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P.240 #27: I don't really understand the question above, but maybe someone can answer me this: At the end of the evaluation of the H(s) functions through partial fractions, shouldn't L^-1{<math> \frac{-1}{s^2+3^2}\ </math>} =(-1/3)sin(3t)? Yet in the solution the 1/3 has disappeared. I've spent hours trying to track down where I went wrong and I can't find where I am missing a 3 to cancel that term. Anybody have an idea what I'm missing here? | P.240 #27: I don't really understand the question above, but maybe someone can answer me this: At the end of the evaluation of the H(s) functions through partial fractions, shouldn't L^-1{<math> \frac{-1}{s^2+3^2}\ </math>} =(-1/3)sin(3t)? Yet in the solution the 1/3 has disappeared. I've spent hours trying to track down where I went wrong and I can't find where I am missing a 3 to cancel that term. Anybody have an idea what I'm missing here? | ||
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| + | Answer: You should have a 4/3 sin(3t) component as part of your y(t). This accounts for the "missing" -1/3 sin(3t) component. Think of it as y(t) = sin(t) -1/3 sin(3t) + 4/3 sin(3t) which becomes [sin(t) + 3/3 sin(3t)] or [sin(t) + sin(3t)] for 0 < t < pi. For t > pi, y(t) = sin(t) - 1/3 sin(3t) - sin(t) - (-1/3) sin(3t) + 4/3 sin(3t) which simplifies to 4/3 sin(3t). Hope this helps. | ||
Page 241, Problem 45: Can any of you EE's explain how to get this started? Do I need to go learn the integro-differential equation from sec 2.9? | Page 241, Problem 45: Can any of you EE's explain how to get this started? Do I need to go learn the integro-differential equation from sec 2.9? | ||
Revision as of 04:02, 13 October 2010
Homework 7 collaboration area
Question: What exactly is 6.2, #9 asking when it says to use another method to find the laplace transform for Prob 1? (AM, 07-Oct)
Answer: I think they just want you to show that it can be computed in two ways. In problem 1, you probably used the identity
L[f'] = s F(s) - f(0).
To compute the same Laplace transform a second way, you could integrate directly from the definition of the Laplace transform, or maybe you could use
L[f"] = s^2 F(s) - s f(0) - f'(0)
to get the same answer as problem 1.
For 6.2, #9 (prob 1 part) I followed the book's hint and used shifting. See Theorem 2 pp. 224. Let f(t)=t, and use $ e^{kt} $ for shifting. --Rekblad 06:57, 11 October 2010 (UTC)
Sec6.2 P232 #31: I've factored out the s in the denominator so it looks like
$ \frac{1}{s}\ \frac{5}{s^2-5} $
But I'm not sure how to proceed from there.
Answer: You will need to use the integration formula on p. 239:
$ \mathcal{L}[\ \int_0^t f(\tau)d\tau \ ]=\frac{1}{s}F(s), $
using F(s) = 5/(s^2 - 5). Find f(t) and integrate as shown to find the inverse transform of the given function.
Question continued: I understand the above formula, but why is F(s) = 5/(s^2 - 5) if the quantity we are trying to evaluate is L^-1[5/(s^3 + 5)]? I see that you can factor out the 1/s, but I still would think F(s) = 5/(s^3 + 5)...
And, if you are able to prove to me that I'm just being an idiot, maybe you can do it again: How do you find the inverse Laplace of (1/s)*(5/(s^2 + 5))? The first part is just 1 and the second is sqrt(5)*sin(sqrt(5)t) but I doubt that's the correct way to go about this.
Sec6.3 P240 #8: I have it written out as
f(t)=[u(t-0)-u(t-pi)]*(1-e^(-t)).
I'm stuck on how to work out (1-e^(-t)). In the previous problem, #5, it was easy to make t^2 into
[(t-1)+1]^2 or [(t-2)+2]^2
and essentially not change the function. However, that's not the case with (1-e^(-t)) and I don't know what to do with it.
Answer: Do the same thing:
$ 1-e^{-t}=1-e^{-[(t-\pi)+\pi]}=1-e^{-\pi}e^{-(t-\pi)} $
Sec6.3 #5: Yes, I agree it is easy to make t^2 into [(t-1)+1]^2 or [(t-2)+2]^2. Doing this, the book's solution is 2/s^3 + 2/s^2 + 1/s but I get: ... - 1/s by expanding upon the previous square. Any thoughts as to why the sign difference? Thanks.
Question: Pg. 232 - #9: The back of the book states that (cos a)^2 = 1/2 + 1/2 cos 2a.....where does this come from?
Answer: (cosx)^2=1-(sinx)^2, (sinx)^2=1/2-1/2cos(2x) => (cosx)^2=1-[1/2-1/2cos(2x)]=1/2+1/2cos(2x)
Question: pg247 #7: Does anyone think that there might be a typo in the back of the book? in the "u(t-4)" term, I don't see how they have 1/3sin(3t-12), I think it should be 2/3. The 1/3e^-t*sin3t makes sense because you have a (1-2/3)e^-t*sin33t, but there are no other sine terms to combine with in the "u(t-4" area. This problem was heavy on the book keeping and I might have dropped something, but I can't find it.
Follow Up Question to Above: The (1/3)sin3t term inside u(t-4) multiplier must have an exp(-t+4) multiplied to it, right? I dont see how this has vanished in the Book answer.--Sdhar 23:04, 12 October 2010 (UTC)
Follow Up Answer: You are right...the (1/3)sin3t term does have an exp(-t+4) component. Look at the book answer again. You will see that the exp(-t+4) term was factored out of both the cos and sine terms for the u(t-4) part.
Answer: I actually made the same mistake my first go around. The problem I encountered was after solving for the partial fractions of the "u(t-4)" term, I had:
$ \frac{-1}{s}\ + \frac{s+2}{(s+1)^2+9}\ $
This should expand to:
$ \frac{-1}{s}\ + \frac{s+1}{(s+1)^2+9}\ + \frac{1}{(s+1)^2+9}\ $
This will give you the answer the book has. Our error came from violating the s-shifting rules and incorrectly solving for the inverse Laplace using the following:
$ \frac{-1}{s}\ + \frac{s}{(s+1)^2+9}\ + \frac{2}{(s+1)^2+9}\ $
As you can see, the inverse laplace cannot be taken of the second term.... Hope this helps you.
It does, I can't believe we both made the same mistake. Thank you.
What do we do with the first 10 on the right hand side of the problem statement? Do we multiply through or is it better to work the problem with the 10 factored out?
Page 240, Problem 16. I don't understand what to do with the e^-s term. Any guidance on this problem?
Answer: the e^t is what allows you to use the theorem suggested in the title above the problem: "inverse transforms by the second shifting theorem."
P.240 #27: Given the piecewise function r(t), I converted this into a function of u(t-pi) and input that as the RHS of the system. I go through and solve for the answer that matches the back of the book, but I notice that it is only valid between 0 and pi and that there is a different answer when t > pi. I thought that by creating the function 8sin(t)*(1-u(t-pi)) we took care of the fact that the function is zero after pi... I am assuming that we obtain the second solution by setting the RHS to zero, but can anyone explain why we have to do this after building the piecewise function? Thanks!
P.240 #27: I don't really understand the question above, but maybe someone can answer me this: At the end of the evaluation of the H(s) functions through partial fractions, shouldn't L^-1{$ \frac{-1}{s^2+3^2}\ $} =(-1/3)sin(3t)? Yet in the solution the 1/3 has disappeared. I've spent hours trying to track down where I went wrong and I can't find where I am missing a 3 to cancel that term. Anybody have an idea what I'm missing here?
Answer: You should have a 4/3 sin(3t) component as part of your y(t). This accounts for the "missing" -1/3 sin(3t) component. Think of it as y(t) = sin(t) -1/3 sin(3t) + 4/3 sin(3t) which becomes [sin(t) + 3/3 sin(3t)] or [sin(t) + sin(3t)] for 0 < t < pi. For t > pi, y(t) = sin(t) - 1/3 sin(3t) - sin(t) - (-1/3) sin(3t) + 4/3 sin(3t) which simplifies to 4/3 sin(3t). Hope this helps.
Page 241, Problem 45: Can any of you EE's explain how to get this started? Do I need to go learn the integro-differential equation from sec 2.9?
Answer: check out example 2 on page 132. The switch in our problem is like the battery and switch in the example. Reply: Thanks. Yeah, I just needed to follow that walk-through.
p. 240:
QUESTION: I'm confused by the directions. It says that what is given is the Laplace transform of f(t) (that is, L[f]). But it looks to me what is given is L[(f)*u(t-a)]), not just L[f]. Can someone please clarify what is going on here? Furthermore, the back of the book gives answers for f(t) for various intervals, but for #21, it doesn't. Please help!! Thanks.
From Bell: Here is a little
about 247:16.
p. 240, #45: Since 1 kV = 1000 V, shouldn't there be a factor of 1000 somewhere in the answer? Thank you.
Re: I agree. Since the answer in the back does not have any units associated with it, I left the factor of 1000 in my i(t) and put [A] at the end. Figured that would cover my bases ---Rayala 20:42, 12 October 2010 (UTC)
