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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 <math>e^{kt}</math> for shifting. --[[User:Rekblad|Rekblad]] 06:57, 11 October 2010 (UTC) | 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 <math>e^{kt}</math> for shifting. --[[User:Rekblad|Rekblad]] 06:57, 11 October 2010 (UTC) | ||
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| + | p. 232, #9, part II: What does the problem mean? It says to express | ||
| + | cos^2(.5t) in terms of cosine and by using problem 3. What does this mean? | ||
| + | |||
| + | Answer: I think it means to use | ||
| + | |||
| + | <math>\cos^2\theta=\frac{1}{2}(1+\cos(2\theta))</math> | ||
| + | |||
| + | to write | ||
| + | |||
| + | <math>\cos^2(t/2)=\frac{1}{2}(1+\cos t))</math>. | ||
| + | |||
| + | Then use that to compute the Laplace Transform. Another way | ||
| + | to compute the same thing, is to use | ||
| + | |||
| + | <math>\cos^2\theta+\sin^2\theta =1</math> | ||
| + | |||
| + | to get | ||
| + | |||
| + | <math>\cos^2(t/2)=1-\sin^2(t/2)</math>, | ||
| + | |||
| + | and now problem 3 can be used to compute the Laplace Transform. | ||
Sec6.2 P232 #31: I've factored out the s in the denominator so it looks like | Sec6.2 P232 #31: I've factored out the s in the denominator so it looks like | ||
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the inverse transform of the given function. | the inverse transform of the given function. | ||
| − | Question | + | Question: How do you find the inverse Laplace of (5/(s^2 - 5))? |
| − | + | Answer: There are two ways to do that. One way is to use the table on | |
| + | page 224 to get | ||
| + | |||
| + | <math>\sqrt{5}\sinh(\sqrt{5}\, t)</math>. | ||
| + | |||
| + | The other way is to use partial fractions to get | ||
| + | |||
| + | <math>\frac{5}{s^2-5}=\frac{A}{s-\sqrt{5}} + \frac{B}{s+\sqrt{5}}</math> | ||
| + | |||
| + | and take it from there. | ||
Sec6.3 P240 #8: I have it written out as | Sec6.3 P240 #8: I have it written out as | ||
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It does, I can't believe we both made the same mistake. Thank you. | It does, I can't believe we both made the same mistake. Thank you. | ||
| + | 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 | |
| − | + | <math>\mathcal{L}^{-1}[ \frac{-1}{s^2+3^2}\ ] = -\frac{1}{3}sin(3t) </math>? | |
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | 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. | 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. | ||
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From Bell: Here is a little | From Bell: Here is a little | ||
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[http://www.math.purdue.edu/~bell/MA527/jing/periodic.swf Flash video hint] | [http://www.math.purdue.edu/~bell/MA527/jing/periodic.swf Flash video hint] | ||
| − | |||
about 247:16. | about 247:16. | ||
| + | And here is a | ||
| + | [http://www.math.purdue.edu/~bell/MA527/jing/circuit.swf Flash video hint] | ||
| + | about 240: 45. | ||
| + | |||
| + | You can find PDFs of the aftermath of the videos at | ||
| + | [http://www.math.purdue.edu/~bell/MA527/jing/ Bell's Jing things] | ||
p. 240, #45: Since 1 kV = 1000 V, shouldn't there be a factor of 1000 somewhere in the answer? Thank you. | p. 240, #45: Since 1 kV = 1000 V, shouldn't there be a factor of 1000 somewhere in the answer? Thank you. | ||
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Re: I was thinking the same. I believe they probably factored out the 1000 and their current is in terms of mA instead of A. -Artesha | Re: I was thinking the same. I believe they probably factored out the 1000 and their current is in terms of mA instead of A. -Artesha | ||
| + | |||
| + | From Bell: Yes, I agree with Artesha. Don't sweat about that factor of 1000. | ||
[[2010 MA 527 Bell|Back to the MA 527 start page]] | [[2010 MA 527 Bell|Back to the MA 527 start page]] | ||
Latest revision as of 16:16, 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)
p. 232, #9, part II: What does the problem mean? It says to express cos^2(.5t) in terms of cosine and by using problem 3. What does this mean?
Answer: I think it means to use
$ \cos^2\theta=\frac{1}{2}(1+\cos(2\theta)) $
to write
$ \cos^2(t/2)=\frac{1}{2}(1+\cos t)) $.
Then use that to compute the Laplace Transform. Another way to compute the same thing, is to use
$ \cos^2\theta+\sin^2\theta =1 $
to get
$ \cos^2(t/2)=1-\sin^2(t/2) $,
and now problem 3 can be used to compute the Laplace Transform.
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: How do you find the inverse Laplace of (5/(s^2 - 5))?
Answer: There are two ways to do that. One way is to use the table on page 224 to get
$ \sqrt{5}\sinh(\sqrt{5}\, t) $.
The other way is to use partial fractions to get
$ \frac{5}{s^2-5}=\frac{A}{s-\sqrt{5}} + \frac{B}{s+\sqrt{5}} $
and take it from there.
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.
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
$ \mathcal{L}^{-1}[ \frac{-1}{s^2+3^2}\ ] = -\frac{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 Flash video hint about 247:16.
And here is a Flash video hint about 240: 45.
You can find PDFs of the aftermath of the videos at Bell's Jing things
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)
Re: I was thinking the same. I believe they probably factored out the 1000 and their current is in terms of mA instead of A. -Artesha
From Bell: Yes, I agree with Artesha. Don't sweat about that factor of 1000.
