Добавил:
ivanov666
Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз:
Предмет:
Файл:Theoretical fundamentals of electrical engineering. P.II. Transients in linear electrical circuits with lamped parameters. A collection of advanced pr
.pdf
61
3.5..Determining the degree of the characteristic polynomial.
Problem 3.56.
Determine the degree of n characteristic polynomials of differential equations describing
transients in the electrical circuits shown in Fig. 3.56.
a
b
c
d
e
f
g
h
62
i
j
k
l
m
Fig. 3.56
Answer:
a b c d e f G h i j k l m 2 0 3 2 4 4 1 0 5 6 5 4
4
63
3.6. Duhamel’s integral method
If the passive electric circuit with zero initial conditions at time t = 0 is connected to the ideal
e.m.f. source e(t), then the current i(t) and voltage u(t) in any section of the circuit are determined
by the formulas (Duhamel integral):
t
0
i(t) e(0) g(t) e (τ) g(t - τ) dτ
= +
;
t
0
u(t)=e(0) y(t)+ e (τ) y(t-τ) dτ
.
Here:
g(t) is the transient conductivity, Ω
-1
;
y(t) is a transient function, dimensionless.
The transient functions g(t) and y(t) are defined as the ratios of the transient current and the
transient voltage in the selected circuit section to the voltage of the constant source e.m.f. E,
connected to the circuit at the moment t = 0.
The transient functions g(t) and y(t) are numerically equal to the current and voltage at the
selected circuit section when the circuit is connected to unit e.m.f.:
0, t<0;
e(t)=1(t)=
1V, t 0.
The transient functions g(t) and y(t) depend on the passive electrical circuit scheme and on
the parameters of its elements.
If the e.m.f. source e(t) has piecewise analytic form of the curve e(t), then the above Duhamel
integral formulas are valid at the first time interval
( )
1
0,t
. At subsequent time intervals
( )
k k+1
t ,t
,
the result of integration in the previous time interval
( )
k-1 k
t t ,t
with the integration interval
( )
k-1 k
t ,t
should be taken into account as a summand.
At the time interval boundary for example
k
t
, the function of input signal e(t) may have type
I jumps
k k+1 k k k
E e (t ) e (t ) = −
. The reaction of the circuit to the jump is taken into account as a
summand
kk
E g(t t )−
or
kk
E y(t t )−
.
Example.
Let g(t) be the transient conductivity of the circuit, e(t) be the input signal having the
following form:
11
2 1 2
32
e (t), 0 t t ;
e(t) = e (t), t t t ;
e (t), t t .
Then, at three time intervals, the transient current will be:
In the first interval
1
0 t t
:
t
1 1 1
0
i (t) e (0) g(t) e (τ) g(t τ) dτ
= + −
;
In the second interval
12
t t t
:
1
t
2 1 1
0
i (t) e (0) g(t) e (τ) g(t τ) dτ
= + − +
64
1
t
2 1 1 1 1 2
t
e (t ) e (t ) g(t t )+ e (τ) g(t τ) dτ
+ − − −
;
In the third interval
2
tt
:
1
t
3 1 1 2 1 1 1 1
0
i (t) e (0) g(t) e (τ) g(t τ) dτ e (t ) e (t ) g(t t )
= + − + − − +
2
12
t
t
2 3 2 2 2 2 3
tt
e(τ) g(t τ) dτ e (t ) e (t ) g(t t )+ e (τ) g(t τ) dτ
+ − + − − −
.
The recommended sequence of application of the Duhamel integral in calculating the
transient.
1. . Calculation of transient functions g(t) and y(t).
In this step, the transient current i(t) and the voltage u(t) for the selected electrical circuit
section are calculated by the classical or operator method. In this case, the transient process is
calculated when the circuit is connected to a source of unit e.m.f..
2. . Calculation of the desired transients.
At this stage, Duhamel integral formulas are used to calculate the transient process when
connecting the circuit to a given e.m.f. source e(t). At the same time, in the case of the piecewise
analytic form of the e.m.f. source, the integration interval must be divided into intervals of
continuity e(t) and, when writing the transient variable using the Duhamel integral, follow the
above instructions.
3. Recording the answer.
4. . Plotting transient variable time diagrams.
Problem 3.57.
For electrical circuits shown in the Figures 3.57.1 - 3.57.6 calculate the transient conductivity
and transient function.
Fig. 3.57.1
Answer:
r
t
L
g(t) 1 e S
−
=−
;
r
t
L
L
y (t) 1 e
−
=
.
65
Fig. 3.57.2
Answer:
1
t
rC
g(t) 1 e S
−
=
;
1
t
rC
C
y (t) 1 e
−
=−
.
Fig. 3.57.3
Answer:
r
t
2L
11
g(t) e S
r 2r
−
= −
;
r
t
2L
L
1
y (t) e
2
−
=
.
Fig. 3.57.4
Answer:
2
t
rC
11
g(t) e S
2r 2r
−
= +
;
2
t
rC
C
11
y (t) e
22
−
= −
.
66
Fig. 3.57.5
Answer:
3
t
rC
12
g(t) e S
3r 3r
−
= +
;
3
t
rC
C
11
y (t) e
33
−
= −
.
Fig. 3.57.6
Answer:
r
t
3L
12
g(t) e S
r 3r
−
= −
;
r
t
3L
L
1
y (t) e
3
−
=
.
Problem 3.58.
Voltage pulse
u(t)
acts on the input of passive one-port with transient conductivity
g(t)
(Fig.3.58).
Calculate the transient current
i(t)
at time intervals
1
0 t t
;
1
tt
.
67
Fig. 3.58
Answer:
10
0 t t : i(t) U g(t) =
;
1 0 0 1
t t : i(t) U g(t) U g(t t ) = − −
.
Problem 3.59.
Voltage pulse
u(t)
acts on the input of passive one-port with transient conductivity
g(t)
(Fig. 3.59).
Calculate transient current
i(t)
at time intervals
1
0 t t
;
1
tt
.
Fig. 3.59
Answer:
t
0
10
1
0
3U
0 t t : i(t) U g(t) g(t τ) dτ
t
= − + −
;
1
t
0
1 0 0 1
1
0
3U
t t : i(t) U g(t) g(t τ) dτ 2U g(t t )
t
= − + − − −
.
68
Problem 3.60.
Voltage pulse
u(t)
acts on the electric circuit input (Fig.3.60).
Calculate the transient current
i(t)
at time intervals
1
0 t t
;
1
tt
.
Fig. 3.60
Answer:
1
t
0
rC
1
U
0 t t : i(t) e
r
−
=
;
1
11
t (t t )
00
r C r C
1
UU
t t : i(t) e e
rr
− − −
= −
.
Problem 3.61.
Voltage pulse
u(t)
acts on the input of passive one-port with transient conductivity
g(t)
(Fig.3.61).
Calculate the transient current
i(t)
at time intervals
1
0 t t
;
11
t t 2t
;
1
t 2t
.
Fig. 3.61
Answer:
t
0
1
1
0
U
0 t t : i(t) g(t ) d
t
= −
;
69
1
1
t
t
00
12
11
0t
UU
t t t : i(t) g(t ) d g(t ) d
tt
= − − −
;
11
1
t 2t
00
1
11
0t
UU
t t : i(t) g(t ) d g(t ) d
tt
= − − −
.
Problem 3.62.
Voltage pulse acts on the electric circuit input
1
u (t)
(Fig.3.62).
Calculate the transient voltage
2
u (t)
at time intervals
1
0 t t
;
1
tt
.
Fig. 3.62
Answer:
r
t
L
1 2 0
M
0 t t : u (t) U e
L
−
=
;
1
rr
t (t t )
LL
1 2 0 0
MM
t t : u (t) U e U e
LL
− − −
= −
.
Problem 3.63.
Voltage pulse acts on the electric circuit input
1
u (t)
(Fig.3.63).
Calculate the transient voltage
2
u (t)
at time intervals
1
0 t t
;
1
tt
.
70
Fig. 3.63
Answer:
1
t
rC
1 2 0
0 t t : u (t) U e
−
=
;
1
11
t (t t )
r C r C
1 2 0 0
t t : u (t) U e U e
− − −
= −
.
Problem 3.64.
For the circuit shown in Fig. 3.64a, it is known:
r 10 =
;
C=200 F
;
voltage pulse acts on the circuit input (Fig. 3.64b)
11
1
2 1 2
32
U
e (t)= t; t 0,t ;
t
e(t) e (t)=U; t t ,t ;
e (t)=0; t t ;
=
U=50 V
;
1
t =2 ms
;
2
t =5 ms
.
Calculate the transient current.
Fig. 3.64
Answer:
At the first time interval
1
0 t t
:
Соседние файлы в предмете [НЕСОРТИРОВАННОЕ]