Question regarding Fourier Transform duality

[diredragon]

Homework Statement

Given the Fourier transformation pair $f(t) \implies F(jw)$ where
$f(t) = e^{-|t|}$ and $F(jw)=\frac{2}{w^2+1}$ find and make a graph of the Fourier transform of the following functions:
a) $g(t)=\frac{2}{t^2+1}$
b) $h(t) = \frac{2}{t^2+1}\cos (w_ot)$

Homework Equations

3. The Attempt at a Solution [/B]
We have just recently started learning about Fourier Transforms and last class we mentioned the duality principle of the Fourier transformations and i didn't quite understand what it was all about. I suppose this is what should be applied here. I made some calculations and wanted to check if they are correct.
So the principle should be:
If $f(t) \implies F(jw)$ then $F(t) \implies 2\pi f(-w)$

a) $g(t)=\frac{2}{t^2+1}$
$g(t)$ is clearly $F(t)$ so $F_g(jw)$ should be:
$F_g(jw) = 2\pi e^{-|w|}$ because of the duality.

b)$h(t) = \frac{2}{t^2+1}\cos (w_ot)$
This one is a little trickier but it boils down to $h(t)=g(t)\cos (w_ot)$ and since $f(t)e^{-jw_ot} \implies F(j(w-w_o))$ then it's clearly:
$F_h(t) = \frac{1}{2}(F_g(w-w_o)+F_g(w+w_o))$

Is this correct? Could you also refer me to a source which explains this principle more in depth?

 

[collinsmark]

Homework Statement

Given the Fourier transformation pair $f(t) \implies F(jw)$ where
$f(t) = e^{-|t|}$ and $F(jw)=\frac{2}{w^2+1}$ find and make a graph of the Fourier transform of the following functions:
a) $g(t)=\frac{2}{t^2+1}$
b) $h(t) = \frac{2}{t^2+1}\cos (w_ot)$

Homework Equations

3. The Attempt at a Solution [/B]
We have just recently started learning about Fourier Transforms and last class we mentioned the duality principle of the Fourier transformations and i didn't quite understand what it was all about. I suppose this is what should be applied here. I made some calculations and wanted to check if they are correct.
So the principle should be:
If $f(t) \implies F(jw)$ then $F(t) \implies 2\pi f(-w)$

a) $g(t)=\frac{2}{t^2+1}$
$g(t)$ is clearly $F(t)$ so $F_g(jw)$ should be:
$F_g(jw) = 2\pi e^{-|w|}$ because of the duality.

b)$h(t) = \frac{2}{t^2+1}\cos (w_ot)$
This one is a little trickier but it boils down to $h(t)=g(t)\cos (w_ot)$ and since $f(t)e^{-jw_ot} \implies F(j(w-w_o))$ then it's clearly:
$F_h(t) = \frac{1}{2}(F_g(w-w_o)+F_g(w+w_o))$

Is this correct? Could you also refer me to a source which explains this principle more in depth?

That looks pretty close to me. Or at least almost correct. The good news is that you seem to have correctly caught on to the tricky bits.

But I do feel the need to comment on the "$2 \pi$" that comes up in some of your answers. Different textbooks/instructors handle this differently. There's isn't necessarily one right answer, but you do need to be consistent with your particular textbook/instructor.

Some textbooks define the Fourier transform as
$F(k) = \int_{- \infty}^\infty f(x) e^{-2 \pi j k x} \ dx$
$f(x) = \int_{- \infty}^\infty F(k) e^{2 \pi j k x} \ d k$

while other textbooks define it as
$F(\omega) = \int_{- \infty}^\infty f(t) e^{-j \omega t} \ dt$
$f(t) = \frac{1}{2 \pi} \int_{- \infty}^\infty F(\omega) e^{j \omega t} \ d \omega$

while still others define it as
$F(\omega) = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^\infty f(t) e^{-j \omega t} \ dt$
$f(t) = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^\infty F(\omega) e^{j \omega t} \ d \omega$

None of them are "wrong," so-to-speak, but they will give you different answers. Specifically, the answers will differ regarding whether or not a " $2 \pi$," or "$\sqrt{2 \pi}$" ends up in your answer, and if it does, where.

Some of your answers have a "$2 \pi$" in them, but without more information regarding your definition of the transforms, I'm not able to comment on whether that aspect of your answer is correct or not. Your instructor should be able to clarify the definition you are supposed to use.

All that said, with the possible exception of a constant (e.g., $2 \pi$), your answers look pretty good.

[Edit: replaced $i$ with $j$, since you are using $j$ as the unit imaginary number.]

 

[diredragon]

That looks pretty close to me. Or at least almost correct. The good news is that you seem to have correctly caught on to the tricky bits.

But I do feel the need to comment on the "$2 \pi$" that comes up in some of your answers. Different textbooks/instructors handle this differently. There's isn't necessarily one right answer, but you do need to be consistent with your particular textbook/instructor.

Some textbooks define the Fourier transform as
$F(k) = \int_{- \infty}^\infty f(x) e^{-2 \pi j k x} \ dx$
$f(x) = \int_{- \infty}^\infty F(k) e^{2 \pi j k x} \ d k$

while other textbooks define it as
$F(\omega) = \int_{- \infty}^\infty f(t) e^{-j \omega t} \ dt$
$f(t) = \frac{1}{2 \pi} \int_{- \infty}^\infty F(\omega) e^{j \omega t} \ d \omega$

while still others define it as
$F(\omega) = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^\infty f(t) e^{-j \omega t} \ dt$
$f(t) = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^\infty F(\omega) e^{j \omega t} \ d \omega$

None of them are "wrong," so-to-speak, but they will give you different answers. Specifically, the answers will differ regarding whether or not a " $2 \pi$," or "$\sqrt{2 \pi}$" ends up in your answer, and if it does, where.

Some of your answers have a "$2 \pi$" in them, but without more information regarding your definition of the transforms, I'm not able to comment on whether that aspect of your answer is correct or not. Your instructor should be able to clarify the definition you are supposed to use.

All that said, with the possible exception of a constant (e.g., $2 \pi$), your answers look pretty good.

[Edit: replaced $i$ with $j$, since you are using $j$ as the unit imaginary number.]

They were using a $2\pi$ in the formulas and i used $j$ since it's like an engineering practise in our school not to use $i$ since it's used to denote current :).