Time
Response Of Electrical Circuits  2nd Order Systems
Introduction
Circuit With
No Input (Initial Condition Response)
Using
Current In An RLC Circuit
Using
Capacitor Voltage In An RLC Circuit
There are numerous physical systems that exhibit second order response
 they satisfy a second order differential equation. In this module
you will examine how those systems respond in different situations including
situations in which initial conditions produce the response and situations
in which inputs to the circuit or system produce the response.
Systems
of the type we will be considering will have a diffential equation like
that at the right. Important terms in this differential equation
are:

u(t)
The
Input or the Driving Function

x(t) The
Response of the system or circuit
There will be several different situations that
are important.

u(t) = 0
Here the response is due to initial conditions

u(t) = Constant
An
important special case
And there are several other special cases like
sinusoidal and exponential input signals that we will not discuss in this
module.
We will begin by examining a situation with no input, i.e. u(t) = 0.
Zero
Input Response
Example Circuit: A
RLC Circuit With a Capacitor that is Initially Charged.
Imagine that the switch in this circuit is closed at some time. We
will call the instant that the switch closes t =
0. When the switch closes the capacitor has some charge on
it and consequently it has some initial voltage. We will call that
initial voltage v_{c}(0).
That initial voltage on the capacitor causes a voltage to appear across
the inductor after the switch is closed. Initially the current, i(t),
through the inductor will be zero. However, the voltage from the
capacitor appears across the inductor (there's no current through the resistor,
so the voltage across the resistor is zero initially.) causing the current
to begin to increase, and it takes off from there.
How it takes off, and what it does requires solving a differential equation.
To get that differential equation, we write KCL around the loop.
At the right voltage polarities are defined for the various components.
If we write KVL after the switch has closed, we will have:
v_{C}(t)
= v_{L}(t) + v_{R}(t)
The capacitor voltage is defined as shown,
and we get the opposite sign there because of the way the capacitor voltage
is defined.
Now, we need to write the differential equation that describes this circuit.
Although it is not immediatelly apparent, we have two choices for the variable
to use. We can write the differential equation in terms of the current
in the circuit, i(t), or we can write it in terms of the capacitor voltage,
v_{c}(t). We'll set it up both ways so you can see
how it goes. Click here to skip
the section on using current.
Writing
The Differential Equation Using Current
First, we
set up the differential equation using current. Rewriting KVL (above)
in terms of current, we get:
Note that we need to compute the capacitor voltage
in terms of the current, and we must integrate the current and divide by
the capacitance to get the capacitor voltage. You may not know how
to solve the integrodifferential equation above, but you can put it into
the usual form of a differential equation by differentiating. If
you do that, and put everything on the same side of the resulting equation
you should be able to get:
Now, you need to consider solution methods.
Since there is no driving function term on the right hand side of this
equation, the equation is already a homogeneous equation. With a
driving function on the right hand side you would have had to consider
a solution that was a sum of a homogeneous solution and a particular solution.
Here the particular solution is zero.
To solve the differential equation you must assume that the solution is
of the form C_{h}e^{st}.
Doing that, we get:
There is a common factor, C_{h}e^{st},
which can be removed, giving:
This equation  since it is a quadratic  has
two possible values for the variable s (call them s_{1} and s_{2}).
Not only are there two values, but there are
three different cases for the solution. We can have:

Two real  and different
 roots.

Two real  and equal 
roots.

A pair of complexconjugate
roots.
The case of two real and equal roots presents
some complications we will not consider here. However, the other
two cases both have the same general form:
i(t) = C_{h,1}e^{s1t}
+ C_{h,2}e^{s2t}
In those cases the solution is composed of
two terms with  as yet  undetermined constants, C_{h,1}
and C_{h,2}^{. }We need to compute those undetermined
constants, and to do that we need initial conditions. We already
have one initial condition because we know that the voltage across the
capacitor has some predetermined value at t = 0. That doesn't seem
to help much however. We need to think this through somehow.
Here is the circuit again. We know that when we close the switch
there cannot be any current in the inductor since there is no closed path
for current to flow before the switch is closed. So, we can deduce
that:
The second condition is
even harder to deduce. Let's look at what happens when the switch
is closed. At that time we know that the current is zero. If
the current is zero, then the voltage across the resistor must also be
zero.

We know that the current
is zero, i.e. i(0) = 0

If the current is zero,
then the voltage across the resistor is also zero.

If the voltage across
the resistor is zero, then KVL tells us that the initial voltage across
the capacitor is applied across the inductor at the instant the switch
is closed.

We know that the voltage
across the inductor is v_{L}(t) = L di/dt.

So, at the instant the
switch is closed, we must have:
v_{C}(0)
= L di/dt_{t=0}

So, the initial condition
we get from this is:
di/dt_{t=0}
= v_{C}(0)/L
Finally, this gives us what we need. If
we have the current and the derivative of the current at the instant of
switch closing (t = 0), then we can solve for the yet undetermined constants
in the solution of the differential equation.
We set the expression for the current and the derivative of the current
to the values we have found, and that gives us two linear equations in
the two unknown coefficients. Here are those equations.
i(0) = C_{1}e^{s1*0}
+ C_{2}e^{s2*0} = C_{1} + C_{2}
and
di/dt_{t=0}
= v_{C}(0)/L = s_{1}C_{1}e^{s1*0}
+ s_{2}C_{2}e^{s2*0}
and these reduce to:
C_{1}
+ C_{2} = i(0) = 0
and
s_{1}C_{1}
+ s_{2}C_{2} = v_{C}(0)/L
From here, you can get the solutions for
the constants:

C_{1} =
 v_{C}(0)/[L (s_{2}  s_{1})]

C_{2} =
+ v_{C}(0)/[L (s_{2}  s_{1})]
With these solutions it is possible to write
out the complete solution for the current. We still have to worry
about whether we have two complex conjugate s values or whether we have
to real, distinct values for s.
We are going to defer consideration of how to interpret the solution in
those two cases until after we have examined how to solve for the solution
in terms of the capacitor voltage.
Writing
The Differential Equation Using Capacitor Voltage
Remember KVL after the switch has closed gives us this equation:
v_{C}(t)
= v_{L}(t) + v_{R}(t)
Then, we know that the current is given by:
i(t) = C dv_{c}(t)/dt
That allows us to compute the voltage across
the resistor:
v_{R}(t)
= R_{L}C dv_{c}(t)/dt
and the voltage across the inductor can also
be computed after differentiating the current.
v_{L}(t)
= LC (d^{2}v_{c}(t)/dt^{2})
Putting it all together gives us the differential
equation we want:
LC (d^{2}v_{c}(t)/dt^{2})
+ R_{L}C (dv_{c}(t)/dt) + v_{c}(t)
= 0
If you worked through the section on using
current as the variable, you will realize that this is exactly the same
differential equation that we had before. (Click
here to look at the derivation using current as the variable.)
Now, you need to consider solution methods. Since there is no driving
function term on the right hand side of this equation, the equation is
already a homogeneous equation. With a driving function on the right
hand side you would have had to consider a solution that was a sum of a
homogeneous solution and a particular solution. Here the particular
solution is zero.
To solve the differential equation you must assume that the solution is
of the form C_{h}e^{st}.
(Note that when using current we get exactly the same form, but the coefficients
will work out differently!) Doing that, we get:
There is a common factor, C_{h}e^{st},
which can be removed, giving:
This equation  since it is a quadratic  has
two possible values for the variable s (call them s_{1} and s_{2}).
Not only are there two values, but there are
three different cases for the solution. We can have:

Two real  and different
 roots.

Two real  and equal 
roots.

A pair of complexconjugate
roots.
The case of two real and equal roots presents
some complications we will not consider here. However, the other
two cases both have the same general form:
v_{c}(t)
= C_{1}e^{s1t} + C_{2}e^{s2t}
Now, if we can compute the constants we have
the solution we want. We need to use initial conditions to compute
those constants. We already have one. We know v_{c}(0).
It's given.

v_{c}(0)
= Some known voltage
The
second initial condition is not quite as easy to determine. However,
note that the current through the inductor is initially zero. We
know that when we close the switch there cannot be any current in the inductor
since there is no closed path for current to flow before the switch is
closed. So, we can deduce that:
But, that doesn't help us much. We need
a condition on the capacitor voltage. Notice, however, that the capacitor
voltage is related to the current, since:
That gives us the condition we want, because
if the current is zero, then the first derivative of capacitor voltage
(with respect to time) is also zero. (And, if current were nonzero,
then we would have to compute the initial value of that derivative.)
Anyhow, we have:
That gives us the following two equations that
we can use to solve for the undetermined constants:

C_{1} +
C_{2} = v_{c}(0)

s_{1} C_{1}
+ s_{2} C_{2} = 0
There
are numerous ways to solve this set of equations for the undetermined constants
(including using determinants).
However you solve this set of equations, you should find:

C_{1} =
s_{2} v_{c}(0)/(s_{2}  s_{1})

C_{2} =
s_{1} v_{c}(0)/(s_{2}  s_{1})
That's it. Those are the values of the
constants. Now, we have what we need to know to determine the capacitor
voltage. Let's go back to the expression for the voltage on the capacitor:

v_{c}(t)
= C_{1}e^{s1t} + C_{2}e^{s2t}
We can insert the expressions above for
the constants to get:

v_{c}(t)
= [s_{2} v_{c}(0)/(s_{2}
 s_{1})]e^{s1t}  [s_{1} v_{c}(0)/(s_{2}
 s_{1})]e^{s2t}
We can use either form from this point on.
However, to keep things simple we will not use the complete form with the
constants expanded. Rather we will work on trying to make sense of
the unexpanded form. Consider:

v_{c}(t)
= C_{1}e^{s1t} + C_{2}e^{s2t}
We will assume that the roots, s_{1}and
s_{2}, are complex conjugates. Then C_{1}
and C_{2} must be complex conjugates if the capacitor voltage
is a real function of time. Since e^{s1t}
and e^{s2t} are complex conjugates (for all values
of t) the only way the capacitor voltage can be a real value for all values
of t is if C_{1} and C_{2} are complex conjugates.
This form does not give a lot of insight into what is actually happening.
If the roots are complex, then the solution will be a decaying sinusoid.
We need to get some feel for what the solution actually looks like and
we can recast the solution into that form. Let:
C_{1}
= C_{real} + jC_{imag}
and C_{2} = C_{real}
 jC_{imag}
expressing the roots in terms of real and
imaginary parts, C_{real} and C_{imag}.
or:
C_{1}
= C_{M}e^{jb} and
C_{2}
= C_{M}e^{jb}
expressing the roots in terms of
a magnitude, C_{M}, and angle, b.
Then let:
s_{1}
= r + jw
and s_{2} = r  jw
Then:
v_{c}(t)
= C_{1}e^{s1t} + C_{2}e^{s2t}
Now, use the magnitude/angle representation
for C_{1} and C_{2}.
v_{c}(t)
= C_{M}e^{jb}e^{(r + j}^{w)t}
+ C_{M}e^{jb}e^{(r  j}^{w)t}
v_{c}(t)
= C_{M}[e^{rt}(e^{j(}^{wt+b)}
 e^{(j}^{wt+b)})]
v_{c}(t)
= 2 C_{M} e^{rt }[cos (wt+b)
]
And, in the end, the voltage across the capacitor
is a decaying sinusoid. It's not obvious that it decays, but we can
see that if we examine the values for r, w and b.
Let's go back to the definitions of those variables.
We had two roots
with values given to the right. We had also broken these roots into
real and imaginary parts:
s_{1}
= r + jw
and s_{2} = r  jw
So, we have:
r = R_{L}/2L
w =
[  (R_{L}/2L)^{2} + 1/LC]^{1/2}
Where:

r determines that rate
of decay of the sinusoid.

w
determines the frequency of the sinusoidal oscillations.

It's the angular frequency,
and:f = 2p/w
Here, the real part of
the roots (r) is negative so the oscillations decay  and do not grow.
The frequency, w, is the angular frequency of the oscillations.
But, we need to compute C_{M} and b. We have:

C_{1} =
C_{M}e^{jb} = s_{2}
v_{c}(0)/(s_{2}  s_{1}) =
(r  jw)
v_{c}(0)/( 2 jw)
and we have:

C_{M} =
(r^{2} +w^{2})^{1/2}
v_{c}(0)/( 2w)

b = tan^{1}(w/r)
 90^{o}
Now, let us examine this in more detail.
First focus on the frequency of the oscillations.

w
= [  (R_{L}/2L)^{2} + 1/LC]^{1/2}
This is the angular frequency of the oscillations
in the decay. Any time you have the opportunity to look at extreme
or limiting cases that's the first thing to do. Here, you can ask
what happens in the resistance is zero. In that case:

w=
[1/LC]^{1/2} = 1/ [LC]^{1/2}
Note that this implies:

If either the inductor
or capacitor is made smaller, the frequency will increase.
Practically, in many cases the frequency will
be very close to the limiting (zero resistance) case if you observe any
oscillations at all.
We also need to examine the decay rate, r. Above we found:
This decay rate  the real part of the roots
 determines how fast the sinusoid decays. The response is:

v_{c}(t)
= 2 C_{M} e^{rt }[cos (wt+b)
]
So, there is a factor
multiplying the sinsusoid of:
We can calculate how much
decay occurs in one period of the frequency. Since the frequency
is:

w
= [  (R_{L}/2L)^{2} + 1/LC]^{1/2}
The period, T, is given
by:
So, in one period, the decay is:

e^{(R}L^{/2L)2}^{p/w }
= e^{(R}L^{/L)}^{p/w
= }e^{(}^{pR}L^{/}^{wL)}
EXAMPLE
Let us look at the response of our example circuit. The response
is shown below. Note the following about this response.

The parameters are:

R_{L} = 10 W

C = 10 mf

L = 0.1 h (100 mh)

v_{c}(0)
= 10 v

The response starts at
a capacitor voltage of 10 volts  which we should expect.

We can compute the frequency
of the oscillations:

w
= [  (R_{L}/2L)^{2} + 1/LC]^{1/2}

w
= [  (10/.2)^{2} + 1/(.1 x 10^{5})]^{1/2}

w
= [  (50)^{2} + 10^{6}]^{1/2}

w
= [  2500 + 10^{6}]^{1/2}

w
~= 10^{3}

f = 159 Hz

Note that the frequency
is pretty much what it would have been with R_{L} = 0.

The decay factor for one
period is:

e^{(}^{pR}L^{/}^{wL)}
= e^{(}^{p10/1000
x .1)}

= e^{(}^{p/10)}
= 0.73

Examine the response.
After one cycle of the oscillations it looks to be 7.3 volts, which is
0.73 x 10 v.

In the second cycle, the
response drops from 7.3 volts to 7.3 x 0.73 = 5.33 v, which is about what
we have on the graph. Here's the graph again so that you can check
it.

The frequency can be approximated
by observing that we have eight cycles in .05 seconds. That means:

f = 8/.05 = 160.

That's pretty close.
We'll take it.
Summing Up The Algorithm
Let's consider what the steps were in this solution. Here they are:

Apply KVL or KCL to the
circuit.

Use element laws (v =
L di/dt) to turn the turn the KVLKCL equation into a differential equation.

Solve the differential
equation. (Get an exponential form.)

Determine the initial
conditions.

You will probably have
to use physical reasoning to get them.

Use the initial conditions
to determine the constants in the solution.

Evaluate the solution.
Make sure that the decay rate is right and that the frequency is correct.

Plot the solution to see
that is looks right.
That's it in a nutshell.
Click
here for a problem on the topic.