| (15 intermediate revisions by 5 users not shown) | |||
| Line 1: | Line 1: | ||
| − | = Homework 8 Collaboration Area = | + | == Homework 8 Collaboration Area == |
| − | + | Question on problem 15 in Sec 6.6: | |
| + | |||
| + | I tried to obtain the expression for | ||
| + | |||
| + | s/(s + 1) * 1/(s+1) | ||
| + | |||
| + | but am not getting the correct result in the Laplace table of | ||
| + | |||
| + | t sin t. | ||
| + | |||
| + | I am using the convolution of | ||
| + | cos(tau)*sin(t-tau). There is no t term in sight. Is it okay to read off the | ||
| + | table? Even if it is, shouldn't the result be the same? | ||
| + | |||
| + | Answer: | ||
| + | |||
| + | To find the inverse Laplace transform of | ||
| + | |||
| + | s/(s + 1) * 1/(s+1) | ||
| + | |||
| + | via convolution, you'll need to compute the convolution integral: | ||
| + | |||
| + | <math>\int_0^t \cos(\tau)\sin(t-\tau)\ d\tau.</math> | ||
| + | |||
| + | You'll have to use a formula for the sine of the | ||
| + | product of cosine and sine. Here it is: | ||
| + | |||
| + | (cos A)(sin B) = (1/2)sin(A+B) + (1/2)sin(B-A). | ||
| + | |||
| + | You'll need to be very careful. | ||
| + | Remember, t acts like a constant in the integrals. | ||
| + | |||
| + | There is only one correct answer, so you should get | ||
| + | it that way. (If it looks different than the back | ||
| + | of the book, a trig identity might be at fault.) | ||
| + | |||
| + | Another, and perhaps better, way to solve this problem is to recognize that the given expression is the derivative of | ||
| + | |||
| + | 1 / [(s+2)^2 + 1]] | ||
| + | |||
| + | ...therefore greatly simplifying the solution (no trig identities required). | ||
| + | |||
| + | (You'll need to use the formula | ||
| + | |||
| + | L[ t f(t) ] = -F'(s) | ||
| + | |||
| + | to get the inverse transform.) | ||
| + | |||
| + | P.257 #8: I'm a bit confused on how to get this problem started... Based on other problems in this section, we can factor out a t such that the function inside the transform is t times f(t) where f(t)=t^(n-1)e^(kt), but I don't see how that helps us much. | ||
| + | |||
| + | Answer from Bell: You'll need to use mathematical induction for this problem. Start with n=1. Then use the formula | ||
| + | |||
| + | L[ t f(t) ] = -F'(s) | ||
| + | |||
| + | to reduce the n=2 case to the n=1 case, etc. | ||
| + | |||
| + | (AJ) Also, there's a better table of Laplace transforms at the end of chapter 6 that has the transform for f(t) = t^(n-1)e^(kt) | ||
| + | |||
| + | Yes, but if you wanted to compute the transform the easy way instead of using the differentiation formula from the section, you would probably use | ||
| + | |||
| + | L[ f(t)e^{at} ] = F(s-a) | ||
| + | |||
| + | to get | ||
| + | |||
| + | L[ t^n e^{kt} ] = n!/ (s-k)^{n+1} | ||
| + | |||
| + | because | ||
| + | |||
| + | L[ t^n ] = n!/s^{n+1}. | ||
| + | |||
| + | (That's how they got that formula in the table in an earlier section.) | ||
| + | |||
| + | [[2010 MA 527 Bell|Back to the MA 527 start page]] | ||
| + | |||
| + | [[Course List|To Rhea Course List]] | ||
| + | |||
| + | [[Category:MA5272010Bell]] | ||
Latest revision as of 14:21, 19 October 2010
Homework 8 Collaboration Area
Question on problem 15 in Sec 6.6:
I tried to obtain the expression for
s/(s + 1) * 1/(s+1)
but am not getting the correct result in the Laplace table of
t sin t.
I am using the convolution of cos(tau)*sin(t-tau). There is no t term in sight. Is it okay to read off the table? Even if it is, shouldn't the result be the same?
Answer:
To find the inverse Laplace transform of
s/(s + 1) * 1/(s+1)
via convolution, you'll need to compute the convolution integral:
$ \int_0^t \cos(\tau)\sin(t-\tau)\ d\tau. $
You'll have to use a formula for the sine of the product of cosine and sine. Here it is:
(cos A)(sin B) = (1/2)sin(A+B) + (1/2)sin(B-A).
You'll need to be very careful. Remember, t acts like a constant in the integrals.
There is only one correct answer, so you should get it that way. (If it looks different than the back of the book, a trig identity might be at fault.)
Another, and perhaps better, way to solve this problem is to recognize that the given expression is the derivative of
1 / [(s+2)^2 + 1]]
...therefore greatly simplifying the solution (no trig identities required).
(You'll need to use the formula
L[ t f(t) ] = -F'(s)
to get the inverse transform.)
P.257 #8: I'm a bit confused on how to get this problem started... Based on other problems in this section, we can factor out a t such that the function inside the transform is t times f(t) where f(t)=t^(n-1)e^(kt), but I don't see how that helps us much.
Answer from Bell: You'll need to use mathematical induction for this problem. Start with n=1. Then use the formula
L[ t f(t) ] = -F'(s)
to reduce the n=2 case to the n=1 case, etc.
(AJ) Also, there's a better table of Laplace transforms at the end of chapter 6 that has the transform for f(t) = t^(n-1)e^(kt)
Yes, but if you wanted to compute the transform the easy way instead of using the differentiation formula from the section, you would probably use
L[ f(t)e^{at} ] = F(s-a)
to get
L[ t^n e^{kt} ] = n!/ (s-k)^{n+1}
because
L[ t^n ] = n!/s^{n+1}.
(That's how they got that formula in the table in an earlier section.)
