Unit 3
Response of Second-Order Systems
In this unit, we consider the natural and step responses of simple series and parallel
circuits containing inductors, capacitors and resistors. The equations which result
from this analysis will, in general, be second-order ordinary differential equations
(O.D.E.’s) with constant coefficients. Before introducing a particular circuit example,
we will consider how to determine the solution of such an equation. The solution of
the general second-order O.D.E. with constant coefficients may then be applied to
any circuit whose equation fits that form.
3.1
3.1.1
Natural Response of a Parallel RLC Circuit
Review of Second-Order O.D.E. with Constant Coefficients
Consider the following second-oder O.D.E. in which x is a function of time and α and
ω0 are constants. The middle term is written with a factor of 2 for reasons that will
become obvious shortly.
. . . (1)
When we get to the actual circuits, x will be current i or voltage v and the constants
will depend on the the elements R, L and C and their particular configuration in the
circuit.
The typical approach in solving (1) is to assume that a solution of the form
x = Aest
. . . (2)
exists. Here, A and s are unknown constants. To determine A and s we begin by
substituting (2) into (1) to get
1
which implies
Aest s2 + (2α)s + ω02 = 0 . . . (3)
From (3) we see that either A = 0 – which is a trivial but useless case since our
voltages and currents won’t be zero for all time – or
s2 + (2α)s + ω02 = 0 . . . (4)
which implies
. . . (5)
Equation (4) is the so-called characteristic equation because its roots (solutions) determine the character of x. These roots may be written explicitly as
s1 =
. . . (6a)
s2 =
. . . (6b)
Since, from (3) we know that both (6a) and (6b) are solutions to the charactersitic
equation regardless of the value of A, equation (2) indicates that both
x1 =
and
x2 =
are solutions of (1). Since x1 and x2 are linearly independent solutions of (1), their
combination given by
x = x1 + x2 =
. . . (7)
is also a solution. As we shall see, s1 and s2 will depend on the circuit parameters R,
L and C, while A1 and A2 will be determined from the initial conditions.
Equation (6) suggests three possible “kinds” of solution:
(1) α2 > ω02 – both characteristic roots are real [overdamped system]
(2) α2 < ω02 – roots are complex conjugates[underdamped system]
(3) α2 = ω02 – both roots are real and equal [critically damped system]
The meanings of these system responses will be considered shortly, as will the significance of α (the so-called neper frequency of the system) and ω0 (the so-called resonant
radian frequency of the system).
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3.1.2
The Parallel RLC Circuit – Natural Response
Consider the following parallel combination of R, L and C.
Since the voltage is the same across each component we will seek that first. Applying KCL to the top node of the circuit, we have
. . . (8)
where we have allowed for an initial (constant) current I0 through the inductor.
Differentiating this equation gives
. . . (9)
and dividing by C gives
d2 v
1 dv
1
+
+
v = 0 . . . (10)
2
dt
RC dt LC
Comparing (10) with the general form of the second-order O.D.E. in (1), we immediately identify
α=
1
2RC
and ω0 = √
1
. . . (11)
LC
and therefore s1 and s2 can be immediately found from (6a) and (6b). From (7) we
see that we must still find A1 and A2 . In fact, while we proceed slightly differently
for the three classes of systems mentioned above, the general procedure for finding v
does not change fundamentally:
(1) Find s1 and s2 using R, L, and C.
(2) Determine the initial voltage across the elements (which is the initial voltage capacitor) – i.e. v(0+ ) = V0 .
iC (0+ )
dv(0+ )
=
.
dt
C
(4) Use (2) and (3) to find the remaining constants required in the solution for v.
(3) Find the initial rate of change of the voltage
This last step will vary in application depending on the type of system as we will
develop below.
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(5) Once the voltage is established, we can easily determine the branch currents.
Overdamped Voltage Response (α2 > ω02 )
For this case, s1 and s2 are real and may be found using (6a), (6b) and (11) (we
won’t carry out this step until we use a particular example).
Next, from (7),
v(0+ ) = A1 + A2
. . . (12)
while
dv(0+ )
iC (0+ )
=
= s 1 A1 + s 2 A 2
dt
C
. . . (13)
From (12) and (13), A1 and A2 may be readily found so that equation (7) is completely specified. Of course, in general, iC (0+ ) will have to be determined by circuit
analysis (or be otherwise specified). See example 8.2 of the text, pages 338-339.
Underdamped Voltage Response (α2 < ω02 )
In this case, s1 and s2 are complex conjugates and may be written as
s1 =
=
⇒ s1 =
. . . (14)
where
ωd =
q
ω02 − α2
. . . (15)
is referred to as the damped radian frequency to be explained shortly. Also,
s2 = −α − jωd
. . . (16)
Substituting these s values into (7), we get (recalling that here x = v(t)),
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It is possible to show (see Problem Set 4 later) that A1 and A2 are complex
conjugates so that B1 and B2 are real, as is required to obtain a real voltage.
We note the following about equation (17):
(1) The sine and cosine functions indicate an oscillatory response. The rate of the
oscillation is determined by ωd , the damped radian frequency.
(2) If there is a resistance, there will be an α and therefore a decay of the voltage
over time by virtue of the e−αt . For this reason, α is often referred to as the damping
coefficient. Thus, for an underdamped RLC parallel circuit, the voltage response will
be as shown below.
Illustration:
The oscillatory behavior occurs because of the presence of two energy storage elements
(the inductor and capacitor) which alternately store and release energy – howbeit in
a decaying fashion if there is a resistance present. Notice that the oscillation occurs
about a final value – this phenomenon is referred to as ringing and it will be a very
important consideration in later courses (eg., control systems and filters).
Critically Damped Voltage Response (α2 = ω02 )
Finally, we consider what happens to the voltage response when s1 = s2 = −α (i.e.,
the characteristic roots are equal). There is now a fundamental difference because
the solution to the general equation changes from that given in (7) to
x = D1 te−αt + D2 e−αt
. . . (18)
where D1 and D2 are constants – this is easily shown to be a solution of (1). In
determining x (or voltage v for the case being discussed) we must find D1 and D2
from the initial conditions as before. That is, we use
v(0+ ) = V0 = D2 (from (18))
and
dv(0+ )
ic (0+ )
=
= D1 − αD2 .
dt
C
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The critically damped response indicates a smooth approach to the final value as
depicted.
Illustration:
Before considering the step response of the system for which we have just completed the natural response, we address the following three examples.
Overdamped Response Drill exercise 8.3, page 340 of the text. Consider the
following circuit in which L = 250 mH, C = 10 nF and R = 2 kΩ. The initial current
I0 in the inductor is -4 A while the initial voltage on the capacitor is 0 V. The output
signal is the voltage v. Determine (a) iR (0+ ); (b) iC (0+ ); (c) dv(0+ )/dt; (d) A1 ; (e)
A2 ; and (f) v(t) for t ≥ 0.
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Underdamped Response Drill exercise 8.4, page 344 of the text. Consider the
following circuit in which L = 10 mH and C = 1 µF. R is adjusted so that the roots
of the characteristic equation are −8000 ± j6000 rad/s. The initial current I0 in the
inductor is 80 mA while the initial voltage on the capacitor is 10 V. Determine (a)
R; (b) dv(0+ )/dt; (c) B1 and B2 in the solution for v; and (d) iL (t).
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Critically Damped Response Drill exercise 8.5, page 346 of the text. Suppose
that the resistance in the previous example is adjusted for critical damping. The
inductance L = 0.4 H and the capacitance C = 10 µF. The initial energy stored in
the circuit is 25 mJ and is distributed equally between the inductor and capacitor.
Determine (a) R; (b) V0 ; (c) I0 ; (d) D1 and D2 in the solution for v; and (e) iR for
t ≥ 0+ .
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3.2
Step Response of a Parallel RLC Circuit
Consider the following circuit in which a step input is initiated by opening the switch
at t = 0. For now we consider that there is no energy initially stored in the system.
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When the switch is opened, the circuit is “forced” by the current I (the forcing
function) from the ideal current source. We will concentrate first on determining the
expression for iL , the current in the inductor. Using KCL,
i L + iR + iC = I
. . . (1)
which, recalling the relationships between currents and voltages in the three components, we can write in terms of voltage and iL as
. . . (2)
Since, here, the voltage across each element of the parallel section is the same (for
example, v = vL = L(diL /dt)) (2) may be divided by LC and written as
d2 iL
1 diL
iL
I
+
+
=
. . . (3)
2
dt
RC dt
LC
LC
This is clearly a second-order inhomogeneous D.E. – see Term 2 Math Course –
whose solution is the sum of the the solution to the homogeneous part (i.e. the the
complementary function) plus any one solution (i.e. the particular solution). The
homogeneous part of (3) looks just like equation (10) on page 3 of this unit with v
there replaced by iL here. Furthermore,
iL = I
is obviously a solution (a “particular” solution) to (3). Thus, considering equations
(7), (17) and (18) of Section 3.1, with the time functions replaced by iL here, the
solution to (3) may be written as
iL = I + A1 es1 t + A2 es2 t [overdamping] . . . (4)
iL = I + e−αt [B1 cos ωd t + B2 sin ωd t] [underdamping] . . . (5)
iL = I + D1 te−αt + D2 e−αt [critical damping] . . . (6)
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Of course, the A’s, B’s and D’s here are arbitrary constants whose values won’t be
the same as those in the voltage equations we developed earlier. They may be found,
in each case here from simultaneously solving the system which results from iL (0)
and diL (0)/dt. If there is an initial stored energy – that is, if there is in the above
analysis a non-zero initial value for iL – this will affect the solutions for the constants.
See Example 8.10, page 354, of the text.
By observation, we see that equations (4), (5) and (6) take the general form
function of the same form
g(t) = Gf +
. . . (7)
as the natural response
where g(t) may be current or voltage and Gf is the final value of the current or voltage
response.
Step Response Example:(Drill exercise 8.6, page 355 of text.) In the circuit shown,
R = 500 Ω, L = 0.64 H, C = 1 µF, I0 = 0.5 A (this is the initial current through
the inductor), V0 = 40 V (this is the initial voltage), and I = −1 A. We find various
circuit parameters and responses below.
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3.3
Natural and Step Response of a Series RLC
Circuit
Natural Response
In this section there are no new mathematical forms – the equations look similar
to those already seen. Consider the following series RLC circuit.
At the instant the analysis begins, the initial current through the inductor is I0 and
the initial voltage across the capacitor is V0 . In general, the current in the circuit is
i. Applying KVL around the path we have
Z
di
1 t
Ri + L +
idt + V0 = 0 . . . (1)
dt C 0
Differentiating again and dividing by L gives
. . . (2)
This looks the same as equation (10) page 3 of this unit with v replaced by i and
1/RC replaced by R/L. The solution to equation (2) is therefore of the form
i(t) = Aest
and
α=
. . . (3)
ω0 =
. . . (4)
while
As before,
α2 − ω02 . . . (5a)
q
= −α − α2 − ω02 . . . (5b)
s1 = −α +
s1
q
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Note that the natural frequency of the series RLC circuit is the same as that for the
parallel RLC circuit.The three possible solutions again take the form
i = A1 es1 t + A2 es2 t [overdamping] . . . (6)
i = e−αt [B1 cos ωd t + B2 sin ωd t] [underdamping] . . . (7)
i = D1 te−αt + D2 e−αt [critical damping] . . . (8)
Once i is known it is straightforward to calculate the voltage v across any element
since
vR = iR ; vL = L
dvc
di
; i=C
. . . (9a,b,c).
dt
dt
We’ll consider an example following the step response analysis below.
Step Response
Consider the following circuit in which the switch has been open for a long time
and then is abruptly closed at t = 0. The source voltage is V .
Applying KVL around the loop gives
V =
. . . (10)
Since the current is related to the capacitor voltage via (9c), equation (10) becomes
and dividing by LC gives
d2 vc R dvc
vc
V
+
+
=
. . . (11)
2
dt
L dt
LC
LC
Equation (11) has exactly the same form as equation (3) of Section 3.2 with vc replacing iL , R/L replacing 1/RC, and V replacing I. It therefore follows that the three
possible solutions are:
vc = V + A1 es1 t + A2 es2 t [overdamping] . . . (4)
vc = V + e−αt [B1 cos ωd t + B2 sin ωd t] [underdamping] . . . (5)
vc = V + D1 te−αt + D2 e−αt [critical damping] . . . (6)
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Note that in all cases, the solutions take the general form given in equation (7) on
page 10 of this unit. V is simply the final voltage across the capacitor here.
Be sure to consider all text examples in Chapter 8 up to page 360.
Example for Natural Response of a Series RLC Circuit: Problem 8.35, page
374 of the text.
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