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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.

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  • primosaur ultimate style avatar for user Raul Lara
    What does Phasor notation have to do with Euler's?
    (3 votes)
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    • spunky sam orange style avatar for user Willy McAllister
      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)
  • blobby green style avatar for user Ta Ay
    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)
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    • old spice man green style avatar for user Willy McAllister
      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)
  • male robot donald style avatar for user Bryan
    What do the sines and cosines do?
    (1 vote)
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    • duskpin ultimate style avatar for user Didi
      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.