Need Help with Fourier Transforms? Urgent Assistance Available!

[jose_peeter]

Fourier transform help urgent?

dear friends,

sorry to bug you all with things that are lengthy and rather tedious. please help me solve these questions if possible.
i have tried them but i just can't do the integrals. please help

FIND THE FOURIER TRANSFORMS OF THE FOLLOWING.

f(t) = ( 1 - 1/t ) 0 <= t <= 2
( 1 + 1/t ) -2 <= t <= 0
( 0 ) otherwise

f(t) = ( e^-t cost ) t >= 0
( 0 ) t < 0

f(t) = e^at -2 <= t <= 2
0 elsewhere

 

[vela]

dear friends,

sorry to bug you all with things that are lengthy and rather tedious. please help me solve these questions if possible.
i have tried them but i just can't do the integrals. please help

FIND THE FOURIER TRANSFORMS OF THE FOLLOWING.

f(t) = ( 1 - 1/t ) 0 <= t <= 2
( 1 + 1/t ) -2 <= t <= 0
( 0 ) otherwise

You can't do this one because 1/0 isn't defined.

f(t) = ( e^-t cost ) t >= 0
( 0 ) t < 0

f(t) = e^at -2 <= t <= 2
0 elsewhere

Show us the integrals and your work so far in evaluating them.

 

[jose_peeter]

dear vela,
thanks you for your prompt response.

the question

f(t) = ( 1 - 1/t ) 0 <= t <= 2
( 1 + 1/t ) -2 <= t <= 0
( 0 ) otherwise

is doable and has the following answer.

( 1 - cos(2ω) ) / ω^2

next MY ATTEMPT on the first two questions.

f(t) = ( e^-t cos t ) >= 0
( 0 ) t < 0

0 - ∞∫e^-t cos t e^-jωt dt

= ∫ e^(-1-jω)tcos t dt * my problem is purely integration. you can use integration by parts but then they just keep repeating without end. SO HOW TO DO?*

the second question was

f(t) = e^at -2 <= t <= 2, where a is a positive constant
0 elsewhere

applying F TRANSFORM.

∫e^(a-jω)t dt

= [ -1/jω * e^(a-jω)t ] 2 < -> (-2)

= [ -1/jω * e^2(a-jω)] - [ -1/jω * e^-2(a-jω)]

= now notice the different SIGNS on the two constant a's, that is the problem

= [ (e^-2a * e^2jω) - (e^2a * e^-2jω) ] / jω

= I cannot factor out the constant e^2a and e^-2a to make (sin jω). so what do i do ? ? ? ? ?

= i am stuck here so please help me ASAP. thank you so much.
just help me get these three questions cleared.

 

[vela]

dear vela,
thanks you for your prompt response.

the question

f(t) = ( 1 - 1/t ) 0 <= t <= 2
( 1 + 1/t ) -2 <= t <= 0
( 0 ) otherwise

is doable and has the following answer.

( 1 - cos(2ω) ) / ω^2

As you've written f(t), it's not because f(0) doesn't exist. 0 can't be part of the domain. But what's more problematic is that the transform of the supposed answer doesn't look anything like your f(t).

next MY ATTEMPT on the first two questions.

f(t) = ( e^-t cos t ) >= 0
( 0 ) t < 0

0 - ∞∫e^-t cos t e^-jωt dt

= ∫ e^(-1-jω)tcos t dt * my problem is purely integration. you can use integration by parts but then they just keep repeating without end. SO HOW TO DO?*

Let $I=\int_0^\infty \cos t\exp[-t(1+j\omega)]\,dt$. At some point, you end up with something of the form I = (stuff)+(constant)I. Just solve for I.

the second question was

f(t) = e^at -2 <= t <= 2, where a is a positive constant
0 elsewhere

applying F TRANSFORM.

∫e^(a-jω)t dt

= [ -1/jω * e^(a-jω)t ] 2 < -> (-2)

You've already made a mistake by this point. The constant out front isn't correct. You need to fix that first.

= [ -1/jω * e^2(a-jω)] - [ -1/jω * e^-2(a-jω)]

= now notice the different SIGNS on the two constant a's, that is the problem

= [ (e^-2a * e^2jω) - (e^2a * e^-2jω) ] / jω

= I cannot factor out the constant e^2a and e^-2a to make (sin jω). so what do i do ? ? ? ? ?

= i am stuck here so please help me ASAP. thank you so much.
just help me get these three questions cleared.

 

[Ray Vickson]

dear friends,

sorry to bug you all with things that are lengthy and rather tedious. please help me solve these questions if possible.
i have tried them but i just can't do the integrals. please help

FIND THE FOURIER TRANSFORMS OF THE FOLLOWING.

f(t) = ( 1 - 1/t ) 0 <= t <= 2
( 1 + 1/t ) -2 <= t <= 0
( 0 ) otherwise

f(t) = ( e^-t cost ) t >= 0
( 0 ) t < 0

f(t) = e^at -2 <= t <= 2
0 elsewhere

No matter what I do, I get a divergent result for the FT of your first function (essentially due to the 1/|t| behaviour near t = 0). Direct computation of the integral from -2 to 2 yields a divergent result; indirect computation by combining the integral from -2 to -a and from a to 2, then letting a --> 0+ also yields divergence. So, show us your work on the first example.

RGV

 

[HACR]

f(t) = ( e^-t cos t ) >= 0
( 0 ) t < 0

0 - ∞∫e^-t cos t e^-jωt dt

For this question I ended up with \frac {1+ix}{(1+ix)^2+1}

the second question was

f(t) = e^at -2 <= t <= 2, where a is a positive constant
0 elsewhere

applying F TRANSFORM.

∫e^(a-jω)t dt

= [ -1/jω * e^(a-jω)t ] 2 < -> (-2)

For this one above, the \Phi (f(t))= \frac {1}{a-ix} (sinh(2a)cos(-2x)+i*cosh(2a)sin(-2x) from the identity, sinh(z)=sinh(x)cos(y)+i cosh(x)sin(y).