What is the significance of negative frequency in Fourier transform? Why we include the band widths of the negative frequency also while calculating band width of the signal.
Fourier Transform – What Is the Significance of Negative Frequency in Fourier Transform?
fourier transform
Related Solutions
A very brief answer: The Fourier transform as used in quantum physics interprets a point in one set of three coordinates as a broad spectrum in another set of three coordinates, and vice-versa. This is the uncertainty relationship, since it means that when the object is point-like in one set of coordinates (space, for example) it become vast and diffuse in the other (momentum, for example).
To make that a bit more precise you need to stop thinking of Fourier transforms as flat sinusoidal graphs, and instead think of them as having complex values -- that is, as having phase values in a plane (two more axes really) that is orthogonal to the other three. In this form, an object that is stretched out along one axis -- say X -- looks more like a stretched-out slinky if you add the two complex coordinates and move along X. The various continuity equations of physics simply say that the coils of any such slinky like to stay as smooth and unkinked as possible. (The slinkies also rotate around their axis based on how massive the object is, but that's a different issue).
If such a slinky is infinitely long in X and perfectly smooth at some frequency of coiling, it represents a particle that whose wave function is infinitely long in X, and whose probability of finding it at any one spot is essentially zero. That's position uncertainty, the worst possible case of it.
But it's the frequency of coiling that's more important for uncertainty, because that frequency just happens to represent the momentum of the particle parallel to the X axis. If the coiling is perfectly regular, there's only one such frequency, and you could map that on frequency out (radio-dial style) into another triplet of axes labeled $p_x$, $p_y$, and $p_z$. That's momentum space, the other side of the Fourier coin.
In that space, the particle has a very precise location indeed, that being the location of that particular frequency along the momentum axis $p_x$.
But what if you try it the other way around? That is, what if you construct a similar helix using the same rules of adding a complex plane perpendicular to $p_x$ and creating a very long, very regular coil there. (There's a beautiful symmetry in physics here, because it turns out that momentum space has continuity and smoothness rules very much like those of regular XYZ space, despite some different meanings of them.)
Well, pretty much exactly the same thing happens, only in reverse: The long, precise coil in momentum space also has a precise frequency, which also has an interpretation over in XYZ space as a precise location. And again, having the coil stretched out in momentum space means that it's hard to find the particle there; it could with equal likelihood be almost anywhere along the coil, meaning it could have almost any momentum. So, in this case, being "uncertain" about the location of the particle in momentum space means that it has a very precise location in regular space.
The symmetry is simply gorgeous, and has many real implications in physics. Metals, for example, are substances in which pairs of electrons exist more in momentum space than in regular space, causing some of them to have very high momenta (energies) and all of them to be distributed rather oddly over a metal crystal. The high energy ones make the metal reflective, the movement (momenta) of the electrons make it conductive, and the delocalization of the electrons combine with charge cancellation to create rather remarkable properties such as tensile strength (it sticks together tightly!) and ductility (it's easier to bend when half the charges are always in motion).
But while the Fourier part of this symmetry between the XYZ and momentum spaces is exact, the meanings of frequencies in the two spaces is anything but. The biggest difference is that while the size of an object in XYZ space has no huge energy implications, size in momentum space has huge implications, because higher momentum means higher energy. So, the longer a slinky (more properly called a wave function) becomes in momentum space, the higher in energy the particle becomes.
And, since the length of the slinky in momentum space determines through the Fourier transform how precisely it can be positioned in regular XYZ space, extremely precise positions come at a cost -- a very high cost. The Stanford linear accelerator, for example, has to accelerate electrons to extremely high velocities (momentums) because it's the only way to make electron locations precise enough to probe the innards of particles such as neutrons and protons.
Oddly, no such cost is accrued when particles become lost in regular XYZ space. So, an electron wandering through the cosmos can in principle be represented by a very large slinky with a very precise frequency, if something (e.g. diffraction) encourages it to form that way. But even though the energy cost is low, such wave functions are unstable for a very different reason: Any kind of information-imparting interaction between them and other matter causes the wave function to be irrelevant (yes, I'm trying to avoid getting into schools of quantum doctrine with that careful wording) and for all practical purpose a vastly smaller wave function. Just not too small, since only the energy of the interaction is available to alter the original diffuse form.
And again, to bring all that back home: It's the Fourier transform and its ability to say how well each frequency "matches" a slinky-form in either of the spaces that creates this interpretation back-and-forth between the two, and thereby creates quantum uncertainty.
Consider a single value of $m$. The Fourier series for just that $m$ gives $$a_m \cos(2\pi m t / T) + b_m \sin(2\pi m t / T) \, .$$ This can be rewritten as $$M_m \cos (2\pi m t / T + \phi_m)$$ where $$M_m = \sqrt{a_m^2 + b_m^2} \qquad \text{and} \qquad \phi_m = \operatorname{atan2}(-b_m,\,a_m) \, .$$ So, you can see that $a_m$ and $b_m$ are just the cartesian coordinate equivalent of the signal's amplitude and phase.
When you have the full series with all the terms it's just a sum of many sinusoids each with their own amplitude and phase. In other words, we could equivalently write $$\frac{a_0}{2} + \sum_{m=1}^\infty M_m \cos(2\pi m t / T + \phi_m)$$ using the relations above.
Anyway, the point is that the physical meaning of the coefficients in the Fourier series is that they tell you the amplitude and phase at each frequency.
Best Answer
If you have a real function of time $x(t)$, which may represent, for example, a modulated carrier, you can, if you wish, avoid negative frequencies by using the Fourier Sine Transform $$X_s(\omega)=\int_{-\infty}^{\infty}{x(t)\sin\omega t dt} $$ and Fourier Cosine Transform $$X_c(\omega)=\int_{-\infty}^{\infty}{x(t)\cos\omega t dt} $$ because then, the inverse transforms can be written (ignoring factors of 2s and $\pi$s): $$x(t)=\int_{0}^{\infty}X_c(\omega)\cos \omega t + \int_{0}^{\infty}X_s(\omega)\sin \omega t $$ i.e no negative frequencies are present.
However, it's much neater to work with the complex exponential representation of modulated signals. A generic modulated signal, in the time domain will have its amplitude and phase varying as a function of time $$x(t)=A(t)\cos(\omega_c t+\phi(t)) $$ We can write this as the real part of a complex signal $$ x(t)=\Re\{A(t)\cos(\omega_c t+\phi(t))+jA(t)\sin(\omega_c t+\phi(t))\}$$ Now I could have written any old rubbish in the imaginary part, and the formula would still have been correct, but the choice I made makes it extremely convenient - mainly because I can write it in exponential form $$ x(t)=\Re\{A(t)\exp(j(\omega_c t+\phi(t)))\}$$ and this makes it really easy for me to "factor out" the carrier frequency $$ x(t)=\Re\{X_B(t)\exp(j(\omega_c t))\} $$ where $X_B(t)$ is the "baseband equivalent" signal. "Baseband" meaning "effectively at 0Hz carrier frequency". The baseband equivalent signal is just $$X_B(t)=A(t)\exp(j(\phi (t))) $$ and is complex and contains the "interesting" information contained in the modulation (the carrier itself being not so interesting!).
Working with these exponential forms, it now seems much more natural to define the Fourier transform in terms of exponentials in the normal way: $$X(\omega) = \int_{-\infty}^{\infty}{x(t)\exp(-j\omega t)}dt $$ together with the inverse transform $$x(t) = \int_{-\infty}^{\infty}{X(\omega)\exp(j\omega t)}d\omega $$ The Fourier transform of the baseband equivalent signal $X_B(t)$ is not symmetric about zero - you can't specify it by giving its values for positive frequencies and invoking symmetry. So if you want to work with this formalism then negative frequencies are fundamental.
The actual spectrum of the signal centred at $+\omega_c$ is the one you'd get if you swept a very narrowband filter over the frequency range. It's not necessarily symmetric about $+\omega_c$ and this gets translated to the complex baseband one (shown dotted) which is now not symmetric about zero, so negative frequencies are needed.