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Chapter 8

Second-Order Circuit

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A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements.

What is second-order circuit?

Typical examples of second-order circuits: a) series RLC circuit, b) parallel RLC circuit, c) RL circuit, d) RC circuit

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1. The Series RLC Circuit

2. The Parallel RLC Circuit

3. Second-Order Circuit Complete Response

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1. The Series RLC CircuitFORMULATING SERIES RLC CIRCUIT EQUATIONS

Eq.(7-33)

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The initial conditions

To solve second-order equation, there must be two initial values.

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ZERO-INPUT RESPONSE OF THE SERIES RLC CIRCUIT

With VT=0(zero-input) Eq.(7-33) becomes

Eq.(3-37)

try a solution of the form

then

Eq.(7-39)

characteristic equation

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In general, a quadratic characteristic equation has two roots:

Eq.(7-40)

three distinct possibilities:

Case A: If                               , there are two real, unequal roots

                                          Case B: If                               , there are two real, equal roots

                           Case C: If                               , there are two complex conjugate

roots                                                          

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A source-free series RLC circuit

Special case: Vc(0)=V0, IL(0)=0

V(t)V0

I(t)

tM

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tM>t>0 t > tM

What happens when R=0?

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Second Order Circuit with no Forcing Function

vc(0) = Vo , iL(0) = Io.

I. OVER DAMPED:

R=2 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A

iL(t) = -0.7 e -0.354t +2.7 e -5.646t A

vc(t) = 1.318 e -0.354t -0.318 e -5.646t V

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II. CRITICALLY DAMPED:

R=0.943 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A

iL(t) = 2e -1.414t -5.83t e -1.414t A

vc(t) = e -1.414t+ 2.75 t e -1.414t V

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III. UNDER DAMPED:

R=0.5 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A

iL(t) =4.25 e -0.75t cos(1.2t + 1.081) A

vc(t) = 2 e -0.75t cos(1.2t - 1.047) V

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IV. UNDAMPED:

R=0 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A

iL(t) =2.915 cos(1.414t + 0.815) A

vc(t) =1.374 cos(1.414t - 0.756) V

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EXAMPLE 7-14A series RLC circuit has C=0.25uF and L=1H. Find the roots of the characteristic equation for RT=8.5kohm, 4kohm and 1kohm

SOLUTION: For RT=8.5kohm, the characteristic equation is                                                         whose roots are

                                                               These roots illustrate case A. The quantity under the radical is positive, and there are two real, unequal roots at S1=-500 and S2=-8000.

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For RT=4kohm, the characteristic equation is

                                            whose roots are                                                                This is an example of case B. The quantity under the radical is zero, and there are two real, equal roots at S1=S2=-2000.For RT=1kohm the characteristic equation is

                                               whose roots are

                                   The quantity under the radical is negative, illustrating case C.                                              In case C the two roots are complex conjugates.

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In case A the two roots are real and unequal                                     and the zero-input response is the sum of two exponentials of the form

Eq.(7-48a)

In case B the two roots are real and equal                       and the zero-input response is the sum of an exponential and a damped ramp.

Eq.(7-48b)tt

C teKeKtV 21)(In case C the two roots are complex conjugates                                            and the zero-input response is the sum of a damped cosine and a damped sine.

)()( 21 dtSinKdtCosKetV tC Eq.(7-48c)

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EXAMPLE 7-15The circuit of Figure 7-31 has C=0.25uF and L=1H. The switch has been open for a long time and is closed at t=0. Find the capacitor voltage for t  0 for (a) R=8.5k ohm, (b) R=4k ohm, and (c) R=1k ohm. The initial conditions are Io=0 and

Vo=15V. SOLUTION:

Fig. 7-31

•(a) In Example 7-14 the value R=8.5kohm yields case A with roots S1=-500 and S2=-8000. The corres

ponding zero-input solution takes the form in Eq.(7-48a).

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The initial conditions yield two equations in the constants K1 and K2:

Solving these equations yields K1=16 and K2 =-1, so that the

zero-input response is

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The initial conditions yield two equations in the constants K1 and K2:

Solving these equations yields K1=15 and K2= 2000 x 15,

so the zero-input response is

•(b) In Example 7-14 the value R=4kohm yields case B with roots S1=S2=-2000. The corresponding zero-input response takes

the form in Eq.(7-48b):

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•c) In Example 7-14 the value R=1k ohm yields case C with roots                                  . The corresponding zero-input response takes the form in Eq.(7-48c):

The initial conditions yield two equations in the constants K1

and K2:

Solving these equations yields K1=15 and K2=( ) , so the

zero-input response is

)()( 21 dtSinKdtCosKetV tC

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Fig. 7-32

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In general, a quadratic characteristic equation has two roots:

Eq.(7-40)

three distinct possibilities:

Case A: If                               , there are two real, unequal roots

                                          Case B: If                               , there are two real, equal roots

                           Case C: If                               , there are two complex conjugate

roots                                                          

Overdamped situation

Ciritically damped situation

Underdamped situation

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2. The Parallel RLC Circuit

FORMULATING PARALLEL RLC CIRCUIT EQUATIONS

Eq. 7-55

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Equation(7-55) is second-order linear differential equation of the same form as the series RLC circuit equation in Eq.(7-33). In fact, if we interchange the following quantities:

we change one equation into the other. The two circuits are duals, which means that the results developed for the series case apply to the parallel circuit with the preceding duality interchanges.

The initial conditions

iL(0)=Io and

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set iN=0 in Eq.(7-55) and obtain a homogeneous equation in the i

nductor current:

A trial solution of the form IL=Kest leads to the characteristic eq

uation

Eq. 7-56

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There are three distinct cases:

Case A: If (GNL)2-4LC>0, there are two unequal real roots a

nd the zero-input response is the overdamped form

Case B: (GNL)2-4LC=0, there are two real equal roots and

the zero-input response is the critically damped form

Case C:(GNL)2-4LC<0, there are two complex, conjugate r

oots and the zero-input response is the underdamped form

0)()(

,

21

21

ttSinKtCosKeti

jsst

L

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EXAMPLE 7-16

In a parallel RLC circuit RT=1/GN=500ohm, C=1uF, L=0.2H. T

he initial conditions are Io=50 mA and Vo=0. Find the zero-input

response of inductor current, resistor current, and capacitor voltage

SOLUTION: From Eq.(7-56) the circuit characteristic equation is

                                                                                           

The roots of the characteristic equation are

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Evaluating this expression at t=0 yields

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EXAMPLE 7-17

The switch in Figure 7-34 has been open for a long time and is closed at t=0

(a) Find the initial conditions at t=0 (b) Find the inductor current for t0 (c) Find the capacitor voltage and current through the switch for t  0

Fig. 7-34

SOLUTION:

(a) For t<0 the circuit is in the dc steady state                   

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(b) For t 0 the circuit is a zero-input parallel RLC circuit with initial conditions found in (a). The circuit characteristic equation is

The roots of this equation are

The circuit is overdamped (case A), The general form of the inductor current zero-input response is

using the initial conditions

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The initial capacitor voltage establishes an initial condition on the derivative of the inductor current since

The derivative of the inductor response at t=0 is

The initial conditions on inductor current and capacitor voltage produce two equations in the unknown constants K1 and K2:

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Solving these equations yields K1=30.3 mA and K2=-0.309 ma

The zero-input response of the inductor current is

(c) Given the inductor current in (b), the capacitor voltage is

For t 0 the current isw(t) is the current through the 50 ohm re

sistor plus the current through the 250 ohm resistor

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3. Second-order Circuit Complete Response

The general second-order linear differential equation with a step function input has the form Eq. 7-60  The complete response can be found by partitioning y(t) into forced and natural components:

Eq. 7-61

yN(t) --- general solution of the homogeneous equation (input set

to zero), yF(t) is a particular solution of the equation

                                                                    ∴ yF=A/ao

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Combining the forced and natural responses

Eq. 7-67

EXAMPLE 7-18 The series RLC circuit in Figure 7-35 is driven by a step function and is in the zero state at t=0. Find the capacitor voltage for t 0.

Fig. 7-35

SOLUTION:

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By inspection, the forced response is vCF=10V. In standard for

mat the homogeneous equation is

the natural response is underdamped (case C)

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The constants K1 and K2 are determined by the initial conditions.

These equations yield K1= -10 and K2= -2.58. The complete

response of the capacitor voltage step response is

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General second-order circuit

Steps:

1. Set a second-order differential equation

2. Find the natural response yN(t) from the homogeneous

equation (input set to zero)

3. Find a particular solution yF(t) of the equation

4. Determine K1 and K2 by the initial conditions

5. Yield the required response

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Summary•Circuits containing linear resistors and the equivalent of two energy storage elements are described by second-order differential equations in which the dependent variable is one of the state variables. The initial conditions are the values of the two state variables at t=0.

•The zero-input response of a second-order circuit takes different forms depending on the roots of the characteristic equation. Unequal real roots produce the overdamped response, equal real roots produce the critically damped response, and complex conjugate roots produce underdamped responses.

•Computer-aided circuit analysis programs can generate numerical solutions for circuit transient responses. Some knowledge of analytical methods and an estimate of the general form of the expected response are necessary to use these analysis tools.