Linear Phase Terms

The reason $e^{-j \omega_k \Delta}$ is called a linear phase term is that its phase is a linear function of frequency:

$$\angle e^{-j \omega_k \Delta} = - \Delta \cdot \omega_k$$

Thus, the slope of the phase versus radian frequency is $-\Delta$. In general, the time delay in samples equals minus the slope of the linear phase term. If we express the original spectrum in polar form as

$$X(k) = G(k) e^{j\Theta(k)},$$

where $G$ and $\Theta$ are the magnitude and phase of $X$, respectively (both real), we can see that a linear phase term only modifies the spectral phase $\Theta(k)$:

$$e^{-j \omega_k \Delta} X(k) \isdef e^{-j \omega_k \Delta} G(k) e^{j\Theta(k)} = G(k) e^{j[\Theta(k)-\omega_k\Delta]}$$

where $\omega_k \isdeftext 2\pi k/N$. A positive time delay (waveform shift to the right) adds a negatively sloped linear phase to the original spectral phase. A negative time delay (waveform shift to the left) adds a positively sloped linear phase to the original spectral phase. If we seem to be belaboring this relationship, it is because it is one of the most useful in practice.

Definition: A signal is said to be a linear phase signal if its phase is of the form

$$\Theta(\omega_k)= \pm \Delta \cdot \omega_k\pm \pi I(\omega_k),$$

where $\pm\Delta$ is any real constant, and $I(\omega_k)$ is an indicator function which takes on the values 0 or $1$ over the points $\omega_k$, $k=0,1,2,\ldots,N-1$.

A zero-phase signal is thus a linear phase signal for which the phase-slope $\Delta$ is zero.