DFT Phase

Discover how the phase plot of a DFT is generated.

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The magnitude plot is drawn by taking the absolute values of the complex DFT output. It might seem that the phase plot is not very useful. This is not true.

  • The magnitude plot tells us about the level of contribution in the signal from each participating sinusoid.
  • The phase plot tells us about the alignment of those sinusoids with respect to the origin.

Time shift

Let’s consider a rectangular signal in which L=3L=3 and N=16N=16. This signal has the following samples:

x1[n]={1index}x_1[n]=\{\underbrace{1}_{\text{index}~0}. 1. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.\}

Now recall the following two facts:

  • Due to DFT periodicity in both the input and output, the last N/2N/2 samples (from N/2N/2 to N1N-1) are the same as the samples from N/2-N/2 to 1-1. These are actually N/2N/2 samples to the left of zero.
  • In this kind of setup, the signal above is simply a right shift of the sequence below:

x0[n]={1index}x_0[n]=\{\underbrace{1}_{\text{index}~0}. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1.\}

From the two expressions above, we can write the following:

x1[n]=x0[(n1) mod 16]x_1[n]=x_0[(n-1)~\text{mod}~ 16]

Phase plot

The phase plot of x0[n]x_0[n] above should be 00 due to symmetry around zero. However, x1[n]x_1[n] is a right circular shift by 11 of x0[n]x_0[n]. Therefore, the phase at each bin for a shift of n0n_0 should become:

ϕk=2πkNn0\phi_k = -2\pi \frac{k}{N}n_0

We will derive this property of the DFT in the next chapter.

For the example above, where N=16N=16 and n0=1n_0=1, the phase at bin k=1k=1 should be:

ϕ1=2π116(1)180π=22.5\phi_1 = -2\pi \frac{1}{16}(1)\cdot \frac{180^\circ}{\pi}=-22.5^\circ

This is visible in the phase plot of x1[n]x_1[n] below. The second subplot is a zoomed-in version of the first.

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