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 vc(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:

vC(t) = vL(t) + vR(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, vc(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 integro-differential 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 Chest.  Doing that, we get:

There is a common factor, Chest, which can be removed, giving:

This equation - since it is a quadratic - has two possible values for the variable s (call them s1 and s2).
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 complex-conjugate 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) = Ch,1es1t + Ch,2es2t

In those cases the solution is composed of two terms with - as yet - undetermined constants, Ch,1 and Ch,2We 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:

• i(0) = 0
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 vL(t) = L di/dt.
• So, at the instant the switch is closed, we must have:
vC(0) = L di/dt|t=0
• So, the initial condition we get from this is:
• di/dt|t=0 = vC(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) = C1es1*0 + C2es2*0 = C1 + C2

and
di/dt|t=0 = vC(0)/L = s1C1es1*0 + s2C2es2*0
and these reduce to:

C1 + C2 = i(0) = 0

and
s1C1 + s2C2 = vC(0)/L

From here, you can get the solutions for the constants:

• C1 = - vC(0)/[L (s2 - s1)]
• C2 = + vC(0)/[L (s2 - s1)]
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:

vC(t) = vL(t) + vR(t)

Then, we know that the current is given by:

i(t) = C dvc(t)/dt

That allows us to compute the voltage across the resistor:

vR(t) = RLC dvc(t)/dt

and the voltage across the inductor can also be computed after differentiating the current.

vL(t) = LC (d2vc(t)/dt2)

Putting it all together gives us the differential equation we want:

LC (d2vc(t)/dt2) + RLC (dvc(t)/dt) + vc(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 Chest.  (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, Chest, which can be removed, giving:

This equation - since it is a quadratic - has two possible values for the variable s (call them s1 and s2).
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 complex-conjugate 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:

vc(t) = C1es1t + C2es2t

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 vc(0).  It's given.

• vc(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:
• i(0) = 0
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:
• i(t) = C dvc(t)/dt
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 non-zero, then we would have to compute the initial value of that derivative.)  Anyhow, we have:
• dvc(t)/dt|t=0 = 0
That gives us the following two equations that we can use to solve for the undetermined constants:
• C1 + C2 = vc(0)
• s1 C1 + s2 C2 = 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:
• C1 = s2 vc(0)/(s2 - s1)
• C2 = -s1 vc(0)/(s2 - s1)
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:
• vc(t) = C1es1t + C2es2t

We can insert the expressions above for the constants to get:

• vc(t) =  [s2 vc(0)/(s2 - s1)]es1t - [s1 vc(0)/(s2 - s1)]es2t
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:
• vc(t) = C1es1t + C2es2t
We will assume that the roots, s1and s2, are complex conjugates.  Then C1 and C2 must be complex conjugates if the capacitor voltage is a real function of time.  Since es1t and es2t 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 C1 and C2 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:

C1 = Creal + jCimag    and C2 = Creal - jCimag
expressing the roots in terms of real and imaginary parts, Creal and  Cimag.
or:

C1 = CMejb  and C2 = CMe-jb

expressing the roots in terms of a magnitude, CM, and angle, b.

Then let:

s1 = r + jw    and  s2 = r - jw

Then:
vc(t) = C1es1t + C2es2t
Now, use the magnitude/angle representation for C1 and C2.
vc(t) = CMejbe(r + jw)t + CMe-jbe(r - jw)t

vc(t) = CM[ert(ej(wt+b) - e-(jwt+b))]

vc(t) = 2 CM ert [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:

s1 = r + jw    and  s2 = r - jw
So, we have:
r = -RL/2L
w = [ - (RL/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 CM and b.  We have:

• C1 = CMejb =  s2 vc(0)/(s2 - s1) = (r - jw) vc(0)/( -2 jw)
and we have:
• CM = (r2 +w2)1/2 vc(0)/( 2w)
• b = tan-1(w/r) - 90o
Now, let us examine this in more detail.  First focus on the frequency of the oscillations.
• w = [ - (RL/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:

• r = -RL/2L
This decay rate - the real part of the roots - determines how fast the sinusoid decays.  The response is:
• vc(t) = 2 CM ert [cos (wt+b) ]
So, there is a factor multiplying the sinsusoid of:
• e-(RL/2L)t
We can calculate how much decay occurs in one period of the frequency.  Since the frequency is:
• w = [ - (RL/2L)2 + 1/LC]1/2
The period, T, is given by:
• T = (1/f) = 2p/w
So, in one period, the decay is:
• e-(RL/2L)2p/w  = e-(RL/L)p/w = e-(pRL/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:
• RL = 10 W
• C = 10 mf
• L = 0.1 h (100 mh)
• vc(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 = [ - (RL/2L)2 + 1/LC]1/2
• w = [ - (10/.2)2 + 1/(.1 x 10-5)]1/2
• w = [ - (50)2 + 106]1/2
• w = [ - 2500 + 106]1/2
• w ~=  103
• f = 159 Hz
• Note that the frequency is pretty much what it would have been with RL = 0.
• The decay factor for one period is:
• e-(pRL/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 KVL-KCL 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.