EGR 2201 Unit 10
Second-Order Circuits



Read Alexander & Sadiku, Chapter 8.
Homework #10 and Lab #10 due next
week.
Quiz next week.
Review: Four Kinds of First-Order
Circuits


The circuits we studied last week are called
first-order circuits because they are
described mathematically by first-order
differential equations.
We studied four kinds of first-order circuits:
 Source-free RC circuits

Source-free RL circuits

RC circuits with sources

RL circuits with sources
Review: A General Approach for
First-Order Circuits (1 of 3)

1.
2.
3.
4.
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
Review: A General Approach for
First-Order Circuits (2 of 3)

The graph of the previous equation,
𝑡
−𝜏


𝑥 𝑡 = 𝑥(∞) + (𝑥(0) − 𝑥(∞))𝑒
is 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().
Review: A General Approach for
First-Order Circuits (3 of 3)

Recall also that we can think of the
complete response as being the sum of a
transient response that dies away with
time and a steady-state response that is
constant and remains after the transient
has died away:
𝑡
𝑥 𝑡 = 𝑥(∞) + (𝑥(0) − 𝑥(∞))𝑒
Steady-state
response
Transient
response
−𝜏
Transient Analysis with Multisim

The textbook’s Sections 7.8 and 8.9 discuss
using PSpice simulation software to perform
transient analysis of first-order and secondorder circuits.

We can also do
this with Multisim,
as shown here.
Our Goal: A General Approach for
Second-Order Circuits


Next we will develop a general approach
for analyzing more complicated circuits
called second-order circuits.
Unfortunately the general approach for
second-order circuits is quite a bit more
complicated than the one for first-order
circuits.
Second-Order Circuits


The circuits we’ll study are called
second-order circuits because they
are described mathematically by
second-order differential equations.
Whereas first-order circuits contain a
single energy-storing element (capacitor
or inductor), second-order circuits
contain two energy-storing elements.

These two elements could both be
capacitors or both be inductors, but we’ll
focus on circuits containing one capacitor
and one inductor.
Four Kinds of Second-Order
Circuits

We’ll study four kinds of second-order circuits:
 Source-free series RLC circuits

Source-free parallel RLC circuits

Series RLC circuits with sources

Parallel RLC circuits with sources
Natural Response and Step
Response



Recall that the term
natural response
refers to the behavior
of source-free circuits.
And the term step
response refers 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 RLC circuits, and to
understand the step response of RLC
circuits with sources.
Redraw, Redraw, Redraw!

Our procedure will usually require us
to find values of voltages or currents
at the following three times:




At t = 0, just before a switch is opened or
closed.
At t = 0+, just after a switch is opened or
closed.
As t  , a long time after a switch is
opened or closed.
Usually the circuit looks different at
these three times, so you’ll want to
redraw the circuit for each of these
times.
Finding Initial Values

To completely solve a first-order
differential equation, you need one
initial condition, usually either:



An initial inductor current i(0+), or
An initial capacitor voltage v(0+).
To completely solve a second-order
differential equation, you need two
initial conditions, usually either:


An initial inductor current i(0+) and its
derivative di(0+)/dt, or
An initial capacitor voltage v(0+) and its
derivative dv(0+)/dt.
Finding Initial Derivative Values

To find initial derivative values such as
dv(0+)/dt, we’ll rely on the basic
relationships for capacitors and
inductors:
𝑑𝑣
𝑖=𝐶
𝑑𝑡

𝑑𝑖
𝑣=𝐿
𝑑𝑡
For example, if we know a capacitor’s
initial current i(0+), then we can use
the left-hand equation above to find
the initial derivative of that capacitor’s
voltage, dv(0+)/dt.
Quantities that Cannot Change
Abruptly

We’ll also rely on the fact that a
capacitor’s voltage and an inductor’s
current cannot change abruptly.


Example: In this
circuit, i(0+) must be
equal to i(0), and
v(0+) must be
equal to v(0).
Since these values
must be equal, we don’t really need to
distinguish between their values at time
t = 0 and at time t = 0+. So we could just
write i(0) instead of i(0+) and i(0).
Caution: Some Quantities Can
Change Abruptly

Don’t assume that every quantity
has the same value at times t = 0
and t = 0+.


Example: In the
same circuit, iC(t)
changes abruptly
from 0 A to 2 A
at time t = 0.
So we must distinguish between iC(0) and
iC(0+):
 iC(0) = 0 A
 iC(0+) = 2 A
 iC(0) is undefined.
Finding Final Values

Our procedure will sometimes also
require us to find final or “steady-state”
values, such as:



A final inductor current i()
A final capacitor voltage v().
Usually these final values are easier to
find than initial values, because:
1.
2.
We don’t have to worry about abrupt
changes as t  , so we never need to
distinguish between t   and t  +.
We don’t have to find derivatives of
currents or voltages as t  .
Natural Response of Source-Free
Series RLC Circuit (1 of 2)



Consider the circuit shown. Assume
that at time t=0, the inductor has
initial current I0, and the capacitor has
initial voltage V0.
As time passes, the
initial energy in the
capacitor and inductor
will dissipate as current
flows through the
resistor.
This results in changing current i(t),
which we wish to calculate.
Natural Response of Source-Free
Series RLC Circuit (2 of 2)



Applying KVL,
𝑑𝑖 1 𝑡
𝑅𝑖 + 𝐿 +
𝑖 𝜏 𝑑𝜏 = 0
𝑑𝑡 𝐶 −∞
A standard trick for such
integro-differential equations is to
take the derivative of both sides:
𝑑𝑖
𝑑2 𝑖 1
𝑅 +𝐿 2+ 𝑖 =0
𝑑𝑡
𝑑𝑡
𝐶
Now divide by L and rearrange terms:
𝑑2 𝑖 𝑅 𝑑𝑖
1
+
+
𝑖=0
2
𝑑𝑡
𝐿 𝑑𝑡 𝐿𝐶
A Closer Look at Our
Differential Equation


Our equation,
𝑑2𝑖
𝑑𝑡 2
+
𝑅 𝑑𝑖
𝐿 𝑑𝑡
1
+ 𝑖
𝐿𝐶
= 0, is a
second-order linear differential
equation with constant
coefficients.
Such equations have been studied
extensively, and we’ll outline the
standard solution.
Solving Our Differential
Equation (1 of 4)



𝑑2𝑖
𝑑𝑡 2
𝑅 𝑑𝑖
+
𝐿 𝑑𝑡
1
𝑖
𝐿𝐶
To solve
+
= 0, first
assume that the solution is of the form
𝑖 𝑡 = 𝐴𝑒 𝑠𝑡
where A and s are constants to be found.
Plugging this into the equation yields:
𝑅
1
2 𝑠𝑡
𝑠𝑡
𝐴𝑠 𝑒 + 𝐴𝑠𝑒 +
𝐴𝑒 𝑠𝑡 = 0
𝐿
𝐿𝐶
Factoring out the common term,
𝑅
1
𝐴𝑒 (𝑠 + 𝑠 + ) = 0
𝐿
𝐿𝐶
𝑠𝑡
2
Solving Our Differential
Equation (2 of 4)

𝑠𝑡
2
From 𝐴𝑒 (𝑠 +

+
1
)
𝐿𝐶
= 0 it follows that:
𝑅
1
+ 𝑠+
=0
𝐿
𝐿𝐶
This is called the characteristic equation of
our original differential equation. Note that it
is purely algebraic (no derivatives or
integrals).
Now we can use the quadratic formula to
solve for s. But first….
𝑠2

𝑅
𝑠
𝐿
Solving Our Differential
Equation (3 of 4)

For convenience we introduce two new
symbols:
𝛼=



𝑅
2𝐿
𝜔0 =
and
1
𝐿𝐶
We call  the neper frequency, and we call
0 the undamped natural frequency.
2
Then we can rewrite 𝑠 +
𝑅
𝑠
𝐿
+
𝑠 2 + 2𝛼𝑠 + 𝜔02 = 0
Now use the quadratic formula….
1
𝐿𝐶
= 0 as:
Solving Our Differential
Equation (4 of 4)

Applying the quadratic formula to
2
2
𝑠 + 2𝛼𝑠 + 𝜔0 = 0 gives two solutions for
s, which we call the natural frequencies:
𝑠1 = −𝛼 + 𝛼 2 − 𝜔02 ,

𝑠2 = −𝛼 − 𝛼 2 − 𝜔02
Because of the square root, we must
distinguish three cases:



 > 0, called the overdamped case.
 = 0, called the critically damped case.
 < 0, called the underdamped case.
The Overdamped Case ( > 0)

If  > 0, our two solutions for s are distinct
negative real numbers:
Real number
𝑠1 = −𝛼 + 𝛼 2 − 𝜔02 ,


𝑠2 = −𝛼 − 𝛼 2 − 𝜔02
In this case, the solution to our differential
equation is
𝑖 𝑡 = 𝐴1 𝑒 𝑠1 𝑡 + 𝐴2 𝑒 𝑠2 𝑡
The initial conditions determine A1 and A2:
𝐴1 + 𝐴2 = 𝑖(0+ ),
𝑠1 𝐴1 + 𝑠2 𝐴2 =
𝑑𝑖(0+ )
𝑑𝑡
The Critically Damped Case ( = 0)

If  = 0, our two solutions for s are equal
to each other and to :
𝑠1 = −𝛼 + 𝛼 2 − 𝜔02 = −𝛼 ,


Zero
𝑠2 = −𝛼 − 𝛼 2 − 𝜔02 = −𝛼
In this case, the solution to our differential
equation is
𝑖 𝑡 = (𝐴2 + 𝐴1 𝑡)𝑒 −𝛼𝑡
The initial conditions determine A1 and A2:
𝐴2 = 𝑖(0+ ),
𝐴1 =
𝑑𝑖(0+ )
𝑑𝑡
+ 𝛼𝐴2
The Underdamped Case ( < 0)

If  < 0, our two solutions for s are complex numbers:
𝑠1 = −𝛼 + 𝛼 2 − 𝜔02 = −𝛼 + 𝑗𝜔𝑑
Imaginary number
𝑠2 = −𝛼 − 𝛼 2 − 𝜔02 = −𝛼 − 𝑗𝜔𝑑

Here j is the imaginary unit, 𝑗 = −1, and d is called
the damped natural frequency, 𝜔𝑑 = 𝜔02 − 𝛼 2 .

In this case, the solution to our differential equation is
𝑖 𝑡 = 𝑒 −𝛼𝑡 (𝐵1 cos 𝜔𝑑 𝑡 + 𝐵2 sin 𝜔𝑑 𝑡)

The initial conditions determine B1 and B2:
𝐵1 = 𝑖(0+ ),
𝐵2 =
𝑑𝑖(0+ )
+ 𝛼𝐵1
𝑑𝑡
𝜔𝑑
Graphs of the Three Cases


Details will differ based on initial conditions and
element values, but the shapes shown here are typical.
Note the oscillation in the underdamped case.
Typing Equations in Word 2013
1.
2.
3.
Select Insert > Equation on Word’s menu bar.
Use the toolbar’s Structures section to create
fractions, exponents, square roots, and more.
Use the toolbar’s Symbols section to insert basic
math symbols, Greek letters, special operators, and
more.
Oscilloscope

Looking ahead, we’ll use an
oscilloscope to display and measure
fast-changing
voltages,
including
transients.
Oscilloscope Challenge Game


The oscilloscope is a complex
instrument that you must learn to use.
To learn the basics, play my
Oscilloscope Challenge game at
http://people.sinclair.edu/nickreeder/flashgames.htm.
Natural Response of Source-Free
Parallel RLC Circuit (1 of 2)

The math for a
source-free parallel
circuit is almost the
same as the math
on the previous
slides, except that:
1. Now the variable in our differential
equation is v(t) instead of i(t).
2. We define  (the neper frequency)
1
𝑅
to be equal to
instead of .
2𝑅𝐶
2𝐿
Natural Response of Source-Free
Parallel RLC Circuit (2 of 2)

By applying KCL and some
algebra, we get:
𝑑2 𝑣
1 𝑑𝑣
1
+
+
𝑣=0
2
𝑑𝑡
𝑅𝐶 𝑑𝑡 𝐿𝐶
By assuming a solution of the form
𝑣 𝑡 = 𝐴𝑒 𝑠𝑡
we can derive the characteristic
equation
1
1
2
𝑠 +
𝑠+
=0
𝑅𝐶
𝐿𝐶

Solving Our Differential
Equation

For convenience we define:
𝛼=


1
2𝑅𝐶
and
𝜔0 =
1
𝐿𝐶
Note that 0 (the undamped natural
frequency) is the same as for series
circuits, but  (the neper frequency) is not.
Therefore, just as for series circuits,
𝑠 2 + 2𝛼𝑠 + 𝜔02 = 0

We get the same three cases as before
(overdamped, critically damped, and
underdamped)….
The Overdamped Case ( > 0)

If  > 0, our two solutions for s are distinct
negative real numbers
𝑠1 = −𝛼 + 𝛼 2 − 𝜔02 ,


𝑠2 = −𝛼 − 𝛼 2 − 𝜔02
In this case, the solution to our differential
equation is
𝑣 𝑡 = 𝐴1 𝑒 𝑠1 𝑡 + 𝐴2 𝑒 𝑠2 𝑡
The initial conditions determine A1 and A2:
𝐴1 + 𝐴2 = 𝑣(0+ ),
𝑠1 𝐴1 + 𝑠2 𝐴2 =
𝑑𝑣(0+ )
𝑑𝑡
The Critically Damped Case ( = 0)

If  = 0, our two solutions for s are equal
to each other and to :
𝑠1 = −𝛼 + 𝛼 2 − 𝜔02 = −𝛼 ,


𝑠2 = −𝛼 − 𝛼 2 − 𝜔02 = −𝛼
In this case, the solution to our differential
equation is
𝑣 𝑡 = (𝐴2 + 𝐴1 𝑡)𝑒 −𝛼𝑡
The initial conditions determine A1 and A2:
𝐴2 = 𝑣(0+ ),
𝐴1 =
𝑑𝑣(0+ )
+
𝑑𝑡
𝛼𝐴2
The Underdamped Case ( < 0)

If  < 0, our two solutions for s are complex numbers:
𝑠1 = −𝛼 + 𝛼 2 − 𝜔02 = −𝛼 + 𝑗𝜔𝑑
𝑠2 = −𝛼 − 𝛼 2 − 𝜔02 = −𝛼 − 𝑗𝜔𝑑

Here j is the imaginary unit, 𝑗 = −1, and d is called
the damped natural frequency, 𝜔𝑑 = 𝜔02 − 𝛼 2 .

In this case, the solution to our differential equation is
𝑣 𝑡 = 𝑒 −𝛼𝑡 (𝐵1 cos 𝜔𝑑 𝑡 + 𝐵2 sin 𝜔𝑑 𝑡)

The initial conditions determine B1 and B2:
𝐵1 = 𝑣(0+ ),
𝐵2 =
𝑑𝑣(0+ )
+ 𝛼𝐵1
𝑑𝑡
𝜔𝑑
Graphs of the Three Cases


Details will differ based on initial conditions and
element values, but the shapes shown here are typical.
Note the oscillation in the underdamped case.
A General Approach for Source-Free Series
or Parallel RLC Circuits (1 of 3)

1.
To find x(t) in a source-free series or parallel RLC circuit,
where x could be any current or voltage:
Find the quantity’s initial value 𝑥(0+ ).
2.
Find the quantity’s initial derivative
3.
Find the neper frequency:
4.

𝛼=

𝛼=
𝑅
for a series RLC circuit.
2𝐿
1
for a parallel RLC circuit.
2𝑅𝐶
Find the undamped natural frequency:

5.
𝑑𝑥(0+ )
.
𝑑𝑡
𝜔0 =
1
𝐿𝐶
Compare  to 0 to see whether circuit is
overdamped, critically damped, or underdamped…
A General Approach for Source-Free Series
or Parallel RLC Circuits (2 of 3)
6.
If it’s overdamped ( > 0), then:
𝑠1 = −𝛼 + 𝛼 2 − 𝜔02 ,
Solve for A1 and A2:
𝐴1 + 𝐴2 =
𝑠2 = −𝛼 − 𝛼 2 − 𝜔02
𝑥(0+ ),
𝑠1 𝐴1 + 𝑠2 𝐴2 =
𝑥 𝑡 = 𝐴1 𝑒 𝑠1 𝑡 + 𝐴2 𝑒 𝑠2 𝑡
7.
If it’s critically damped ( = 0), then:
𝐴2 =
𝑥(0+ ),
𝐴1 =
𝑑𝑥(0+ )
+
𝑑𝑡
𝑥 𝑡 = (𝐴2 + 𝐴1 𝑡)𝑒 −𝛼𝑡
𝛼𝐴2
𝑑𝑥(0+ )
𝑑𝑡
A General Approach for Source-Free Series
or Parallel RLC Circuits (3 of 3)
8.
If it’s underdamped ( < 0), then:
𝜔𝑑 =
𝐵1 = 𝑥(0+ ),
𝜔02 − 𝛼 2
𝐵2 =
𝑥 𝑡 = 𝑒 −𝛼𝑡 (𝐵1 cos 𝜔𝑑 𝑡
𝑑𝑥(0+ )
+ 𝛼𝐵1
𝑑𝑡
𝜔𝑑
+ 𝐵2 sin 𝜔𝑑 𝑡)