# Why does this FFT based signal deliver a sinusoidal result?

I have noticed that there is an identical pattern between two signals:

one is a signal of n identically spaced samples over a single period of a sinusoid

the other is a signal constructed of the real component of the second (i.e. first periodic) Fourier bin of n circular shifts of the signal, so pseudocode would be:

allocate a final result, size n
loop from 1 to n
newSignal = circularly shift signal one step to the left
newValue = take the FFT, second (first periodic) bin, real component
finalResult(n) = newValue

The only difference between these two is the scaling. If one's input is known to be a single cycle of a sinusoid, this is a nice way to remove all noise without knowing the phase of the input. Obviously it works in part because the second Fourier bin corresponds to the frequency of the input. But that's as far as my intuition gets me. In terms of theory, I don't understand why a signal composed of the real component of the second Fourier bins of the circular shift of each step of the signal looks the same as the original signal. Can anyone shed insight?

Let me see if I understood you correctly. You have one period of a sinusoidal signal:

$$x[n]=\sin\left (\frac{2\pi n}{N}\right ),\quad n=0,1,\ldots N-1$$

Its DFT is given by

$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}$$

For $k=1$ we have

$$X[1]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi n/N}$$

For the given signal $x[n]$ we get

$$X[1]=\frac{N}{2j}$$

For the left-shifted versions of $x[n]$ we get

$$X_l[1]=\sum_{n=0}^{N-1}x[n+l]e^{-j2\pi n/N}=e^{j2\pi l/N}\sum_{n=0}^{N-1}x[n]e^{-j2\pi n/N}=e^{j2\pi l/N}\frac{N}{2j}$$

So we have

$$\Re\{X_l[1]\}=\Re\left\{e^{j2\pi l/N}\frac{N}{2j}\right\}=\frac{N}{2}\sin\left (\frac{2\pi l}{N}\right )=\frac{N}{2}x[l],\quad l=0,1,\ldots, N-1$$

Q.E.D.

• That's great thanks, and my scaling is indeed N/2. I do not see how you got to N / 2j with a j in the denominator. Are there trig identies involved? – barnhillec Mar 28 '14 at 15:10
• If you write the sine function as $(e^{j2\pi n/N}-e^{-j2\pi n/n})/2j$ then you see that you get the $2j$ in the denominator. You get the $N$ because you simply sum $N$ times the number 1. – Matt L. Mar 28 '14 at 15:12