Step Response of a Series RLC Circuit

8.5 Step Response of a Series RLC Circuit

Figure 8.17

As we learned in the preceding chapter, the step response is obtained For Practice Prob. 8.6. by the sudden application of a dc source. Consider the series RLC cir-

cuit shown in Fig. 8.18. Applying KVL around the loop for t 70 ,

i⫽C Figure 8.18

Chapter 8

Second-Order Circuits

Substituting for i in Eq. (8.39) and rearranging terms,

d 2 v R dv

LC which has the same form as Eq. (8.4). More specifically, the coeffi-

dt L dt

LC

cients are the same (and that is important in determining the frequency parameters) but the variable is different. (Likewise, see Eq. (8.47).) Hence, the characteristic equation for the series RLC circuit is not affected by the presence of the dc source.

The solution to Eq. (8.40) has two components: the transient response v t (t) and the steady-state response v ss (t); that is,

v (t) ⫽ v t (t) ⫹ v ss (t) (8.41)

The transient response v t (t) is the component of the total response that dies out with time. The form of the transient response is the same as the form of the solution obtained in Section 8.3 for the source-free circuit, given by Eqs. (8.14), (8.21), and (8.26). Therefore, the transient response

v t (t) for the overdamped, underdamped, and critically damped cases are:

v (t) ⫽ A e s 1 t

t 2 1 ⫹A s t

2 e (Overdamped)

(8.42a)

v t (t) ⫽ (A 1 ⫹A 2 t )e ⫺at (Critically damped)

(Underdamped) (8.42c)

The steady-state response is the final value of v (t) . In the circuit in Fig. 8.18, the final value of the capacitor voltage is the same as the

source voltage . Hence, V s

v ss (t) ⫽ v(⬁) ⫽ V s

Thus, the complete solutions for the overdamped, underdamped, and critically damped cases are:

v (t) ⫽ V s ⫹A

1 e s 1 t ⫹A 2 e t (Overdamped)

(Critically damped)

(Underdamped) (8.44c)

The values of the constants A 1 and A 2 are obtained from the initial con- ditions: v (0) and dv (0) 兾dt. Keep in mind that v and i are, respectively, the voltage across the capacitor and the current through the inductor. Therefore, Eq. (8.44) only applies for finding v. But once the capaci-

tor voltage v C ⫽v is known, we can determine i⫽C dv 兾dt, which is the same current through the capacitor, inductor, and resistor. Hence, the voltage across the resistor is v R ⫽ iR , while the inductor voltage is v L ⫽L di 兾dt.

Alternatively, the complete response for any variable x (t) can be found directly, because it has the general form

x (t) ⫽ x ss (t) ⫹ x t (t)

where the x ss ⫽x (⬁) is the final value and x t (t) is the transient response. The final value is found as in Section 8.2. The transient response has the same form as in Eq. (8.42), and the associated constants are deter- mined from Eq. (8.44) based on the values of x (0) and dx(0) 兾dt.

8.5 Step Response of a Series RLC Circuit

For the circuit in Fig. 8.19, find v (t) and i (t) for t 70 . Consider these

Example 8.7

cases: , R⫽ 5 ⍀, R ⫽ 4 ⍀ and R⫽ 1⍀ .

0.25 F v 1Ω

■ CASE 1 When R⫽ 5 ⍀. For t 6 0, the switch is closed for a

long time. The capacitor behaves like an open circuit while the

inductor acts like a short circuit. The initial current through the Figure 8.19

inductor is For Example 8.7.

and the initial voltage across the capacitor is the same as the voltage across the 1-⍀ resistor; that is,

v (0) ⫽ 1i(0) ⫽ 4 V

For t 7 0, the switch is opened, so that we have the 1-⍀ resistor disconnected. What remains is the series RLC circuit with the voltage source. The characteristic roots are determined as follows:

Since a 7␻ 0 , we have the overdamped natural response. The total response is therefore

where v ss is the steady-state response. It is the final value of the

capacitor voltage. In Fig. 8.19, v f ⫽

24 V. Thus,

(t) ⫽ 24 ⫹ (A ⫺ e ⫺t ⫹A 4t

We now need to find A 1 and A 2 using the initial conditions. v (0) ⫽ 4 ⫽ 24 ⫹ A 1 ⫹A 2

or

⫺ 20 ⫽ A 1 ⫹A 2 (8.7.2)

The current through the inductor cannot change abruptly and is the

same current through the capacitor at ⫹ t⫽ 0 because the inductor and

capacitor are now in series. Hence, dv (0)

dv (0)

i (0) ⫽ C ⫽ 4 ⫽ ⫽

1 ⫽ dt 16 C 0.25

dt

Before we use this condition, we need to take the derivative of v in Eq. (8.7.1).

16 ⫽ ⫺A 1 ⫺ 4A 2 (8.7.4)

Chapter 8

Second-Order Circuits

From Eqs. (8.7.2) and (8.7.4), A 1 ⫽⫺ 64 兾3 and A 2 ⫽ 4 兾3. Substituting

A 1 and A 2 in Eq. (8.7.1), we get

(t) ⫽ 24 ⫹ (⫺16e ⫺t

⫹e ⫺ 4t )V

Since the inductor and capacitor are in series for t 7 0, the inductor current is the same as the capacitor current. Hence,

dv

i (t) ⫽ C dt

Multiplying Eq. (8.7.3) by C⫽ 0.25 and substituting the values of A 1

and gives A 2

i (t) ⫽ (4e ⫺t ⫺e 4t )A

3 Note that i (0) ⫽ 4 A, as expected.

■ CASE 2 When R⫽ 4 ⍀. Again, the initial current through the inductor is

4⫹1 and the initial capacitor voltage is

v (0) ⫽ 1i(0) ⫽ 4.8 V

For the characteristic roots,

a⫽

2L 2⫻1

while ␻ 0 ⫽ 2 remains the same. In this case, s 1 ⫽s 2 ⫽ ⫺a ⫽ ⫺ 2, and we have the critically damped natural response. The total response is therefore

and, as before v ss ⫽

To find A 1 and A 2 , we use the initial conditions. We write

v (0) ⫽ 4.8 ⫽ 24 ⫹ A 1 1 A 1 ⫽⫺ 19.2 (8.7.8)

Since or i (0) ⫽ C dv(0) 兾dt ⫽ 4.8 dv (0) ⫽ 4.8 ⫽

dt

From Eq. (8.7.7),

dv ⫽

(⫺2A ⫺ 2t

19.2 ⫽ ⫺2A 1 ⫹A 2 (8.7.10)

8.5 Step Response of a Series RLC Circuit

From Eqs. (8.7.8) and (8.7.10), A 1 ⫽⫺ 19.2 and A 2 ⫽⫺ 19.2 . Thus,

Eq. (8.7.7) becomes v

(t) ⫽ 24 ⫺ 19.2(1 ⫹ t)e ⫺ 2t V (8.7.11)

The inductor current is the same as the capacitor current; that is,

dv

i (t) ⫽ C dt

Multiplying Eq. (8.7.9) by C⫽ 0.25 and substituting the values of A 1

and gives A 2

(t) ⫽ (4.8 ⫹ 9.6t)e ⫺ i 2t A (8.7.12)

Note that i (0) ⫽ 4.8 A, as expected.

■ CASE 3 When R⫽ 1⍀ . The initial inductor current is

and the initial voltage across the capacitor is the same as the voltage across the 1-⍀ resistor,

v (0) ⫽ 1i(0) ⫽ 12 V

a⫽

2L 2⫻1 Since a⫽ 0.5 6␻ 0 ⫽ 2 , we have the underdamped response

s 1,2 ⫽ ⫺a ⫾ 2a 2 ⫺␻ 2 0 ⫽⫺

0.5 ⫾ j1.936

The total response is therefore v

(t) ⫽ 24 ⫹ (A ⫺ 0.5t

1 cos 1.936t ⫹ A 2 sin 1.936t)e

We now determine A 1 and A 2 . We write

v (0) ⫽ 12 ⫽ 24 ⫹ A 1 1 A 1 ⫽⫺ 12 (8.7.14)

Since i (0) ⫽ C dv(0) 兾dt ⫽ 12,

⫽e ⫺ 0.5t (⫺1.936A

1 sin 1.936t ⫹ 1.936 A 2 cos 1.936t)

dt

⫺ 0.5e ⫺ 0.5t (A

1 cos 1.936t ⫹ A 2 sin 1.936t)

At t⫽ 0, dv (0) ⫽

48 ⫽ (⫺0 ⫹ 1.936 A 2 ) ⫺ 0.5(A 1 ⫹ 0)

dt

Substituting A 1 ⫽⫺ 12 gives A 2 ⫽ 21.694, and Eq. (8.7.13) becomes

(t) ⫽ 24 ⫹ (21.694 sin 1.936t ⫺ 12 cos 1.936t)e ⫺ 0.5t V (8.7.17)

The inductor current is

dv

i (t) ⫽ C

Chapter 8

Second-Order Circuits

Multiplying Eq. (8.7.16) by C⫽ 0.25 and substituting the values of A 1

and gives A 2

i (t) ⫽ (3.1 sin 1.936t ⫹ 12 cos 1.936t)e ⫺ 0.5t A (8.7.18)

Note that i (0) ⫽ 12 A, as expected.

Figure 8.20 plots the responses for the three cases. From this figure, we observe that the critically damped response approaches the step input of 24 V the fastest.

v (t) V 40

Underdamped 35

30 Critically damped

For Example 8.7: response for three degrees of damping.

Practice Problem 8.7

Having been in position a for a long time, the switch in Fig. 8.21 is moved to position b at Find t⫽ 0. v (t) and for v R (t) t 7 0.

For Practice Prob. 8.7.

Answer: ⫺ 15 ⫺ (1.7321 sin 3.464t ⫹ 3 cos 3.464t)e 2t V, 3.464e ⫺ 2t sin 3.464t V.