Unit 2
Response of First-Order Systems
In this unit, we consider inductors or capacitors (but not both together) as elements
in circuits containing resistors. That is, we wish to examine resistor-inductor (RL)
circuits and resistor-capacitor (RC) circuits. These circuits are referred to as firstorder circuits because their voltages and currents can be described by first-order
differential equations. If a circuit can be reduced to a Thévenin or Norton equivalent
connected to an equivalent inductor or capacitor, then the circuit will be first-order.
The four possible combinations are depicted below.
In the first two sections of this unit we will consider
1. natural response – that is, what voltages and currents arise when a dc source
is suddenly disconnected from an RL or RC combination so that energy is released from the “storage” element (i.e. the inductor or capacitor) to the resistive
part of the network
1
and
2. step response – that is, what voltages and currents arise when a dc source is
suddenly connected to an RL or RC combination so that energy is acquired by
the “storage” element (i.e. the inductor or capacitor).
2.1
Natural and Step Response of RL Circuits
2.1.1
Natural Response of a RL Circuit
The following circuit will be useful in this analysis.
We note:
(1) Before our analysis begins, the switch has been closed for a long time (we’ll
quantify this better later).
(2) Thus, for t < 0 the inductor acts as a short circuit and I0 flows through L but
not through R0 or R.
(3) At time t = 0 the switch is opened (source I0 is removed from the circuit) and
the circuit reduces to:
Our analysis involves finding v(t) and i(t) for t ≥ 0.
Current Equation
Using the passive sign convention and applying KVL to the circuit,
vL + vR = 0
so that
. . . (1)
2
Since R and L are constants, the current is obviously modelled by a first-order differential equation with constant coefficients. From (1)
. . . (2)
Taking the initial time as t0 = 0 so that i(t0 ) = i(0), equation (2) may be integrated
as
Z
i(t)
i(0)
di
R
=−
i
L
Z
t
dt
0
from which
and
i(t) = i(0)e−(R/L)t
. . . (3)
Since, as discussed in the previous unit, there cannot be an instantaneous change in
the current (I0 ) in the inductor, then
lim i(t) = lim+ i(t) = I0 .
t→0−
t→0
Therefore, we may finally write equation (3) as
i(t) = I0 e−(R/L)t = I0 e−t/τ
. . . (3)
where τ = L/R and is referred to as the time constant of the circuit.
The time constant is a very important descriptor of the circuit. One time constant
is the time necessary for the current to be reduced to e−1 (or 36.788%) of its original
value when the switch was first opened. By the time t = 5τ , the current is reduced
to less than 1% of its original value. This tells us that the existence of the current in
the circuit after the switch is opened is a momentary or transient effect and its nature
is referred to as the transient response of the circuit. The response that remains a
long time after the switching occurs is referred to as the steady-state response. Read
pages 281 and 282 of the text for further insight into the time constant.
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Voltage Equation
We have noticed that before the switch is opened in the circuit shown on page
2, the current is I0 , a constant. Since, for the inductor, the voltage depends on the
derivative of the current, the voltage for t < 0 must be 0. Also, since before the
switch is opened, the voltage v across the resistor is v = vL ,
v(0− ) = 0 . . . (4)
while after the switch is opened we get using equation (3)
v = iR =
. . . (5)
From this we see that
v(0+ ) = I0 R . . . (6)
so that from (4) and (6) together we see that there is an instanantaneous change in
the voltage when the switch is opened.
Power and Energy Equations
Whether we express power dissipated in the resistor as p = vi or p = i2 R or p =
v 2 /R, after the switch is opened the deliberations above lead to
p = I02 Re−2t/τ . . . (7)
Then, the energy at time t ≥ 0 after the switch is closed becomes
Z t
w =
pdt =
0
=
=
. . . (8)
What does equation (8) tell us about energy conservation?
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2.1.2
Step Response of an RL Circuit
We next examine what happens when a voltage is suddenly impressed on an RL
circuit – i.e., what is the step response? Consider the following circuit in which at
time t = 0 the switch is suddenly closed.
Illustration:
Current Equation To keep our analysis as general as possible, at the outset we will
imagine that at time t = 0 the inductor is already energized by a current I0 . Applying
KVL after the switch is closed, we have
Vs = vR + vL =
from which
Z
i(t)
I0
di
R
=−
i − (Vs /R)
L
Z
t
dt
0
Carrying out the integration,
– the final result becomes
Vs
Vs −t/τ
+ I0 −
e
. . . (9)
i(t) =
R
R
where we have used τ = L/R. I0 is positive if it is in the same direction as i, otherwise
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it is negative. Clearly, if the initial current is zero, equation (9) takes the form
i(t) =
Vs
1 − e−t/τ . . . (10)
R
Equation (10) indicates that after a period of time equal to one time constant, the
current will have reached about 63% of its final value of Vs /R. These results for
I0 = 0 are depicted below.
Voltage Equation for the Inductor
di
Given that vL = L , we have, using equation (9) after the switch is closed, that
dt
vL =
. . . (11)
or for an initial current of 0,
vL = Vs e−t/τ . . . (12)
From both (11) and (12) we see that there is a “jump” in the voltage when the switch
is closed. Convince yourself of the size of this “jump”.
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2.2
Natural and Step Response of RC Circuits
2.2.1
Natural Response of a RC Circuit
The following circuit will be useful in this analysis.
We note:
(1) Before our analysis begins, the switch has been closed for a long time (–i.e., the
capacitor is fully charged and the voltage across it is V0 .
(2) For t < 0 the capacitor acts as an open circuit in the presence of a constant
voltage.
(3) At time t = 0 the switch is moved from position a to position b. There can be no
instantaneous change in the voltage on the capacitor (recall i depends on dv/dt and
the latter quantity cannot physically be infinite). For t ≥ 0, the circuit above thus
reduces to
Our analysis involves finding v(t) and i(t) for t ≥ 0.
Voltage Equation
Using KCL at the upper node between R and C we may write
iC + IR = 0
where the subscripts on the i’s indicate the element with which the current is associated. This may be written as
C
dv
v
+ = 0 . . . (1)
dt R
which is identical in form to the inductor result [equation (1) of Section 2.1]. We
notice here that C, v and 1/R take the place of L, i and R, respectively, of the
corresponding inductor equation. Thus, we may immediately write the solution as
v(t) = V0 e−(1/RC)t = V0 e−t/τ , t ≥ 0, . . . (2)
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where the time constant for the RC circuit is now given by τ = (RC). A sketch of v
versus t for the capacitor corresponding to that for i versus t for the natural response
of the inductor [see sketch on page 3 of this unit] is therefore
Current, Power and Energy
Considering equation (2) and the circuit on the previous page we immediately
have for the current
i(t) =
v(t)
V0
= e−t/τ , t ≥ 0+ . . . (3).
R
R
From (2) and (3) the power is
p=
. . . (4)
while the energy dissipation is given by
w=
. . . (5)
A nice summary of what’s involved in calculating the natural response of an RC
circuit is found just prior to Example 7.3 on page 288 of the text.
2.2.2
Step Response of an RC Circuit
Consider the following circuit in which at time t = 0 the switch is suddenly closed.
Using KCL at the top node, while recalling that the current through the capacitor is
given by C(dvc /dt), we get (when the switch is closed)
. . . (6)
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which is identical in mathematical form with the initial step response equation written
di
for the inductor on page 5 (i.e. L + Ri = Vs ). Thus, replacing Vs , i, I0 , R and L of
dt
equation (9), page 5, with Is , vc , V0 , 1/R and C, respectively, we write the solution
to (6) above as
vc = Is R + (V0 − Is R)e−t/τ , t ≥ 0 . . . (7)
[YOU SHOULD VERIFY THIS.] Using (7), the current may be deduced as
Interesting (and maybe) useful summary note:
The results of Sections 2.1 and 2.2 may be summarized as follows:
Natural Response of RL circuit (current): i(t) = I0 e−(R/L)t = I0 e−t/τ
Natural Response of RL circuit (voltage): v(t) = I0
Re−(R/L)t= I0 Re−t/τ
Vs −t/τ
Vs
Step Response of RL circuit (current): i(t) =
+ I0 −
e
R
R
Step Response of RL circuit (voltage): vL = (Vs − I0 R) e−t/τ
Natural Response of RC circuit (voltage): v(t) = V0 e−(1/RC)t = V0 e−t/τ
V0
Natural Response of RC circuit (current): i(t) = e−t/τ
R
−t/τ
Step Response of RC circuit (voltage): vc = Is
R + (V0 −
Is R)e
V0 −t/τ
Step Response of RC circuit (current): i(t) = Is −
e
R
In all cases the form is
function of time = final value + (initial value − final value) e−
(elapsed time after switching)
(time constant)
This can easily be deduced by writing any one of the first-order differential equations
we encountered in this unit in the general form of
dx x
+ =K
dt
τ
where x is the variable and K is a constant which is allowed to have a value of 0 when
required (see “natural response” sections).
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2.3
Sequential Switching
Sequential switching as the name suggests refers to an orderly opening and closing of
a multiple-position switch or a series of switches which may be opened and closed in
sequence. In analyzing such a circuit it is critically important to
(1) know the initial values of the variables (i.e. currents and/or voltages) at each
stage of the switching process and
(2) remember that the time variable in the exponentials of the equations encountered
previously may be read as “elapsed time” – thus, for example, if a switching process
occurs at t1 (not equal to 0), then the time t in the exponentials must be replaced by
t − t1 .
We illustrate with an example: Problem 7.71, page 340 of the text.
Initially, switch A has been open and switch B has been closed for a long time.
At t = 0, switch A closes. One second after switch A closes, switch B opens. Find
iL (t) for t ≥ 0.
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•
•
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2.4
Unbounded Response
It is possible that a circuit response may grow, rather than decay, with time. This
is referred to as an unbounded response – of course, in a practical sense there will be
a physical limit to the growth because circuit elements will eventually break down
or saturate, thus limiting further increase in voltages or currents. The circuits where
this phenomenon is possible contain “dependent sources” (or elements that are acting
like dependent sources). In the case of unbounded response, the Thévenin resistance
will be negative and the exponent on the exponentials which we encountered in the
previous sections will be positive (thus the exponential growth). In these cases, rather
than drawing upon the general solution for step-responses in first-order circuits, we
develop and solve (by separation of variables) the equations from first principles.
We illustrate with an example: Problem 7.78, page 342 of the text.
The capacitor in the circuit shown below is charged to 25 V at the time the switch is
closed. If the capacitor ruptures when its terminal voltage equals or exceeds 50 kV,
how long after the switch closes does the capacitor rupture?
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