Take the 2-minute tour ×
Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free, no registration required.

In the field of Digital signal processing I have seen people using words

Complex signals and Negative frequencies. For eg. in FFT Spectrum.

Does it really have significant meaning in the time domain or is just a part of the mathematical symmetry.

How do you visualize negative Frequency in Time domain ?

share|improve this question
Please, have a look at this DSP SE question - dsp.stackexchange.com/questions/431/… –  yuvi Mar 11 '14 at 12:36
This question is much easier when you have a solid grasp of complex (I/Q) representation of signals. See Constellations in Digital Communication and What are the I and Q in quadrature sampling?. –  Phil Frost Mar 11 '14 at 12:55

4 Answers 4

up vote 20 down vote accepted

FFTs work by treating signals as 2-dimensional -- with real and imaginary parts. Remember the unit circle? Positive frequencies are when the phasor spins counter-clockwise, and negative frequencies are when the phasor spins clockwise.

If you throw away the imaginary part of the signal, the distinction between positive and negative frequencies will be lost.

For example (source):

Phasor spinning

If you were to plot the imaginary part of the signal, you would get another sinusoid, phase shifted with regards to the real part. Notice how if the phasor were spinning the other way, the top signal would be exactly the same but the phase relationship of the imaginary part to the real part would be different. By throwing away the imaginary part of the signal you have no way of knowing if a frequency is positive or negative.

share|improve this answer
Very good illustration. I think worth underscoring that if you only think of frequencies as sinusoidal waves, then you can't have negative frequencies, because if you spin the other way, the top half of the illustration looks the same. This is also why when you do a FFT of real signals (by arbitrarily setting the complex part to 0), the negative frequencies in the result are a mirror of the positive frequencies. –  Phil Frost Mar 11 '14 at 12:58
Also a good follow-up question for anyone who wanted to ask it: "Why does the FFT treat signals as 2-dimensional?" –  Phil Frost Mar 11 '14 at 13:09
@PhilFrost Thanks, updated! –  sbell Mar 11 '14 at 13:16
Well, Lets say I have a sine wave signal(freq = F) sampled at frequency Fs. How can I obtain the real & Imaginary part out of it ? Does it has to do anything with phase shifted Current or Voltage ? I may be totally wrong at this point ...but I need more inputs to get it straight & practically clear in sense! –  rahulb Mar 11 '14 at 13:45
@rahulb If you don't have the imaginary part, you can make it with the Hilbert transform. –  Phil Frost Mar 12 '14 at 0:36

In the time domain, a negative frequency is represented by a phase reversal.

For a cosine wave, it doesn't make any difference, since it is symmetrical around zero time anyway. It starts out at 1 and falls to zero in either direction.

$$cos(t) = cos(-t)$$

However, a sine wave starts with a value of zero at zero time and rises in the positive direction, but falls in the negative direction.

$$sin(t) = -sin(-t)$$

share|improve this answer
I can't argue with the math, so this isn't wrong per se, but I think it misses addressing what is likely the knowledge lacking in the question: quadrature, complex representation of signals. In practice, we deal with signals with arbitrary phase offsets anyway, and in that case, simply reversing the phase (such as by swapping the feed polarity on an antenna) most absolutely doesn't get you negative frequencies. –  Phil Frost Mar 11 '14 at 13:02
I think this answer captures it correctly. I just wanted to comment that the problem isn't that you simplify sine by phase-shifting. The problem is that you can't simplify the pair (cosine, sine) by phase-shifting. –  SomeEE Mar 11 '14 at 15:48

Here is a slightly different approach. Let's see which periodic function has Fourier transform exactly with frequency \$-1\$.

It is the function \$t \mapsto e^{-2\pi \mathrm{i} t} = \cos(-2\pi t) + \mathrm{i}\sin(-2\pi t) = \cos(2\pi t) - \mathrm{i}\sin(2\pi t) \$ for \$ t \in [0,1]\$.

Notice that this function has the same real part as the function \$t \mapsto e^{2\pi \mathrm{i}t}\$. This latter function has only a single frequency component - the frequency \$1\$.

The reason these negative frequencies show up when considering only real signals is because they give an easier way to describe strictly complex eigenvalues of the action of the unit circle on its function space.

Edit: To expand upon the last comment, in order to do frequency analysis what we really wished to do is take the space of real valued functions on \$[0,1]\$, \$F([0,1], \mathbb{R})\$ and be able to express any function \$f \in F([0,1], \mathbb{R})\$ in terms of some natural basis of \$F([0,1], \mathbb{R})\$. We agree that it doesn't really that much if we start our period is \$0\$ to \$1\$ or \$1/2\$ to \$3/2\$ so we really would desire that this basis behave well with respect to the shift operator \$f(x) \mapsto f(a+x)\$.

The problem is, with appropriate adjectives, \$F([0,1], \mathbb{R})\$ not a direct sum of functions that behave well with respect to shifting. It is a (completed) direct sum of two dimensional vector spaces which behave well with respect to the shift operator. This is because the matrix representing the map \$f(x) \mapsto f(a+x)\$ has complex eigenvalues. These matrices will be diagonal (in an appropriate basis) if we complexify the situation. That is why we study \$F([0,1], \mathbb{C})\$ instead. Introducing complex numbers has a penalty though - we obtain a concept of negative frequencies.

This is all a bit abstract but to see concretely what I am talking about consider my two favorite functions: $$\cos(2\pi t) = \frac{1}{2}(e^{2\pi \mathrm{i} t} + e^{-2\pi \mathrm{i} t})$$ $$\sin(2\pi t) = \frac{1}{2 \mathrm{i}}(e^{2\pi \mathrm{i} t} - e^{-2\pi \mathrm{i} t})$$

Consider the shift by \$\frac{1}{4}\$, \$s(f(x)) = f(x+\frac{1}{4})\$. $$s(\cos(2\pi t)) = -\sin(2 \pi t)$$ $$s(\sin(2\pi t)) = \cos(2 \pi t)$$
The real vector space span of \$\cos(2 \pi t)\$ and \$\sin(2 \pi t)\$ is a two dimensional vector space of functions which is preserved by \$s\$. We can see that \$s^2 = -1\$ so \$s\$ has eigenvalues \$\pm \mathrm{i}\$

This two dimensional space of functions cannot be decomposed into eigenspaces for \$s\$ unless we complexify it. In this case the eigenvectors will be \$e^{2\pi \mathrm{i} t}\$ and \$e^{-2 \pi \mathrm{i}t}\$.

To recap, we started with two positive frequencies but in order to diagonalize the action of \$s\$ we had to add in the negative frequency function \$e^{-2 \pi \mathrm{i} t}\$.

share|improve this answer

A great way of visualizing negative frequencies is to modulate the original signal. Say you have a sine wave with frequency \$\omega_0\$ (in radians):


The spectrum of this signal has a peak at \$\omega=\omega_0\$ and one at the negative frequency \$\omega=-\omega_0\$.

By modulating the signal \$x(t)\$ you basically shift the original spectrum by the carrier frequency \$\omega_c>\omega_0\$:


Now the original negative peak at \$-\omega_0\$ has become visible after shifting it up by \$\omega_c\$. It is now at \$\omega=\omega_c-\omega_0\$. The peak at positive frequencies is not at \$\omega=\omega_c+\omega_0\$.

share|improve this answer
The OP specifically asked about visualization in the time domain, but you talk only about the frequency domain and the spectrum of the signal. –  Joe Hass Mar 11 '14 at 12:20
@JoeHass Well, the signal \$y(t)\$ is in the time domain, and here you can see both frequency components. –  Matt L. Mar 11 '14 at 12:24
I think you are missing the point. All I see is an equation where one of the terms may have a negative frequency. I think the OP is wondering what a negative frequency would look like on an oscilloscope. –  Joe Hass Mar 11 '14 at 12:53
Maybe it would be helpful if you could submit an answer to this question, as you seem to understand what the OP is wondering about. –  Matt L. Mar 11 '14 at 13:03
No, I can't submit an answer because I am also confused by this topic. However, I do understand the question. I think Dave Tweed came as close as anyone in describing "negative" frequency as being a phase reversal. –  Joe Hass Mar 11 '14 at 15:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.