On Oct 1, 10:01�am, dbell <bellda2...@cox.net> wrote:> On Oct 1, 1:15�am, vv <vanam...@netzero.net> wrote: > > > On Oct 1, 1:18 am, Cary <c...@domain.invalid> wrote: > > > > To the OP: �Post the exact statement of the problem from the textbook > > > you referenced so we can consider what the author had in mind. > > > I don't have the book handy, but it is along the following lines: > > Knowing the transform of x(t) = 1 for t >= 0 to be 1/s, use the > > integration property to obtain the transform of x(t)/t. �This leads to > > Int[s..oo] du/u, which I suspected didn't exist. �I guess the textbook > > author was not too careful when framing this problem. �If someone has > > the Solutions Manual to this book and post the solution given there, > > maybe it will throw more light on what the author had in mind. > > > --vv > > Are you looking for something that works for all s? �If you consider > integral(1/t) for t = 0 to 1 is infinite (=-ln(0)), it looks like with > s not in the left half plane you are guaranteed a problem. > > DirkI am sorry, I take my comment back. Left hal plane part is definitely not correct, I need to rethink the rest. Dirk

# Unilateral Laplace transform of 1/t

Started by ●September 30, 2008

Reply by ●October 1, 20082008-10-01

Reply by ●October 1, 20082008-10-01

On Tue, 30 Sep 2008 22:15:42 -0700 (PDT), vv <vanamali@netzero.net> wrote:>On Oct 1, 1:18 am, Cary <c...@domain.invalid> wrote: >> To the OP: Post the exact statement of the problem from the textbook >> you referenced so we can consider what the author had in mind. > >I don't have the book handy, but it is along the following lines: >Knowing the transform of x(t) = 1 for t >= 0 to be 1/s, use the >integration property to obtain the transform of x(t)/t. This leads to >Int[s..oo] du/u, which I suspected didn't exist. I guess the textbook >author was not too careful when framing this problem. If someone has >the Solutions Manual to this book and post the solution given there, >maybe it will throw more light on what the author had in mind. > >--vvI believe the book you first made reference to was "Linear Systems and Signals" by B. P. Lathi, and I'll guess the 2nd ed. I don't have the book, but after some digging I may have found the exercise in question, which I've reproduced below. The problem begins by referring to the Laplace transform X(s) of signal x(t) = 1/t u(t) That looks to be a mistake (typo?). I don't think the author meant to write u(t), which most often refers to the "unit step function", also known as the Heaviside step function. From what appears in the problem below, that line should probably read x(t) = 1/t y(t) It appears the point of the exercise is to derive the "division by t" property. If the exercise below is the one you had in mind, look at it again and see what you think. Page 480, exercise 4.2-9: ------------------------ It is difficult to compute the Laplace transform X(s) of signal x(t) = 1/t u(t) by using direct integration. Instead, properties provide a simpler method. (a) Use Laplace transform properties to express the Laplace transform of t x(t) in terms of the unknown quantity X(s). (b) Use the definition to determine the Laplace transform of y(t) = t x(t). (c) Solve for X(s) by using the two pieces from a and b. Simplify your answer.

Reply by ●October 1, 20082008-10-01

On Oct 1, 8:16�pm, Cary <c...@domain.invalid> wrote:> I believe the book you first made reference to was "Linear Systems and > Signals" by B. P. Lathi, and I'll guess the 2nd ed.Yes, the above book is the one in question; I inadvertently called it "Signals and Linear Systems" when I first posted the query.> �Page 480, exercise 4.2-9: > �------------------------ > > It is difficult to compute the Laplace transform X(s) of signal > > � �x(t) = 1/t u(t) > > by using direct integration. �Instead, properties provide a simpler > method.If the author meant y(t)/t, and that the u(t) is a typo, then that clears the air. The problem as printed was the cause for confusion. Thanks. --vv

Reply by ●October 1, 20082008-10-01

On Oct 1, 1:15�am, vv <vanam...@netzero.net> wrote:> On Oct 1, 1:18 am, Cary <c...@domain.invalid> wrote: > > > To the OP: �Post the exact statement of the problem from the textbook > > you referenced so we can consider what the author had in mind. > > I don't have the book handy, but it is along the following lines: > Knowing the transform of x(t) = 1 for t >= 0 to be 1/s, use the > integration property to obtain the transform of x(t)/t. �This leads to > Int[s..oo] du/u, which I suspected didn't exist. �I guess the textbook > author was not too careful when framing this problem. �If someone has > the Solutions Manual to this book and post the solution given there, > maybe it will throw more light on what the author had in mind. > > --vvw, (hopefully not THE 'w', on Pennsylvania Ave) Since your Laplace transform is unilateral, consider the possibility that the Laplace transform does not converge (i.e., does not exist). Unilateral LT(1/t)=integral(1/t*exp(-s*t)) integrated over t=[0, inf). Consider that there is an 'a' such that Re(exp(-s*t))>=a>0 for t in the interval [0,1], with s fixed, but unspecified. Also consider that integral(1/t) integrated over [0,1] = ( -ln(0) ), which is infinite; this follows from the definition of ln(t). Then it follows that Re( integral(1/t*exp(-s*t) ) integrated over [0,1] is also infinite. As the upper limit of integration is increased, the real part inside the integral stays strictly positive, so increasing the upper limit on integral(1/t*exp(-s*t)) strictly adds to Re( integral(1/t*exp(- s*t)) ). Therefore the real part of the unilateral Laplace transform of 1/t is unbounded for every s. So the unilateral Laplace transform of 1/t does not exit. Dirk P.S. The Fourier transform of this does not exist either when the function is considered to be 0 for t<0. The Fourier transform that you see computed is for ( 1/t, where t is from (-inf, inf), usually with the function redefined to be 0 for t=0).

Reply by ●October 1, 20082008-10-01

On Oct 1, 1:45�pm, dbell <bellda2...@cox.net> wrote:> On Oct 1, 1:15�am, vv <vanam...@netzero.net> wrote: > > > On Oct 1, 1:18 am, Cary <c...@domain.invalid> wrote: > > > > To the OP: �Post the exact statement of the problem from the textbook > > > you referenced so we can consider what the author had in mind. > > > I don't have the book handy, but it is along the following lines: > > Knowing the transform of x(t) = 1 for t >= 0 to be 1/s, use the > > integration property to obtain the transform of x(t)/t. �This leads to > > Int[s..oo] du/u, which I suspected didn't exist. �I guess the textbook > > author was not too careful when framing this problem. �If someone has > > the Solutions Manual to this book and post the solution given there, > > maybe it will throw more light on what the author had in mind. > > > --vv > > w, �(hopefully not THE 'w', on Pennsylvania Ave) > > Since your Laplace transform is unilateral, consider the possibility > that the Laplace transform does not converge (i.e., does not exist). > > � � Unilateral LT(1/t)=integral(1/t*exp(-s*t)) integrated over t=[0, > inf). > > Consider that there is an 'a' such that Re(exp(-s*t))>=a>0 for t in > the interval [0,1], with s fixed, but unspecified. Also consider that > integral(1/t) integrated over [0,1] = ( -ln(0) ), which is infinite; > this follows from the definition of ln(t). Then it follows that > Re( integral(1/t*exp(-s*t) ) integrated over [0,1] is also infinite. > > As the upper limit of integration is increased, the real part inside > the integral stays strictly positive, so increasing the upper limit on > integral(1/t*exp(-s*t)) strictly adds to Re( integral(1/t*exp(- > s*t)) ). > > Therefore the real part of the unilateral Laplace transform of 1/t is > unbounded for every s. So the unilateral Laplace transform of 1/t does > not exit. > > Dirk > > P.S. The Fourier transform of this does not exist either when the > function is considered to be 0 for t<0. �The Fourier transform that > you see computed is for ( 1/t, where t is from (-inf, inf), usually > with the function redefined to be 0 for t=0).Does not "exist", not Does not "exit". Can't type. Dirk

Reply by ●October 1, 20082008-10-01

On Oct 1, 11:16�am, Cary <c...@domain.invalid> wrote:> On Tue, 30 Sep 2008 22:15:42 -0700 (PDT), vv <vanam...@netzero.net> > wrote: > > >On Oct 1, 1:18 am, Cary <c...@domain.invalid> wrote: > >> To the OP: �Post the exact statement of the problem from the textbook > >> you referenced so we can consider what the author had in mind. > > >I don't have the book handy, but it is along the following lines: > >Knowing the transform of x(t) = 1 for t >= 0 to be 1/s, use the > >integration property to obtain the transform of x(t)/t. �This leads to > >Int[s..oo] du/u, which I suspected didn't exist. �I guess the textbook > >author was not too careful when framing this problem. �If someone has > >the Solutions Manual to this book and post the solution given there, > >maybe it will throw more light on what the author had in mind. > > >--vv > > I believe the book you first made reference to was "Linear Systems and > Signals" by B. P. Lathi, and I'll guess the 2nd ed. > > I don't have the book, but after some digging I may have found the > exercise in question, which I've reproduced below. > > The problem begins by referring to the Laplace transform X(s) of > signal > > � �x(t) = 1/t u(t) > > That looks to be a mistake (typo?). �I don't think the author meant to > write u(t), which most often refers to the "unit step function", also > known as the Heaviside step function. �From what appears in the > problem below, that line should probably read > > � �x(t) = 1/t y(t) > > It appears the point of the exercise is to derive the "division by t" > property. �If the exercise below is the one you had in mind, look at > it again and see what you think. > > �Page 480, exercise 4.2-9: > �------------------------ > > It is difficult to compute the Laplace transform X(s) of signal > > � �x(t) = 1/t u(t) > > by using direct integration. �Instead, properties provide a simpler > method. > > (a) Use Laplace transform properties to express the Laplace transform > � � of �t x(t) �in terms of the unknown quantity X(s). > > (b) Use the definition to determine the Laplace transform of > � � y(t) = t x(t). > > (c) Solve for X(s) by using the two pieces from a and b. �Simplify > � � your answer.Checking CRC tables, combining results for unit step function u(t) and 1/t*F(t) (F(t) some function) you get Laplace transform (1/t*u(t)) =integral (1/x dx) from s to inf, where s is the Laplace variable. Using definition of ln(), integral (1/x dx) from a to b = ln(b)- ln(a). Then integral (1/x dx) for x from s to inf = lim (as b=>inf) of [ln(b)- ln(s)], which is unbounded for any s. The unilateral Laplace transform of 1/t does not exist. Dirk

Reply by ●October 1, 20082008-10-01

Cary wrote: (snip)> The problem begins by referring to the Laplace transform X(s) of > signal> x(t) = 1/t u(t)> That looks to be a mistake (typo?). I don't think the author meant to > write u(t), which most often refers to the "unit step function", also > known as the Heaviside step function. From what appears in the > problem below, that line should probably read> x(t) = 1/t y(t)u is the variable most commonly used for change of variables problems in calculus. (v is second.) In any case, I agree it doesn't sound like the unit step function. (That was H in the books I remember, but others may use u.) -- glen

Reply by ●October 1, 20082008-10-01

On Wed, 01 Oct 2008 12:47:50 -0800, glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:>Cary wrote: >(snip) > >> The problem begins by referring to the Laplace transform X(s) of >> signal > >> x(t) = 1/t u(t) > >> That looks to be a mistake (typo?). I don't think the author meant to >> write u(t), which most often refers to the "unit step function", also >> known as the Heaviside step function. From what appears in the >> problem below, that line should probably read > >> x(t) = 1/t y(t) > >u is the variable most commonly used for change of variables >problems in calculus. (v is second.) > >In any case, I agree it doesn't sound like the unit step function. >(That was H in the books I remember, but others may use u.)Indeed, H is still widely used. Some authors use the even longer name 'Heaviside unit step function'. MathWorld, Wikipedia, and PlanetMath all use the subject title 'Heaviside Step Function'. Abramowitz and Stegun use 'unit step function' and the notation u(t). Authors differ in their definition of the function. Simple. 8-) Refs: <http://mathworld.wolfram.com/HeavisideStepFunction.html> <http://en.wikipedia.org/wiki/Unit_step_function> <http://planetmath.org/encyclopedia/HeavisideStepFunction.html>> >-- glen