 4 \times 10^3 \times 0.5 \times 10^{-3} = 2 \text{ s}$$

Since the capacitor acts like an open circuit to dc at steady state, v(∞) = 30 V. Thus,

$$v(t) = v(\infty) + [v(0) - v(\infty)]e^{-t/\tau}$$

= 30 + (15 - 30)e^{-t/2} = (30 - 15e^{-0.5t}) V

At t = 1,

$$v(1) = 30 - 15e^{-0.5} = 20.9 \text{ V}$$

At t = 4,

$$v(4) = 30 - 15e^{-2} = 27.97$$

V

Example 7.10

Practice Problem 7.10

Find v(t) for t > 0 in the circuit of Fig. 7.44. Assume the switch has been open for a long time and is closed at t = 0. Calculate v(t) at t = 0.5.

Answer: (9.375 + 5.625e⁻2*^t* ) V for all t > 0, 11.444 V.

Example 7.11 In Fig. 7.45, the switch has been closed for a long time and is opened at t = 0. Find i and v for all time.

Solution:

The resistor current i can be discontinuous at t = 0, while the capacitor voltage v cannot. Hence, it is always better to find v and then obtain i from v.

By definition of the unit step function,

$$30u(t) = \begin{cases} 0, & t < 0\\ 30, & t > 0 \end{cases}$$

For t < 0, the switch is closed and 30 u(t) = 0, so that the 30u(t) voltage source is replaced by a short circuit and should be regarded as contributing nothing to v. Since the switch has been closed for a long time, the capacitor voltage has reached steady state and the capacitor acts like an open circuit. Hence, the circuit becomes that shown in Fig. 7.46(a) for t < 0. From this circuit we obtain

$$v = 10 \text{ V}, \qquad i = -\frac{v}{10} = -1 \text{ A}$$

Since the capacitor voltage cannot change instantaneously,

$$v(0) = v(0^-) = 10 \text{ V}$$

For t > 0, the switch is opened and the 10-V voltage source is disconnected from the circuit. The 30 u(t) voltage source is now operative, so the circuit becomes that shown in Fig. 7.46(b). After a long time, the circuit reaches steady state and the capacitor acts like an open circuit again. We obtain v(∞) by using voltage division, writing

$$v(\infty) = \frac{20}{20 + 10}(30) = 20 \text{ V}$$

Figure 7.46 Solution of Example 7.11: (a) for t < 0, (b) for t > 0.

The Thevenin resistance at the capacitor terminals is

$$R_{\text{Th}} = 10||20 = \frac{10 \times 20}{30} = \frac{20}{3}\Omega$$

and the time constant is

$$\tau = R_{\text{Th}} C = \frac{20}{3} \cdot \frac{1}{4} = \frac{5}{3} \text{ s}$$

Thus,

$$v(t) = v(\infty) + [v(0) - v(\infty)]e^{-t/\tau}$$

= 20 + (10 - 20)e^{-(3/5)t} = (20 - 10e^{-0.6t}) V

To obtain i, we notice from Fig. 7.46(b) that i is the sum of the currents through the 20-Ω resistor and the capacitor; that is,

$$i = \frac{v}{20} + C\frac{dv}{dt}$$

= 1 - 0.5e^-0.6t + 0.25(-0.6)(-10)e^-0.6t = (1 + e^-0.6t) A

Notice from Fig. 7.46(b) that v + 10i = 30 is satisfied, as expected. Hence,

$$v = \begin{cases} 10 \text{ V}, & t < 0 \\ (20 - 10e^{-0.6t}) \text{ V}, & t \ge 0 \end{cases}$$ $$i = \begin{cases} -1 \text{ A}, & t < 0 \\ (1 + e^{-0.6t}) \text{ A}, & t > 0 \end{cases}$$

Notice that the capacitor voltage is continuous while the resistor current is not.

The switch in Fig. 7.47 is closed at t = 0. Find i(t) and v(t) for all time. Note that u(−t) = 1 for t < 0 and 0 for t > 0. Also, u(−t) = 1 − u(t). Practice Problem 7.11

7.6 Step Response of an RL Circuit

Consider the RL circuit in Fig. 7.48(a), which may be replaced by the circuit in Fig. 7.48(b). Again, our goal is to find the inductor current i as the circuit response. Rather than apply Kirchhoff’s laws, we will use the simple technique in Eqs. (7.50) through (7.53). Let the response be the sum of the transient response and the steady-state response,

$$i = i_t + i_{ss} \tag{7.55}$$

We know that the transient response is al ways a decaying e xponential, that is,

$$i_t = Ae^{-t/\tau}, \qquad \tau = \frac{L}{R} \tag{7.56}$$

where A is a constant to be determined.

The steady-state response is the value of the current a long time after the switch in Fig. 7.48(a) is closed. We know that the transient response essentially dies out after five time constants. At that time, the inductor becomes a short circuit, and the v oltage across it is zero. The entire source v oltage Vs appears across R. Thus, the steady-state response is

$$i_{ss} = \frac{V_s}{R}$$

(7.57)

Substituting Eqs. (7.56) and (7.57) into Eq. (7.55) gives

$$i = Ae^{-t/\tau} + \frac{V_s}{R}$$

\n

$$(7.58)$$

We now determine the constant A from the initial v alue of i. Let I0 be the initial current through the inductor, which may come from a source other than Vs. Since the current through the inductor cannot change instantaneously,

$$i(0^+) = i(0^-) = I_0 \tag{7.59}$$

Thus, at t = 0, Eq. (7.58) becomes

$$I_0 = A + \frac{V_s}{R}$$

From this, we obtain A as

$$A = I_0 - \frac{V_s}{R}$$

Substituting for A in Eq. (7.58), we get