Negative frequency

The concept of negative and positive frequency can be as simple as a wheel rotating one way or the other way. A signed value of frequency indicates both the rate and direction of rotation. The rate is expressed in units such as revolutions (aka cycles) per second (hertz) or radians/second (where 1 cycle corresponds to 2π radians).

Sinusoids

A sinusoid is a function of an angular argument, and its amplitude varies cyclically as the angle (aka phase) steadily increases or decreases. When the angle is a function of time, the concept of negative frequency is sometimes used to distinguish a decreasing angle from an increasing one. But sinusoids are not monotonic functions. Consequently, $\cos(\omega t + \theta)\,$ does not preserve the sign of $\omega \,$ , just as $f(x)=x^2\,$ does not preserve the sign of $x\,$ . note that $\theta \,$ represents a usually unknown, random phase offset. In most cases of dealing with a single, real-valued sinusoid, it is sufficient to assume that $\omega \,$ is positive. It represents the frequency, in units of radians/sec.

Sometimes there are two sinusoids with the same frequency, and a known phase difference, for instance:

R(t) = \cos(\omega t + \theta)\,

and

I(t) = \cos(\omega t + \theta -\begin{matrix}\frac{\pi }{2} \end{matrix}) = \sin(\omega t + \theta )\,

When $\omega > 0\,$ , $R(t)\,$ appears to lead $I(t)\,$ by $\begin{matrix} \frac{1}{4} \end{matrix}\,$ cycle ( $=\begin{matrix}\frac{\pi }{2} \end{matrix}\,$ radians). But when $\omega < 0\,$ , the roles are reversed. So in that case it is possible to distinguish negative and positive frequencies. The diagram depicts a negative frequency. $R(t)\,$ and $I(t)\,$ are referred to as real and imaginary, respectively. And $\theta = 0\,$ .

A parametric plot of imaginary vs real would trace a circular path (like the rotating wheel). The addition of a time dimension creates a corkscrew pattern. A negative frequency (decreasing phase) causes a clockwise rotation in a right hand coordinate system as time increases:

Complex sinusoids

The complex function: $\cos(\omega t) + j\cdot \sin(\omega t)\,$ facilitates many kinds of mathematical operations involving $\cos(\omega t)\,$ , due in large part to Euler's simplification:

e^{j \omega t} = \cos(\omega t) + j\cdot \sin(\omega t)\,

This very useful form is often referred to as a complex sinusoid. And of course it preserves the distinction between positive and negative $\omega \,$ .

For positive values, it is also called the analytic representation of $\cos(\omega t)\,$ .

The Fourier transform of $e^{j \omega t}\,$ produces a non-zero response only at frequency $\omega \,$ .

The transform of \cos(\omega t)\, has responses at both \omega \, and -\omega \,, which reflects the fact that \cos(\omega t)\, is insufficient to determine the sign of \omega \,.
- So just as $\sqrt{4} = \pm 2 \,$ , the interpretation of the ambiguity often depends on collateral information.
- An alternative, and surprisingly useful, viewpoint is that both frequencies are present, as implied by the inverse of Euler's formula: $\cos(\omega t) = \begin{matrix} \frac{1}{2} \end{matrix}(e^{j \omega t}+e^{-j \omega t})\,$ .

Sampling of positive and negative frequencies and aliasing

When a complex sinusoid is sampled at regular intervals, its frequency becomes indistinguishable from certain other frequencies, including negative ones (referred to as aliasing). The adjacent figure illustrates this effect for several cases. The red indicates 0 Hz (aka DC). Successively higher frequencies are indicated by orange, blue, purple, violet, black, and blue. Note that some frames depict "R" and "I" for the same frequency, and others depict the "I" samples of different frequencies that are aliases of each other.

For instance, the fourth frame (purple and green) compares samples of the imaginary component of the fractional frequency + $\begin{matrix}\frac{5}{8}\end{matrix}$ with those of negative frequency $-\begin{matrix}\frac{3}{8}\end{matrix}$ , to illustrate that they are indistinguishable. Or in other words: $e^{j2\pi \left(+\begin{matrix}\frac{5}{8}\end{matrix}\right) n} = e^{j2\pi \left(-\begin{matrix}\frac{3}{8}\end{matrix}\right) n}\,$ for integer values of n, representing the sample number. The underlying waveforms are just the imaginary components of: $e^{j 2 \pi \left(+\begin{matrix}\frac{5}{8}\end{matrix}\right) F_s t}\,$ and $e^{j 2 \pi \left(-\begin{matrix}\frac{3}{8}\end{matrix}\right) F_s t}\,$ , where $F_s \,$ is the sample rate (samples/sec).

Likewise + $\begin{matrix}\frac{7}{8}\end{matrix}$ is indistinguishable from $-\begin{matrix}\frac{1}{8}\end{matrix}$ . And $\begin{matrix}\frac{8}{8}\end{matrix}$ (last plot) is indistinguishable from $\begin{matrix}\frac{0}{8}\end{matrix}$ (first plot).