Hello, Please can you advise me on the below question from my text book. Since the text book only provides answers for odd numbered questions, I am not able to check my answer while I am studying Laplace. I typed the question out in MS WORD and did a screen shot. Must I expand the function then apply the Laplace or do I just apply Laplace with expanding? Thank you. Find the Laplace transform of the following piece-wise function: f(t) = (t+2)^2 , t > -2 0 , t < -2
First, I think you should check the condition when the shift theorem works (when and how it can be applied). Second, you can expand the function and compute the Laplace transform of each. However, your final result is wrong.
Hi Anhnha, Thank you, but please can you tell me how I would properly use the shift theorem. My text book does not clearly provide many examples. So if I expand the function and compute the Laplace transform I get the below. This is the answer that I now worked out, is this correct? I am getting confused with regards the Heaviside Laplace problems and those that contain the period restrictions. Please can some explain to me where will the t > -2 and t <-2 come into my equation or will it ever come into my equation?
I don't think you can apply the theorem directly as you did above. For the theorem to work the function should be zero when t < 0. However, your function is not. I think you can modify the theorem somehow to apply for the function above. However, I have not done it yet. From the proof of the theorem here, you can see that the step with red arrow will only be right if a > 0. Yes, it is correct. I am not sure what you mean by this also.
I can't see how your function can be transformed by using shifting theorem now. However, in general the information t>a and t< a with a>0 is used in unit step u(t-a) in the proof of the theorem. I am also not an expert just trying to help. To better understand when to use it, I think you should read or prove the therem yourself. Also, when you want to check the result, you can calculate Laplace transform by definition then compare your between them.
If you are using the single-sided Laplace Transform (which is almost certainly what you are using), then your function must be zero for t<0 otherwise the transform does not exist. If you look at most decent transform tables, they make this requirement explicitly clear by showing the transforms for f(t)·u(t). Double-sided Laplace Transforms are rarely encountered in signals and systems type work; when functions spanning t<0 are needed, the Fourier Transform is usually used.
Thanks Anhnha and WBahn, I understand. I am going to go over more examples to help me master this section of work. I will post up more questions which I feel is very difficult and "confusing" to me.
Hola naickej4, From my understanding on Laplace Transforms, I am very certain beyond a shadow of a doubt that your first answer is correct. I am also a student and will not be able to explain to you the finer details of my argument, since my theoretical knowledge is limited pertaining to this topic, but I can provide you with my worked solution and provide you a picture of my text if that may help you to briefly understand this theorem ( Second translation or shifting property). I am open to any correction of my evaluation from the lecturers (experts). gracias A picture from my textbook:
Hi Josh007, Thank you very much. This does make sense to me but why the p? (sounds like stupid question). I am going to find more examples on the internet for this theorem. My textbook does not have much. thanks
Hi naickej4, The author of the textbook prefers to use "p" instead of "s" to depict the transformed Laplace expression. I have not the faintest idea why he did this, but it can become very confusing when relating to other solutions that is found on other textbooks. regards Joshy
Hello there, When all else fails, read the instructions First, to apply the time shift theorem (not that it should be applied directly here though) we dont time shift the 'entire' function which here is (t+2)^2. The function that gets time shifted is t^2, and that is because that is the function that is time shifted not (t+2)^2. If we really wanted to time shift (t+2)^2 by 3 more units we would do this: ((t+3)+2)^2 and to time shift that by a delay of 1 unit we would do this: (((t-1)+3)+2)^2 and of course that simplifies to (t+4)^2. So (t+2)^2 is the time function t^2 shifted forward in time by two units. Most of the time however we work with delays, so we would have for example (t-2)^2, and that would be the same result as below but with a negative exponent. The time shift theorem may not work for all functions, but it might work for this one. The determining condition i think is that the integral has to exist. Since it exists, i think the result should be: F(s)=(2*e^(2*s))/s^3 and that of course is recognized as the time shifted function of t^2, but this is not the delayed function. Does it really exist? We might say that we cant have continuous functions that exist in future time, but i myself am not entirely clear on this either. That's because when this idea came about in history there were no computers with memory that could work in what appears to be future time. We now have such things. If the book gives that result then the book assumes we can, but if it doesnt then the book assumes we cant. Some of the better books have an 'answers' edition to supplement the main text. If you could find such an edition for your main text that could help you find answers to other questions too, if they provide them of course as some 'answers' editions only show answers to odd questions too. Consider yourself lucky if you got ANY answers in the book itself
Hi Mr AI, thank you sir for the explanation. So this mean what Josh007 says is correct and the theorem? Thank you
Hi, Well if that result matches my result (which you can tell by inspection) then all i can say is that if there is an answer then that is it. The ultimate decider is the book from which it came. If it is a more recent book it is more likely that they agree with that, but if a very old book then it is less likely. But again ultimately the book sets the precedence so that's the best resource. If you could find another similar example in the book you could find out from there if they allow that solution or they would rather call it 'undefined'. I have to admit i dont come across this too much because usually we work with delays which would mean the transformation t=>t-a not t+a. But if you apply the full definition of the Laplace Transform and not any short cuts, you get the result i showed, therefore i tend to believe that's the best solution. The only other choice is "undefined", and that doesnt seem right because then any question of this nature that has a positive time shift would simply be deemed undefined, so it would be a stupid question to ask. For example, how about this: K1*(t+a)^9+K2*(t+a)^8+K3*(t+a)^7+... we would think that was silly because of the t+a if we had no way to define functions with t+a in them. I'll check some other references in a little while and see what i can find. If i find anything interesting i'll make another post in this thread.
Hi, All i found was what i remembered, if t<0 then you have to use the full definition of the LT and that means many things that deal with the one sided LT are not the same.
Hi MrAl: What you did is the double-sided Laplace Transforms, so I think to know what is right, we should make clear what is meant in the original post (one sided LT or double sided LT).
Hello all, I think I'm getting confused now. Please may I know what is one sided LT and double sided LT accompanied with a small example for each. (forgive me if this seems like a silly question but I started studying Laplace from Saturday) thank you
The one sided transform has a lower integration limit of t=0- (i.e., an infinitesimal amount of time before t=0 so that impulse events at t=0 can be included, which are usually used to take care of initial conditions, which is a way of saying that they deal with the net effect of everything for all time prior to t=0). The two sided transform has a lower limit of negative infinity. The problem with the two sided transform is that the exponential in the transform integral blows up as t get more negative, so the transform integral itself is less likely to converge.