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

raffle_snaffle
 one year ago
Best ResponseYou've already chosen the best response.0dw:1401497886718:dw

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

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1I'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
 one year ago
Best ResponseYou've already chosen the best response.0This 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
 one year ago
Best ResponseYou've already chosen the best response.0But that is what i did, not sure how to proceed after defining E(t) as the piecewise function

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

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

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

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

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

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1Actually, 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).

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

raffle_snaffle
 one year ago
Best ResponseYou've already chosen the best response.0I 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

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1Yeah 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
 one year ago
Best ResponseYou've already chosen the best response.0yeah we can do that later.

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1I think I got a different result.

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

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

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1i also did it without laplace

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

myininaya
 one year ago
Best ResponseYou've already chosen the best response.1We 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

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

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

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

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

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.0okay thank you. Please show me how to extract the unit step function from the info given

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1Here'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

raffle_snaffle
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1Sure, take your time.

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

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

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1The 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
 one year ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles

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

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

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

raffle_snaffle
 one year ago
Best ResponseYou've already chosen the best response.0img 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
 one year ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles

raffle_snaffle
 one year ago
Best ResponseYou've already chosen the best response.0nvm 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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