2

Wolfram alpha gives the answer to be

$$F(\omega)=\sqrt{2\pi}\delta(\omega)$$

Does that mean that the function is valued $\sqrt{2\pi}$ at all points in the frequency domain? I think this is reasonable because such function i.e. $f(t)=1$ in the time domain would be sum of all the harmonics of a sinusoid and hence would contain all the frequencies. Maybe no, the function isn't varying at all and hence the frequency is $0$. But then the Fourier transform should have been $\delta(0)$ instead of $\delta(\omega)$.

Someone please shed some light on this!

Chirag Arora
  • 233
  • "Does that mean that the function is valued 2π−−√2π at all points in the frequency domain?" No, completely different. – Cardinal May 06 '17 at 19:26
  • Okay then what is the right answer? – Chirag Arora May 06 '17 at 19:27
  • If you think that dirac or delta function is discrete, an imaginary situation, then it will have a value on 0 and it will be zero for the other points. In fact, the continuous form of the delta function has no definite value at zero. I think you may want to have a look on delta function definition. – Cardinal May 06 '17 at 19:29
  • I agree on what you are saying but all I want to know is what would DiracDelta(w) look like? – Chirag Arora May 06 '17 at 19:31
  • It would look like an impulse. Consider a pulse, a step function, with a very short width. https://en.wikipedia.org/wiki/Dirac_delta_function – Cardinal May 06 '17 at 19:32
  • Rule of thumb: extension in time domain will translate to contraction in frequency domain and vice versa. – Cardinal May 06 '17 at 19:34
  • It is a function of frequency, w. So if you are plotting it against w, wouldn't it take all values of w? – Chirag Arora May 06 '17 at 19:35
  • Yes, but it will have a value on zero. For the rest of the points, it will be zero. – Cardinal May 06 '17 at 19:36
  • $\delta(w)$ is not, technically, a function. It can be thought of as a "distribution." Or as a kind of abstract limit of spike functions at $0$. – Thomas Andrews May 06 '17 at 19:37
  • @ThomasAndrews I just wanted to make it more simple. – Cardinal May 06 '17 at 19:38
  • Wasn't replying to you, just the post. @Cardinal – Thomas Andrews May 06 '17 at 19:38
  • I am not able to digest the fact that it will be 0 for every point except w=0. Then why isn't the Fourier transform of 1 simply DiracDelta(0). Why put a variable inside it if it can only be spiked at a single point? – Chirag Arora May 06 '17 at 19:41
  • Did you read the wikipedia page that I posted above? – Cardinal May 06 '17 at 19:43
  • 1
    There is something mentioned as "Dirac Comb" which represents the way I think DiracDelta(w) should be represented. The only difference is that it is defined at discrete points. Now this has made me even more confused ;___; – Chirag Arora May 06 '17 at 19:50
  • "Why put a variable inside it if it can only be spiked at a single point?" Imagine the function $d(x) = 1$ if $x = 0$, $d(x) = 0$ otherwise. Then the function $d(x)$ will be 0 for every point except $x = 0$ (because that's how we defined it). But if we define the function $f(x) = d(0)$, this function is 1 everywhere. We have to put a variable inside it because we only want it to spike at a single point. – Sophie Swett May 06 '17 at 21:38
  • 1
    Here's another way of putting it. The Dirac delta distribution $\delta(x)$ has a spike whenever $x = 0$. That means that $\delta(\omega)$ has a spike only at the single point $\omega = 0$. On the other hand, $\delta(0)$ (if there were such a thing) would have a spike whenever $0 = 0$; it would have a spike everywhere! We want it to spike only at $\omega = 0$, so we use $\delta(\omega)$ instead of $\delta(0)$. – Sophie Swett May 06 '17 at 21:41

4 Answers4

4

I think the clearest way to see this is by noting that we have (depending on your convention for the placement of $2 \pi$ in Fourier transforms) that $$\mathcal{F}(\mathcal{F}(f(x))) = 2 \pi f(-x)$$ Taking the convention that $$\tilde{f}(k) = \int_{-\infty}^\infty e^{-ikx} f(x) \; dx$$ so $$f(x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{ikx} \tilde{f}(k) \; dk$$ We get $$f(-x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{-ikx} \tilde{f}(k) \; dk = \frac{1}{2\pi} \mathcal{F(\tilde{f}(k))} = \frac{1}{2\pi}\mathcal{F}(\mathcal{F}(f(x)))$$ Note we have $$\mathcal{F}(\delta(x)) = \int_{-\infty}^\infty e^{-ikx} \delta(x) \; dx = 1$$ So then $$\mathcal{F}(1) = \mathcal{F}(\mathcal{F}(\delta(x))) = 2 \pi \delta(-x) = 2 \pi \delta(x)$$ For other constants, by linearity, we have $$\mathcal{F}(c) = c \mathcal{F}(1) = 2 \pi c \delta(x)$$

Vivek Thazhathattil
  • 35
LtSten
  • 1,704
3

You can derive the answer very easily with the general formula for the fourier series of a complex exponential:

$\mathcal F(e^{jw_0 t}) = 2\pi \delta(w-w_0)$

This identity is very intuitive: Since a complex exponential only has one frequency ($w_0$), its fourier transform only has one pulse at that frequency[1].

Set $w_0 = 0$ and you get:

$\mathcal F(e^{0}) = \mathcal F(1) = 2\pi \delta(w-0) = 2\pi \delta(w)$

[1] Here is a proof: https://staff.fnwi.uva.nl/r.vandenboomgaard/SignalProcessing/FrequencyDomain/CTNP.html#complex-exponential

alvitawa
  • 131
1

I know that this has been answered, but it's worth noting that the confusion between factors of $2\pi$ and $\sqrt{2\pi}$ is likely to do with how you define the Fourier transform in the first place.

Wolfram Alpha seems to be taking the definition that involves placing a factor of $1/\sqrt{2\pi}$ on both the transform and the inverse transform (as is sometimes done in physics), rather than placing the factor $1/2\pi$ entirely on the inverse transform.

If you take the former approach, I believe you end up with the desired result, $\mathcal{F}(1)=\sqrt{2\pi}\delta(k)$, since (compare LtSten's answer) then $\hat{f}(k)$ will originally contain a factor of $1/\sqrt{2\pi}$ (i.e. $\mathcal{F}(\delta(x))=1/\sqrt{2\pi}$), meaning that $\mathcal{F}(1)=\sqrt{2\pi}\delta(x)$ is consistent with $\mathcal{F}(\mathcal{F}(f(-x)))=2\pi f(x)$.

Erik Baxter
  • 21
0

Just use the definition. $F(w) = \int_{-\infty}^\infty e^{-iwt}dt =\delta(w)$ by how the dirac delta is defined. This is normalized to 1at zero and 0 otherwise.

JMJ
  • 4,866
  • 14
  • 31
  • 2
    I don't think anyone defines $\delta(\omega) $ as the limit in the sense of (tempered) distributions $\lim_{T \to \infty}\frac1{2\pi} \int_{-T}^T e^{-i\omega t}dt=\lim_{T \to \infty} \frac{\sin(\omega T)}{\pi \omega }$, which is a theorem equivalent to the Fourier inversion theorem for tempered distributions. – reuns Aug 11 '19 at 14:06
  • 1
    See https://math.stackexchange.com/questions/1343859/why-does-integrating-a-complex-exponential-give-the-delta-function – alvitawa Nov 15 '20 at 17:43
  • Yes, all of this can be made very rigorous using distribution theory. But most people who use the FT in the "real world" (whatever that actually is) don't bother with that much rigor. Statics has a good answer on the linked question which explains the reasoning behind this definition in a simple way. – JMJ Nov 16 '20 at 18:33
  • This is a math forum ... people come here to understand what is the good rigorous explanation – LL 3.14 Nov 03 '23 at 23:58
  • Not necessarily. Many people come just wanting the answer to or a way to think about a math problem they don't understand, and a rigorous answer by current standards might just make things more obscure. Many of the most useful results in mathematics were not discovered or first presented in a "rigorous" way. In the current case, an actually rigorous answer would require the first few chapters of a distribution theory book, which is probably not what the asker was looking for. – JMJ Nov 04 '23 at 00:44