# Continuous Fourier transform and properties for a simple function

This was an exercise to reconcile properties of a discrete and continuous Fourier transform.

The continuous function to be approximated is $$f(t)=t$$ on the closed interval $$[0,1].$$ The discrete approximation is $$f(n)=n/N$$ with $$n=1,2,...,N.$$ Let $$N=10,000$$ for purposes of this question.

For the real side, the discrete and continuous properties are as expected.

$$(1) \hspace{10mm}\int_0^1|t|^2dt=1/3,~~(1/N)\sum_n |n/N|^2=0.333383.$$

The quantity that most interests me (in both t and $$\omega$$ domains) is:

$$W_f=\frac{\int_0^1 |t\cdot t|^2 dt}{\int_0^1 |t^2|dt}=3/5.$$

For the discrete version on the real side,

$$\frac{(1/N)\sum_n[|(n/N\cdot n/N|^2]}{(1/N)\sum_n{|n/N|^2}}=0.60006$$

In the $$\omega$$ domain I have the following:

For the discrete transform I allowed Mathematica to find the discrete transform of the sequence $$f(n)=\{1/N,2/N,..,n/N,..,1\},$$ giving $$\hat{f}(n).$$ I then calculated

$$(1/N)\sum_n |\hat{f}(n)|^2=0.333383$$

which as a casual check means that with (1) above Parseval's relation is satisfied. As for

$$W_{\hat{f}}=\frac{(1/N\sum_n)|\hat{f}(n)\cdot \hat{f}(n)|^2}{(1/N)\sum_n |\hat{f}(n)|^2}=1917.129,$$

and I noticed that as N grows the result of this calculation divided by 2N seems to approach a constant, 0.09585...(if there is a simple explanation for this I'd be interested).

My question is how to do the continuous part of the transformed function and evaluate these quantities.

Using the definition and Mathematica I find that

$$F(f(t))=\frac{1}{\sqrt{2\pi}}\int_R t\cdot \exp(-i\omega t)dt$$ and since $$f(t)$$ vanishes outside $$[0,1]$$ I integrated from $$0$$ to $$1$$ giving

$$F(f(t))=\hat{f}(\omega)=\frac{e^{-i\omega} (1+i\omega)-1}{\sqrt{2\pi}\omega^2}$$

Because Plancherel's relation is not satisfied as a threshold matter I know this is not correct.

Edit: I verified the Mathematica result by hand. Does $$\omega$$ in the denominator prevent Plancherel's relation from holding? Mathematica's numerical integration does give the anticipated value if I find for large-ish $$\Omega$$ and small-ish $$0+$$

$$2 \int_{0+}^{\Omega} |\hat{f(\omega)}|^2d\omega$$

So maybe this is the conditionally valid answer. Appreciate any discussion or correction.

• It may be that the weight functions $W_{\hat{f}}$ above and the continuous cognate do not agree because they both diverge but not in tandem. – daniel Mar 3 at 10:31