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Homework 5, due February 18, 2009.
Posted: Thu Feb 12, 2009 3:46 pm
by goodwine
Reading: Chapter 8 from the course text.
Problems: 8.1, 8.2, 8.4, 8.7, 8.8, 8.12, 8.13 and 8.14.
Re: Homework 5, due February 18, 2009.
Posted: Sat Feb 14, 2009 7:18 pm
by goodwine
For problem number 8.13, you have that xDdot+25x = t when t < t < 1 is the range supposed to be from 0 to 1 or something else?. Also, for the solutions to some of these problems it seems like it is possible to get more than one solution depending on how you combine the terms, is this ok or is there a specific way to try to combine the terms? For example, in the book on pg 209 you pulled the 1/s out of the equations w/ the e^... and ended up with the solution shown in ( 8.8 ). If I had done that problem I would have probably left it as the [1(t-2)-1(t-3)] and just multiplied it by e^-t would that also be an acceptable answer or is that not the right way to do it?
It should be 0<t<1
These equations have unique solutions. They may be in different forms, but the solution is unique, so if you plot them they should be the same.
If I understand the latter part, what you are saying is wrong. It seems like you are taking the inverse Laplace transform of part of it but not another? I suspect I misunderstand what you are asking.
FWIW, I'm going to do a full example on Monday for these types of problems.
Re: Homework 5, due February 18, 2009.
Posted: Wed Feb 18, 2009 7:08 am
by goodwine
I was wondering if you can multiply F(s) terms to turn them back into f(t) terms. For example, can
(s/(s^2+25)) * (1 / (s^2 + 1))
be simply transformed into cos(5t) * sin (t), or would I have to combine them so that it becomes the sum of two F(s) terms, not the product?
No, the inverse Laplace transform of a product is not the product of the inverse Laplace transforms of the individual terms. That is why you have to do partial fractions.
It is linear, however, so the inverse Laplace transform of a sum or difference is the sum or difference of the inverse Laplace transforms of the individual terms.
Re: Homework 5, due February 18, 2009.
Posted: Thu Feb 19, 2009 8:42 pm
by goodwine
have a couple of questions on this weeks homework. First, how do you do the first problem? When I tried a partial fraction expansion with denominators of S^3 and (S+a) or S^2 and S(S+a) it made things worse looking. Any hints?
My second question regards problem 8.12. I know the solution should be [(1/3)cos(t) - (1/3)cos(2t)]*[1-step(t-pi)]. That answer makes sense and it it was my ODE45 approximation shows. Yet I keep getting a positive step function to be added: [(1/3)cos(t) - (1/3)cos(2t)]*[1+step(t-pi)] because of the fact that cos(t-pi) = -cos(t). What am I doing wrong?
For the first one you definitely want to split it into two denominators: s^3 and (s+a). It may be worse looking, but at least you can look the stuff up on the table.
I think you are not using the theorem correctly for the second term. If there is a step(t-pi) wherever t appears in the function should also be replaced by (t-pi). Or perhaps I'm misunderstanding the question.
Re: Homework 5, due February 18, 2009.
Posted: Thu Feb 19, 2009 8:43 pm
by goodwine
I noticed that if F(s) is 1/s, then f(t) can be equal to step(t) or
equal to 1. How do we know which to use when doing the inverse of the
Laplace transform?
Since we don't know anything before t=0, 1 and 1(t) are equivalent, so you can use either one. Actually 1(t) is more correct, but if you just use 1, that's fine for this class.
Re: Homework 5, due February 18, 2009.
Posted: Thu Feb 19, 2009 9:14 pm
by wstaruk
In 8.12 the equation the needs to be transformed is 1(t)cos(t) - 1(t-pi)cos(t), but because the step function and the cosine function have different arguments you can't transform it yet. In example 8.4.4 you have a similar situation with cos(2t)1(t-7pi/2). There, to get the arguments the same, you use the fact that cos(2(t-7pi/2))=-cos(2t). When I try using the fact that cos(t-pi) = -cos(t) on problem 8.12, however, I get the wrong answer.
Re: Homework 5, due February 18, 2009.
Posted: Thu Feb 19, 2009 9:20 pm
by goodwine
wstaruk wrote:In 8.12 the equation the needs to be transformed is 1(t)cos(t) - 1(t-pi)cos(t), but because the step function and the cosine function have different arguments you can't transform it yet. In example 8.4.4 you have a similar situation with cos(2t)1(t-7pi/2). There, to get the arguments the same, you use the fact that cos(2(t-7pi/2))=-cos(2t). When I try using the fact that cos(t-pi) = -cos(t) on problem 8.12, however, I get the wrong answer.
You are using the right approach. You need to replace
1(t-pi)*cos(t)
with
-1(t-pi)*cos(t-pi).
I don't think that's your error. Something else is probably giving you the wrong answer.
Re: Homework 5, due February 18, 2009.
Posted: Thu Feb 19, 2009 11:07 pm
by sagarwal
Sorry this is so late, but could you post the MATLAB syntax for problems 12-14 on here? Thanks
Re: Homework 5, due February 18, 2009.
Posted: Fri Feb 20, 2009 11:21 am
by goodwine
sagarwal wrote:Sorry this is so late, but could you post the MATLAB syntax for problems 12-14 on here? Thanks
To plot
x(t) = sin(t) + t^2/2 + 1(t-3)[t sin(t-3)+cos(4 t)]
do
Code: Select all
t=linspace(0,20,1000);
plot(t,sin(t)+t.^2/2+(t>3).*(sin(t-3).*t+cos(t*t));
Basically, it just boils down to using (t>3) for the time-shifted step function and making sure you use .* in all the right places.