Anyone feel like helping me with a Laplace transform problem? It is summer break so I bought a 1000 page book on Engineering Mathematics and I am slowly working through it so I will have some background when I start this year. Anyway, I have made it to Ch. 5 of Erwin Kreyszig's Advanced Engineering Mathematics which is titled "Laplace Transforms". I understand the simple examples that the book gives (like find the Laplace transfrom of f(t) = 1 or f(t) = e^at blah, blah, blah) but there is a question in the problems at the end which is giving me a hard time. Here it is... Find the Laplace transform of sin[pi]t. I get L(f) = S e^-st*sin[pi]t = 1/-s*e^-st*sin[pi]t |oo,0 + 1/s S e^-st*cos[pi]t. Where "S" is the improper integral between infinity and zero. and |oo,0 means the integral has been done and is evaluated between infinity and zero. I am very doubtful that this is correct could anyone please tell me where I have gone wrong! The thing is, I can guess the answer! It is simply [pi]/(s^2 + [pi]^2). But how do I get this. (clue: do I have to use the fact that e^iwt (where i = sqrt(-1)) is coswt + isinwt? I probably do don't I?) PS. What is the point of the shifting theorem? Thanks.
The Laplace transform of f(x) is the function F(s) = integral (from 0 to inf) f(t)e^(-st)dt. In order for that to exist, f must be "of exponential type"- increases no worse than e^x (a pretty large group; exponentials increase rapidly) and cos(x) certainly is of that type. The simplest way to integrate e^(-st)cos(t)dt is to use integration by parts (that's generally true of Laplace transforms). Let u be e^(-st) and dv= cos(t)dt. The du= -se^(-st)dt and v= sin(t). The integral becomes {e^(-st)(sin(t)) evaluated between 0 and infinity} + (s) integral (from 0 to infinity) e^(-st) sin(t) dt. {e^(-st)(sin(t)) evaluated between 0 and infinity} is 0 because sin(t) is 0 at 0 and e^(-st) goes to 0 as t goes to 0 so the Laplace transform is simply (s) integral (from 0 to infinity) e^(-st) sin(t) dt. Now, do the integration by parts again. Let u= e^(-st) again and dv= sin(t) so that du= -se^(-st)dt and v= -cos(t). Now the integral is s({-e^(-st)cos(t) evaluated between 0 and infinity)}- s integral (from 0 to infinity) e^(-st)cos(t)dt. e^(-st) is still 0 at infinity but now, at t=0, e^(-st)cos(t) is -e^(0)cos(0)= -1. We have integral (from 0 to infinity)e^(-st)cos(t)dt= -s- s^2 integral (from 0 to infinity) e^(-st)cos(t)dt. We are right back where we started! Adding s^2 integral (from 0 to infinity) e^(-st)cos(t)dt to both sides: (1+s^2) integral (from 0 to infinity) e^(-st).(t)dt= -s so the Laplace transform of cos(x) is -s/(1+s^2). If by "shifting theorem", you mean what I think you do, then the point is this: since the Laplace transform is an integral from 0 to infinity, it doesn't matter what the function is for x< 0 so we often take it to be 0 there. If we have a function that is 0 for x< a (for some a>0) then we can take the integral from a to infintiy. Now make a change of variable: let u= t-a so that when t=a, u=0. integral (from t=a to infinity) e^(-st)f(t)dt= integral(from u= 0 to infinity) e^(-s(u+a))f(u+a)du= e^(-sa)*Laplace transform of f "shifted" so that it now starts at 0.
Let sin(pi)t = Im exp(i(pi)t) then evaluate the integral. To clear up my notation Im exp means take only the imaginary component of complex exponential. By the way, the real part of the answer will give you the Laplace transform of cos(pi)t. You do not need to shift or integrate by parts or anything because now you simply have the integral of an exponential.
Thanks for the help guys! I can do it now! Here is my next question: When solving second-order nonhomogeneous equations with constant coefficients... y'' + ay' + by = r There are three steps to follow... 1. Form the Subsidiary Equation by letting y'' = s^2Y - sy(0) - y'(0), y' = sY - y(0), y = Y and r + R. 2. Then rearrange and solve for Y. Multiply both sides by the Transfer Function so you have your equation in terms of Y. 3. [Here is my problem] "Reduce equation to a sum of terms" whose inverses can be found on a Laplace transform table. The part in inverted commas is my problem. I can do most of the problems up until I have to play with the fractions. Can anyone give me an idea on how they do this bit (methods, hints, guides...). Thanks alot. Ben.
The "reduce to a sum of terms" is typically the "partial fractions" you learned in calculus. That is: 1/(x-a)(x-b)= A/(x-a)+ B(x-b), etc. You can find the coefficients, A, B, etc. by multiplying both sides by the denominator on the left side, then (i) substituting values of x that will simplify the equation or (ii) equating coefficients of like powers of x. (i) is simpler when it works. (ii) always works.
There are a few tricks to know when doing partial fractions which come only once you easily recognise the form. Indeed the method which HallsofIvy outlined is correct, but if there is higher order terms in the denominator you can get into trouble, as shown below 1/(s^2(s+1)= A/(s^2)+ B/(s+1), Implies that A=B=0 which is clearly not true. In fact the correct form is 1/(s^2(s+1)=(As+B)/(s^2)+C/(s+1)
Thanks for clearing that up ryans. I had to go back to my first-year calculus textbook and re-learn the section on integrating partial fractions. This has helped.
I'd like to bring to your attention a simple, beautiful, but rather unusual exposition of Laplace Transforms. It can be found here: http://www.mathpages.com/home/kmath508.htm Where Msr. Brown says: "Now, the key to the usefulness of Laplace transforms arises from the following indefinite integral..." The equation that follows can be derived by the use of repeated Integration by Parts or by the use of Kronecker's Formula. I hope you enjoy it, Michael
Then you should be able to help me even more!!!! Please Register or Log in to view the hidden image! The unit step function is causing me hassles! (also called Heaviside function in some texts) My main question is: What is the reason we use it? What is it good for? What can it do? The example question shows how to find the transform of f(t) which is defined in three parts: f(t) = 2 for 0 < t < pi f(t) = 0 for pi < t < 2pi f(t) = sin(t) if t > 2pi The first part of the function (0 < t < pi) is a straight line and has a unitstep function of 2u(t). Now the book does not say how it got this! But my guess would be to use the formula: xu(t-a) where x=2 and a=0. For pi < t < 2pi the book says the unitstep function is 2u(t-pi). Once again, the book does not show how it got to this conclusion, instead leaving it up to me to guess. I guessed that we now are jumping from f(t) = 2 to f(t) = 0 so we take 2u(t) (as before) and have a = pi. Leaving 2u(t-pi) (in thinking about this, we are SUBTRACTING the f(t) = 0 from f(t) = 2). Following in this fashion I figured that the third unitstep function goes from f(t) = 0 into a sinusoidal f(t) = sint. So we take the 2u(t-pi) and "stick" sint to it. Giving 2u(t-pi)sint. (so the f(t) = 0 line is all-of-a-sudden a sin wave) If this is a correct way of thinking about it then YAY. But it probably isn't. If so, can someone give me their opinion on solving these problems. Anyway, so the function in terms of its unitstep functions is: f(t) = 2u(t) - 2u(t-pi) + 2u(t-pi)sint The Laplace transform of this is of course: L(f) = (2/s) - (2exp(-[pi]s))/s + (exp(-2[pi]s))/(s^2 + 1) Thanks for any help on this matter. Cheers. Ben.
The low down on step functions is that they are very useful, as they act like a mathematical switch which turns things like currents, forces and the such on in our equations. For a Laplace transform, the step function basically changes the range of integration. Say we have a function f(t) multiplied by H(t-a) (step function with H(t-a)=0 for t<a and H(t-a)=1 for t>a). The integral limits for the Laplace transform of f(t)H(t-a) is then inf / | f(t)H(t-a)exp(-st)dt = / 0 inf / | f(t)exp(-st)dt / a That's basically it. A point of interest maybe that the derivative of the step function is the dirac delta function, and thus, by using the dirac delta function as a "test" force for say a harmonic oscillator, Laplace transforms maybe used to find what is known as the Green's function of the problem, meaning that we can reconstruct difficult differential equations into simpler form. Do no underestimate the step function.
Thankyou ryans for your motivation. I think I am starting to like unit step functions. A neat trick indeed! A few more questions. 1. What is the Dirac delta function. From what I understand it is similar to the unitstep function but more of an impulsive nature. How is it applied to problems and what does it teach us? 2. I am moving through the section entitled "integration of transforms". The book is telling me that "integration of the transform of a function f(t) corresponds to the division of the function by t. So say you had f(t) then L{(f(t)/t} = int[s -> oo] F(s)ds. ...(1) Now applying this fact to the example problem... Find the inverse transform of the function ln[1 + w[sup]2[/sup] / s[sup]2[/sup] ] ...(2) From (1) it is equivalent to say that L[sup]-1[/sup] {int[s->oo]F(s)ds} = f(t)/t ...(3) So if (2) is going to be our function inside the brackets in (3) then what we have already is the integrated function. We need f(t) so we differentiate with respect to s. Which leads to... F = 2/s - s(s/s[sup]2[/sup] + w[sup]2[/sup] ) f(t) = L[sup]-1[/sup] {F} = L[sup]-1[/sup] {2/s - s(s/s[sup]2[/sup] + w[sup]2[/sup] )} = 2 - 2coswt Summary: I was given a function and was required to find the inverse Laplace transform. I found the inverse Laplace transform by rearranging (1) which was possible due to the linearity of the transform. I needed to find F(s). I was given the integral of F(s) so I differentiated. From this answer I substituted into (3) and after using a table of Laplace transforms I found the answer. Could someone please read this to check my reasoning. Thanks.
If you had to find the Laplace transform of something like 5e^2t sinh(2t). Would you use the fact that L{sinh(2t)} = 0.5 L{e^2t} + 0.5 L{e^-2t}? If so then L{sinh(2t)} = 2/(s^2 - 4) And L{5e^2t} = 5/(2 - s) Then is it safe to say that L{5e^2t sinh(2t)} = (2/(s^2 - 4)) + (5/(2 - s)) = 10/((2-s)(s^2 - f))? Can someone help me with this. Thanks.
The dirac delta is, very meaningfully, the Fourier transform of the constant function. The Fourier transform is rather similar to the Laplace transform, so it has a related notion in terms of the Laplace transform. Basically, the dirac delta is a "spike." It is defined in terms of an integral as pulling the value of the function out of the integral. It also has the requirement that it vanishes at every point in its domain except the point at which its arguement vanishes, at which point it is undefined (not infinite, though it acts a lot like infinity at this point.) It is the derivative of the unit step function. It can be used as an alternate definition for the unite step: u(t-&tauPlease Register or Log in to view the hidden image! = int[-inf -> t]δ(t'-&tauPlease Register or Log in to view the hidden image!. It is useful as a direct consequence of its definition as a selector function. It is the fundamental kernel of the transfer function to characterize the response of a system. I don't think linearity applies to this justification. At least, I don't think that is the only property that is important in this case. I have forgotten most of these theorm type solutions, though. Did you mean to say "duality" instead of "linearity?" I wouldn't do it that way. I would start with your first step, but then diverge from there. Recognize that the product of two exponentials is the exponential of the sum of the two exponents. Then, you have the Laplace transform of the sum of two exponentials. I looks like you're using some rule: L{f(t)g(t)} = F(s) + G(s)? This is invalid, so, if that's what you were doing, then don't. The Laplace transform of a product results in what is called a convolution. Coincidently, it is a bit more complicated that a simple summation: L{f(t)g(t)} = F*G(s) = int[0 -> inf]F(σ-s)G(&sigmaPlease Register or Log in to view the hidden image!. Anyway, I get a transform that disagrees with yours, even in order: 10/[s(s-4)]. Do you know that yours is correct?
Hello errandir, The answer in the book is 10/((s-2)^2 - 4) = 10/((s-2)(s-2) - 4) = 10/(s^2 - 4s + 4 - 4) = 10/(s^2 - 4s) = 10/(s(s - 4)) *writing what I am thinking* Your answer is identical. The problem is that I am not up to convolution yet so the only tool I have to work with is s-shifting. L{5e^at f(t)} = F(s-a) So L{5e^at sinh(2t)} can be rearranged as L{e^at 5sinh(2t)} where f(t) = 5sinh(2t), F(s) = 10/(s^2 - 4) F(s-a) = 10/((s-a)^2 - 4) Is this correct reasoning? 2. Find the Laplace transform of sinh(t)cos(t)? Using the shifting theorem. I cannot seem to get it into the correct form! Thankyou for your description of the Dirac delta function.
It seems to be correct. Of course, you don't have to rearrange the "5" in that way. Since the transform is linear, you can just pull it out front as a constant multiplier (just like you can in an integral, and, of course, it really <i>is</i> an integral anyway). If a = 2, then you would get the correct answer, if I'm not mistaken. (Of course, you probably well know that a correct answer certainly does <i>not</i> invariably indicate a correct technique/procedure.) Why can you not express the function in terms of exponentials and then look up the transform in a table from there? You probably are expected to express the sinh in terms of exponentials and then look up the transform of cos in a table. You will of course have two transforms, one shifted the negative amount of the other.
5e^at sinh(2t) = 5(e^at (1/2e^at + 1/2e^-at)) = 5(1/2e^(at + at) + 1/2e^(at-at)) = 5(1/2e^2at + 1/2) = 5/2e^2at + 5/2 Is this what you mean errandir?
Yes, except I think sinh is a difference instead of a sum? BTW, please quote whenever possible so that I know to what you are referring.
