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.
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.
$$sin(t) = -sin(-t)$$
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\$.
"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.
