# Circular Time Delay in signal, effect on phase spectra of DFT

If you circularly shift a signal x[n]= [-3 -2 -1 0 1 2 3 2 1 0 -1 -2 ] to the right by M=1 compared to M=2, the phase spectra has a lot many zeros when M=2 compared to M=1.

here, I am talking about Circular Time shifting.

I know that if x[n] is even, then DFT is purely real and hence phase is zero. However, why are there way many more zeros when M=2 compared to M=1?

Thanks

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If by phase spectrum you mean the sequence $$\bigr(\angle X[0], \angle X[1], \angle X[2], \ldots, \angle X[N-1]\bigr),$$ then remember that a circular shift by $M$ in the time domain results in each $X[n]$ being multiplied by $\exp(-j2\pi Mn/N)$, and thus causes a change in $\angle X[n]$; in particular, $\angle X[n]$ decreases by $2\pi Mn/N$. For the case $M=1$, the multipliers are distinct $N$-th roots of unity, that is, the changes in phase are

$$\Bigr(0, 2\pi\frac{1}{N}, 2\pi\frac{1}{N}, \ldots, 2\pi\frac{N-1}{N}\Bigr)$$

all of which are different numbers. Thus, each entry in the phase spectrum changes (except for the entry for $n=0$). For the case $M=2$ and $N=12$, the multipliers are $\exp(-j2\pi 2n/12) = \exp(-j2\pi n/6)$ and so the changes in phase are

$$\Bigr(0, 2\pi\frac{1}{6}, 2\pi\frac{2}{6}, \ldots, 2\pi\frac{5}{6}, 2\pi\frac{6}{6}, 2\pi\frac{7}{6}, \ldots 2\pi\frac{11}{6}\Bigr)$$

But a phase change of $2\pi\frac{6}{6} = 2\pi$ is the same as no phase change, a phase change of $2\pi\frac{7}{6}$ is the same as a phase change of $2\pi\frac{1}{6}$, and so on. So the displayed vector above of phase changes is actually

$$\Bigr(0, 2\pi\frac{1}{6}, 2\pi\frac{2}{6}, \ldots, 2\pi\frac{5}{6}, 0, 2\pi\frac{1}{6}, \ldots 2\pi\frac{5}{6}\Bigr)$$

which is two periods of a sequence of period $6$. Note each entry in the phase spectrum changes except for $n=0$ and $n=6$ which do not change.

Now suppose that each $X[n]$ is a real number and thus the phase spectrum is $0$ everywhere. Then, if you circularly shift by one place ($M=1$ in the time domain, all entries in the phase spectrum will be nonzero except for the $n=0$ entry. If instead you circularly shift by two places ($M=2$) in the time domain, the phase spectrum will have zeroes in both the $n=0$ and $n=6$ entries. Note also that $\exp(-j\pi = -1$, and so for $M=1$, $X[0]$ and $X[6]$ will still be real-valued (though there will have been a phase change of $\pi$ in $X[6]$) and similarly, for $M=2$, $X[0], X[3], X[6]$, and $X[9]$ will still remain real-valued though there will have been a phase change of $\pi$ in $X[3]$ and $X[9]$.

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I am really new to this; hence I could not understand what you meant by "For the case M=1, the multipliers are distinct N-th roots of unity and each"--esp, the multipliers being distinct N-roots of unity. By unity, do you mean the unit circle with the multiplier being the $exp(−j2πMn/N)$? If you could kindly break it down for me, it help make sense, especially how you are accounting for the periods of the multiplier sequence. – dsp_ent Nov 4 '12 at 18:56
@dsp_ent See edited answer. – Dilip Sarwate Nov 4 '12 at 22:15