Fourier transform of the sawtooth function

[jashua]

Let the sawtooth function be defined as follows:

h(t) = t, 0<t<1,
h(t) = 0, elsewhere

The problem is two explain the reason for difference between the following two forms of the Fourier transform of h(t), which is denoted as H(f).

First method is straightforward, i.e., use the Fourier transform. Hence, we have:

H(f)= int (-inf to inf) t*exp(-j*2*pi*f*t) dt
H(f)= int (0 to 1) t*exp(-j*2*pi*f*t) dt

Then, using integration by part, we get (skipping some tedious steps)

H(f)= 1/(j*2*pi*f) * [sinc(f)*exp(-j*pi*f) - exp(-j*2*pi*f)].

On the other hand, if we think the sawtooth function h(t) as the integral of a rectangular pulse g(t), which is given as:

g(t)=1, 0<t<1,
g(t)=0, elsewhere,

such that h(t)=int (-inf to t) g(t) dt, then, using Fourier property (integration in the time domain), we get the following result:

H(f)=1/(j*2*pi*f) * sinc(f)*exp(-j*pi*f) + 1/2*dirac(f).

As you see these two results are different. Why?.. What is the mistake? Any help will be appreciated.

 

[vela]

Hint: Is

$$\int_{-\infty}^2 g(t)\,dt$$

equal to h(2)?

 

[jashua]

Ops..! got it :)

So,

g(t) = rect(t-1/2) - dirac(t-1).

many thanks vela.

 

[Ray Vickson]

Let the sawtooth function be defined as follows:

h(t) = t, 0<t<1,
h(t) = 0, elsewhere

The problem is two explain the reason for difference between the following two forms of the Fourier transform of h(t), which is denoted as H(f).

First method is straightforward, i.e., use the Fourier transform. Hence, we have:

H(f)= int (-inf to inf) t*exp(-j*2*pi*f*t) dt
H(f)= int (0 to 1) t*exp(-j*2*pi*f*t) dt

Then, using integration by part, we get (skipping some tedious steps)

H(f)= 1/(j*2*pi*f) * [sinc(f)*exp(-j*pi*f) - exp(-j*2*pi*f)].

On the other hand, if we think the sawtooth function h(t) as the integral of a rectangular pulse g(t), which is given as:

g(t)=1, 0<t<1,
g(t)=0, elsewhere,

such that h(t)=int (-inf to t) g(t) dt, then, using Fourier property (integration in the time domain), we get the following result:

H(f)=1/(j*2*pi*f) * sinc(f)*exp(-j*pi*f) + 1/2*dirac(f).

As you see these two results are different. Why?.. What is the mistake? Any help will be appreciated.

Your first H(f) = int (-inf to inf) t*exp(-j*2*pi*f*t) dt is not the F.T of the sawtooth function h(t); it is the F.T of the function k(t) = t for all t in R. Of course, your second form H(f)= int (0 to 1) t*exp(-j*2*pi*f*t) dt is correct for h(t).

RGV

 

[jashua]

thank you very much for your correction in my question.