Revision as of 16:11, 11 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.

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.

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!

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?

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