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### 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 3

We use Euler's Formula to change our complex exponential solution into a solution expressed in terms of sines and cosines. Created by Willy McAllister.

## Want to join the conversation?

- What does Phasor notation have to do with Euler's?(3 votes)
- Euler's equation says we can disassemble a sine or cosine into the sum of two complex exponential terms. We do this disassembly because it is simpler to solve a circuit with exponential inputs compared to sine or cosine inputs. The word phasor refers to the exponential term, specifically when there is a phase shift represented by angle phi: e^j(wt + phi) = e^jwt * e^jphi. We call the term e^jphi a Phasor.(4 votes)

- do i need to constantly come up with a solution or do i just use the solution presented in this video? Or does it vary?(1 vote)
- The solution for LC Natural Response is derived independent of the value of L and C, so it is a general solution for the circuit LC configuration presented. Likewise with the RC and RL Natural Responses.

The next more complex circuit is R L and C together. You start again from scratch to derive the general solution(s) for that new circuit. The RLC solution is quite rich and varied. I consider it the "Queen of Analog Circuits" with wide application to most analog systems.

Anything more complex than RLC is best viewed as a "filter". We don't use the same solution method as for simpler circuits, but rather move on/up to Laplace Transform Theory.(2 votes)

- What do the sines and cosines do?(1 vote)
- Sines and cosines are needed to solve the exponentials for current, i = K_1*e^j⍵0t + K_2*e^-j⍵0t. Applying Euler's formula is necessary to solve for K_1 and K_2. Current for an LC circuit natural response is:

I = sqrt(C/L) * V_o sin(ω0t), ω0 = sqrt(1/LC)(1 vote)

## Video transcript

- [Voiceover] In the last
video we took a guess at what the solution was for our
differential equation and we came up with an exponential as
our guess and as we did the analysis we developed a
characteristic equation. We ended up with a complex
answer for one of the adjustable parameters, the
natural frequency of our circuit. So here's the form of
our proposed solution, now this is getting kinda
complicated but I promise you we're gonna simplify this pretty soon. We have two solutions for s, s
one and s two that we plugged in as plus and minus j omega naught, and we have more adjustable
amplitude parameters that we have to figure out. So now in this video we're
gonna continue on solving our differential equation, this
is our proposed solution, this is a pretty complicated
looking expression and what we're gonna call on
here now is something really important in electronics and
in general and it's called Euler's Identity. And we're gonna use this
identity to figure out what to do with these complex exponential terms. So if you search on this term
on Kahn Academy you'll find an explanation of where
these identities come from but I'm just gonna state 'em right here. And what it says is that
Euler's identitiy is e to the j x anything up
there equals cosign x plus j times sine x, that's one of the identities and the other identity is
e to the minus j x equals cosign of x minus j sine x. So these are useful because
we have this exponential function with the complex
unit inside of it, when we go over to this side, this is sort of a normal complex number, cosign x is some number
between plus and minus one, sine x is some number
between plus and minus one, and it's just a normal complex number so this may help us
simplify our life here as we move forward. So that's a really important
identity that we get to use to solve our LC circuit. Now I'm gonna go back and
rewrite these two exponential's using Euler's Identity here. This is gonna get big but it'll
collapse down pretty soon. Okay, let me move over here,
i equals k one times e to the plus j omega t let's use this
one here so that would be cosign x is omega naught t
plus j sine x is omega naught t and we use the first one. Okay, now the second term is
plus k two times now we have the negative up in the
exponent so we use this one cosign omega naught t minus
j sine omega naught t. Now we have our solution spread
all out across the screen and let's see if we can
tidy things up here. So what I'm gonna do is I'm
gonna gather all the cosign terms together, this one and this one and then I'll gather
the sine terms together. Okay so i equals cosign omega
naught t, and the cosign is multiplied by k one
and plus k two alright. Now let's add to that we have j sine and j sine so
I'll write sine omega naught t over here and j times what,
j times k one and this time we have this minus sign
that makes it minus k two, alright so now, current is
some number times cosign plus j times some number times sine. Now these are two arbitrary
constants and I'm just gonna make up another one, I'm
just call this one a one we'll call this one a
two and we'll call that, I'll use the j, k one minus k two, now I can rewrite this as i equals a one cos mega naught t plus a two sine omega naught t, good. So from now on we're
gonna work with these a's and if I ever want to know
what the original k's were I would just come back
to these equations here, once I figure out a, I can
figure out both the k's. So let's keep pressing on, how do we figure out a one and a two? Well to do that we're gonna
use the initial conditions, the ic's. And if we think back we
remember that in our original schematic we had some q
here which means we had a plus or minus v zero, v
naught, and we said the current through here started at zero. So that's our initial conditions, v of time equals zero equals v naught and current of time
equals zero equals zero. Let's use these two values
to help us figure out what a one and a two are. And we'll do that in the
next video as we continue the derivation of the natural response.