|
10.3.2
complementary Function and particular Integral
The governing equation for inductor current in a series RL circuit with unit step voltage input is
di
dt
i
u t
L
L
+
=
a
b
( )
(10.3-3)
where
a
=
R/L and
b
=
1/L. There is a jump discontinuity from 0 to 1V at t
=
0 in the applied voltage.
This jump in applied voltage travels straight to the inductor and appears as a jump in voltage across
it at t
=
0.
Particular Integral of a differential equation is the solution term due to the forcing function. The
domain of particular integral is same as the domain of the input function. In this case, the input is 1V
in t
∈
(0,
∞
). Hence, the particular integral will be valid for t
∈
(0,
∞
). The particular integral in the
present case is obtained by solving the following equation in the interval t
∈
(0,
∞
):
di
dt
i
t
L
L
for all
+
=
∈ ∞
a
b
( , )
0
(10.3-4)
The differential equation in Eqn. 10.3-4 is to be true for all t
∈
(0,
∞
). This is possible only if i
L
is such a function that its first derivative along with its own copy becomes a constant for all t. The
only such function is i
L
=
c, a constant. Substituting this trial solution in Eqn. 10.3-4, we get that the
particular integral is
b
/
a
A for t
∈
(0,
∞
).
What about particular integral for t
∈
(
-∞
,0)? The voltage applied in this interval was 0 V.
Therefore, the particular integral is 0 for t
∈
(
-∞
,0).
Therefore, the particular integral of the circuit changes from 0 to
b
/
a
A at t
=
0. That is, there
is a step discontinuity in the particular solution at t
=
0. However, such an instantaneous change in
inductor current is not permitted in an RL circuit unless impulse voltage is applied across the inductor.
The inductor current has to be continuous at t
=
0.
Hence, we need another term in the circuit solution which will force the total circuit solution to
satisfy the inductor current continuity requirement at t
=
0.
This additional function has to satisfy a constraint. The solution
i t
R
t
t
L
A ;
A ;
( )
[ , )
(
,
]
=
=
∈
∞
∈ −∞
+
−
b
a
1
0
0
0
satisfies the differential equation in Eqn. 10.3-4 at all t
∈
0. Therefore, the additional function we are
to going to add to this solution to enforce compliance with continuity requirement at t
=
0 should not
add anything to the right side of differential equation. This implies that it has to be a function that will
satisfy the following differential equation for all t excluding t
=
0.
di
dt
i
t
L
L
+
=
≠
a
0
0
;
(10.3-5)
Step Response of
RL
Circuit by Solving Differential Equation
10.11
A trivial solution to this equation is i
L
=
0. We are not interested in that. We recast the above
equation as
di
dt
i
t
L
L
= −
≠
a
;
0
and note the fact that if i
L
is a function of time, then both sides of this equation will be functions of
time. Two functions of time can be equal to each other over their entire domain if and only if they are
same kind of functions – they must have same shape when plotted. Hence, we look for functions which
produce a copy (probably scaled versions) of themselves on differentiation.
Sinusoidal function comes to our mind first. But then, we remember that sinusoidal function can
be covered by exponential function with imaginary exponent by Euler’s formula. Hence, e
g
t
where
g
can be a complex number is a function with the desired property. So we try out Ae
g
t
, where A is an
arbitrary constant, as a possible solution in Eqn. 10.3-5. We get
A e
A e
t
Ae
t
t
t
t
g
a
g a
g
g
g
+
=
+
=
0
0
for all in any interval
for all
(
)
iin any interval
A
=
0 is a trivial solution. e
g
t
=
0 cannot be a solution since the equation has to be true for all t.
Therefore,
g
has to be equal to
-
a
. Thus, A e
-
a
t
is the solution for the Eqn. 10.3-5 in any given
interval. The differential equation cannot help us in deciding the value of A. It decides only the value
of exponent in the exponential function. This solution we have arrived at i.e., A e
-
a
t
is called the
complementary function of the differential equation in Eqn. 10.3-5 and the differential equation with
zero forcing function is called the homogeneous differential equation.
Now that we have got the function required to enforce compliance with the current continuity
requirement in the circuit, let us proceed to form the total solution for i
L
in series RL circuit with
step input. There are two intervals over which the particular integral is known. They are (
-∞
,0
-
] and
[0
+
,
∞
). Let the complementary solution that we accept for t
∈
(
-∞
,0
-
] be A
-
e
-
a
t
and the complementary
solution that we accept for t
∈
(
-∞
,0
-
] be A
+
e
-
a
t
.
i t
A e
R
t
i
t
L
L
for
and
( )
( )
=
+
≥
=
+
−
+
+
a
1
0
0
0
(10.3-6)
The voltage applied to the circuit was zero right from t
= -∞
up to t
=
0
-
. Hence, the circuit current
has to remain zero over (
-∞
,0
-
]. Therefore, the amplitude of complementary solution for t
∈
(
-∞
,0
-
]
has to be chosen as zero, i.e., A
-
=
0. However, this will mean that there will be a step discontinuity at
t
=
0 in the complementary solution for any non-zero A
+.
That is precisely what we want. We want the
discontinuity in complementary solution to cancel the discontinuity in particular integral at t
=
0 and
thereby make inductor current continuous at t
=
0. Obviously, A
+
must be chosen as negative of the
size of discontinuous jump in a particular integral. This gives us A
+
= -
1/R. Hence, the final solution is
i t
R
e
t
t
L
R
t
L
where
( )
(
);
;
/
=
−
≥
≤
=
=
−
+
−
1
1
0
0
0
1
t
t
a
(10.3-7)
Equation 10.3-7 gives the unit step response (abbreviated as step response) of current in a series
RL circuit.
10.12
First-Order
RL
Circuits
Do'stlaringiz bilan baham: |
