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raffle_snaffle
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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
 one month ago
 one month ago
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
 one month ago
 one month ago

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Loser66 Group TitleBest ResponseYou've already chosen the best response.0
@SithsAndGiggles
 one month ago

Loser66 Group TitleBest ResponseYou've already chosen the best response.0
to make sure you get the correct explanation.
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
dw:1401497886718:dw
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
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)
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
Do we use this piecewise function to determine e(t) dw:1401498254051:dw
 one month ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
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?
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
This looks very familiar. We did one of these RLC circuits in class together. Looking back at my notes I am a bit lost...
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
But that is what i did, not sure how to proceed after defining E(t) as the piecewise function
 one month ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
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 \[\large \mathcal{L}\left\{E(t)\right\}=100\int_0^{2\pi} e^{st}~dt=100\cdot\frac{1e^{2\pi s}}{s}\]
 one month ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
You were on the right track with the step function idea  you can definitely express \(E(t)\) in terms of \(u(tc)\): \[E(t)=100u(2\pit)\]
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
we have to change that 2pit because we can't take the LT
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
so would it be 100u(t2pi)
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
But if I was to use the unit step function how to identify the equation you have there E(t)=100u(2pit)?
 one month ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
Actually, if you want to write it in terms of \(t2\pi\), the change would be \[E(t)=100100u(t2\pi)\] 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).
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
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.
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
Can we solve this using the definition of the Laplace transform? What you showed me fist?
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
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
 one month ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
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.
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
yeah we can do that later.
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
I think I got a different result.
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
I based my result off these notes: http://calculus7.com/sitebuildercontent/sitebuilderfiles/rlc.rtf
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
After I found Q Then I used Q'=i to use my condition i(0)=0
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
i also did it without laplace
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
which you can
 one month ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
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: \[\begin{align*} \mathcal{L}\left\{\frac{dI}{dt}+10^4\int_0^t I(\tau)~d\tau\right\}&=\frac{100(1e^{2\pi s})}{s}\\ \left(sF(s)I(0)\right)+10^4\frac{F(s)}{s}&=\cdots\\ \left(s+\frac{10^4}{s}\right)F(s)&=\frac{100(1e^{2\pi s})}{s}\\ \left(s^2+10^4\right)F(s)&=100(1e^{2\pi s})\\ F(s)&=\frac{100(1e^{2\pi s})}{s^2+10^4} \end{align*}\]
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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
 one month ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
^^ yup that method works too. I assumed this was a LT question based on previous questions.
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
okay thanks @SithsAndGiggles that makes a lot of sense.
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
If I am not able to identify the equation using the unit step function I can always use the definition of the laplace transform?
 one month ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Yea there looks like a ton of laplace going on in those notes I posted
 one month ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
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.
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
okay thank you. Please show me how to extract the unit step function from the info given
 one month ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
Here's a plot of \(E(t)\): dw:1401500411049:dw So the regular step function is defined as \[u(tc)=\begin{cases}1&\text{for }tc\ge 0\\0&\text{for }tc<0\end{cases}~~~~~~~~\left[\iff~~u(tc)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t<c\end{cases}\right]\] 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 nonzero and vice versa. In order to do that, we negate the argument, i.e. use \(u((tc))=u(ct)\). From the function's definition, we get \[u((tc))\begin{cases}1&\text{for }(tc)\ge 0\\0&\text{for }(tc)<0\end{cases}\] or \[u(ct)\begin{cases}1&\text{for }t\le c\\0&\text{for }t>c\end{cases}\] 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 \(t2\pi\), and to do that you would negate this function and its argument and add 100: dw:1401501104472:dw
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
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.
 one month ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
Sure, take your time.
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
So what piecewise function do we use to obtain the unit step function? @SithsAndGiggles
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
Do we use dw:1401581129762:dw or dw:1401581158916:dw
 one month ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
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.
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
Okay thanks.
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
@SithsAndGiggles
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
How does my work look @SithsAndGiggles ?
 one month ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
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.
 one month ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.1
*That last \(F(s)\) should be \(f(t)\).
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
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?
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
@SithsAndGiggles
 one month ago

raffle_snaffle Group TitleBest ResponseYou've already chosen the best response.0
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.
 one month ago
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