Hi there, Suppose I have a function f(t) which I knew its laplace and fourier transform. What is the laplace and fourier transform of the following: exp(a*f(t)) ??? ---------------------------- Is there a way to evaluate the laplace transform of heaviside((exp(x)-a)), where the heaviside function is also called step function, heaviside(x) = 1, when x>0, and =0, when x<0, and it has a jump from 0 to 1 at x=0... --------------------------- Please give me some pointers! thanks a lot!

# how to compute the laplace and fourier transform of this function?

Started by ●August 20, 2006

Reply by ●August 20, 20062006-08-20

In article <1156111223.392350.157580@75g2000cwc.googlegroups.com>, Luna Moon <lunamoonmoon@gmail.com> wrote:>Hi there, > >Suppose I have a function f(t) which I knew its laplace and fourier >transform. > >What is the laplace and fourier transform of the following: > >exp(a*f(t)) > >???No nice formula. exp(a f(t)) = sum_{n=0}^infty a^n f(t)^n/n! Now the Fourier transform of a power of f is a "convolution power" of the Fourier transform of f. But in general that's making matters more complicated rather than less.>---------------------------- > >Is there a way to evaluate the laplace transform of > >heaviside((exp(x)-a)), > >where the heaviside function is also called step function, > >heaviside(x) = 1, when x>0, and =0, when x<0, and it has a jump from 0 >to 1 at x=0...Yes, that's easy. Hint: when does exp(x) - a change sign? Do the cases a <= 0 and a > 0 separately. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada

Reply by ●August 20, 20062006-08-20

On 20 Aug 2006 15:00:23 -0700, "Luna Moon" <lunamoonmoon@gmail.com> wrote:>Hi there, > >Suppose I have a function f(t) which I knew its laplace and fourier >transform. > >What is the laplace and fourier transform of the following: > >exp(a*f(t)) > >??? > >---------------------------- > >Is there a way to evaluate the laplace transform of > >heaviside((exp(x)-a)), > >where the heaviside function is also called step function, > >heaviside(x) = 1, when x>0, and =0, when x<0, and it has a jump from 0 >to 1 at x=0... > >--------------------------- > >Please give me some pointers! thanks a lot!The Heaviside transform is just like the Laplace except with the Heaviside, the step function is already built in, as I recall. So the inverse of the Heaviside is the response to a unit step. Laplace requires you to specify the input function so the Laplace transform is s times the Heaviside. See if that doesn't work. John Polasek John Polasek http://www.dualspace.net

Reply by ●August 21, 20062006-08-21

On 20 Aug 2006 15:00:23 -0700, "Luna Moon" <lunamoonmoon@gmail.com> wrote:>Hi there, > >Suppose I have a function f(t) which I knew its laplace and fourier >transform. > >What is the laplace and fourier transform of the following: > >exp(a*f(t)) > >??? > >---------------------------- > >Is there a way to evaluate the laplace transform of > >heaviside((exp(x)-a)), > >where the heaviside function is also called step function, > >heaviside(x) = 1, when x>0, and =0, when x<0, and it has a jump from 0 >to 1 at x=0... > >--------------------------- > >Please give me some pointers! thanks a lot!(This message did not appear when I posted it last night) The Heaviside transform is just like the Laplace except with the Heaviside, the step function is already built in, as I recall. So the inverse of the Heaviside is the response to a unit step. Laplace requires you to specify the input function so the Laplace transform is s times the Heaviside. See if that doesn't work. John Polasek John Polasek http://www.dualspace.net

Reply by ●August 22, 20062006-08-22

Luna Moon wrote:> > Suppose I have a function f(t) which I knew its laplace and fourier > transform. > > What is the laplace and fourier transform of the following: > > exp(a*f(t)) > > ???there is no theorem that will do this nicely. you gotta plug in f(t) and see what you get.> Is there a way to evaluate the laplace transform of > > heaviside((exp(x)-a)), > > where the heaviside function is also called step function, > > heaviside(x) = 1, when x>0, and =0, when x<0, and it has a jump from 0 > to 1 at x=0...this one can be simplified because of the nature of heaviside(x) h(x) = heaviside( exp(x) - a ) = heaviside( x - log(a) ) the Laplace transform is H(s) = Laplace{ h(x) } = 1/s * exp(-s*log(a)) = 1/(s*a^s) r b-j