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Electrical engineering
Course: Electrical engineering > Unit 2
Lesson 4: Natural and forced response- Capacitor i-v equations
- A capacitor integrates current
- Capacitor i-v equation in action
- Inductor equations
- Inductor kickback (1 of 2)
- Inductor kickback (2 of 2)
- Inductor i-v equation in action
- RC natural response - intuition
- RC natural response - derivation
- RC natural response - example
- RC natural response
- RC step response - intuition
- RC step response setup (1 of 3)
- RC step response solve (2 of 3)
- RC step response example (3 of 3)
- RC step response
- RL natural response
- Sketching exponentials
- Sketching exponentials - examples
- LC natural response intuition 1
- LC natural response intuition 2
- LC natural response derivation 1
- LC natural response derivation 2
- LC natural response derivation 3
- LC natural response derivation 4
- LC natural response example
- LC natural response
- LC natural response - derivation
- RLC natural response - intuition
- RLC natural response - derivation
- RLC natural response - variations
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RLC natural response - derivation
A formal derivation of the natural response of the RLC circuit. Written by Willy McAllister.
Introduction
Let's take a deep look at the natural response of a resistor-inductor-capacitor circuit left parenthesis, start text, R, L, C, right parenthesis, end text. This is the last circuit we'll analyze with the full differential equation treatment.
The start text, R, L, C, end text circuit is representative of real life circuits we can actually build, since every real circuit has some finite resistance. This circuit has a rich and complex behavior that finds application in many areas of electrical engineering.
What we're building to
We will model the start text, R, L, C, end text circuit with a 2nd-order linear differential equation with current, i, as the independent variable:
The resulting characteristic equation is:
We will solve for the roots of the characteristic equation using the quadratic formula:
By substituting variables alpha and omega, start subscript, o, end subscript we can write s a little simpler as:
where
alpha, equals, start fraction, start text, R, end text, divided by, 2, start text, L, end text, end fraction, and omega, start subscript, o, end subscript, equals, start fraction, 1, divided by, square root of, start text, L, C, end text, end square root, end fraction
alpha is called the damping factor and omega, start subscript, o, end subscript is the resonant frequency.
We will solve an example start text, R, L, C, end text circuit with specific component values and discover what the current and voltages look like.
Strategy
We follow the same line of reasoning we used for solving the second-order LC circuit in an earlier article.
- Create a second-order differential equation based on the i-v equations for the start text, R, end text, start text, L, end text, and start text, C, end text components. We will use Kirchhoff's Voltage Law to build the equation.
- Make an informed guess at a solution. As usual, our guess will be an exponential function of the form K, e, start superscript, s, t, end superscript.
- Insert the proposed solution into the differential equation. The exponential terms will factor out and leave us with a characteristic equation in variable s.
- Find the roots of the characteristic equation. This time we will need to use the quadratic formula to solve for the roots.
- Find the constants by accounting for the initial conditions.
- Celebrate the solution.
Model the circuit with a differential equation
When the switch closes, the circuit looks like this (now with voltage labels on the inductor and resistor, v, start subscript, start text, L, end text, end subscript and v, start subscript, start text, R, end text, end subscript).
For each individual element, we can write i-v equations.
We can write Kirchhoff's Voltage Law (KVL) starting in the lower left corner and summing voltages going around the loop clockwise. The inductor has a voltage rise, while the resistor and capacitor have voltage drops.
Replacing the v terms with the corresponding i terms gives us:
If we wanted to, we could attack this equation and try to solve it, but the integral term is awkward to deal with. We can retire the integral if we take the derivative of the entire equation.
This gives us the following equation with a second derivative term, a first derivative term, and a plain i term, all still equal to 0.
This is called a homogeneous second-order ordinary differential equation. It is homogeneous because every term is related to i and its derivatives. It is second order because the highest derivative is a second derivative. It is ordinary because there is only one independent variable (no partial derivatives). Now we set about solving our differential equation.
Propose a solution
Just like we did with previous natural response problems ( RC, RL, LC), we assume a solution with an exponential form. Exponential functions have the wondrous property that the derivatives look a lot like the original function. When you have multiple derivatives participating in a differential equation, it is really nice when they look alike. We assume a solution with this form:
K is an adjustable parameter representing the amplitude of the current.
s is up in the exponent next to t, so it must represent some kind of frequency (s has to have units of 1, slash, t). We call this the natural frequency.
Try the proposed solution
Next, substitute the proposed solution into the differential equation. If the equation turns out to be true, then our solution is a winner.
Now let's work on the terms with derivatives.
Middle term: The first derivative of the start text, R, end text term is
Leading term: We take the derivative if the leading start text, K, end text, e, start superscript, s, t, end superscript term two times:
so the leading term becomes:
Plug these back into the differential equation:
Now we can factor out the common K, e, start superscript, s, t, end superscript term:
Make the equation true
Now let's figure out how many ways we can make this equation true.
We could set K equal to 0. That means i, equals, 0 and we are putting nothing into the circuit and getting nothing out. Pretty boring.
The term e, start superscript, s, t, end superscript never becomes 0 unless we wait until t goes to infinity. That's a long time from now. That leaves us with one interesting way to make the equation true: if the term with all the s's is zero.
This is called the characteristic equation of the start text, L, R, C, end text circuit.
Solve for the roots of the characteristic equation
Let's find values of s that make the characteristic equation true. (We want to find the roots of the characteristic equation.)
We have exactly the right tool for this, the quadratic formula:
For any quadratic equation: a, x, squared, plus, b, x, plus, c, equals, 0,
the quadratic formula gives us the roots (the zero-crossings):
Looking back at the characteristic equation, we can plug in our circuit component values to get the roots. a, equals, start text, L, end text, b, equals, start text, R, end text, and c, equals, 1, slash, start text, C, end text.
That's the answer for s, the natural frequency. We need to break this down a little more to get a feeling for the meaning of this solution.
We can make the notation a little more compact by replacing parts of the expression with two new variables, alpha and omega, start subscript, o, end subscript.
Let me write the characteristic equation this way left parenthesisdividing through by start text, L, end text, right parenthesis:
If we use alpha and omega, start subscript, o, end subscript, the characteristic equation can be written:
We can revise the quadratic formula by mashing the 2, start text, L, end text denominator up into each term of the numerator:
The second term under the square root reduces to:
And this lets us write s in terms of alpha and omega, start subscript, o, end subscript as:
We know s is some sort of frequency (it has to have units of 1, slash, t). That means the two terms making up s are also some sort of frequency.
- alpha is called the damping factor. It will determine how quickly the overall signal fades to zero.
- omega, start subscript, o, end subscript is called the resonant frequency. It will determine how fast the system swings back and forth. This is the same resonant frequency we found in the start text, L, C, end text natural response.
Proposed solution, upgraded
The quadratic formula gave us two solutions for s, we'll call these s, start subscript, 1, end subscript and s, start subscript, 2, end subscript. We need to include both of these in the proposed solution, so we update our proposed solution to be a linear combination (the superposition) of two separate exponential terms with four adjustable parameters:
s, start subscript, 1, end subscript and s, start subscript, 2, end subscript are natural frequencies,
K, start subscript, 1, end subscript and K, start subscript, 2, end subscript are amplitude terms.
K, start subscript, 1, end subscript and K, start subscript, 2, end subscript are amplitude terms.
Example circuit
At this point it's helpful to do a specific example with some actual component values, to see how one particular solution plays out. Here is our example circuit:
The differential equation for the start text, R, L, C, end text circuit is
With real component values it becomes:
As we always do, assume a solution of the form: i, left parenthesis, t, right parenthesis, equals, K, e, start superscript, s, t, end superscript
We perform the analysis we did above, which results in this characteristic equation:
Solving for the roots of the characteristic equation with the quadratic formula:
With real component values:
(Electrical engineers use the letter j for the imaginary square root of, minus, 1, end square root, since we use i as the symbol for current.)
We get a complex answer, just as we did with the start text, L, C, end text natural response, only this time the complex answer includes both a real part and an imaginary part.
Solving for the roots of the characteristic equation gave us two possible answers for s, so the proposed solution for i is now written as the superposition of two different exponential terms:
The terms up in the exponents are complex conjugates. Let's fuss around with the way this is written. We can tease apart the real and imaginary parts of the exponents:
and factor out the common e, start superscript, minus, 1, t, end superscript term:
Notice how the real part of s came through the factoring process to give us the leading term, a decaying exponential, e, start superscript, minus, t, end superscript.
The terms in the parentheses are a sum of two imaginary exponentials where the exponents are complex conjugates. This looks just like what we saw in the start text, L, C, end text natural response. As we did then, we call on Euler's formula to help us with these terms.
Euler's formula
Using Maclaurin series expansions for e, start superscript, j, x, end superscript, sine, j, x, and cosine, j, x, it is possible to derive Euler's formula:
and
In the linked video, whenever Sal says i, we say j.
These formulas let us turn e, start superscript, i, m, a, g, i, n, a, r, y, end superscript into a normal complex number.
Use Euler's formula
We can use Euler's formula to transform the sum
into
Multiply through the constants K, start subscript, 1, end subscript and K, start subscript, 2, end subscript:
and gather the cosine terms and sine terms:
Without messing up the equation, we can simplify how this appears if we replace the unknown K's with different unknown A's. Let A, start subscript, 1, end subscript, equals, left parenthesis, K, start subscript, 1, end subscript, plus, K, start subscript, 2, end subscript, right parenthesis, and A, start subscript, 2, end subscript, equals, j, left parenthesis, K, start subscript, 1, end subscript, minus, K, start subscript, 2, end subscript).
The previous expression becomes:
And now put this back into our proposed solution:
So far so good. Next, we need to figure out A, start subscript, 1, end subscript and A, start subscript, 2, end subscript using the initial conditions.
Find the initial conditions
For a second-order equation, you need two initial conditions to get a complete solution: one for the independent variable, i, and another for its first derivative, d, i, slash, d, t.
If we can figure out i and d, i, slash, d, t at a specific time, we can find A, start subscript, 1, end subscript and A, start subscript, 2, end subscript.
Finding the initial conditions for start text, R, L, C, end text is pretty much the same as for the LC circuit. We just have account for the resistor.
Here's what we know about t, equals, 0, start superscript, minus, end superscript (the moment before the switch closes):
- The switch is open, so i, left parenthesis, 0, start superscript, minus, end superscript, right parenthesis, equals, 0
- The starting capacitor voltage is specified: v, start subscript, start text, C, end text, end subscript, left parenthesis, 0, start superscript, minus, end superscript, right parenthesis, equals, start text, V, end text, start subscript, 0, end subscript
If t, equals, 0, start superscript, plus, end superscript is the moment just after the switch closes, our goal is to find i, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis and d, i, slash, d, t, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis.
We know some properties of inductors and capacitors that tell us what happens when the switch closes, going from t, equals, 0, start superscript, minus, end superscript to t, equals, 0, start superscript, plus, end superscript:
- Inductor current does not change instantly, so i, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis, equals, i, left parenthesis, 0, start superscript, minus, end superscript, right parenthesis, equals, 0
- Capacitor voltage does not change instantly, so v, start subscript, start text, C, end text, end subscript, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis, equals, v, start subscript, start text, C, end text, end subscript, left parenthesis, 0, start superscript, minus, end superscript, right parenthesis, equals, start text, V, end text, start subscript, 0, end subscript
Now we know one initial condition, i, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis, equals, 0, and we know something about the voltage, but we don't know d, i, slash, d, t, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis.
Let's go after the second initial condition, d, i, slash, d, t, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis. Any time I see d, i, slash, d, t it makes me think of the inductor i-v equation.
If we can figure out the voltage across the inductor, we can figure out d, i, slash, d, t. Let's do that by process of elimination.
Kirchhoff's Voltage Law around the loop is:
Since i, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis, equals, 0, that means the voltage across the resistor, v, start subscript, start text, R, end text, end subscript, has to be 0. We also know the voltage across the capacitor is v, start subscript, start text, C, end text, end subscript, equals, start text, V, end text, start subscript, 0, end subscript. Let's fill in the KVL equation with these values:
And now we know the voltage across the inductor at t, equals, 0, start superscript, plus, end superscript:
We can use this to derive d, i, slash, d, t using the inductor i-v equation.
The moment just after the switch closes, the current in the inductor has an initial slope of 10 amperes per second.
Find constants A, start subscript, 1, end subscript and A, start subscript, 2, end subscript using the initial conditions
As a reminder, our proposed solution is:
and the initial conditions are:
If we evaluate i at t, equals, 0, we can find one of the A constants. Insert t, equals, 0 and i, equals, 0 into the proposed solution:
A, start subscript, 1, end subscript, equals, 0, so the cosine term drops out of the solution. One down, one to go. Our proposed solution now looks like:
Let's chase down A, start subscript, 2, end subscript using the second initial condition:
We need an equation for the derivative of i. Where might we get such a thing? How about we take the derivative of the proposed solution?
The proposed solution is the product of two functions. To take its derivative we use the product rule:
Identify the two parts of the product and their derivatives:
Assemble the parts according to the product rule:
We can evaluate this expression at t, equals, 0, start superscript, plus, end superscript:
Solution for current
And finally, after a bunch of hard work, the solution for current is:
The graph of i as a function of time looks like this:
When the switch closes, the current takes a big surge upwards and takes on the shape of the first hump of a sine wave. The sine wave quickly fades away after a few swings because the energy in the system rapidly dissipates as heat as charge flows back and forth through the resistor.
The role of "friction" in this example, as played by the resistor value, represents a fairly high rate of energy dissipation. The current visibly changes sign only two times before settling to zero.
This is an example of an underdamped solution. We will introduce this descriptive term in the next section.
Solve the voltages
There is only one current in the circuit. Now that we know the natural response of the current, we can find the natural response of the three voltages.
Resistor voltage
We use Ohm's Law to find the resistor voltage: left parenthesisthere's a minus sign because i is backwards relative to v, start subscript, start text, R, end text, end subscript, right parenthesis
Inductor voltage
The inductor voltage emerges from the inductor i-v equation:
Capacitor voltage
To find the capacitor voltage we can use the integral form of the capacitor i-v equation: left parenthesisanother extra minus sign because of the direction of i is reversed relative to v, start subscript, start text, C, end text, end subscript, right parenthesis
Here are all three voltages plotted together:
Summary
The start text, R, L, C, end text circuit is the electronic equivalent of a swinging pendulum with friction.
The circuit can be modeled by this 2nd-order linear differential equation:
The resulting characteristic equation is:
We solved for the roots of the characteristic equation using the quadratic formula:
By substituting variables alpha and omega, start subscript, o, end subscript we wrote s a little simpler as:
where
alpha, equals, start fraction, start text, R, end text, divided by, 2, start text, L, end text, end fraction and omega, start subscript, o, end subscript, equals, start fraction, 1, divided by, square root of, start text, L, C, end text, end square root, end fraction
We finished up by solving an example circuit whose components produced a current (and voltages) that swing back and forth a few times.
The roots of the characteristic equation can take on both real and complex forms, depending on the relative size of alpha and omega, start subscript, o, end subscript. In the next article, we will describe these three forms in additional detail:
- overdamped, alpha, is greater than, omega, start subscript, 0, end subscript, leads to the sum of two decaying exponentials
- critically damped, alpha, equals, omega, start subscript, 0, end subscript, leads to t, dot decaying exponential
- underdamped, alpha, is less than, omega, start subscript, 0, end subscript, leads to a decaying sine
Want to join the conversation?
You guys should really make a video on this topic! Also, it is much easier to find v(0+), i(0+), dv(0+)/dt, di(0+)/dt, v(infinity), i(infinity) right off the bat so you don't have to pound through the problem the long way.
Oh, and the alpha and omega for this only works if they are in series. You need to do a write up to show how to solve if they are in parallel.
Thanks(25 votes)
It's the same process for a parallel, it's just that the differential equation is voltage-based: Cv''(t) + v'(t)/R + v(t)/L = 0.(1 vote)
- Why is the voltage across the capacitor (vc(t)) equals to the integral of the opposite of the current ? where does this minus sign come from ?(5 votes)
Because, in this article, it is assumed that the positive sign of the capacitor is on the top, so, the direction of current that it used is different from the current that supposed to be in the circuit... so that's why it got the minus sign...
If you flip the sign of the capacitor, it could be also worked though... so we don't get the minus sign, but in the KVL equation, it will be regarded as voltage rise, and it will get the plus sign... and in the end... it gets the same equation...(5 votes)
- Hi,
When finding A2 in derivative of the proposed solution, di/dt=d(e^(−t)A2sin2t)/dt
Shouldn't the answer be e^(-t) (4 cos(2 t) - 2 sin(2 t)) since we are using the product rule? With the answer coming out as A2=10/4?(2 votes)
Thanks for pointing how where I skipped the product rule step when taking the derivative. I will ask to have that fixed. Here's how it should go:
di/dt = d/dt ( e^-t A2 sin2t) = e^-t A2 (2cos2t - sin 2t)
Evaluate this at t=0,
10 = e^0 A2 (2cos 0 - sin 0)
10 = 1 . A2 . (2 - 0) = 2 . A2
A2 = 5(1 vote)
- How do you get the simpler s equation (the one with the variables α and ωnaught) from the quadratic formula?(1 vote)
- Review the definitions of alpha and ωo. This happens right after it says, "We can make the notation a little more compact by replacing parts of the expression with two new variables, alpha and ωo:"
Now look at the equation right after the sentence, "We can revise the quadratic formula by mashing the 2L denominator up into each term of the numerator:".
The first part of the equation, before the +/- sign, looks like -alpha.
The second part of the equation, under the square root symbol you find two terms, they look like alpha^2 and ωo^2.
Please let me know if this helped you understand. This is a spot where the article needs to be really clear.(3 votes)
- How can we include imaginary constant (j) in A2 and still get A2 as real (=5)?(2 votes)
- Back in the section Use Euler's Formula, A2 was defined in terms of the K constants.
A2 = j(K1 - K2)
Then we pressed ahead and figured out A2 = 5.
If you go back to the definition of A2 in terms of K's, that means the K's have to be imaginary numbers. They have to include a j term to get rid of the j in A2 = j(K1 - K2).(1 vote)
When you assume a solution in the exponential form in the "Propose a solution" section, that lets you factor out the exponential and set the other, quadratic factor to 0. Then you use the solutions for s in the "Proposed solution, upgraded" section to make a new form for the solution as a sum of 2 exponentials. But why were you able to do that when you assumed the form of the solution to be in the form of a single exponential to begin with? Wouldn't that be contradicting the assumption?(2 votes)
What role does s(natural frequency) play here ? What it determines, can you please explain about it ?(1 vote)- In the RLC derivation we found that s could have real or complex values. We described s in terms of alpha and omega, and found there were three general classes of behavior (under-damped, critically-damped, and over-damped). The solutions are rich and wonderful and varied. You get a drooping behavior from the real parts, and a wiggly behavior from the imaginary parts.
If you want to think about the meaning of s, it is a parameter (a complex number) that captures both the natural droopy and wiggly aspects of current and voltage as they change with time. The units of s are 1/t, which is the unit of frequency. The real part of s specifies the rate of droop while the imaginary part of s tells you the rate of wiggle.(2 votes)
- Why do we have 2L as the denominator in the quadratic formula?
Rachel(1 vote)
Our differential equation for the RLC can be turned into a "characteristic equation". It turns out the characteristic equation follows the general form of a quadratic equation...
ax^2 + bx + c = 0
which can be solved with the quadratic formula...
x = -b +/- sqrt(b^2 -4ac) / 2a
In our case the RLC circuit assigns a, b, and c as...
a = L, b = R, and c = 1/C
The 2 comes from the quadratic formula.(1 vote)
- Please , how do I find the value of R in terms of L and C that will result in repeated roots to be able to demonstrate an underdamped, overdamped and critically damped response in a RLC combination circuit.(1 vote)
- How did you obtained the second order DE for the RLC in series? Did you apply Kirchhoff's Loop Rule and take a total derivative of the function with respect to current?(1 vote)
- The differential equation for the RLC is constructed by applying Kirchhoff's Voltage Law around the loop. You write the voltage across each element in terms of the common current using the individual i-v equations for L, R, and C.(1 vote)