# How do you visualize negative Frequency in Time domain ?

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 ?

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): 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.

• 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. Mar 11, 2014 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?" Mar 11, 2014 at 13:09
• 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! Mar 11, 2014 at 13:45
• Whoever is generating the sine wave is the one responsible for keeping the imaginary part or not. If you only get one sine wave, that means there is no imaginary part. If you do get two separate signals (each a sine wave), you can treat the second wave as the imaginary part of the same signal. Mar 11, 2014 at 14:25
• @rahulb If you don't have the imaginary part, you can make it with the Hilbert transform. Mar 12, 2014 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.

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

• 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. Mar 11, 2014 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. Mar 11, 2014 at 15:48
• "In the time domain, a negative frequency is represented by a phase reversal." And - suddenly - the counting of periodic events per second gives a negative value? I think, this claim is not in accordance with the definition of the term "frequency".
– LvW
Mar 12, 2015 at 10:02
• @LvW: The generalized concept of "frequency" is much broader than the simple counting of discrete periodic events. You can add and subtract frequencies, and when you subtract a large frequency from a small one, you get a negative frequency. In its most general form, frequency is a complex number, and in some cases, the associated time-domain phenomena are not periodic at all! Mar 12, 2015 at 11:16
• @Dave Tweed, yes-I can do all mathematical manipulations (add, subtract) with SIGNALS having different frequencies - however, I wonder how I can identify (measure) negative frequencies in the time domain (and THAT was the questuon).
– LvW
Mar 12, 2015 at 14:57

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 $e^{-2 \pi \mathrm{i} t}$.

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):

$$x(t)=\sin(\omega_0t)$$

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 $\omega_c>\omega_0$:

$$y(t)=x(t)\cos(\omega_ct)=\sin(\omega_0t)\cos(\omega_ct)=\frac{1}{2}[\sin(\omega_c+\omega_0)t-\sin(\omega_c-\omega_0)t]$$

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 $\omega=\omega_c+\omega_0$.

• The OP specifically asked about visualization in the time domain, but you talk only about the frequency domain and the spectrum of the signal. Mar 11, 2014 at 12:20
• @JoeHass Well, the signal $y(t)$ is in the time domain, and here you can see both frequency components. Mar 11, 2014 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. Mar 11, 2014 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. Mar 11, 2014 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. Mar 11, 2014 at 15:01

"How do you visualize negative Frequency in Time domain ?"

I interprete this question as follows: Do negative frequencies exist in reality?

If this interpretation is correct (and meets the core of the question) my answer is simply: NO - they do not exist.

More than that (to be a bit "sophistic") - "frequencies" cannot exist because they are not a physical quantity. Instead, we have sinusoidal waves with some specific properties - and on of these properties is the number of periods per second. And that`s what we call "frequency". And this number cannot be negative.

Hence, the introduction of signals having "negative frequencies" may have a lot of advantages but it is a pure abstract and theoretical "tool" allowing simplifications of mathematical expressions/descriptions.