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raffle_snaffle

  • one year ago

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

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  1. Loser66
    • one year ago
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    @SithsAndGiggles

  2. Loser66
    • one year ago
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    to make sure you get the correct explanation.

  3. raffle_snaffle
    • one year ago
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    |dw:1401497886718:dw|

  4. raffle_snaffle
    • one year ago
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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)

  5. raffle_snaffle
    • one year ago
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    Do we use this piecewise function to determine e(t) |dw:1401498254051:dw|

  6. SithsAndGiggles
    • one year ago
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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?

  7. raffle_snaffle
    • one year ago
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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...

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

  9. SithsAndGiggles
    • one year ago
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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 \[\large \mathcal{L}\left\{E(t)\right\}=100\int_0^{2\pi} e^{-st}~dt=100\cdot\frac{1-e^{-2\pi s}}{s}\]

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

  11. raffle_snaffle
    • one year ago
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    we have to change that 2pi-t because we can't take the LT

  12. raffle_snaffle
    • one year ago
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    so would it be 100-u(t-2pi)

  13. raffle_snaffle
    • one year ago
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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)?

  14. SithsAndGiggles
    • one year ago
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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).

  15. raffle_snaffle
    • one year ago
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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.

  16. raffle_snaffle
    • one year ago
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    Can we solve this using the definition of the Laplace transform? What you showed me fist?

  17. raffle_snaffle
    • one year ago
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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

  18. SithsAndGiggles
    • one year ago
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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.

  19. raffle_snaffle
    • one year ago
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    yeah we can do that later.

  20. myininaya
    • one year ago
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    I think I got a different result.

  21. myininaya
    • one year ago
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    I based my result off these notes: http://calculus7.com/sitebuildercontent/sitebuilderfiles/rlc.rtf

  22. myininaya
    • one year ago
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    After I found Q Then I used Q'=i to use my condition i(0)=0

  23. myininaya
    • one year ago
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    i also did it without laplace

  24. myininaya
    • one year ago
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    which you can

  25. SithsAndGiggles
    • one year ago
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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: \[\begin{align*} \mathcal{L}\left\{\frac{dI}{dt}+10^4\int_0^t I(\tau)~d\tau\right\}&=\frac{100(1-e^{-2\pi s})}{s}\\ \left(sF(s)-I(0)\right)+10^4\frac{F(s)}{s}&=\cdots\\ \left(s+\frac{10^4}{s}\right)F(s)&=\frac{100(1-e^{-2\pi s})}{s}\\ \left(s^2+10^4\right)F(s)&=100(1-e^{-2\pi s})\\ F(s)&=\frac{100(1-e^{-2\pi s})}{s^2+10^4} \end{align*}\]

  26. myininaya
    • one year ago
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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

  27. SithsAndGiggles
    • one year ago
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    ^^ yup that method works too. I assumed this was a LT question based on previous questions.

  28. raffle_snaffle
    • one year ago
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    okay thanks @SithsAndGiggles that makes a lot of sense.

  29. raffle_snaffle
    • one year ago
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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?

  30. myininaya
    • one year ago
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    Yea there looks like a ton of laplace going on in those notes I posted

  31. SithsAndGiggles
    • one year ago
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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.

  32. raffle_snaffle
    • one year ago
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    okay thank you. Please show me how to extract the unit step function from the info given

  33. SithsAndGiggles
    • one year ago
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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|

  34. raffle_snaffle
    • one year ago
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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.

  35. SithsAndGiggles
    • one year ago
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    Sure, take your time.

  36. raffle_snaffle
    • one year ago
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    So what piecewise function do we use to obtain the unit step function? @SithsAndGiggles

  37. raffle_snaffle
    • one year ago
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    Do we use |dw:1401581129762:dw| or |dw:1401581158916:dw|

  38. SithsAndGiggles
    • one year ago
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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.

  39. raffle_snaffle
    • one year ago
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    Okay thanks.

  40. raffle_snaffle
    • one year ago
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    @SithsAndGiggles

  41. raffle_snaffle
    • one year ago
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  42. raffle_snaffle
    • one year ago
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    How does my work look @SithsAndGiggles ?

  43. SithsAndGiggles
    • one year ago
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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.

  44. SithsAndGiggles
    • one year ago
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    *That last \(F(s)\) should be \(f(t)\).

  45. raffle_snaffle
    • one year ago
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  46. raffle_snaffle
    • one year ago
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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?

  47. raffle_snaffle
    • one year ago
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    @SithsAndGiggles

  48. raffle_snaffle
    • one year ago
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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.

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