# Question regarding Fourier Transform duality

• diredragon
In summary, the conversation discusses the duality principle of Fourier transformations and how it applies to finding the Fourier transform of functions. The participant also asks for confirmation on their calculations and sources for further understanding. The expert provides a brief explanation of the different definitions of Fourier transforms and advises the participant to check with their instructor for the specific definition they are using.
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?

diredragon said:

## 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.]

Last edited:
diredragon
collinsmark said:
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 :).

## What is the concept of Fourier Transform duality?

The concept of Fourier Transform duality refers to the relationship between a signal and its Fourier Transform. It states that a signal in the time domain has a corresponding representation in the frequency domain, and vice versa. This means that the characteristics of a signal can be examined and analyzed in both the time and frequency domains.

## What is the mathematical formula for Fourier Transform duality?

The mathematical formula for Fourier Transform duality is given by the following equations:
FT(x) = ∫f(t)e^(-iωt)dt
FT(f) = ∫x(t)e^(iωt)dω

## What are the applications of Fourier Transform duality?

Fourier Transform duality has various applications in fields such as signal processing, image processing, and data analysis. It is used to analyze and manipulate signals and images in the frequency domain, which can provide insights into their characteristics and patterns.

## What is the difference between Fourier Transform and Fourier Transform duality?

The Fourier Transform is a mathematical operation that transforms a signal from the time domain to the frequency domain. On the other hand, Fourier Transform duality refers to the relationship between a signal and its Fourier Transform. While the Fourier Transform transforms a signal from one domain to another, Fourier Transform duality shows the connection between the two domains.

## What are some common misconceptions about Fourier Transform duality?

One common misconception about Fourier Transform duality is that it is a one-to-one mapping between the time and frequency domains. In reality, there can be multiple signals with the same Fourier Transform, and vice versa. Another misconception is that Fourier Transform duality only applies to periodic signals, when in fact it can be applied to non-periodic signals as well.

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