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raffle_snaffle Group Title

The values of the elements of an RLC circuit are given. Solve the initial value problems. L=1, R=0, C=10^-4; e(t)=100 if 0<=t<2pi: e(t)=0 if t>=2pi

  • 6 months ago
  • 6 months ago

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  1. Loser66 Group Title
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    @SithsAndGiggles

    • 6 months ago
  2. Loser66 Group Title
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    to make sure you get the correct explanation.

    • 6 months ago
  3. raffle_snaffle Group Title
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    |dw:1401497886718:dw|

    • 6 months ago
  4. raffle_snaffle Group Title
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    So i was working on this problem about 10 min ago. I am having a hard time determining e(t). Can you please help me find e(t). Once I obtain e(t) I can solve the equations for f(t)

    • 6 months ago
  5. raffle_snaffle Group Title
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    Do we use this piecewise function to determine e(t) |dw:1401498254051:dw|

    • 6 months ago
  6. SithsAndGiggles Group Title
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    I'm at a bit of a disadvantage here with the physics concepts, but I'll give it a shot. As I see it, the diff. eq. is \[\frac{dI}{dt}+10^4\int_0^t I(\tau)~d\tau=E(t)\] and \(E(t)\) is defined as \[E(t)=\begin{cases}100&\text{for }0\le t<2\pi\\0&\text{for }t\ge2\pi\end{cases}\] and with initial value \(I(0)=0\). Have I interpreted this correctly?

    • 6 months ago
  7. raffle_snaffle Group Title
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    This looks very familiar. We did one of these RLC circuits in class together. Looking back at my notes I am a bit lost...

    • 6 months ago
  8. raffle_snaffle Group Title
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    But that is what i did, not sure how to proceed after defining E(t) as the piecewise function

    • 6 months ago
  9. SithsAndGiggles Group Title
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    Taking the Laplace transform of both sides is the way to go here, but let's focus on the RHS first: \[\large\mathcal{L}\left\{E(t)\right\}=\int_0^\infty e^{-st}E(t)~dt\] \(E(t)=0\) for \(t\ge2\pi\) and \(E(t)=100\) for \(t<2\pi\), so we can reduce to \[\large \mathcal{L}\left\{E(t)\right\}=100\int_0^{2\pi} e^{-st}~dt=100\cdot\frac{1-e^{-2\pi s}}{s}\]

    • 6 months ago
  10. SithsAndGiggles Group Title
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    You were on the right track with the step function idea - you can definitely express \(E(t)\) in terms of \(u(t-c)\): \[E(t)=100u(2\pi-t)\]

    • 6 months ago
  11. raffle_snaffle Group Title
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    we have to change that 2pi-t because we can't take the LT

    • 6 months ago
  12. raffle_snaffle Group Title
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    so would it be 100-u(t-2pi)

    • 6 months ago
  13. raffle_snaffle Group Title
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    But if I was to use the unit step function how to identify the equation you have there E(t)=100u(2pi-t)?

    • 6 months ago
  14. SithsAndGiggles Group Title
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    Actually, if you want to write it in terms of \(t-2\pi\), the change would be \[E(t)=100-100u(t-2\pi)\] But that's not that important. Using the definition, you can find the same answer without having to memorize any formulas (other than the integral definition).

    • 6 months ago
  15. raffle_snaffle Group Title
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    I was looking at my notes... my instructor went over the definition of the LT so I understand what you did up top with the integral sign. It's just a way I am not use to when finding the laplace transform.

    • 6 months ago
  16. raffle_snaffle Group Title
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    Can we solve this using the definition of the Laplace transform? What you showed me fist?

    • 6 months ago
  17. raffle_snaffle Group Title
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    I think it would help me understand this LT better by using the definiton first because in all these mass spring system or RLC circuits we been using the unit step function

    • 6 months ago
  18. SithsAndGiggles Group Title
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    Yeah I'll continue with my method, but if you still want to know how to extract the unit step function expression, I can help you figure that out too.

    • 6 months ago
  19. raffle_snaffle Group Title
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    yeah we can do that later.

    • 6 months ago
  20. myininaya Group Title
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    I think I got a different result.

    • 6 months ago
  21. myininaya Group Title
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    I based my result off these notes: http://calculus7.com/sitebuildercontent/sitebuilderfiles/rlc.rtf

    • 6 months ago
  22. myininaya Group Title
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    After I found Q Then I used Q'=i to use my condition i(0)=0

    • 6 months ago
  23. myininaya Group Title
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    i also did it without laplace

    • 6 months ago
  24. myininaya Group Title
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    which you can

    • 6 months ago
  25. SithsAndGiggles Group Title
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    Okay, so we have \[\frac{dI}{dt}+10^4\int_0^t I(\tau)~d\tau=E(t)\] with \[E(t)=\begin{cases}100&\text{for }0\le t<2\pi\\0&\text{for }t\ge2\pi\end{cases}\] and with initial value \(I(0)=0\). Taking the LT of both sides: \[\begin{align*} \mathcal{L}\left\{\frac{dI}{dt}+10^4\int_0^t I(\tau)~d\tau\right\}&=\frac{100(1-e^{-2\pi s})}{s}\\ \left(sF(s)-I(0)\right)+10^4\frac{F(s)}{s}&=\cdots\\ \left(s+\frac{10^4}{s}\right)F(s)&=\frac{100(1-e^{-2\pi s})}{s}\\ \left(s^2+10^4\right)F(s)&=100(1-e^{-2\pi s})\\ F(s)&=\frac{100(1-e^{-2\pi s})}{s^2+10^4} \end{align*}\]

    • 6 months ago
  26. myininaya Group Title
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    We have Q''+10^4Q=100 first find homogenous solution to Q''+10^4Q=0 then find the particular solution I chose to use general form Q=At^2+Bt+C to get there

    • 6 months ago
  27. SithsAndGiggles Group Title
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    ^^ yup that method works too. I assumed this was a LT question based on previous questions.

    • 6 months ago
  28. raffle_snaffle Group Title
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    okay thanks @SithsAndGiggles that makes a lot of sense.

    • 6 months ago
  29. raffle_snaffle Group Title
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    If I am not able to identify the equation using the unit step function I can always use the definition of the laplace transform?

    • 6 months ago
  30. myininaya Group Title
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    Yea there looks like a ton of laplace going on in those notes I posted

    • 6 months ago
  31. SithsAndGiggles Group Title
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    Yes, the integral from the definition would be reduced to a simple calculation by changing the limits of integration based on the the step function you use.

    • 6 months ago
  32. raffle_snaffle Group Title
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    okay thank you. Please show me how to extract the unit step function from the info given

    • 6 months ago
  33. SithsAndGiggles Group Title
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    Here's a plot of \(E(t)\): |dw:1401500411049:dw| So the regular step function is defined as \[u(t-c)=\begin{cases}1&\text{for }t-c\ge 0\\0&\text{for }t-c<0\end{cases}~~~~~~~~\left[\iff~~u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t<c\end{cases}\right]\] with the plot |dw:1401500438474:dw| What we want to do is reflect about \(t=c\) so that the function is is 0 where it used to be non-zero and vice versa. In order to do that, we negate the argument, i.e. use \(u(-(t-c))=u(c-t)\). From the function's definition, we get \[u(-(t-c))-\begin{cases}1&\text{for }-(t-c)\ge 0\\0&\text{for }-(t-c)<0\end{cases}\] or \[u(c-t)-\begin{cases}1&\text{for }t\le c\\0&\text{for }t>c\end{cases}\] which gives the plot |dw:1401500884228:dw| So, multiplying by the scale factor of 100 and letting \(c=2\pi\), we get \(E(t)=100u(2\pi -t)\). However, like you said, you want this in terms of \(t-2\pi\), and to do that you would negate this function and its argument and add 100: |dw:1401501104472:dw|

    • 6 months ago
  34. raffle_snaffle Group Title
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    I got to leave here soon. Let me read this over some more and I will tag you later, k? I just need to study this a bit. If I have any questions I will let you know. For the most part I understand what is going on mathematically.

    • 6 months ago
  35. SithsAndGiggles Group Title
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    Sure, take your time.

    • 6 months ago
  36. raffle_snaffle Group Title
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    So what piecewise function do we use to obtain the unit step function? @SithsAndGiggles

    • 5 months ago
  37. raffle_snaffle Group Title
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    Do we use |dw:1401581129762:dw| or |dw:1401581158916:dw|

    • 5 months ago
  38. SithsAndGiggles Group Title
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    The second one. The first is the unit step function; the second is the unit step function scaled by a factor of 100, which is what you want.

    • 5 months ago
  39. raffle_snaffle Group Title
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    Okay thanks.

    • 5 months ago
  40. raffle_snaffle Group Title
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    @SithsAndGiggles

    • 5 months ago
  41. raffle_snaffle Group Title
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    • 5 months ago
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  42. raffle_snaffle Group Title
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    How does my work look @SithsAndGiggles ?

    • 5 months ago
  43. SithsAndGiggles Group Title
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    I think you skipped a step in the last two lines. Partial fractions first, then take the inverse transform. For example, \[F(s)=\frac{1}{s(s^2+4)}=\frac{A}{s}+\frac{Bs+C}{s^2+4}\] Rearranging factors, you have \[\begin{align*}1&=A(s^2+4)+s(Bs+C)\\ 1&=(A+B)s^2+Cs+4A\end{align*}\] so you have \[\begin{cases}A+B=0\\C=0\\4A=1\end{cases}~~\Rightarrow~~A=\frac{1}{4},~B=-\frac{1}{4},~C=0\] Then \[F(s)=\frac{1}{4}\cdot\frac{1}{s}-\frac{1}{4}\cdot\frac{s}{s^2+4}\] Taking the inverse transform yields \[F(s)=\frac{1}{4}-\frac{1}{4}\cos2t\] The same applies for the \(\dfrac{e^{-\pi s}}{s(s^2+4)}\) term.

    • 5 months ago
  44. SithsAndGiggles Group Title
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    *That last \(F(s)\) should be \(f(t)\).

    • 5 months ago
  45. raffle_snaffle Group Title
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    • 5 months ago
  46. raffle_snaffle Group Title
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    img 102 has my work and img 103 has the "advance" LT's. How would I continue my work without the cheat sheet? If I was only given the basic LT?

    • 5 months ago
  47. raffle_snaffle Group Title
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    @SithsAndGiggles

    • 5 months ago
  48. raffle_snaffle Group Title
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    nvm I got it figured out. Sorry I have not been here to discuss that one problem. I have an exam tomorrow on LTs so I am reviewing material instead of working on hw.

    • 5 months ago
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