In my notes there is the following: $$f(x)=\delta(x-x_0)$$ $$\text{The more closed a function is, the spectrum will be wider.}$$ $$\widetilde{f}(k)=\int_{-\infty}^{+\infty}{\delta(x-x_0)e^{-ikx}}dx$$ $$\widetilde{f}(k)=e^{-ikx_0}(*)$$ $$ |\widetilde{f}(k)|=1$$ $$$$ $$f(x)=\cos{(k_0x)}$$ $$\widetilde{f}(k)=\int_{-\infty}^{+\infty}{\frac{e^{ik_0x}+e^{-ik_0x}}{2}e^{-ikx}}dx=\frac{1}{2}[\int_{-\infty}^{+\infty}{(e^{i(k_0-k)x}+e^{-i(k_0+k)x})}dx]$$ $$\widetilde{f}(k)=\frac{1}{2}[\delta(k+k_0)+\delta(k-k_0)](**)$$ $$$$ From the relation $$\int_{-\infty}^{+\infty}{f(x) \delta(x-a)}dx=f(a)$$ we conlculde to the relation $(*)$.

Could you explain me how we conclude to the fact that $ |\widetilde{f}(k)|=1$ ??

I have also a question about the relation $(**)$:

We know that: $$2 \pi \delta(x)=\int_{-\infty}^{+\infty}{e^{ikx}}dk$$ So shouldn't it be: $$\frac{1}{2}[\int_{-\infty}^{+\infty}{(e^{i(k_0-k)x}+e^{-i(k_0+k)x})}dx]=\frac{1}{2}[ 2 \pi \delta(k_0-k)+2 \pi \delta(-k-k_0)]= \pi [ \delta(k_0-k)+ \delta(-k-k_0)]$$ Or do we conclude to that by an other way??

Last but not least, could you explain me the meaning of the following sentence? $$\text{The more closed a function is, the spectrum will be wider.}$$


1 Answer 1

  1. The absolute value (modulus) of any complex exponential is $1$. The reason is that $|z| = \sqrt{z\bar{z}}$. If $z = e^{ikx_0}$, then $\bar{z} = e^{-ikx_0}$ and so $z\bar{z} = e^{-ikx_0}e^{ikx_0} = e^{-ikx_0+ikx_0} = e^0 = 1$.

  2. These are actually the same. $\delta(x-x_0) = \delta(x_0-x)$ and $\delta(-x-x_0) = \delta(x+x_0)$. You can check these by integrating against functions. The factor of $2\pi$ occurs because you're not being consistent with where you put your $\frac{1}{2\pi}$. Fix this and it will work out the way you want. I'm highly against using a non-unitary version of the Fourier transform for this reason.

Your last sentence really needs some work but it is simply a realization of the Heisenberg uncertainty principle. If you have a narrow function, its Fourier transform is much wider and vice versa.

  • $\begingroup$ 1. I got it!!! $$$$ 2. A ok!!! I use the following version of Fourier transform: $$\text{Fourier transform } : \widetilde{f}(k)=\int_{-\infty}^{+\infty}{f(x)e^{-ikx}}dx$$ $$\text{Inverse Fourier transform } :f(x)=\frac{1}{2 \pi} \int_{- \infty}^{+ \infty}{\widetilde{f}(k) e^{ikx}}dk$$ Where should I have put the term $\frac{1}{2 \pi}$?? $\endgroup$
    – Mary Star
    Commented Apr 16, 2014 at 20:36
  • $\begingroup$ You should have had that $\tilde{f}(k) = 2\pi\left(\frac{1}{2}(\delta(k+k_0)+\delta(-k-k_0))\right)$ according to $2\pi\delta(k) = \int_{-\infty}^{\infty}e^{ikx}dx$. $\endgroup$ Commented Apr 16, 2014 at 21:39
  • $\begingroup$ Yes, but at the relation $(**)$ there is no $\pi$.. $\endgroup$
    – Mary Star
    Commented Apr 16, 2014 at 22:22
  • $\begingroup$ My point is that there should be based on your observation that $2\pi\delta(k) = \int_{-\infty}^{\infty}e^{ikx}dx.$ $\endgroup$ Commented Apr 16, 2014 at 23:15
  • $\begingroup$ But I have found the same result as yours, or not? At the result that is in my notes there is no $\pi$.. Or have I understood it wrong? $\endgroup$
    – Mary Star
    Commented Apr 16, 2014 at 23:18

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