Main content

## 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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# LC natural response derivation 2

Starting from the differential equation, we come up with a proposed exponential solution and plug it into the equation. This gives us a characteristic equation. A "natural frequency" emerges. Created by Willy McAllister.

## Want to join the conversation?

- this is a tiny thing, but do engineers not use i for imaginary number because i is already used for current so often?(3 votes)
- That is exactly right. We (engineers) use i for current and j for the imaginary unit (sqrt(-1)). The i comes from Ampere himself. When he talked about current in his native French he called it "intensité du courant".(5 votes)

- Hi. I'm sorry, can you explain me why is it Vc+ Vl ? Isn't it Ic=IL and Vc=Vl because they are in parallel? if we do Vc+ Vl, instead of Vc - Vl, it doesn't make sense to me. Hoping for your reply, thank you(1 vote)
- This is slightly tricky because of the Sign Convention for Passive Components. You are correct that VC = VL = little v. Little v for both is measured from the bottom to the top (+ is at the top).

BUT, the currents are not identical, it is not the case that IL = IC. They are in opposite directions. If we point the current arrow DOWN for the inductor, it points UP for the capacitor. This introduces a negative sign when applying the I-V equation to the capacitor.

v = L di/dt

-i = C dv/dt

This is a weird case for such a super-simple circuit. It introduces a quirk in the I-V equation for one of them.(2 votes)

- I'm sorry, but how do you get the equation at00:56(eq. with the green color)?(1 vote)
- The green equation at :56 comes from dividing both sides of the previous equation by L.

I just noticed an error in the green equation, the denominator of the left-hand second derivative term should be dt^2, instead of just plain dt.(2 votes)

- Why the K variables in the Superposition of i are K1 and K2? Why not just K?(1 vote)
- As we solve this problem, we came up with an s^2 term. To solve for s we have to take a square root. Square roots always have two possible solutions, one with a plus sign, and one with a negative sign. This is where the terms s1 and s2, and eventually K1 and K2 come from. We can't choose one of the solutions as the right one, so we carry through the rest of the solution with both alternatives.(1 vote)

- @2:40s=1/t, and the unit is rad/sec , but if you multiply rad/sec times sec , the unit is rad and you said st has no units

and what is meant by natural frequency , what is special about it? and why is the characteristics equation named that name?(1 vote)- The Radian is a unit, but it is a "dimensionless" unit. That is 1 Rad = 1. Check out the Wikipedia page for Radian.

We use the term "natural" frequency to describe the way something vibrates when there is no external energy applied (when there is no "driving" force).

A good example is a pendulum. When you pull the pendulum's weight to the side you are initializing it with some starting energy. When you let go of the weight the pendulum swings back and forth with some period and some amplitude. Since you are no longer touching the pendulum there is no additional energy coming in from outside, and it will do whatever it does "naturally". In this state we would call its frequency the "natural frequency". If you grab the weight and move it back and forth at your will, that is no longer its natural frequency but rather its driven frequency.(1 vote)

- What are the other types of frequencies, besides "natural frequency"?(1 vote)
- We use the term 'natural frequency' when focussed on the few components right in front of us, without regard for any signals coming in from outside. 'Natural frequency' is what the LC (or RC, or RL, or RLC) does 'naturally' when nothing else is driving the circuit.

If we attach a signal source it can introduce its own frequency signal to the circuit. We would use the term 'frequency' to talk about that.(1 vote)

- At1:10, the square of t in denominator is missing. Is it ok or a mistake?(1 vote)
- Good eye. I made a notation error by leaving out the ^2 on the bottom of the second derivative term. This is a small flaw, it doesn't harm the derivation.(1 vote)

- 6:55This may be a math question. Why there are two solutions (positive & nagative).

I was not able to see two answers when dealing with imaginary numbers in Sal lectures.(1 vote) - Excuse me. do you know which lesson is Energy Storage in LC Circuits and Electromagnetic Oscillations ? I want to learn this lesson.(1 vote)
- In the last equation, why is it possible to add the two possible solutions for
`s`

?

Shouldn't it be that we attempt`i_1`

and`i_2`

with the respective`s`

's? (And most probably get only one valid answer?)(1 vote)- Good question. Did you try to solve for i_1 by itself? What did you get?

The characteristic equation produced two roots, (s_1 and s_2). We propose a general solution by tossing both of them into the proposed answer. If one of them is not needed to get a valid result (a function that solves the original differential equation) then the useless s will disappear during the math. If both s terms are needed to get a valid answer they will both hang around to the end.

To review, there are two approaches you can take.

1. Try one root, try the other root, try both roots together. Do as many as needed to get a valid answer.

2. Try both roots together. If only one is needed count on the math to make the other one melt away. If both are needed then you've done the math once.(0 votes)

## Video transcript

- [Voiceover] In the last video, we set up this differential equation that described an LC circuit, and now we're gonna go about solving this second order circuit. The technique that works here is the same that worked with first order ordinary differential equations. We're looking for a function of I that makes this whole equation true, and we're gonna do that
by guessing at an I, put it back into the equations, and see if it works. If it works, we win. If it doesn't work, we have
to think of something else, and we keep doing that until we solve it. So, knowing what I know about derivatives, and knowing and looking at this equation, can I guess at a plausible solution? So what we have here is two things. We have two terms that
have to add up to zero for all time, and so that means that
whatever function I pick for I, and there's some scaling factors here. Lemme write this equation like this. So what I have here is
a scaling factor times I has to somehow fully subtract from the second derivative of I. So these two terms have to
somehow look alike for all time. There's one function I
know where its derivatives sort of look like what we started with, and that's the exponential function. So I'm actually gonna make a guess. I'm gonna guess that I of T is something of the form some constant times the exponential of time with some other scale factor. Now, K is an adjustable parameter, and that's an amplitude. So it tells us how big the signal is, and what's S? S is up in the exponent along with T, and we know that by the time we take an exponent of anything that whatever's up here
has to have no units, so that means that ST, S times T, has no units, and that means that S has units of one over time so S is a frequency of some sort. And in particular, it's
gonna be a radian frequency. It's gonna be in radians per second. So, S is gonna be called
a natural frequency. So let's keep working on this. We're gonna plug I back into our equation, and see if it works. So we can plug I straight into here, and we need the second derivative of I. Let's first take the first
derivative D, DT of I equals D, DT of K E to the ST, and we can take that derivative, and that equals K times S times E to the ST. So it's also an exponential, and now we need the second derivative, so we wanna take the
derivative of this guy, so second derivative of
I with respect to time equals the first derivative of S, K, E to the ST, and that equals-- Another S comes down, so it's S squared, K, E to the ST. Good, now we have our second derivative. We can plug that in here, so let's do that. So the equation becomes S squared, K, E to the ST, plus one over LC, times K, E to the ST, equals zero. Let's do a little factoring. There's a common term here. There's a common term. K, E to the ST, K to the E, to the ST, so lemme factor that out. K, E to the ST times, what's left? S squared plus one over LC equals zero. Okay, we have how many
adjustable parameters here? We have K, S. Those are the two. L and C are constants that
are values of our circuit. So, we need to find some values of K and S that make this equation zero. Okay, I can make K equal to zero, and that would mean that zero equals zero, so our amplitude would be zero. So if we put nothing in the circuit, we get nothing out. Okay, that's totally boring. So that's not so
interesting of a solution. Now, does E to the ST ever become zero? E to something, does it ever become zero? It never does. If I let T go to plus infinity, and S is negative, then E to the ST would become zero, but plus infinity time from now is pretty far in the future, and I don't wanna wait that long. So the interesting solution becomes can we make S squared plus one over LC equal to zero? This equation is referred to as the characteristic equation. So let's see what happens
when we try to solve this. Well, the first step of this is I'm gonna get S squared equals minus one over LC, or S equals square root of minus one over LC. Uh oh, look what happens here. We're taking the square
root of a negative number. So what's gonna happen here? We're gonna get an imaginary
number for our answer. I can write this as square root of minus one times square root of one over LC, and that gives me two answers, and I'm gonna call the first answer S one, and that's gonna be equal to J, which is the imaginary number. That's the square root of minus one for electrical engineers, times what? Times square root of one over LC, and the other, S two, is the negative of that, minus J. So these are two possible solutions to our differential equation. Now, I'm gonna give a nickname
to this expression here because I don't wanna write it so much. I'm gonna call this omega knot. So this is a lower case
omega Greek letter. This is the capital omega that we use for Ohms, but little omega is often used as this variable here. So I can say that S one equals plus J, omega knot, and S two equals minus J, omega knot. So we're gonna use this
notation for a little bit, but just remember I made
that simple substitution. We have two different roots that can cause our differential
equation to go to zero. So when we combine these
into a solution for I, we're gonna use a
combination of these two. We don't know which one it is. It could be both. So they could be
superimposed on each other, and this is what superposition is for. So we're gonna now have, we need to have K one. We'll have two constants, E to the S one T plus K two, E to the S two T, and I could also write this as-- Let's fill in some of the numbers here, K one, E to the plus J, omega knot, T, plus K two, E to the minus, J omega knot, T. So that's my proposed solution, and what we need to do now, we found S. We found two values of S. Now we have these two constants. We gotta work those out. So, in the next video, we'll use the initial conditions to figure out what K one and K two are.