Is the Fourier Transform of x(t)=1 equal to X(jω)=2πδ(ω)?

[asmani]

Hi all

I know that the Fourier transform of x(t)=1 is X(jω)=2πδ(ω) by using the duality property.
This implies:
\int_{-\infty }^{+\infty }e^{-j\omega t}dt=2\pi\delta(\omega)
Consequently, for ω≠0:
\int_{-\infty }^{+\infty }e^{-j\omega t}dt=0
And as a result:
\int_{-\infty }^{+\infty }\cos t\: dt=0

Is this result true?!

Thanks in advance

 

[jsgruszynski]

Yes.

But intuitively you should be able to solve this also: the amplitude is constant so this translates to an impulse in the opposite domain which is a single frequency in frequency domain, or a frequency domain impulse function.

Similarly, in the frequency domain, a A(w)=1 corresponds to a time-domain impulse function - all frequencies represented.

These are duals.

 

[rbj]

Hi all

I know that the Fourier transform of x(t)=1 is X(jω)=2πδ(ω) by using the duality property.
This implies:
\int_{-\infty }^{+\infty }e^{-j\omega t}dt=2\pi\delta(\omega)
Consequently, for ω≠0:
\int_{-\infty }^{+\infty }e^{-j\omega t}dt=0
And as a result:
\int_{-\infty }^{+\infty }\cos t\: dt=0

Is this result true?!

it depends on what you really mean by "true".

mathematicians will have trouble with anything you wrote here. the integrals do not converge. you *can* say that

\lim_{B \rightarrow + \infty}\frac{1}{2B} \int_{-B }^{+B }\cos t \ dt = 0even the electrical engineering use of the dirac impulse function, \delta(t) is not kosher, from the POV of strict mathematicians. someday, you might take a course in Real Analysis and you will learn that if

f(t) = g(t)

"almost everywhere" (that is, everywhere except for a countable number of discrete points), then

\int f(t) \ dt = \int g(t) \ dt

but we are saying that \delta(t) = 0 almost everywhere, yet the integral of \delta(t) is 1 and the integral of 0 is 0, not the same.

the mathematicians don't even grant the Dirac delta function the unqualified label "function". they call it a "distribution", party because you cannot have a (true) function that is zero everywhere except for one point and have its integral be anything other than zero.

when i do electrical engineering, i do treat the Dirac impulse function as a function, in the sense we commonly do in electrical engineering (it's zero almost everywhere, but its integral is 1), but i know that this doesn't fly, given the language and definitions that the mathematicians give things.

so, my advice is just to be careful with what you say and whom you say it to.

 

[asmani]

Thanks for the replies.
I just followed the definitions of Oppenheim Signal and Systems.

The Fourier transform pair represented in this book:
x(t)=\frac{1}{2\pi}\int_{-\infty}^{+\infty }X(j\omega)e^{j\omega t}d\omega<br />
<br /> X(j\omega)=\int_{-\infty}^{+\infty }x(t)e^{-j\omega t}dt<br />
Now the question is do x(t)=1 and X(jω)=2πδ(ω) satisfy these equations?

On page 297 of this book (second edition):

To suggest the general result, let us consider a signal x(t) with Fourier transform X(jω) that is a single impulse of area 2π at ω=ω₀; that is X(j\omega)=2\pi\delta(\omega - \omega_{0})
To determine the signal x(t) for which this is the Fourier transform, we can apply the inverse transform relation, eq. (4.8), to obtain x(t)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}2\pi\delta(\omega - \omega_{0})e^{j\omega t}d\omega=e^{j\omega_0 t}

After this, X(jω)=2πδ(ω-ω₀) is considered as the Fourier transform of x(t)=e^jω₀t. Is it a valid argument? This pair only satisfy the first equation.