Chapter 9

(1)

Chapter 9

The Complete Response of Circuits with Two Energy Storage Elements

Seoul National University

School of Electrical Engineering and Computer Science

Prof. SungJune Kim

(2)

School of Electrical Engineering and Computer Science, SNU

Index

 Parallel RLC circuit

 Direct Method

 Operator Method using differential operator s

 Solution: Natural response using characteristic equation

 Undamped, underdamped,critically damped, overdamped cases.

 Solution: Forced response

 Complete response

 State variable approach

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Second-Order Circuit

 A second-order circuit is a circuit that is represented by a second-order differential equation.

x(t): output of the circuit (=response of the circuit) f(t) : input to the circuit

α : damping coefficient ω₀: resonant frequency

) ( )

( )

( 2

)

( ₀²

2 2

t f t

x t

dt x t d

dt x

d +

α

+

ω

=

(4)

Second-Order Circuit

 To find the response of the second-order circuit,

 Represent the circuit by a second-order differential equation.

 Find the general solution of the homogeneous differential equation. This solution is the natural response, x_n(t). The natural response will contain two unknown constants that will be evaluated later.

 Find a particular solution of the differential equation. This solution is the forced response, x_f(t).

 Represent the response of the second-order circuit as x(t)=x_n(t) + x_f(t).

 Use the initial conditions, for example, the initial values of the currents in

inductors and the voltage across capacitors, to evaluate the unknown constants.

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Direct Method

 Let us consider the circuit shown in Figure 9.2-1.

Writing the nodal equation at the top node, we have

We write the equation for the inductor as

Then

This method of obtaining the second-order differential equation may be called the direct method and is

summarized in Table 9.2-1.

S

v dv

i C i

R + + dt =

v L di

= dt

2

2 S

L di d i

i CL i

R dt + + dt =

(6)

Direct Method

 Table 9.2-1

The Direct Method for Obtaining the Second-Order Differential Equation of a Circuit

Step1 Identify the first and second variables, x₁ and x₂. These variables are capacitor voltages and/or inductor currents.

Step2 Write one first-order differential equation, obtaining

Step3 Obtain an additional first-order differential equation in terms of the second variable so that

Step4 Substitute the equation of step3 into the equation of step2, thus obtaining a second-order differential equation in terms of x₂

) , ( ₁ ₂

1 f x x

dx =dt

dt dx x K

dt Kx

dx ₂

1 1

2 1

or =

=

(7)

Direct Method-example

 Let us consider the circuit shown in Figure 9.2-2.

Writing KVL around the loop, we have

We write the equation for the capacitor as

Then

This method is the direct method.

S

L di v Ri v dt + + =

i C dv

= dt

2 2

1 v

_S

d v R dv

dt + L dt + LC v = LC

v

S

di v R

dt + + L L i = L

or

(8)

Operator Method: using differential operator s

 Another method of obtaining the second-order equation describing a circuit is called the operator method.

 Consider the circuit shown in Figure 9.2-3 The mesh equations are

and

Now let us use Then we have

and

1

1 di (1 2) _S

L R i i v

dt + − =

2

2 1 2

( ) di 0

R i i L

− + dt =

1 2

1 , 1 H, and 2 H

R = Ω L = L =

1

1 2 S

di i i v dt + − =

2

2 1 2di 0

i i

− + dt =

(9)

We may rearrange these equations as

The differential operator s, where s=d/dt, is used to transform differential equations into algebraic equations.

We may use Cramer’s rule to solve for i₂, obtaining

Therefore

and the differential equation is

1

1 2 S

di i i v

dt + − = ²

1 2 2 di 0

i i

− + + dt = and

1 2

(s+1)i − =i v_S and − +i₁ (2s +1)i₂ = 0

2

1

( 1)(2 1) 1 vS

i = s s

+ + −

2

(2s +3 )s i2 = v_S

2

2 2

2 d i2 3di _S dt + dt =v

Operator Method: Form 2

^nd

order algebraic equation using

differential operator s, then back to the differential eq.

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see Appendix A.4 “ Cramer’s rule” (page 830)

 Cramer’s rule

 A set of simultaneous equations

can be written in matrix form as

 Cramer’s rule states that the solution for the unknown, x_k, is

where Δis the determinant of A and Δ_k is Δ with the kth column replaced by the column vector b.

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

n n

n n nn n n

a x a x a x b

+ + + =



   



Ax =b

k

xk = ∆

∆

(11)

Operator Method

 Table 9.2-2

Operator Method for obtaining the second-order differential equation of a circuit Step1 Identify the variable x₁ for which the solution is desired

Step2 Write one differential equation in terms of the desired variable x₁ and a second variable x₂

Step3 Obtain an additional equation in terms of the second variable and the first variable

Step4 Use the operator s=d/dt and 1/s=∫dt to obtain two algebraic equations in terms of s and the two variables x₁ and x₂.

Step5 Using Cramer’s rule, solve for the desired variable so that

x₁=f(s,source)=P(s)/Q(s), where P(s) and Q(s) are polynomials in s Step6 Rearrange the equation of step 5 so that Q(s)x₁=P(s)

Step7 Convert the operators back to derivatives for the equation of step 6 to obtain the second-order differential equation

(12)

Operator Method: Example 9.2-1

Representing a Circuit by a Differential Equation

 Find the differential equation for the current i₂ for the circuit of Figure 9.2-4

(13)

 Write the two mesh equations using KVL to obtain

Using the operator s=d/dt, we have

and

Using Cramer’s rule to solve for i₂, we obtain

Rearranging Eq. 9.2-16, we obtain

Therefore, the differential equation for i₂ is

Solution

vs

dt di dt

i₁ + di¹ − ² = 2

0 )

2 3 ( ) 2 (

2 1

= +

+

−

=

− +

i s si

v si i

s _s

0 2

3 ₂ ²

1 + + =

− dt

i di dt

di

(9.2-16)

6 7 )

2 3 )(

2

( ² ²

2 = + +

− +

= +

s s

sv s

s s

i sv^s ^s

svs

i s

s² +7 +6) ₂ = (

dt i dv dt

di dt

i

d²₂² +7 ² +6 ₂ = _s

(14)

Operator Method Example 9.2-2

Representing a Circuit by a Differential Equation

 Find the differential equation for the voltage v for the circuit of Figure 9.2-5

(15)

 The KCL node equation at the upper node

Write the equation for the current through the branch containing the inductor as

Using the operator s=d/dt, we have the two equations

Substituting the parameter values and rearranging, we have

Using Cramer’s rule, solve for v to obtain Therefore,

Solution

0

1

= +

− +

dt C dv R i

v v _s

0

1 1

= +

+

−

= + +

Lsi Ri

v

R i v R Csv

v _s

dt v L di

Ri+ =

0 ) 1 10

(

10 )

10 10

(

3

3 3

3

= + +

−

= + +

−

i s v

v i

v

s _s

3 2

6 1001 1001 10

) 1000 (

10 ) 1000 )(

1 (

) 1000 (

× +

+

= + +

+ +

= +

s s

v s

s s

v

v s ^s ^s

vs

s v s

s 1001 1001 10 ) ( 1000)

( ² + + × ³ = +

dv dv

v d²

(16)

Solution of the Second-Order Differential Equation – The Natural Response

 A circuit with two irreducible energy elements can be represented by a second-order differential equation of the form

where the constants a₂, a₁, a₀ are known and the forcing function f(t) is specified.

 The complete response is given by

where x_n is natural response and x_f is forced response. The natural response satisfies the unforced differential equation when f(t)=0. The forced response x_f satisfies the differential equation with the forcing function present.

2

2 d x2 1 dx 0 ( )

a a a x f t

dt + dt + =

n f

x = + x x

(17)

The Natural Response from Characteristic Equation

 The natural response of a circuit, x_n, will satisfy the equation

 Since x_n and its derivatives must satisfy the equation, we postulate the exponential solution

where A and s are to be determined.

 Substituting Eq. 9.3-3 in Eq. 9.3-2, we have

Since , we may rewrite Eq. 9.3-4 as

 Since we do not accept the trivial solution, it is required that

This equation is called a characteristic equation.

2

2 ^st 1 ^st 0 ^st 0

a As e +a Ase +a Ae =

2

2 d x2 1 dx 0 0

a a a x

dt + dt + =

x = Aest

(

^{a s}² ² ⁺^{a s a x}¹ ⁺ ⁰

)

⁼ ⁰

2

2 1 0 0

a s +a s a+ =

(9.3-3) (9.3-2)

(9.3-4)

(9.3-5)

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The characteristic equation is derived from the governing differential equation for a circuit by setting all independent source to zero value and assuming an exponential solution.

 The solution of Eq. 9.3-5 has two roots, s₁ and s₂, where

 When there are two distinct roots, there are two solutions such that where A1 and A2 are unknown constants that will be evaluated later.

The roots of the characteristic equation contain all the information necessary for determining the character of the natural response.

2 2

1 1 2 0 1 1 2 0

1 2

2 2

4 4

and

2 2

a a a a a a a a

s s

a a

− + − − − −

= =

1 2

n 1 2

s t s t

x = A e + A e

Characteristic Equation

(19)

Example 9.3-1

Natural Response of a Second-Order Circuit

 Find the natural response of the circuit current i₂ shown in Figure 9.3-2. Use operators to formulate the differential equation and obtain the response in terms of two arbitrary constants.

(20)

 Writing the two mesh equations

Using the operator s=d/dt, we obtain

Using Cramer’s rule, solve for i₂

Therefore,

Solution

0 1

4 4

4 2

12

2 2

1

2 1

1

= +

+

−

=

− +

dt i di

i

v dt i

i di _s

16 10

2 32

20 2

4 16

) 4 )(

2 12 (

4

2

2 2 = + +

+

= +

− +

= +

s s

v s

s v s

s

i v^s ^s ^s

0 )

4 ( 4

4 )

2 12 (

2 1

= +

+

−

=

− +

i s i

v i

i

s _s

vs

i s

s 10 16) 2

( ² + + ₂ =

(21)

 Note that is the characteristic equation. Thus, the roots of the characteristic equation are s₁=-2 and s₂=-8. Therefore, the natural response is

where x=i₂. The roots s₁ and s₂ are the characteristic roots and are often called the natural frequencies. The reciprocals of the magnitude of the real characteristic roots are the time constants. The time constants of this circuit are 1/2s and 1/8s.

Solution

t t

n Ae A e

x = ₁ ⁻² + ₂ ⁻⁸

0 ) 16 10

(s² + s+ =

(22)

Natural Response of the Unforced Parallel RLC Circuit

 Consider the circuit shown in Figure 9.4-1 Write the KCL at the node to obtain

Taking the derivative, we have

Using the parameter s, we obtain the characteristic equation

The two roots of the characteristic equation are

1/2 1/2

2 2

1 2

1 1 1 1 1 1

and

2 2 2 2

s s

RC RC LC RC RC LC

     

= − +  −  = − −  − 

2 1 1

0

s s

RC LC

+ + =

2 2

1 1

d v dv 0

C v

dt + R dt + L =

0

1 ^t (0) 0

v dv

vd i C

R + L

∫

τ + + dt =

(9.4-2) (9.4-1)

(23)

Natural Response of the Unforced Parallel RLC Circuit

 When s₁ is not equal to s₂, the solution to the second-order differential Eq. 9.4-2 for t>0 is

 The roots of the characteristic equation may be rewritten as

Where and

 The roots of the characteristic equation assume three possible conditions:

1. Two real and distinct roots when  overdamped

2. Two real equal roots when  critically damped

3. Two complex roots when  underdamped

2 2 2 2

1 0 and 2 0

s = − +α α ω− s = − −α α ω−

1 2

n 1 2

s t s t

v = A e + A e

1/(2RC)

α = ²

0 1/(LC) ω =

2 2

α >ω0

2 2

α =ω0

2 2

α <ω0

(9.4-6)

(24)

Natural Response of the Unforced Parallel RLC Circuit

 Let us determine the natural response for the overdamped RLC circuit of Figure 9.4-1 when the initial conditions are v(0) and i(0) for the capacitor and the inductor, respectively.

At t=0 for Eq. 9.4-6, we have

Since A₁ and A₂ are both unknown, we need one more equation at t=0.

Rewriting Eq. 9.4-1 at t=0, we have

Since i(0) and v(0) are known, we have

n

(0)

1 2

v = + A A

(0) (0)

(0) 0

v dv

i C

R + + dt =

(0) (0) (0)

dv v i

dt = − RC − C

(9.4-9)

(9.4-10)

(25)

Natural Response of the Unforced Parallel RLC Circuit

 Thus, we now know the initial value of the derivative of v in terms of the initial conditions.

Taking the derivative of Eq. 9.4-6 and setting t=0, we obtain

Using Eqs. 9.4-10 and 9.4-11, we obtain a second equation in terms of the two constants as

Using Eqs. 9.4-9 and 9.4-12, we may obtain A₁ and A₂.

n

1 1 2 2

(0)

dv s A s A

dt = + (9.4-11)

1 1 2 2

(0) (0)

v i

s A s A

RC C

+ = − − _(9.4-12)

(26)

Example 9.4-1

Natural Response of an Overdamped Second-Order Circuit

 Find the natural response of v(t) for t>0 for the parallel RLC circuit shown in Figure 9.4-1 when R=2/3Ω, L=1H, C=1/2F, v(0)=10V, and i(0)=2A.

(27)

 Using Eq. 9.4-3, the characteristic equation is

or

Therefore, the roots of the characteristic equation are

Then the natural response is

The initial capacitor voltage is v(0)=10, so we have

or

We use Eq. 9.4-12 to obtain the second equation for the unknown constants.

Solution

0 2 3

1 0 1

2 2

= + +

= +

+ s s

s LC s RC

t t

n Ae A e

v = ₁ ⁻ + ₂ ⁻²

2 and

1 ₂

1 = − s = −

s

2 1

10 ) 0 (

A A

v_n

+

=

+

=

2 10

) 0 ( ) 0 (

2 2 1 1

−

=

−

=

−

=

+ C

i RC A v

s A s

(9.4-13)

(9.4-14)

(28)

 Solving Eqs. 9.4-14 and 9.4-15 simultaneously, we obtain A2=24 and A1=-14.

Therefore, the natural response is

The natural response of the circuit is shown in Figure 9.4-2

Solution

V ) 24

14

( ^t ^2t

n e e

v = − ⁻ + ⁻

(29)

Natural Response of the Critically Damped Unforced Parallel RLC Circuit

 We consider the parallel RLC circuit, and we will determine the

special case when the characteristic equation has two equal real roots.

 Let us assume that s₁=s₂ and proceed to find v_n(t)

Since the two roots are equal, we have only one undetermined

constant, but we still have two initial conditions to satisfy. Clearly, Eq.

9.5-1 is not the total solution.

1 2 1

n 1 2 3

s t s t s t

v = A e + A e = A e _(9.5-1)

(30)

recap:Natural Response of the Unforced Parallel RLC Circuit

Where and

2 2 2 2

1 0 and 2 0

s = − +α α ω− s = − −α α ω−

1 2

n 1 2

s t s t

v = A e + A e

1/(2RC)

α = ²

0 1/(LC) ω =

2 2

α >ω0

2 2

α =ω0

2 2

α <ω0

(9.4-6)

(31)

Natural Response of the Critically Damped Unforced Parallel RLC Circuit

 We try the solution

 Let us consider a parallel RLC circuit where The characteristic equation for the circuit is

or

The two roots are then

Using Eq. 9.5-2 for the natural response, we have

1

n ^{s t}

(

1 2

)

v = e A t + A

(9.5-2)

1 H, 1 , 1/ 4 F, (0) 5 V, and (0) 6 A

L = R = Ω C = v = i = −

2 1 1

0

s s

RC LC

+ + =

2 4 4 0

s + + =s

1 2

2 s = = − s

2

n ^t( 1 2)

v =e⁻ A t + A (9.5-3)

(32)

Natural Response of the Critically Damped Unforced Parallel RLC Circuit

 Since v_n(0)=5, we have at t=0 Differentiate Eq. 9.5-3 to obtain

Evaluating Eq. 9.5-4 at t=0, we have

Again, we may use Eq. 9.4-10 so that

or

Therefore,

2

5 A =

2 2 2

1 1 2

2 ^t ^t 2 ^t

dv A te A e A e

dt

− − −

= − + −

(9.5-4)

1 2

(0) 2

dv A A

dt = −

(0) (0) (0)

dv v i

dt = − RC − C

1 2

5 6

2 4

1/ 4 1/ 4 A − A = − − − =

1 14

A =

(33)

Natural Response of the Critically Damped Unforced Parallel RLC Circuit

 The natural response is

 The critically damped natural response of this RLC circuit is shown in Figure 9.5-1

2

n ^t(14 5) V

v = e⁻ t +

(34)

Natural Response of an Underdamped Unforced Parallel RLC Circuit

 The characteristic equation of the parallel RLC circuit will have two complex conjugate roots when . This condition is met when or when

Recall that where

When we have where

(2 )2

LC < RC

2 2

α <ω0

4 2

L < R C

2 2

1,2 0

s = − ±α α ω−

1 2

n 1 2

s t s t

v = A e + A e

2 2

ω0 >α

2 2

1,2 0

s = − ±α j ω α−

1 j = −

(35)

recap:Natural Response of the Unforced Parallel RLC Circuit

Where and

2 2 2 2

1 0 and 2 0

s = − +α α ω− s = − −α α ω−

1 2

n 1 2

s t s t

v = A e + A e

1/(2RC)

α = ²

0 1/(LC) ω =

2 2

α >ω0

2 2

α =ω0

2 2

α <ω0

(9.4-6)

(36)

Natural Response of the Unforced Parallel RLC Circuit

Critical damping provides the quickest approach to zero amplitude for a damped

oscillator. With less damping (underdamping) it reaches the zero position more quickly, but oscillates around it. With more damping (overdamping), the approach to zero is slower.

Critical damping occurs when the damping coefficient is equal to the undamped resonant

frequency of the oscillator.

(37)

Natural Response of an Underdamped Unforced Parallel RLC Circuit

 The complex roots lead to an oscillatory-type response. We define the square root as , which we will call the damped resonant frequency. The factor , called the damping coefficient, determines how quickly the oscillator subside. Then the roots are

Therefore, the natural response is

or

1,2 d

s = − ± α j ω

d d

n 1 2

j t j t

t t

v = A e

⁻^α

e

^ω

+ A e

⁻^α

e

⁻ ^ω

2 2

ω α0 − ω_d α

(

^d ^d

)

n 1 2

j t j t

v = e⁻^αt A e ^ω + A e⁻ ^ω

(38)

using Euler identity

 Let us use the Euler identity

The natural response can be rewritten as

 We replace (A₁+A₂) and j(A₁–A₂) with new constants B₁ and B₂. A₁ and A₂ must be complex conjugates so that B₁ and B₂ are real numbers.

 The natural underdamped response is oscillatory with a decaying

magnitude. The rapidity of decay depends on , and the frequency of oscillation depends on .^ω^d

α

( )

( ) ( )

n 1 1 2 2

1 2 1 2

cos sin cos sin

cos sin

t

d d d d

t

d d

v e A t jA t A t jA t

e A A t j A A t

α

ω ω ω ω

ω ω

−

= + + −

=  + + − 

cos sin e^±j t^ω =

ω

t ± j

ω

t

( )

n ^t 1

cos

_d 2

sin

_d

v = e

⁻^α

B ω t + B ω t

(9.6-5)

(39)

Back to basics: in Appendix D “ Euler’s Formula”

 Euler’s formula

 Euler’s formula is

An alternative form of Euler’s formula is

To derive Euler’s formula, let Differentiating, we obtain

When , we have θ

θ θ

sin cos j

e^j = +

θ θ

θ cos jsin

e⁻^j = −

θ θ sin

cos j

f = +

if

j j

d j df

=

+

=

+

−

=

) sin (cos

cos sin

θ θ

θ θ θ

θ

ej

f = ^if

ddf = θ

(D-1)

(40)

Finding coefficients using initial conditions

 Let us find the general form of the solution for B₁ and B₂ in terms of the initial conditions when the circuit is unforced.

Then at t=0 we have The derivative of v_n is

at t=0 we obtain

Recall that we found earlier that Eq. 9.4-10 provides dv(0)/dt for the parallel RLC circuit as

Therefore, we use Eqs. 9.6-6 and 9.6-7

n(0) 1

v = B

( ) ( )

n

2 1 cos 1 2 sin

t

d d d d

dv e B B t B B t

dt

α ω α ω ω α ω

= −  − − + 

n

2 1

(0)

d

dv B B

dt =ω −α

(0) (0) (0)

dv v i

dt = − RC − C (9.6-7)

2 1

(0) (0)

d

v i

B B

RC C

ω =α − − _(9.6-8)

(41)

Example 9.6-1

Natural Response of an Underdamped Second-Order Circuit

 Consider the parallel RLC circuit when R=25/3Ω, L=0.1H, C=1mF, v(0)=10V, and i(0)=-0.6A. Find the natural response v_n(t) for t>0

(42)

Solution

 First, we determine and to determine the form of the response.

Consequently we obtain

Therefore, and the natural response is underdamped. The damped resonant frequency is

Hence, the characteristic roots are

and

Consequently, the natural response is obtained as

1/(2RC) 60

α = = ω₀² =1/(LC) 10= ⁴ α2 ω₀²

and

2 2

ω0 >α

(

² ²

) (

^{1/ 2} ⁴ ³

)

^{1/ 2}

d 0 10 3.6 10 80 rad/s

ω = ω α− = − × =

1

2

60 80

d

s j j

s j

α ω

= − + = − +

= − −

60 60

n( ) 1 ^t cos80 2 ^t sin 80 v t = B e⁻ t + B e⁻ t

(43)

Solution

 Since v(0)=10, we have

We can use Eq. 9.7-8 to obtain

 Therefore, the natural response is

 A sketch of this response is shown in Figure 9.6-1.

60

n( ) 10 ^t cos80 V

v t = e⁻ t

1 (0) 10 B =v =

2 1

3

(0) (0)

60 10 10 0.6

80 80 25 / 3000 80 10 7.5 15.0 7.5 0

d d d

v i

B B

RC C

α

ω ω ω

−

= − −

× −

= − −

× ×

= − + =

(44)

Natural Response of an Underdamped Unforced Parallel RLC Circuit

 The period of the oscillation is the time interval, denoted as T_d. expressed as

However, the natural response of an underdamped circuit is not a pure oscillatory response. Thus we may approximate T_d by the period

between the first and third zero crossings, as shown in Figure 9.6-1.

 The frequency in hertz is

 The period of the oscillation of the circuit of example 9.6-1 is 2 s

d

T

π

=

ω

1 Hz

d

f = T

2 79 ms

d 80

T ₌ π ₌

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Forced Response of an RLC Circuit

 The forced response of an RLC circuit described by a second-order differential equation must satisfy the differential equation and no

arbitrary constants. The response to a forcing function will often be of the same form as the forcing function.

 We consider the differential equation for the second-order circuit as

The forced response x_f must satisfy Eq. 9.7-1.

 If the forcing function is a constant, we expect the forced response also to be a constant since the derivatives of a constant are zero.

 If the forcing function is of the form , we expect

2

2 d x2 1 dx 0 ( )

a a a x f t

dt + dt + =

2

f f

2 d x2 1 dx 0 f ( )

a a a x f t

dt + dt + =

( ) ^at f t = Be⁻

f

x = De

⁻at

(9.7-1)

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Forced Response of an RLC Circuit

 If the forcing function is a sinusoidal function, we can expect the forced response to b a sinusoidal function. If , we will try

 Table 9.7-1 summarizes selected forcing functions and their associated assumed solutions.

( ) sin 0

f t = A ω t

f

sin

0

cos

0

sin(

0

)

x = M ω t + N ω t = Q ω t + θ

Forcing Function Assumed solution

K A

Kt At + B

Kt² At² + Bt + C

K sin ωt A sin ωt + B cos ωt

Ke^-at Ae^-at

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Example 9.7-1

Forced Response to an Exponential Input

 Find the forced response for the inductor current i_f for the parallel RCL circuit shown in Figure 9.7-1 when is=8e^-2tA. Let R=6Ω, L=7H, and C=1/42F.

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Solution

 The source current is applied at t=0 as indicated by the unit step function u(t).

The KCL equation at the upper node is

We wish to obtain the second-order differential equation in terms of i.

Substituting the component values and the source i_s, we obtain

S

v dv

i C i

R dt

+ + =

2

and 2

di dv d i

v L L

dt dt dt

= =

2 2

1 1 1

S

d i di

i i

dt + RC dt + LC = LC

2

2 7 6 48 ^t

d i di

i e

dt dt

+ + = −

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Solution

 We wish to obtain the forced response, so we assume that the response will be

where B is to be determined. Substituting the assumed solution into the differential equation, we have

or

Therefore, B=-12 and

2 f

i = Be

⁻ t

2 2 2 2

4Be⁻ ^t + −7( 2Be⁻ ^t) 6+ Be⁻ ^t = 48e⁻ ^t

2 2

(4 14 6)− + Be⁻ ^t = 48e⁻ ^t

2

f 12 ^t

i = − e⁻

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Example 9.7-2

Forced Response to a Constant Input

 Find the forced response i_f of the circuit of Example 9.7-1 when i_s=I₀, where I₀ is a constant.

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Solution

 Since the source is a constant applied at t=0, we expect the forced response to be a constant also.

As a 1^st method, we will use the differential equation to find the forced response.

2^nd method, we will demonstrate the alternative method that uses the steady-state behavior of the circuit to find i_f.

 The differential equation with the constant source is obtained,

Again, we assume that the forced response is i_f=D, a constant,. Since the first and second derivatives of the assumed forced response are zero. We have

or

Therefore,

2 0 2

6 6

7 i I

dt di dt

i

d + + =

6D=6I₀ D=I₀

i_f=I₀

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Solution

 Another approach is to determine the steady-state response if of the circuit of Figure 9.7-1 by drawing the steady-state circuit model.

i_f=I₀

Fig 9.7-1 Fig 9.7-2