EGR 2201 Unit 9
First-Order Circuits



Read Alexander & Sadiku, Chapter 7.
Homework #9 and Lab #9 due next
week.
Quiz next week.
Review: DC Conditions in a Circuit
with Inductors or Capacitors



Recall that when power
is first applied to a
dc circuit with
inductors or
capacitors, voltages
and currents change briefly as the
inductors and capacitors become energized.
But once they are fully energized (i.e.,
“under dc conditions”), all voltages and
currents in the circuit have constant values.
To analyze a circuit under dc conditions,
replace all capacitors with open circuits and
replace all inductors with short circuits.
What About the Time Before DC
Conditions?


We also want to be able to analyze
such circuits during the time while
the voltages and currents are
changing, before dc conditions have
been reached.
This is sometimes called transient
analysis, because the behavior that
we’re looking at is short-lived. It’s
the focus of Chapters 7 and 8 in the
textbook.
Four Kinds of First-Order Circuits


The circuits we’ll study in this unit are called
first-order circuits because they are
described mathematically by first-order
differential equations.
We’ll study four kinds of first-order circuits:
 Source-free RC circuits

Source-free RL circuits

RC circuits with sources

RL circuits with sources
Natural Response and Step
Response



We use the term
natural response to
refer to the behavior
of source-free circuits.
And we use the term
step response to
refer to the behavior
of circuits in which a
source is applied at some time.
So our goal in this unit is to
understand the natural response of
source-free RC and RL circuits, and to
understand the step response of RC
and RL circuits with sources.
Natural Response of Source-Free
RC Circuit (1 of 2)



Consider the circuit shown. Assume
that at time t=0, the capacitor is
charged and has an initial voltage, V0.
As time passes, the
initial charge on the
capacitor will flow
through the resistor,
gradually discharging
the capacitor.
This results in changing
voltage v(t) and currents iC(t) and iR(t),
which we wish to calculate.
Natural Response of Source-Free
RC Circuit (2 of 2)

Applying KCL,
𝑖𝐶 + 𝑖𝑅 = 0

Therefore

Therefore

𝑑𝑣 𝑣
𝐶
+ =0
𝑑𝑡 𝑅
𝑑𝑣
𝑣
+
=0
𝑑𝑡 𝑅𝐶
This equation is an example of a firstorder differential equation. How do
we solve it for v(t)?
Math Detour: Differential Equations

Differential equations arise frequently in
science and engineering. Some examples:
dv
2 v 8
dt
d 2v
dv
 5  6v  3t
2
dt
dt
d 4 v d 3v
dv
3 4  3  8  sin t
dt
dt
dt

A first-order diff. eq.
A second-order diff. eq.
A fourth-order diff. eq.
The equations above are all called linear
ordinary differential equations with
constant coefficients.
Solving Differential Equations

The differential equations on the previous
slide are quite easy to solve. The ones
shown below are more difficult.
dv
2t  7v  0
Non-constant coefficient
dt
2
 dv 
3
 2   (7v)  sin t
 dt 
v
v 2
2 7  t
t
x

Non-linear
A partial differential equation
In a later math course you’ll learn many
techniques for solving such equations.
A Closer Look at Our Differential
Equation
Note that our equation,

𝑑𝑣
𝑑𝑡
+
𝑣
𝑅𝐶
= 0,
contains two constants, R and C.
It also contains two variables, v and t.
Also, t is the independent variable, while v
is the dependent variable. We sometimes
indicate this by writing v(t) instead of just v.
Our goal is to write down an equation that
expresses v(t) in terms of t, such as:





𝑣(𝑡) = 8𝑡
𝑣 𝑡 = 𝑅𝐶 sin 𝑡 (But neither of those is right!)
Solving Our Differential Equation



To solve our equation,
𝑑𝑣
𝑑𝑡
𝑣
+
𝑅𝐶
technique called separation of variables.
First, separate the variables v and t:
𝑑𝑣
𝑑𝑡
=−
𝑣
𝑅𝐶
Then integrate both sides:
𝑡
ln 𝑣 = −
+ ln 𝐴
𝑅𝐶

= 0, use a
Then raise e to both sides:
𝑣(𝑡) = 𝐴𝑒
𝑡
−𝑅𝐶
where A is a constant
Apply the Initial Condition

At this point we have:
𝑡
−𝑅𝐶

𝑣(𝑡) = 𝐴𝑒
The last step is to note that if we set t equal
to 0, we get:
𝑣 0 =𝐴

But we assumed earlier that the initial
voltage is some value that we called V0. So
A must be equal to V0, and therefore:
𝑡
−
𝑣(𝑡) = 𝑉0 𝑒 𝑅𝐶
The Bottom Line


I won’t expect you to be able to reproduce
the derivation on the previous slides.
The important point is
to realize that whenever
we have a circuit like this
the solution for v(t) is:
𝑣(𝑡) =
𝑡
−
𝑉0 𝑒 𝑅𝐶
Graph of Voltage
Versus Time

Here’s a graph of
𝑡
−
𝑣(𝑡) = 𝑉0 𝑒 𝑅𝐶

This curve is called
a decaying
exponential curve.

Note that at first the voltage falls steeply
from its initial value (V0). But as time
passes, the descent becomes less steep.
The Time Constant



The values of R and
C determine how
rapidly the voltage
descends.
The product RC is
given a special name
(the time constant) and symbol ():
𝜏 = 𝑅𝐶
The greater  is, the more slowly the
voltage descends.
Don’t Confuse t and 

Since
𝑣(𝑡) =
𝑡
−
𝑉0 𝑒 𝑅𝐶
and  = RC, we can write
𝑣(𝑡) =


𝑡
−
𝑉0 𝑒 𝜏
Remember that t is a variable (time).
But  is a constant.
Rules of Thumb


After one time
constant (i.e., when
t = ), the voltage
has fallen to about
36.8% of its initial
value.
After five time constants (i.e., when t = 5),
the voltage has fallen to about 0.7% of its
initial value. For most practical purposes we
say that the capacitor is completely
discharged and v = 0 after five time
constants.
Comparing Different
Values of 

The greater  is, the more slowly the voltage
descends, as shown below for a few values
of .
Finding Values of
Other Quantities


𝑡
𝜏
−
We’ve seen that 𝑣(𝑡) = 𝑉0 𝑒
From this equation we can use our prior
knowledge to find equations for other
quantities, such as current, power, and
energy. For example, using Ohm’s law we
find that
𝑉0 −𝑡
𝑖𝑅 (𝑡) = 𝑒 𝜏
𝑅
The Keys to Working with
a Source-Free RC Circuit
1.
2.
3.
Find the initial voltage 𝑣(0) = 𝑉0 across
the capacitor.
Find the time constant 𝜏 = 𝑅𝐶.
Once you know these two items, voltage as
a function of time is: 𝑣(𝑡)
4.
= 𝑉0 𝑒
𝑡
−
𝜏
Once you know the voltage, solve for any
other circuit variables of interest.
More Complicated Cases

A circuit that looks more complicated at
first might be reducible to a simple sourcefree RC circuit by combining resistors.

Example: Here we can combine the three
resistors into a single equivalent resistor, as
seen from the capacitor’s terminals.
Where Did V0 Come From?


In previous examples you’ve simply been
given the capacitor’s initial voltage, V0.
More realistically, you have to find V0 by
considering what happened before t = 0.

Example: Suppose you’re told that the switch in
this circuit has been closed for a long time
before it’s opened
at t = 0. Can you
find the capacitor’s
voltage V0 at time
t = 0?
Natural Response of Source-Free
RL Circuit (1 of 2)



Consider the circuit shown. Assume
that at time t=0, the inductor is
energized and has an initial current, I0.
As time passes, the
inductor’s energy
will gradually dissipate
as current flows
through the resistor.
This results in changing
current i(t) and voltages vL(t) and
vR(t), which we wish to calculate.
Natural Response of Source-Free
RL Circuit (2 of 2)

Applying KVL,
𝑣𝐿 + 𝑣𝑅 = 0

Therefore

Therefore

𝑑𝑖
𝐿 + 𝑖𝑅 = 0
𝑑𝑡
𝑑𝑖 𝑅
+ 𝑖=0
𝑑𝑡 𝐿
This first-order differential equation is
similar to the equation we had for
source-free RC circuits.
Solving Our Differential Equation
𝑑𝑖
𝑑𝑡
𝑅
+ 𝑖
𝐿

To solve our equation,

separate the variables i and t:
𝑑𝑖
𝑅
= − 𝑑𝑡
𝑖
𝐿
Then integrate both sides:
𝑅𝑡
ln 𝑖 = − + ln 𝐴
𝐿

Then raise e to both sides:
𝑖(𝑡) = 𝐴𝑒
𝑅𝑡
−𝐿
= 0, first
Apply the Initial Condition

At this point we have:
𝑅𝑡
−𝐿

𝑖(𝑡) = 𝐴𝑒
The last step is to note that if we set t equal
to 0, we get:
𝑖 0 =𝐴

But we assumed earlier that the initial
current is some value that we called I0. So
A must be equal to I0, and therefore:
𝑅𝑡
−
𝑖(𝑡) = 𝐼0 𝑒 𝐿
The Bottom Line


I won’t expect you to be able to reproduce
the derivation on the previous slides.
The important point is
to realize that whenever
we have a circuit like this
the solution for i(t) is:
𝑖(𝑡) =
𝑅𝑡
−
𝐼0 𝑒 𝐿
The Time Constant




The values of R and L determine how
rapidly the current decreases from its initial
value to 0.
Recall that for RC circuits we defined the
time constant as 𝜏 = 𝑅𝐶.
For RL circuits, we define it as
𝐿
𝜏=
𝑅
The greater  is, the more slowly the
current decreases from its initial value.
Don’t Confuse t and 

Since
𝑖(𝑡) =
𝑅𝑡
−
𝐼0 𝑒 𝐿
and  = L/R, we can write
𝑖(𝑡) =


𝑡
−
𝐼0 𝑒 𝜏
Remember that t is a variable (time).
But  is a constant.
Graph of Current
Versus Time

Here’s a graph of
𝑡
−
𝑖(𝑡) = 𝐼0 𝑒 𝜏

It’s a decaying exponential curve, with the
current falling steeply from its initial value
(I0). But as time passes, the descent
becomes less steep.
Rules of Thumb


After one time
constant (i.e., when
t = ), the current
has fallen to about
36.8% of its initial
value.
After five time constants (i.e., when t = 5),
the current has fallen to about 0.7% of its
initial value. For most practical purposes we
say that the inductor is completely deenergized and i = 0 after five time constants.
Finding Values of
Other Quantities


𝑡
𝜏
−
We’ve seen that 𝑖(𝑡) = 𝐼0 𝑒
From this equation we can use our prior
knowledge to find equations for other
quantities, such as voltage, power, and
energy. For example, using Ohm’s law we
find that
𝑣𝑅 (𝑡) =
𝑡
−
𝐼0 𝑅𝑒 𝜏
The Keys to Working with
a Source-Free RL Circuit
1.
Find the initial current 𝑖(0) = 𝐼0 through
the inductor.
𝐿
.
𝑅
2.
Find the time constant 𝜏 =
3.
Once you know these two items, current as
a function of time is: 𝑖(𝑡)
4.
= 𝐼0 𝑒
𝑡
−
𝜏
Once you know the current, solve for any
other circuit variables of interest.
More Complicated Cases

A circuit that looks more complicated at
first might be reducible to a simple sourcefree RL circuit by combining resistors.

Example: Here we can combine the resistors
into a single equivalent resistor, as seen from
the inductor’s terminals.
Where Did I0 Come From?


In previous examples you were simply
given the inductor’s initial current, I0.
More realistically, you have to find I0 by
considering what happened before t = 0.

Example: Suppose you’re told that the switch in
this circuit has been closed for a long time
before it’s opened
at t = 0. Can you
find the inductor’s
current I0 at time
t = 0?
Where We Are

We’ve looked at:



We still need to look at:



Source-free RC circuits
Source-free RL circuits
RC circuits with sources
RL circuits with sources
Before doing this, we’ll look at some
mathematical functions called
singularity functions (or switching
functions), which are widely used to
model electrical signals that arise
during switching operations.
Three Singularity Functions

The three singularity functions that
we’ll study are:

The unit step function

The unit impulse function

The unit ramp function
The Unit Step Function

The unit step function u(t) is 0 for
negative values of t and 1 for positive
values of t:
0,
𝑢 𝑡 =
1,
𝑡<0
𝑡>0
Shifting and Scaling the Unit Step
Function

We can obtain other step functions by
shifting the unit step function to the
left or right…
0,
𝑢 𝑡−2 =
1,

𝑡 <2s
𝑡 >2s
…or by multiplying the unit step function
by a scaling constant:
0,
3𝑢 𝑡 =
3,
𝑡<0
𝑡>0
Flipping the Unit Step Function
Horizontally

As with any mathematical function, we
can “flip” the unit step function
horizontally by replacing t with t:
1,
𝑢 −𝑡 =
0,
𝑡<0
𝑡>0
Adding Step Functions

By adding two or more step functions
we can obtain more complex “steplike” functions, such as the one shown
below from the book’s Practice Problem
7.6.
Using Step Functions to Model
Switched Sources

Step functions are useful for modeling
sources that are switched on (or off) at
some time:
The Unit Impulse Function


The unit step function’s derivative is the
unit impulse function (t), also called
the delta function.
The unit impulse function is 0 everywhere
except at t =0, where it is undefined:
0,
𝑡<0
𝛿 𝑡 = Undefined, 𝑡 = 0
0,
𝑡>0

It’s useful for modeling “spikes” that can
occur during switching operations.
Shifting and Scaling the Unit
Impulse Function


We can obtain other impulse functions by
shifting the unit impulse function to the
left or right, or by multiplying the unit
impulse function by a scaling constant:
We won’t often use impulse functions in this
course, but they’re used in more advanced
courses.
The Unit Ramp Function


The unit step function’s integral is the
unit ramp function r(t).
The unit ramp function is 0 for negative
values of t and has a slope of 1 for positive
values of t:
0,
𝑟 𝑡 =
𝑡,
𝑡<0
𝑡≥0
Shifting and Scaling the Unit Ramp
Function

We can obtain other ramp functions by
shifting the unit ramp function to the
left or right…
0,
𝑟 𝑡 − 0.2 =
𝑡,

𝑡 < 0.2 s
𝑡 ≥ 0.2 s
…or by multiplying the unit ramp function
by a scaling constant:
0,
4𝑟 𝑡 =
4𝑡,
𝑡<0
𝑡≥0
Adding Step and Ramp Functions

By adding two or more step functions
or ramp functions we can obtain more
complex functions, such as the one
shown below from the book’s Example
7.7.
Step Response of a Circuit


A circuit’s step
response is the
circuit’s behavior due
to a sudden application
of a dc voltage source
or current source.
We can use a step
function to model this
sudden application.
t = 0 versus t = 0+

We distinguish the following two
times:




t = 0 (the instant just
before the switch closes)
t = 0+ (the instant just
after the switch closes)
Since a capacitor’s voltage
cannot change abruptly, we know that
v(0) = v(0+) in this circuit.
But on the other hand, i(0)  i(0+) in
this circuit.
Step Response of RC Circuit (1 of 2)



Assume that in the circuit shown, the
capacitor’s initial voltage is V0 (which
may equal 0 V).
As time passes after the
switch closes, the
capacitor’s voltage
will gradually approach
the source voltage VS.
This results in changing
voltage v(t) and current i(t), which we
wish to calculate.
Step Response of RC Circuit (2 of 2)




Applying KCL, for t>0,
𝑑𝑣 𝑣 − 𝑉𝑆
𝐶
+
=0
𝑑𝑡
𝑅
Therefore
𝑑𝑣
𝑣 − 𝑉𝑆
=−
𝑑𝑡
𝑅𝐶
Separating variables,
𝑑𝑣
𝑑𝑡
=−
𝑣 − 𝑉𝑆
𝑅𝐶
Integrating both sides,
𝑡
ln(𝑣 − 𝑉𝑆 ) = −
+ ln(𝐴)
𝑅𝐶
Solution of Our Differential
Equation

Raising e to both sides and rearranging:
𝑡
−𝑅𝐶

𝑣 𝑡 = 𝑉𝑆 + 𝐴𝑒
Applying the initial condition and letting
 = RC:
𝑣 𝑡 = 𝑉𝑆 + (𝑉0 − 𝑉𝑆 )𝑒

−
𝑡
𝜏
Finally,
𝑣 𝑡 =
𝑉0 ,
𝑉𝑆 + (𝑉0 − 𝑉𝑆
𝑡<0
𝑡
−
)𝑒 𝜏 ,
𝑡≥0
The Bottom Line


I won’t expect you to be able to reproduce
the derivation on the previous slides.
The important point is
to realize that for a
circuit like this:
the solution for v(t) is:
𝑣 𝑡 =
𝑉0 ,
𝑉𝑆 + (𝑉0 − 𝑉𝑆
𝑡<0
𝑡
−
)𝑒 𝜏 ,
𝑡≥0
Graph of Voltage
Versus Time

Here’s a graph of
𝑣 𝑡 =


𝑉0 ,
𝑉𝑆 + (𝑉0 −
𝑡<0
𝑡
−
𝑉𝑆 )𝑒 𝜏 ,
𝑡≥0
(assuming V0< VS).
This is a saturating
exponential curve.
Note that the voltage at first rises steeply
from its initial value (V0), and then gradually
approaches its final value (VS).
Rules of Thumb, etc.

We can repeat many of the same remarks
as for source-free circuits, such as:
 The greater  is, the more
slowly v(t) approaches its
final value.
 For most practical purposes,
v(t) reaches its final value
after 5.
 Knowing v(t), we can use Ohm’s law to
find current, and we can use other familiar
formulas to find power, energy, etc.
Another Way of Looking At It

We can rewrite the complete response
𝑣 𝑡 = 𝑉𝑆 + (𝑉0 −
𝑡
−
𝑉𝑆 )𝑒 𝜏
as:
𝑣 𝑡 = 𝑣(∞) + (𝑣(0) −


𝑡
−
𝑣(∞))𝑒 𝜏
Here v(0) is the initial value and v() is the
final, or steady-state, value.
This same equation works for
source-free RC circuits too, since
setting v() to 0 gives
𝑣 𝑡 = 𝑣(0)𝑒
−
𝑡
𝜏.
The Keys to Finding an
RC Circuit’s Step Response
1.
2.
3.
4.
Find the capacitor’s initial voltage 𝑣(0).
Find the capacitor’s final voltage 𝑣(∞).
Find the time constant 𝜏 = 𝑅𝐶.
Once you know these items, voltage is:
𝑡
−𝜏
5.
𝑣 𝑡 = 𝑣(∞) + (𝑣(0) − 𝑣(∞))𝑒
Once you know the voltage, solve for any
other circuit variables of interest.
More Complicated Cases

A circuit that looks more complicated at
first might be reducible to a simple RC
circuit by combining resistors.

Example: Here, for t > 0 we can combine the
resistors into a single equivalent resistor, as
seen from the capacitor’s terminals.
Two Ways of
Breaking It Down

There are two useful ways of looking
at the response
𝑣 𝑡 = 𝑣(∞) + (𝑣(0) − 𝑣(∞))𝑒
1.
2.
−
𝑡
𝜏
As the sum of a natural response
and a forced response.
As the sum of a transient
response and a steady-state
response. This way is of more
interest to us.
Natural Response Versus
Forced Response

We can think of the complete response
𝑣 𝑡 = 𝑣(∞) + (𝑣(0) −
𝑡
−
𝑣(∞))𝑒 𝜏
as being the sum of:
1.
2.
−
𝑡
𝜏
A natural response 𝑣𝑛 𝑡 = 𝑣(0)𝑒
that depends the capacitor’s initial
charge.
Plus a forced response 𝑣𝑓 𝑡 = 𝑣(∞) −
𝑡
−𝜏
𝑣(∞)𝑒
that depends on the voltage source.
Transient Response Versus
Steady-State Response

We can think of the complete response
𝑣 𝑡 = 𝑣(∞) + (𝑣(0) −
𝑡
−
𝑣(∞))𝑒 𝜏
as being the sum of:
−𝜏𝑡
1.
A transient response 𝑣𝑡 𝑡 = (𝑣(0) − 𝑣(∞))𝑒
that dies away as time passes.
2.
Plus a steady-state response 𝑣𝑠𝑠 𝑡 = 𝑣(∞)
that remains after the transient response
has died away.
“Under DC Conditions” = SteadyState


Recall that earlier we used the term
“under dc conditions” to refer to the
time after an RC or RL circuit’s currents
and voltages have “settled down” to
their final values.
This is just another way of referring to
what we’re now calling steady-state
values.

So way can say that in the steady state,
capacitors look like open circuits and
inductors look like short circuits.
Step Response of RL Circuit (1 of 2)



Assume that in the circuit shown, the
inductor’s initial current is I0 (which
may equal 0 A).
As time passes, the
inductor’s current
will gradually approach
a steady-state value.
This results in changing
current i(t) and voltage v(t), which we
wish to calculate.
Step Response of RL Circuit (2 of 2)

Using the same sort of
math we used previously
for RC circuits, we find
𝐼0 ,
𝑉𝑆 −𝑡
𝑖 𝑡 = 𝑉𝑆
+ 𝐼0 −
𝑒 𝜏,
𝑅
𝑅
𝑡<0
𝑡>0
where  = L/R, just as
for source-free RL
circuits.
Another Way of Looking
At It

We can rewrite
𝑖 𝑡 =
as:
𝑉𝑆
𝑅
𝑖 𝑡 = 𝑖(∞) +


+ 𝐼0 −
𝑉𝑆
𝑅
𝑡
−
𝑒 𝜏
𝑡
−
(𝑖(0) − 𝑖(∞))𝑒 𝜏
Here i(0) is the initial value and i() is the
final, or steady-state, value.
This same equation works for
source-free RL circuits too, since
setting i() to 0 gives
𝑖 𝑡 = 𝑖(0)𝑒
−
𝑡
𝜏.
Graph of Voltage
Versus Time

Here’s a graph of
𝑖 𝑡 = 𝑖(∞) + (𝑖(0) − 𝑖(∞))𝑒
(assuming i(0) > i(∞)).
−
𝑡
𝜏
Rules of Thumb, etc.

We can repeat many of the same remarks
as for previous circuits, such as:
 The greater  is, the more
slowly i(t) approaches its
final value.
 For most practical purposes,
i(t) reaches its final value
after 5.
 Knowing i(t), we can use Ohm’s law to
find voltage, and we can use other familiar
formulas to find power, energy, etc.
The Keys to Finding an
RL Circuit’s Step Response
1.
2.
3.
4.
Find the inductor’s initial current 𝑖(0).
Find the capacitor’s final current 𝑖(∞).
Find the time constant 𝜏 = 𝐿/𝑅.
Once you know these items, current is:
𝑡
−𝜏
5.
𝑖 𝑡 = 𝑖(∞) + (𝑖(0) − 𝑖(∞))𝑒
Once you know the current, solve for any
other circuit variables of interest.
More Complicated Cases

A circuit that looks more complicated at
first might be reducible to a simple RL
circuit by combining resistors.

Example: Here we can combine the resistors
into a single equivalent resistor, as seen from
the inductor’s terminals.
A General Approach for FirstOrder Circuits (1 of 3)

As noted on page 276 of the textbook:
 Once we know 𝑥(0), 𝑥(∞), and ,
almost all the circuit problems in this
chapter can be solved using the
formula
𝑥 𝑡 = 𝑥(∞) + (𝑥(0) − 𝑥(∞))𝑒

𝑡
−𝜏
What is x here? It could be any current
or voltage in a first-order circuit.
A General Approach for FirstOrder Circuits (2 of 3)

1.
2.
3.
4.
So to find x(t) in a first-order circuit, where x
could be any current or voltage:
Find the quantity’s initial value 𝑥(0).
Find the quantity’s final value 𝑥(∞).
Find the time constant:
 𝜏 = 𝑅𝐶 for an RC circuit.
 𝜏 = 𝐿/𝑅 for an RL circuit.
Once you know these items, solution is:
𝑡
𝑥 𝑡 = 𝑥(∞) + (𝑥(0) − 𝑥(∞))𝑒
−𝜏
A General Approach for FirstOrder Circuits (3 of 3)

The equation from the previous slide,
𝑡
−𝜏


𝑥 𝑡 = 𝑥(∞) + (𝑥(0) − 𝑥(∞))𝑒
always graphs as either:
A decaying exponential curve
if the initial value x(0) is greater
than the final value x().
Or a saturating exponential
curve if the initial value x(0)
is less than the final value x().