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)

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

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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

d

s j j

s j

α ω

α ω

= − + = − +

= − −

60 60

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

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

−

= − −

× −

= − −

× ×

= − + =

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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

d

T

π

=

ω

1 Hz

d

d

f = T

2 79 ms

d 80

T ₌ π ₌

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

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

2 7 6 48 ^t

d i di

i e

dt dt

+ + = −

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.

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

Forced Response of an RLC Circuit

 Again, consider the circuit of Example 9.7-1 and 9.7-2 (Figure 9.7-1) when the differential equation is

The characteristic equation of the current is or

Thus, the natural response is

2

2 7 6 6 _S

d i di

i i dt + dt + =

2 7 6 0

( 1)( 6) 0

s s

s s

+ + =

+ + =

6

n 1 2

t t

i = A e⁻ + A e⁻

(9.7-9)

(9.7-10)

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Forced Response of an RLC Circuit

 Consider the special case where .

Then we at first expect the forced response to be

However, the forced response and one component of the natural response would then both have the form .

Let’s try substituting Eq. 9.7-11 into the differential equation (9.7-9). We then obtain

or

which is an impossible solution. Therefore, we need another form of the forced response when one of the natural response terms has the same form as the forcing function.

6

S 3 ^t

i = e⁻

6 f

i = Be

⁻ t

De

⁻6t

(9.7-11)

6 6 6 6

6

36 42 6 18

0 18

t t t t

t

Be Be Be e

e

− − − −

−

− + ≠

=

Forced Response of an RLC Circuit

 Let us try the forced response

 Then, substituting Eq. 9.7-12 into Eq. 9.7-9, we have

where . We have

In general, if the forcing function is of the same form as one of the components of the natural response, x_n1, we will use

where the integer p is selected so that the x_f is not duplicated in the natural response. Use the lowest power, p, of t that is not duplicated in the natural response.

( 6 6 36 ) 7 ( 6 ) 6 18 B − −g g + tg + B g − tg + Btg = g

6 f

i = Bte

⁻ t

( ) 6^t

g = g t =e⁻

6 f

18 18

and

5 5

B = − i = − te⁻ t

f n1

x = t x

p

(9.7-12)

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문서에서 Chapter 9 (페이지 40-56)