raffle_snaffle
  • raffle_snaffle
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
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Loser66
  • Loser66
@SithsAndGiggles
Loser66
  • Loser66
to make sure you get the correct explanation.
raffle_snaffle
  • raffle_snaffle
|dw:1401497886718:dw|

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raffle_snaffle
  • raffle_snaffle
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)
raffle_snaffle
  • raffle_snaffle
Do we use this piecewise function to determine e(t) |dw:1401498254051:dw|
anonymous
  • anonymous
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?
raffle_snaffle
  • raffle_snaffle
This looks very familiar. We did one of these RLC circuits in class together. Looking back at my notes I am a bit lost...
raffle_snaffle
  • raffle_snaffle
But that is what i did, not sure how to proceed after defining E(t) as the piecewise function
anonymous
  • anonymous
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 \[\large \mathcal{L}\left\{E(t)\right\}=100\int_0^{2\pi} e^{-st}~dt=100\cdot\frac{1-e^{-2\pi s}}{s}\]
anonymous
  • anonymous
You were on the right track with the step function idea - you can definitely express \(E(t)\) in terms of \(u(t-c)\): \[E(t)=100u(2\pi-t)\]
raffle_snaffle
  • raffle_snaffle
we have to change that 2pi-t because we can't take the LT
raffle_snaffle
  • raffle_snaffle
so would it be 100-u(t-2pi)
raffle_snaffle
  • raffle_snaffle
But if I was to use the unit step function how to identify the equation you have there E(t)=100u(2pi-t)?
anonymous
  • anonymous
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).
raffle_snaffle
  • raffle_snaffle
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.
raffle_snaffle
  • raffle_snaffle
Can we solve this using the definition of the Laplace transform? What you showed me fist?
raffle_snaffle
  • raffle_snaffle
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
anonymous
  • anonymous
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.
raffle_snaffle
  • raffle_snaffle
yeah we can do that later.
myininaya
  • myininaya
I think I got a different result.
myininaya
  • myininaya
I based my result off these notes: http://calculus7.com/sitebuildercontent/sitebuilderfiles/rlc.rtf
myininaya
  • myininaya
After I found Q Then I used Q'=i to use my condition i(0)=0
myininaya
  • myininaya
i also did it without laplace
myininaya
  • myininaya
which you can
anonymous
  • anonymous
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: \[\begin{align*} \mathcal{L}\left\{\frac{dI}{dt}+10^4\int_0^t I(\tau)~d\tau\right\}&=\frac{100(1-e^{-2\pi s})}{s}\\ \left(sF(s)-I(0)\right)+10^4\frac{F(s)}{s}&=\cdots\\ \left(s+\frac{10^4}{s}\right)F(s)&=\frac{100(1-e^{-2\pi s})}{s}\\ \left(s^2+10^4\right)F(s)&=100(1-e^{-2\pi s})\\ F(s)&=\frac{100(1-e^{-2\pi s})}{s^2+10^4} \end{align*}\]
myininaya
  • myininaya
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
anonymous
  • anonymous
^^ yup that method works too. I assumed this was a LT question based on previous questions.
raffle_snaffle
  • raffle_snaffle
okay thanks @SithsAndGiggles that makes a lot of sense.
raffle_snaffle
  • raffle_snaffle
If I am not able to identify the equation using the unit step function I can always use the definition of the laplace transform?
myininaya
  • myininaya
Yea there looks like a ton of laplace going on in those notes I posted
anonymous
  • anonymous
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.
raffle_snaffle
  • raffle_snaffle
okay thank you. Please show me how to extract the unit step function from the info given
anonymous
  • anonymous
Here's a plot of \(E(t)\): |dw:1401500411049:dw| So the regular step function is defined as \[u(t-c)=\begin{cases}1&\text{for }t-c\ge 0\\0&\text{for }t-c<0\end{cases}~~~~~~~~\left[\iff~~u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }tc\end{cases}\] 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|
raffle_snaffle
  • raffle_snaffle
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.
anonymous
  • anonymous
Sure, take your time.
raffle_snaffle
  • raffle_snaffle
So what piecewise function do we use to obtain the unit step function? @SithsAndGiggles
raffle_snaffle
  • raffle_snaffle
Do we use |dw:1401581129762:dw| or |dw:1401581158916:dw|
anonymous
  • anonymous
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.
raffle_snaffle
  • raffle_snaffle
Okay thanks.
raffle_snaffle
  • raffle_snaffle
@SithsAndGiggles
raffle_snaffle
  • raffle_snaffle
1 Attachment
raffle_snaffle
  • raffle_snaffle
How does my work look @SithsAndGiggles ?
anonymous
  • anonymous
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.
anonymous
  • anonymous
*That last \(F(s)\) should be \(f(t)\).
raffle_snaffle
  • raffle_snaffle
raffle_snaffle
  • raffle_snaffle
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?
raffle_snaffle
  • raffle_snaffle
@SithsAndGiggles
raffle_snaffle
  • raffle_snaffle
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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