Download Step Response

Dave Shattuck
University of Houston
© University of Houston
ECE 2202
Circuit Analysis II
Lecture Set #5
Special Cases and Approaches
with First Order Circuits
Dr. Dave Shattuck
Associate Professor, ECE Dept.
[email protected]
713 743-4422
W326-D3
Lecture Set #5
Special Cases and
Approaches with First Order
Circuits
Dave Shattuck
University of Houston
Overview of this Part
Special Cases and Approaches with First
Order Circuits
© University of Houston
In this part, we will cover the following topics:
• Finding Time Constants
• Initial and Final Values
• Sequential Switching
• Negative Time Constants
• Generalized Solution for First Order Circuits
Dave Shattuck
University of Houston
© University of Houston
Textbook Coverage
This material is introduced in different ways in
different textbooks. Approximately this same
material is covered in your textbook in the
following sections:
• Electric Circuits 7th Edition by Nilsson and
Riedel: Sections 7.5 and 7.6
• Electric Circuits 10th Edition by Nilsson and
Riedel: Sections 7.5 and 7.6
Dave Shattuck
University of Houston
© University of Houston
6 Different First-Order Circuits
There are six different STC circuits.
These are listed below.
• An inductor and a resistance (called
RL Natural Response).
• A capacitor and a resistance (called
RC Natural Response).
• An inductor and a Thévenin
equivalent (called RL Step
Response).
• An inductor and a Norton equivalent
(also called RL Step Response).
• A capacitor and a Thévenin
equivalent (called RC Step
Response).
• A capacitor and a Norton equivalent
(also called RC Step Response).
RX
LX
CX
RX
RX
+
vS
LX
-
iS
RX
iS
RX
LX
RX
+
vS
CX
CX
-
These are the simple, first-order
cases. We have found the
solutions for the inductive
currents and capacitive voltages
in the last two parts of this
module.
Dave Shattuck
University of Houston
© University of Houston
Finding the Three Parameters
In the step response solutions that we found in the
last part, there were three parameters:
1. The time constant
2. The initial condition (at the time of switching)
3. The final value (a long time after the time of
switching)
The key in these solutions is finding these three
items.
Dave Shattuck
University of Houston
© University of Houston
Finding the Time Constant – 1
The time constant is L/R in an RL circuit, and RC in an RC
circuit. If there is more than one inductor, we start by
finding the equivalent inductance. Similarly, if there is more
than capacitor, we start by converting them to a single
capacitor. If we can’t take this step of converting to an
equivalent
The inductors L3
with only one
and L2 are in
parallel, and that
inductor or
combination is in
L1
capacitor, we
series with L1.
can’t use this
t=0
method.
IS1
RS1
L2
L3
RS2
IS2
Dave Shattuck
University of Houston
© University of Houston
Finding the Time Constant – 2
After converting to an equivalent with only one
inductor or capacitor, we need to find the
resistance seen by that inductor or capacitor. This
means finding the Thévenin resistance seen by the
inductor or capacitor.
For t > 0:
IS1
A
RS1
L
RS2
B
We need the resistance seen by the
inductor, at terminals A and B.
A
RS1
RS2
B
IS2
The inductor L sees an
equivalent resistance, or
Thévenin resistance, of RS1
in parallel with RS2. In some
cases there may be
dependent sources present.
If so, we would use our testsource method to find the
Thévenin resistance.
Dave Shattuck
University of Houston
© University of Houston
Finding the Initial Condition – 1
In finding the initial condition, we usually are given
the information that the circuit has been in some
condition “for a long time”. In that case, the
voltages and currents will stop changing. As a
result, the inductor behaves like a short circuit, and
the capacitor behaves like an open circuit.
RS1
t=0
RS2
+
vC
+
VS1
+
C
VS2
-
-
We are making an
assumption here that should
be kept in mind. The
voltages and currents will
only stop changing in the six
circuits we are studying, or
circuits that can be reduced
with equivalent circuits to one
of those six circuits.
-
Dave Shattuck
University of Houston
© University of Houston
Finding the Initial Condition – 2
If we are told that the circuit has been in some
condition “for a long time”, and the voltages and
currents stop changing, then an inductor behaves like
a short circuit, and a capacitor behaves like an open
circuit. We can make this replacement, and then the
circuit becomes a fairly straightforward circuit to solve.
RS1
t=0
RS2
+
vC
+
VS1
+
C
VS2
-
-
In the circuit shown here, we
are told that it has been in this
condition for a long time before
t = 0. Therefore, for t < 0, the
switch is closed, and we would
replace the capacitor with an
open circuit, and solve for vC.
Note that this will have to be
equal to vC(0), since it was valid
just before zero, and can’t
change instantaneously.
-
Dave Shattuck
University of Houston
© University of Houston
Finding the Initial Condition – 3
In this example, we have replaced the
capacitor with an open circuit for t < 0. We
solve then for vC(0).
In order to find vC(0), we first
solve for the current iX, which is
VS 1  VS 2
iX 
.
RS 1  RS 2
For t < 0:
RS1
Using this, we can get vC(0),
which is
iX
RS2
+
vC(0)
VS2
-
+
 VS1  VS 2 
vC (0)  VS1  
 RS1.
 RS1  RS 2 
-
VS1
+
vC (0)  VS1  iX RS1 
-
Dave Shattuck
University of Houston
© University of Houston
Finding the Initial Condition – 4
It should be noted that the initial condition cannot always be
found this way, since the circuit may not have been in a
given condition for a long time. However, there must be
some way that the initial condition can be found, if we are to
find the solution. In some cases, an expression for the
capacitive voltage or inductive current may be known, in
which case we simply evaluate at the time of switching.
Let’s imagine that in this circuit, we knew that
[V]; for  100[ms]  t  0.
RS1
We would plug in to get
VS1
C
VS2
-
-
[V]  1.07[V].
RS2
+
vC
+
vC (0)  30e
 100[ms] 


 30[ms] 
t=0
+
vC (t )  30e
 t 100[ms] 


 30[ms] 
-
Dave Shattuck
University of Houston
© University of Houston
Finding the Final Value – 1
In finding the final value, we are looking for the
value after “a long time”. Similar to the situation in
finding the initial condition, this will be when the
voltages and currents stop changing. As a result,
again, this is when the inductor behaves like a
short circuit, and the capacitor behaves like an
open circuit.
RS1
t=0
RS2
+
vC
+
VS1
+
C
VS2
-
-
We are making an
assumption here that should
be kept in mind. The
voltages and currents will
only stop changing in the six
circuits we are studying, or
circuits that can be reduced
with equivalent circuits to one
of those six circuits.
-
Dave Shattuck
University of Houston
© University of Houston
Finding the Final Value – 2
In this example, we have replaced the
capacitor with an open circuit for a long time
after the switching, or t >> 0. We might also
refer to this as t = .
For t = :
RS1
In this circuit, the value of vC()
is clear from the fact that the
current through RS1 is zero, and
+
VS1
vC()
-
vC ()  VS1.
+
-
Sequential Switching – 1
Dave Shattuck
University of Houston
© University of Houston
Now, we will consider the case where there is more than one
switching event, which occur at more than one point in time.
Remember that in general we define the time t = 0 as the time
when the switching takes place. However, when there is more
than one time involved, we can’t make time equal to zero at two
different times in the same problem. Usually, we assign the time
t = 0 to the first event, and then have all other events at times
relative to that point in time.
The question, then, is what to do with the general solution
equation when this occurs.
+
vC
RS2=8[W]
t=0
C=2[F]
-
SWA
VS1=
9[V]
SWB
t = 1[s]
+
RS1=5[W]
-
VS2=
-4[V]
+
-
Sequential Switching – 2
Dave Shattuck
University of Houston
© University of Houston
The question, then, is what to do with the general solution
equation when there is more than one switching event, which
occur at more than one point in time. Conceptually, what we do
is to apply a transformation of variables to the time variable.
Specifically, we take a solution that would be valid with t' = 0 for
that switching event, and then translate it back to the time
variable t, which has another origin.
In the example time line below, note that the time of the second
switching was t = 2[s]. If we had a solution that would work for
t' = 0, we can transform it to the time variable t, by replacing
t' with t – 2[s], since t' = t – 2[s].
time of
first
switching
0
1
time of
second
switching
0
1
2
t', in [s]
2
3
4
t, in [s]
Sequential Switching – 3
Dave Shattuck
University of Houston
© University of Houston
The general solution equation when there is more than one
switching event, which occur at more than one point in time,
applies a transformation of variables. The general solution has
the quantity t – t0, wherever the variable t would otherwise occur.
The other key is that the initial condition for this time period
occurs at the time of the second switching, and holds for the time
after the second switching. The general solution then becomes
x(t )  x f   x(t0 )  x f  e
time of
first
switching
0
 t t 
 0 
  
; for t  t0 .
time of
second
switching
0
t0
t'
t0
2t0
t
Sequential Switching – 4
Dave Shattuck
University of Houston
© University of Houston
Note that this new general solution includes, as a
special case, the solution when the switching occurs at
t = 0. In this case, t0 = 0, and this solution reduces to
the solution that we had previously.
x(t )  x f   x(t0 )  x f  e
time of
first
switching
0
 t t 
 0 
  
; for t  t0 .
time of
second
switching
0
t0
t'
t0
2t0
t
Dave Shattuck
University of Houston
© University of Houston
Sequential Switching – Example – 1
Let's solve this example problem, below.
The problem has two switches, called SWA and SWB.
We are told that switch SWA was closed for a long time,
and switch SWB was open for a long time before t = 0.
Then, at t = 0, switch SWB closed. Then, at t = 1[s],
switch SWA opened. We wish to find the voltage
across the capacitor, vC(t), for t  0.
+
vC
RS2=8[W]
t=0
C=2[F]
-
SWA
VS1=
9[V]
SWB
t = 1[s]
+
RS1=5[W]
-
VS2=
-4[V]
+
-
Dave Shattuck
University of Houston
© University of Houston
Sequential Switching – Example – 2
The problem has two switches, called SWA and SWB. We are told that
switch SWA was closed for a long time, and switch SWB was open for a
long time before t = 0. Then, at t = 0, switch SWB closed. Then, at
t = 1[s], switch SWA opened. We wish to find the voltage across the
capacitor, vC(t), for t  0.
Our first step is to redraw the circuit for t < 0. We replace the capacitor
with an open circuit, since it has been in the condition for a long time, and
the voltages and currents have stopped changing. We close switch SWA,
and open switch SWB, since that was where they were for t < 0. We have
the circuit that follows.
SWB
VS1=
9[V]
+
vC(0)
-
SWA
RS2=8[W]
+
for t < 0:
RS1=5[W]
-
VS2=
-4[V]
Clearly, the
voltage
vC(0) = VS1
= 9[V].
+
-
Dave Shattuck
University of Houston
© University of Houston
Sequential Switching – Example – 3
The problem has two switches, called SWA and SWB. We are told that
switch SWA was closed for a long time, and switch SWB was open for a
long time before t = 0. Then, at t = 0, switch SWB closed. Then, at
t = 1[s], switch SWA opened. We wish to find the voltage across the
capacitor, vC(t), for t  0.
The next step is to redraw the circuit for 0 < t < 1[s]. This is the time period
until the next switching occurs. During this time, the capacitor acts like a
capacitor; we do not replace it with anything else. We close switch SWA,
and close switch SWB. We have the circuit that follows.
SWB
VS1=
9[V]
+
vC
C=2[F]
-
SWA
RS2=8[W]
+
for 0 < t < 1[s]:
RS1=5[W]
-
VS2=
-4[V]
We want the
time
constant,
which means
we need the
equivalent
resistance
seen by the
capacitor.
+
-
Dave Shattuck
University of Houston
© University of Houston
Sequential Switching – Example – 4
The problem has two switches, called SWA and SWB. We are told that
switch SWA was closed for a long time, and switch SWB was open for a
long time before t = 0. Then, at t = 0, switch SWB closed. Then, at
t = 1[s], switch SWA opened. We wish to find the voltage across the
capacitor, vC(t), for t  0.
To find the equivalent resistance seen by the capacitor, we remove the
capacitor, and find the resistance seen by the two terminals of the
capacitor, which we will name A and B. We next set the independent
sources equal to zero, which in this case means that they act like short
circuits. We have the circuit that follows.
The equivalent resistance
for 0 < t < 1[s]:
seen by the capacitor,
RS1=5[W]
RS2=8[W]
that is, seen by terminals
SWB
A
A and B, will be RS1||RS2,
SWA
or 5[W]||8[W] which is
3.1[W]. This gives us the
time constant for this time
period,
 = REQC = 6.2[s] for
B
0 < t < 1[s].
Sequential Switching – Example – 5
Dave Shattuck
University of Houston
© University of Houston
The problem has two switches, called SWA and SWB. We are told that
switch SWA was closed for a long time, and switch SWB was open for a
long time before t = 0. Then, at t = 0, switch SWB closed. Then, at
t = 1[s], switch SWA opened. We wish to find the voltage across the
capacitor, vC(t), for t  0.
The next step find the final value for the circuit for 0  t  1[s]. This will be
the circuit valid when the voltages and currents would have stopped
changing. Note that this will not happen for this time period, since it is less
than a time constant. The voltage never reaches this final value.
However, we need it to get the equation for the solution. We have the
circuit that follows.
Final value for t = 
RS1=5[W]
i
SWB
X
+
vC()
-
+
VS1=
9[V]
RS2=8[W]
+
SWA
We solve for this voltage
in two parts. First, we
find the current, iX,
-

Note that the voltage does not actually reach this final value.
VS2=
-4[V]
iX 
VS 1  VS 2 13[V]

 1[A].
RS 1  RS 2 13[W]
Then, vC (* )  VS1  iX RS1 
vC (* )  9[V]  1[A]5[W]  4[V].
Sequential Switching – Example – 6
Dave Shattuck
University of Houston
© University of Houston
The problem has two switches, called SWA and SWB. We are told that
switch SWA was closed for a long time, and switch SWB was open for a
long time before t = 0. Then, at t = 0, switch SWB closed. Then, at
t = 1[s], switch SWA opened. We wish to find the voltage across the
capacitor, vC(t), for t  0.
The final value for the circuit for 0  t  1[s] is the circuit when the voltages
and currents would have stopped changing. We call this the steady-state
value, which is what is needed for the equation. It is better to use this
terminology. We have the circuit that follows.
Steady State value for 0 < t < 1[s]
RS1=5[W]
RS2=8[W]
i
SWB
X
vC,SS
-
+
VS1=
9[V]
+
+
SWA
We solve for this voltage
in two parts. First, we
find the current, iX,
-

Note that the voltage does not actually reach this final value.
VS2=
-4[V]
iX 
VS 1  VS 2 13[V]

 1[A].
RS 1  RS 2 13[W]
Then, vC ,SS  VS1  iX RS1 
vC , SS  9[V]  1[A]5[W]  4[V].
Sequential Switching – Example – 7
Dave Shattuck
University of Houston
© University of Houston
The problem has two switches, called SWA and SWB. We are told that
switch SWA was closed for a long time, and switch SWB was open for a
long time before t = 0. Then, at t = 0, switch SWB closed. Then, at
t = 1[s], switch SWA opened. We wish to find the voltage across the
capacitor, vC(t), for t  0.
Now we can write the expression for the voltage vC(t) for 0  t  1[s]. We
have found the initial condition, the time constant, and the steady-state
value. We can write the solution as
vC (t )  4   9  4  e
SWB
VS1=
9[V]
+
vC
C=2[F]
-
SWA
RS2=8[W]
+
for 0 < t < 1[s]:
RS1=5[W]
-
 t 


 6.2[s] 
VS2=
-4[V]
[V]; for 0  t  1[s].
+
-
Dave Shattuck
University of Houston
© University of Houston
Sequential Switching – Example – 8
The problem has two switches, called SWA and SWB. We are told that
switch SWA was closed for a long time, and switch SWB was open for a
long time before t = 0. Then, at t = 0, switch SWB closed. Then, at
t = 1[s], switch SWA opened. We wish to find the voltage across the
capacitor, vC(t), for t  0.
Now we want to find the solution for t  1[s]. To do this, we need to again
find the initial condition, time constant, and final value, this time for the new
time period, during which switch SWA is open. That circuit is shown below.
SWB
VS1=
9[V]
+
vC(t)
C=2[F]
-
SWA
RS2=8[W]
+
For t > 1[s]:
RS1=5[W]
-
VS2=
-4[V]
The portion of
the circuit to
the left of the
capacitor does
not affect the
rest of this
solution, and
we will remove
it.
+
-
Dave Shattuck
University of Houston
© University of Houston
Sequential Switching – Example – 9
The problem has two switches, called SWA and SWB. We are told that
switch SWA was closed for a long time, and switch SWB was open for a
long time before t = 0. Then, at t = 0, switch SWB closed. Then, at
t = 1[s], switch SWA opened. We wish to find the voltage across the
capacitor, vC(t), for t  0.
Now we want to find the initial condition of the solution for t  1[s]. Here,
the solution for 0  t  1[s] has not had time to reach a steady state, thus
the voltages and currents do not stop changing. However, we can get the
initial value for the capacitive voltage, since it cannot change
instantaneously. Substituting the time t = 1[s] into the solution for
0  t  1[s], we get
For t > 1[s]:
RS2=8[W]
SWB
C=2[F]
vC (1[s])  4   5  e
+
+
vC(t)
-
-
VS2=
-4[V]
 1[s] 


 6.2[s] 
vC (1[s])  8.25[V].
[V] 
Dave Shattuck
University of Houston
© University of Houston
Sequential Switching – Example – 10
The problem has two switches, called SWA and SWB. We are told that
switch SWA was closed for a long time, and switch SWB was open for a
long time before t = 0. Then, at t = 0, switch SWB closed. Then, at t =
1[s], switch SWA opened. We wish to find the voltage across the
capacitor, vC(t), for t  0.
Next, we want to find the time constant of the solution for t  1[s]. Here,
the equivalent resistance seems fairly clear, and we can see that the
equivalent resistance will be 8[W], after we have set the independent
source equal to zero. The time constant will be RS2C, which will be
For t > 1[s]:
RS2=8[W]
  2[F]8[W]  16[s].
SWB
C=2[F]
+
+
vC(t)
-
-
VS2=
-4[V]
Dave Shattuck
University of Houston
© University of Houston
Sequential Switching – Example – 11
The problem has two switches, called SWA and SWB. We are told that
switch SWA was closed for a long time, and switch SWB was open for a
long time before t = 0. Then, at t = 0, switch SWB closed. Then, at
t = 1[s], switch SWA opened. We wish to find the voltage across the
capacitor, vC(t), for t  0.
Next, we want to find the final value of the solution for t  1[s]. Here, the
solution seems fairly clear. For t = , the final value can be obtained when
the capacitor is replaced by an open circuit, since the voltages and
currents will have stopped changing. We get the circuit below, and we can
solve, to get
For t = :
+
vC()
vC ()  4[V].
+
SWB
RS2=8[W]
-
-
VS2=
-4[V]
Dave Shattuck
University of Houston
© University of Houston
Sequential Switching – Example – 12
The problem has two switches, called SWA and SWB. We are told that
switch SWA was closed for a long time, and switch SWB was open for a
long time before t = 0. Then, at t = 0, switch SWB closed. Then, at
t = 1[s], switch SWA opened. We wish to find the voltage across the
capacitor, vC(t), for t  0.
Now we can write the expression for the voltage vC(t) for t  1[s]. We have found
the initial condition, the time constant, and the final value. We can write the
solution as
For t = :
+
vC()
vC (t )  4   8.25  4  e
RS2=8[W]
+
SWB
-
-
VS2=
-4[V]
vC (t )  4  12.25  e
 t 1[s] 


 16[s] 
 t 1[s] 


 16[s] 
[V]; for t  1[s], or
[V]; for t  1[s]
Dave Shattuck
University of Houston
© University of Houston
Sequential Switching – Example – 13
The problem has two switches, called SWA and SWB. We are told that
switch SWA was closed for a long time, and switch SWB was open for a
long time before t = 0. Then, at t = 0, switch SWB closed. Then, at t =
1[s], switch SWA opened. We wish to find the voltage across the
capacitor, vC(t), for t  0.
So, our complete solution, which has two parts, is the following,
vC (t )  4   5  e
+
vC
C=2[F]
-
[V]; for 0  t  1[s], and
 t 1[s] 


 16[s] 
t=0
-
SWA
VS1=
9[V]
SWB
t = 1[s]
vC (t )  4  12.25 e
RS2=8[W]
+
RS1=5[W]
 t 


 6.2[s] 
VS2=
-4[V]
[V]; for t  1[s].
+
-
Note 1 – Sequential
Switching
Dave Shattuck
University of Houston
© University of Houston
Notice, as you go through the previous slides in this example, that we
switched back and forth between < and >, and  and . This has been
done carefully. Note that when we talk about the capacitive voltage, vC, we
used  and . When we talk about anything else, such as the circuit as a
whole, we use < and >, since there can be step changes in any other
quantity.
 t 
vC (t )  4   5  e


 6.2[s] 
vC (t )  4  12.25 e
+
vC
RS2=8[W]
t=0
C=2[F]
-
SWA
VS1=
9[V]
SWB
t = 1[s]
+
RS1=5[W]
-
VS2=
-4[V]
[V]; for 0  t  1[s], and
 t 1[s] 


 16[s] 
[V]; for t  1[s].
+
-
Dave Shattuck
University of Houston
© University of Houston
Negative Time Constants – 1
A circuit that illustrates what happens with a
negative time constant is shown below. The switch
was closed for a long time, and then opened at
t = 0. We will solve this circuit to find vX(t) for t > 0.
Before we start, though, let us consider what we
should expect from a negative-valued time
constant.
+
vX
10[W]
1[mA]
30[W]
-
4[mH]
t=0
0.2[S]vX
50[W]
Negative Time Constants – 2
Dave Shattuck
University of Houston
© University of Houston
Let's start by remembering what happens when we have a positivevalued time constant. We get a solution that moves from an initial
condition, towards a final value. It moves exponentially, meaning that it
approaches the final value, getting closer and closer but never actually
reaching the final value, which occurs when things stop changing. The
final value is an steady-state value, and is found with all derivatives
equal to zero, which means that inductors become short circuits, and
capacitors become open circuits.
15
10
5
Current in [mA]
This was for
a positive
time
constant.
0
0
50
100
150
-5
-10
-15
-20
time (t) in [ms]
200
250
300
Negative Time Constants – 3
Dave Shattuck
University of Houston
© University of Houston
Now, notice what happens when we extend the range of
time for this solution to t < 0. This is the same solution, but
plotted for negative values of time. Note that the time
constant is still positive. The solution is decreasing rapidly
as time goes more and more negative, that is, moving to the
left.
Positive Time Constant
25
0
-100
-50
0
50
100
current, in [mA]
-25
-50
-75
-100
-125
-150
-175
time, in [ms]
150
200
250
300
Negative Time Constants – 4
Dave Shattuck
University of Houston
© University of Houston
Next, see what happens when we change the time constant
to a negative value. We have also changed the time period
to that from –300[ms] to +100[ms]. This looks just like our
last plot, but rotated around the ordinate (vertical axis).
Note in particular that we have the same value at t = 0, and
we have the same steady state value, which is 10[mA].
Positive Time Constant
25
25
0
0
-25
0
100
200
-300
300
-200
-100
-50
-75
-100
0
-25
current in [mA]
current, in [mA]
-100
Negative Time Constant
-50
-75
-100
-125
-125
-150
-150
-175
-175
time, in [ms]
time in [ms]
100
Negative Time Constants – 5
Dave Shattuck
University of Houston
© University of Houston
The key, then, is that when we have a negative time constant, our
solution technique really does not change. The final value is no longer
a final value, but it is still the steady-state value, and is found in the
same way as with a positive time constant. That is, we replace the
inductor with a short circuit, and the capacitor with an open circuit. We
should really refer to the final value as the steady-state value, which
happens to occur at t =  for the case where the time constant is
positive, and happens to occur at t = - for the case where the time
constant is negative.
Positive Time Constant
Negative Time Constant
25
25
0
-25
0
0
100
200
300
-300
-200
-100
current in [mA]
current, in [mA]
-100
-50
-75
-100
-25
-50
-75
-100
-125
-125
-150
-150
-175
-175
time, in [ms]
time in [ms]
0
100
Dave Shattuck
University of Houston
© University of Houston
Negative Time Constant Example – 1
Let's solve this circuit, which will have a negative-valued time constant.
The switch was closed for a long time, and then opened at t = 0. We
will solve this circuit to find vX(t) for t > 0.
We will start, as always, by defining the inductive current. Then, our next
step is to find the initial condition. For this circuit, that means drawing the
circuit for t < 0. Remember that the circuit had been in the given condition
for a long time.
+
vX
10[W]
1[mA]
30[W]
iL
-
L=
4[mH]
t=0
0.2[S]vX
50[W]
Dave Shattuck
University of Houston
Negative Time Constant Example – 2
© University of Houston
Let's solve this circuit, which will have a negative-valued time constant.
The switch was closed for a long time, and then opened at t = 0. We
will solve this circuit to find vX(t) for t > 0.
We start by finding the initial condition. We redraw the circuit for t < 0. The
switch was closed. Remember that the circuit had been in the condition for
a long time, so we replace the inductor with a short circuit. When we see
this circuit, it is clear that vX is zero. In addition, there is no current through
the 30[W] and 50[W] resistors. Thus, we can say that
iL (0)  1[mA].
For t < 0:
+
vX
10[W]
1[mA]
30[W]
iL(0)
-
0.2[S]vX
50[W]
Dave Shattuck
University of Houston
Negative Time Constant Example – 3
© University of Houston
Let's solve this circuit, which will have a negative-valued time constant.
The switch was closed for a long time, and then opened at t = 0. We
will solve this circuit to find vX(t) for t > 0.
Next, we need to find the time constant. We redraw the circuit for t > 0.
The switch is now open.
For t > 0:
+
vX
10[W]
1[mA]
30[W]
A -
L=
4[mH]
iL
B
0.2[S]vX
50[W]
Dave Shattuck
University of Houston
Negative Time Constant Example – 4
© University of Houston
Let's solve this circuit, which will have a negative-valued time constant.
The switch was closed for a long time, and then opened at t = 0. We
will solve this circuit to find vX(t) for t > 0.
With the circuit redrawn for t > 0, we can find the time constant. We want
the equivalent resistance seen by the inductor. Thus, we remove the
inductor, and then set the independent source equal to zero. Then, since
we have a dependent source present, we apply a test source to the
terminals of the inductor, A and B.
For t > 0:
+
vX
+
We have
v X  1[A]10[W]  10[V].
10[W]
30[W]
A +
vT
vD
0.2[S]vX
50[W]
Writing KCL at the top
node, we get
1[A]
B
-
0.2[S]  10[V] 
vD
v
 1[ A]  D  0.
50[W]
30[W]
Dave Shattuck
University of Houston
Negative Time Constant Example – 5
© University of Houston
Let's solve this circuit, which will have a negative-valued time constant.
The switch was closed for a long time, and then opened at t = 0. We
will solve this circuit to find vX(t) for t > 0.
We want the equivalent resistance seen by the inductor. Solving for vD, we get
vD
v
 1[ A]  D  0. Solving for vD yields
50[W]
30[W]
vD  18.75[V].
2[A] 
+
vT  10[V]  18.75[V] 
vT  8.75[V].
10[W]
30[W]
A +
vT
vD
1[A]
B
vX  vD  vT  0. Thus, we have
vT  v X  vD 
For t > 0:
+
vX
Then, we write KVL around
the loop, and we get
-
0.2[S]vX
50[W]
The equivalent resistance is
v
REQ  T . Thus, we have
1[A]
8.75[V]
REQ 
 8.75[W].
1[A]
Dave Shattuck
University of Houston
Negative Time Constant Example – 6
© University of Houston
Let's solve this circuit, which will have a negative-valued time constant.
The switch was closed for a long time, and then opened at t = 0. We
will solve this circuit to find vX(t) for t > 0.
From this result, we can find the time constant, which is
L
4[mH]


 460[ s].
REQ 8.75[W]
Next, we need to find the steadystate value of the current through
the inductor.
For t > 0:
+
vX
10[W]
1[mA]
30[W]
A -
L=
4[mH]
iL
B
0.2[S]vX
50[W]
Dave Shattuck
University of Houston
Negative Time Constant Example – 7
© University of Houston
Let's solve this circuit, which will have a negative-valued time constant.
The switch was closed for a long time, and then opened at t = 0. We
will solve this circuit to find vX(t) for t > 0.
The steady-state value of the inductive current can be found by redrawing
the circuit, for the switches in the t > 0 positions, and the inductor replaced
by a short circuit. We have the circuit shown below.
vX
v
v
 X  X  1[mA]  0.2[S]vX  0.
50[W] 10[W] 30[W]
For steady state:
Solving, we have
+
vX
46.7[mS]vX  1[mA], or
10[W]
1[mA]
30[W]
A -
0.2[S]vX
v X  21.4[mV]. Thus,
50[W]
iL , SS
iL,SS
B
vX

 2.14[mA].
10[W]
Dave Shattuck
University of Houston
Negative Time Constant Example – 8
© University of Houston
Let's solve this circuit, which will have a negative-valued time constant.
The switch was closed for a long time, and then opened at t = 0. We
will solve this circuit to find vX(t) for t > 0.
Thus, our solution for the inductive current is
iL (t )  2.14  3.14e
iL (t )  2.14  3.14e


t


 460[  s] 


t


 460[  s] 
[mA]; for t  0. This simplifies to
[mA]; for t  0.
For t > 0:
+
vX
10[W]
1[mA]
30[W]
A -
4[mH]
iL
B
0.2[S]vX
50[W]
Since this is the
current through the
10[W] resistor, we can
say that
vX  iL (t )10[W], or
vX  21.4  31.4e


t


 460[  s] 
[mV]; for t  0.
Dave Shattuck
University of Houston
© University of Houston
Note 1 – Negative
Time Constants
Note that the voltage vX made a step jump in
this problem. Just before t = 0 (at t = 0-) the
voltage vX was zero, because of short across it.
Just after t = 0 (at t = 0+) the voltage vX was
10[mV]. This is one of the reasons why it is a
good idea to solve for the inductive currents
and capacitive voltages first.
vX (t )  21.4  31.4e


t


 460[  s] 
[mV]; for t  0.
Dave Shattuck
University of Houston
© University of Houston
Note 2 – Negative
Time Constants
Some students are concerned about the practical
implications of a voltage which increases exponentially with
time. This concern is reasonable. In practice, such a circuit
cannot continue like this for a very long time. Note also that
this circuit is an exception to our rule that the six circuits we
are studying always move to steady-state conditions “after a
long time”.
Such circuits are useful, however. They require an
amplifier (the dependent source) to be able to provide the
energy required. This kind of situation typically results from
positive feedback, which is considered “unstable”. More about
all of this may be covered in your future electronics courses.
vX (t )  21.4  31.4e


t


 460[  s] 
[mV]; for t  0.
Dave Shattuck
University of Houston
© University of Houston
Complete Generalized
Solution
We are now ready for our most generalized version of the
solution of first order circuits. We use xss for the steady-state
value, which is also the final value when the time constant is
positive. We use t0 for the time of switching, so that it can be
at times other than t = 0. Using all this, we get the following
general solution,
x(t )  xSS   x(t0 )  xSS  e
 t t 
 0 
  
; for t  t0 .
In this expression, we should note that  is L/R in the RL
case, and that  is RC in the RC case.
The expression for greater-than-or-equal-to () is only used
for inductive currents and capacitive voltages.
Any other variables in the circuits can be found from these.
Dave Shattuck
University of Houston
© University of Houston
Generalized Solution Technique
– Step Response
To find the value of any variable in a Step
Response circuit, we can use the following general
solution,
 t t0 
x(t )  xSS   x(t0 )  xSS  e


  
; for t  t0 .
Our steps will be:
1) Define the inductive current iL, or the capacitive voltage vC.
2) Find the initial condition, iL(t0), or vC(t0).
3) Find the time constant, L/REQ or REQC. In general the REQ is the
equivalent resistance as seen by the inductor or capacitor, and found
through Thévenin’s Theorem.
4) Find the steady-state value, iL,SS, or vC ,SS.
5) Write the solution for inductive current or capacitive voltage using the
general solution.
6) Solve for any other variable of interest using the general solution found
in step 5).
Dave Shattuck
University of Houston
© University of Houston
6 Different First-Order Circuits
The generalized solution that we just
found applies only to the six different
STC circuits.
1. An inductor and a resistance (called
RL Natural Response).
2. A capacitor and a resistance (called
RC Natural Response).
3. An inductor and a Thévenin
equivalent (called RL Step
Response).
4. An inductor and a Norton equivalent
(also called RL Step Response).
5. A capacitor and a Thévenin
equivalent (called RC Step
Response).
6. A capacitor and a Norton equivalent
(also called RC Step Response).
x(t )  xSS   x(t0 )  xSS  e
 t t 
 0 
  
; for t  t0 .
RX
LX
CX
RX
RX
+
vS
LX
-
iS
RX
iS
RX
LX
RX
+
vS
CX
-
If we cannot reduce the circuit,
for t > 0, to one of these six
cases, we cannot solve using
this approach.
CX
Generalized Solution Technique
– Example 1
To illustrate these steps, let’s work Assessment
Problem 7.8, from page 240 of the 10th Edition.
Dave Shattuck
University of Houston
© University of Houston
Our steps will be:
1) Define the inductive current iL, or the capacitive voltage vC.
2) Find the initial condition, iL(0), or vC(0).
3) Find the time constant, L/R or RC. In general the R is the equivalent
resistance, REQ, as seen by the inductor or capacitor, and found through
Thévenin’s Theorem.
4) Find the steady-state value, iL,SS, or vC ,SS.
5) Write the solution for inductive current or capacitive voltage using the
general solution.
6) Solve for any other variable of interest using the general solution found
in step 5).
t t
x(t )  xSS   x(t0 )  xSS  e


 0 
  
; for t  t0 .
Problem 7.8 from page 240
Switch a in the circuit shown has been
in the 10th Edition
Dave Shattuck
University of Houston
© University of Houston
open for a long time, and switch b has
been closed for a long time. Switch a is
closed at t = 0 and, after remaining closed
for 1[s], is opened again. Switch b is
opened simultaneously, and both switches
remain open indefinitely. Determine the
expression for the inductor current iL that
is valid when (a) 0 < t < 1[s]; and
(b) t > 1[s].
Dave Shattuck
University of Houston
© University of Houston
Generalized Solution Technique
– Example 2
To illustrate these steps, let’s work another
problem, on the board.
Solution: iX(200[ms]) = -2.70[mA] (Quiz 5, Spring 2003)
Generalized Solution Technique
– Example 3
Both switches in this circuit had been in position a for a
long time before t = 0. At t = 0, switch SW2 moved to
position b, and switch SW1 moved to position b 1[ms]
later.
a) Find iR(2[ms]).
b) Find the energy stored in the 7.2[μF] capacitor at
t = 2[ms].
Dave Shattuck
University of Houston
© University of Houston
Solution: iR(2[ms]) = -2.327[mA]; wSTO,7.2(2[ms])=55.6[J]
1.5[kΩ]
a
SW1
b
2.2[kΩ]
t=1[ms]
3[mA]
4.7[kΩ]
SW2
a
7.5[kΩ]
3.3[kΩ]
t=0
25[V]
2.7[kΩ]
5iX
7.2[μF]
iX
iR
b
5.6[μF]
+
39[kΩ]
_
Generalized Solution Technique
– Example 4
Dave Shattuck
University of Houston
© University of Houston
For the circuit shown below, switches SW1 and
SW2 have been in position a for a long time. At
t = 0, both switches are moved instantaneously and
simultaneously to position b and remain there.
Calculate the numerical value of the total energy
stored in the capacitors at t = .
Solution: 25[mJ]
C2 = 2[F]
SW1
a
+
vS1 = C =
1
100[V] 3[F]
t=0
b
t=0
b SW2
R2 = 5[kW]
a
R3 = 40[kW]
vS2 =
200[V]
+
-
R1 = 10[kW]
-
Dave Shattuck
University of Houston
© University of Houston
Isn’t this situation pretty rare?
• This is a good question. Yes, it would seem to
be a pretty special case, until you realize that
with Thévenin’s Theorem, many more circuits
can be considered to be equivalent to these
special cases.
• In fact, we can say that the RL technique will
apply whenever we have only one inductor, or
inductors that can be combined into a single
equivalent inductor, and no capacitors.
• A similar rule holds for the RC technique.
Many circuits fall into one of these two groups.
• Note that the Natural Response is
simply a special case of the Step
Go back to
Response, with a final value of zero.
Overview
slide.