what is negative frequency(fourier transform)by ankities Tags: frequencyfourier, negative, transform 

#1
Dec2012, 06:00 AM

P: 9

in fourier transforms of normal baseband sigal , spectral components are replicated on both +ve and ve sides of frequency axis.
i know that both ve and +ve frequency components contribute to the total power of the signal but i dont know the physical significance of the ve frequencies used? are these ve frequencies just the mathematical imaginary tool ? 



#2
Dec2012, 07:09 AM

P: 134

There is no physical reality of the negative frequency. It is, as you say, a mathmatical tool.
However, the negative frequencies only emerge because we wish to simplify the fourier transform. It is perfectly possible to have a Fourier Transform without any imaginary and negative components. If you look into Fourier Series (from which Fourier Transforms are developed), you will see that it can be represented as [itex]f(t) = a_{0} + \sum_{n}a_{n} cos(n\omega t + \theta_{n}) + \sum_{m}b_{m} sin(m\omega t + \theta_{m})[/itex] It is only because we wish to simplify this that we make use of eulers identity that e^{±iθ} = cos(θ) ± i sin(θ) When substituting this you will get "negative frequencies" when deriving all the formulas. 



#3
Dec2012, 07:54 AM

P: 580

[itex]cos(ωt) = \frac{e^{+jωt} + e^{jωt}}{2}[/itex] On the other hand there is a physical reality to complex frequencies when we use them to describe modulation. Specifically, they represent a *pair* of real modulation signals. There are two independent degrees of freedom when modulating a sinusoid; phase/amplitude in polar coordinates, I/Q in rectangular coordinates. We can incorporate both of these independent signals into our single complex frequency expression. In other words just as complex numbers can represent a pair of real numbers on an Argand diagram, complex frequencies can represent a pair of orthogonal modulation components of a real sinusoid. 


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