电路chapter6

--- damped radian frequency
v(t ) A1e A1e
( j d ) t t
A2 e
( j d ) t
e
jd t
A2 e
t
A2 e
jd t
e j cos j sin
---The Euler identity
t ) 14e 5000 t 26e 20000 t (V), t 0
v(t ) 14e 5000 t 26e 20000 t (V), t 0
v(t ) i R (t ) 70e 5000 t 130e 20000 t (mA), t 0 R
Hence
v ( t ) e t ( A1 cos d t jA1 sin d t A2 cos d t jA2 sin d t ) e t [( A1 A2 ) cos d t j( A1 A2 ) sind t ]
Assume
Therefore v ( t ) B1e t cos d t B2 e t sin d t
0
6.2 the forms of the natural response of a parallel RLC circuit
Completing the description of the natural response requires finding two unknown coefficients, such as A1, A2. the method used to find two unknown coefficients is based on matching the solution for the natural response to the initial conditions imposed by the circuit, which are the initial value of the current (or voltage) and the initial value of the first derivative of the current (or voltage).
s 1 Ae ( s ) 0 Hence: s 2 s 1 0 RC LC RC LC
s 1 s 0 RC LC
2
Therefore, characteristic roots (特征根) are
1 1 2 1 s1 ( ) 2 RC 2 RC LC

I
t0
ic
C
iL vo L
iR
R
v

Form KCL, we have
iL iR i C I
d i dv L 2L dt dt
2
d 2iL dt 2

i 1 diL I L RC dt LC LC
The indirect approach We can solve for iL indirectly by fist finding the voltage v.
t
sind t
v0
0
t
3. The critically damped voltage response
1 Hence s1 s2 2 RC In the case of a repeated root, the solution is
v(t ) D1tet D2e t
1 dv v d 2v c 0 R dt L dt
d 2v 1 dv v 0 dt RC dt LC
The general solution of the second-order differential equation
d 2v 1 dv v 0 dt RC dt LC
2. The underdamped voltage response
(oscillatory discharge process)
2 if 0 2 , then
2 s1 j 0 2 jd
In which
2 d 0 2
2 s2 j 0 2 jd
v( t )
2 0 2
According to the initial values
v (0 ) v 0 D2
dv(0 ) ic (0 ) D1 D2 dt C
O
t
6.3 the step response of a parallel RLC circuit
v(t ) A1e s1t A2e s2t v(t ) B1e t cosd t B2e t sind t v(t ) D1te t D2e t
e s1t A2 e s2 t i L (t ) I A1 e t cosd t B2 e t sind t i L (t ) I B1 te t D2 e t i L (t ) I D1

iL ic vo 0.2 F
Io
iR 200 v

50mH
Solution:
1 1 2 1 s1 ( ) 2 RC 2 RC LC 1.25 104 1.5625 108 108 12500 7500 500rad / s
s2 12500 7500 20000 rad / s
1 1 2 1 s2 ( ) 2 RC 2 RC LC
Hence, the natural response of the parallel RLC circuit is of the form
v A1e s1t A2e s2t
Assume
1 2 RC 1 0 LC
B1 A1 A2 B2 j ( A1 A2 )
Ae t sin( d t )
Initial conditions
v(0 ) B1 dv(0 ) ic (0 ) B1 d B2 dt C
v(t ) B1e
v(t)
t
cosd t B2e
According to the initial conditions, we get
ic ( 0 ) i L ( 0 ) i R ( 0 ) 12 3 30 10 90mA 200
A1 A2 vc (0 )
dv(0 ) s1 A1 s 2 A2 dt i c (0 ) 90 10 3 C 0.2 10 6 -450 103 (kV/s)
Assume the solution is of exponential form, that is , to assume that the voltage is of the st form v Ae
We get
st 2
As st A st As e e e 0 RC LC
2 st
Processes of finding the overdamped response v(t):
1. Find the roots of the characteristics equation, s1 and s2 , using the values of R, L and C.
2. Find v(0+) and d v(0+) /dt using the circuit analysis. 3. Finding the values of A1 and A2 by solving the equations bellow simultaneously. v(0 ) A1 A2
6.1 introduction to the natural response of a parallel RLC circuit
Applying KCL :
v 1 dv vd I 0 c 0 R L 0 dt

ic
C

t
iL vo L
Io R
iR
v

Second-order circuits Differentiate the equation once with respect to t, we get
iL iR i C I
1 t v dv vd C I L 0 R dt

Differentiating the equation once with respect to t. thus
d 2v 1 dv v 0 2 RC dt LC dt
Caution: because there is a source in the circuit for t>0, you must take into account the value of the source current at t=0+ when you evaluate the coefficients in equations. Therefore, the three solutions for iL will be
The direct approach
neper frequency（奈培频率 ） Resonant radian frequency （角频率）
2 s1 2 0
2 s2 2 0
Three possible outcomes
2 0 2 Overdamped（过阻尼） 2 0 2 Underdamped（欠阻尼） 2 2 critically damped（临界阻尼）
dv( t ) ic ( t ) C dt 0.2 10 6 (5000 14 e 5000 t 26 20000 e 20000 t ) 14e 5000 t 104e 20000 t (mA), t 0
i L ( t ) i R ( t ) iC ( t ) t 56e 5000 t 26e 20000 t mA , t 0
1. The overdamped voltage response
( non-oscillatory discharge process)
s1,2
2 0
2
2 0

ic
C

2
iL vo L
Io R
iR
v

The roots of the characteristics equation are real and distinct. 1
v A1e
s1t
A2e
s2 t
v(0 ) A1 A2
dv(0 ) s1 A1 s2 A2 dt
d ic (0 )dt ic (0 ) dv(0 ) c dt dt c

v 0 i c (0 ) I0 R
dv(0 ) ic (0 ) s1 A1 s2 A2 dt c
dv(0 ) ic (0 ) s1 A1 s2 A2 dt c
4. Find the expression of v(t) for t≥0.
v(t ) U0
uc
t
t
Example: for the circuit as shown in the figure, v(0+)=12V, and iL(0+)=30mA. (a) Find the expression for v(t). (b)Derive the expressions of iR(t), iL(t), ic(t).