# Phase Shifts in Fourier Space

In wireless radio networks the data is modulated onto electromagnetic signals for transmission. According to Jean Baptiste Joseph Fourier these signals can be represented as sums of simple harmonic waves, like sines and cosines. Harmonic waves are characterized by their harmonic oscillatory nature. A wave can be thoroughly described by providing amplitude, frequency and a constant or time dependant phase term for every such simple harmonic wave. In the simple case of a periodic signal, this leads to a sum of a fundamental wave of the circular frequency $$\omega$$ and a set of harmonics, whose circular frequencies are integer multiples of the fundamental's. We can represent such a signal as \begin{equation} \label{eq:realsign} x_k = \sum \limits_{m = -\frac{M-1}{2}}^{\frac{M-1}{2}} A_m \cdot \cos(\omega m k + \varphi_m(k)) \end{equation} where $$k \text{ is time, } m$$ is the integer factor denoting the harmonic, and as mentioned above, $$\omega = 2 \pi f$$ is the circular frequency, $$\frac{M-1}{2}$$ is the number of basic harmonic waves and $$\varphi_m(k)$$ is a time dependent phase delay. $$m = 0$$ denotes the direct current component. The Fourier transform is a mapping between a time domain, showing the signal as a time dependent function and a frequency domain, where the contained frequencies show up as peaks. In replacement to the time domain we can as well map from the frequency domain to the space domain for space dependency rather than time. A single peak of amplitude $$A = 1$$ at a frequency $$\omega$$ maps under the Fourier transform to a complex function of the form \begin{equation} \label{eq:simplest} x_k = A \cdot e^{j \omega k} \end{equation}. where $$j$$ is the imaginary unit. We can now depict an advancing wave in the time domain as a moving radius, which we will denote as a phasor in the circle of unity. Such a phasor will move with the velocity of the circular frequency $$\omega$$ in an always changing angle. After a time of $$k = \frac{1}{\omega} s$$ it will complete one cycle. In the frequency domain the frequency component of a phasor which moves counterclockwise around the circle of unity is denoted by a peak in the positive half-plane, which we call upper sideband (USB), while frequency components of phasors which move clockwise show up in the negative half-plane of the frequency domain which we call lower sideband (LSB). This is true as long as we describe signals in the baseband. When baseband signals are modulated onto a carrier frequency, the LSB and USB are understood in relation to the carrier frequency rather than the point of origin. The well known Euler's formula states that \begin{equation} \label{eq:euler} e^{j \omega k} = \cos(\omega k) + j \cdot \sin(\omega k) \end{equation} From this, we obtain \begin{equation*} \sin(\omega k) = \frac{1}{2j} \left( e^{j \omega k} - e^{-j \omega k} \right) \end{equation*} \begin{equation*} \cos(\omega k) = \frac{1}{2} \left( e^{j \omega k} + e^{-j \omega k} \right) \end{equation*} The formula for the cosine tells us, that the cosine appears in the frequency domain as 2 peaks, one at $$\omega \text{ and one at } -\omega$$. The cosine is an even function, its frequency spectrum is axially symmetrical to a vertical line through the point of origin in baseband or in the passband at the carrier frequency respectively. We denote this kind of spectrum simply as symmetrical. A real signal can be described as a sum of weighted cosine terms as in (\ref{eq:realsign}). Since the cosines all have a symmetrical spectrum, every real signal has a symmetrical spectrum as well. Now we know why our signal was said to have $$\frac{M-1}{2}$$ components plus direct current at $$m = 0$$, while the equation had $$M$$ summands. It is because every component but the direct current component is composed of 2 such summands. Using the formula for the sine on the imaginary part of (\ref{eq:simplest}), we obtain \begin{equation*} j \sin(\omega k) = \frac{1}{2} \left( e^{j \omega k} - e^{-j \omega k} \right) \end{equation*} The sine is an odd function, its frequency spectrum is point symmetrical to the point of origin. The positive and negative frequency components differ by their sign, the LSB component is negative, the USB component positive. When we add the real and imaginary terms together, the positive frequency components add up to a total amplitude of 1, whereas the negative frequency components cancel out. A complex signal, also called analytical signal, has no negative frequency components. Since a real signal is symmetrical, all the information is already contained in the USB. Therefore we model baseband signals as complex signals \cite{babe} for simplicity. Nevertheless, the complex signal representation is only a theoretical concept of signal processing. All existing electromagnetic signals are real. When sampling a real signal at the receiver followed by a Fourier transform, the phase informations in $$\varphi_m(k)$$ are lost, when we regard the amplitudes as real valued. These phase informations can be preserved by modifying our signal representation by deriving complex amplitudes. \begin{equation*} x_k = \sum \limits_{m = -\frac{M-1}{2}}^{\frac{M-1}{2}} A_m \cdot \cos(m \omega k + \varphi_m(k)) \end{equation*} From the trigonometric identities of addition follows \begin{equation*} x_k = \sum \limits_{m = -\frac{M-1}{2}}^{\frac{M-1}{2}} A_m \cdot (\cos(m \omega k) \cos(\varphi_m(k)) - \sin(m \omega k) \sin(\varphi_m(k))) \end{equation*} \begin{equation*} = \sum \limits_{m = -\frac{M-1}{2}}^{\frac{M-1}{2}} \underbrace{A_m \cos(\varphi_m(k))}_{Inphase component (I_m)} \cdot \cos(m \omega k) - \underbrace{A_m \sin(\varphi_m(k))}_{Quadrature component (Q_m)} \cdot \sin(m \omega k) \end{equation*} \begin{equation*} = \sum \limits_{m = -\frac{M-1}{2}}^{\frac{M-1}{2}} I_m \cdot {1 \over 2} (e^{j m \omega k} + e^{-j m \omega k}) - Q_m \cdot {1 \over 2 j} (e^{j m \omega k} - e^{-j m \omega k}) \end{equation*} From (\ref{eq:euler}) follows for $$\omega k = 1$$ \begin{equation*} j = e^{j \frac{\pi}{2}} \end{equation*} which leads to $${1 \over j} = \frac{1}{e^{j\frac{\pi}{2}}} = e^{-j\frac{\pi}{2}} = -j$$ resulting in \begin{equation*} x_k = \sum \limits_{m = -\frac{M-1}{2}}^{\frac{M-1}{2}} {1 \over 2} (\underbrace{(I_m + j Q_m)}_{X_m} e^{j m \omega k} + \underbrace{(I_m - j Q_m)}_{X_m^*} e^{-j m \omega k}) \end{equation*} Here $$x_m$$ is the complex Fourier coefficient for the nth harmonic to the fundamental frequency $$\omega$$ and $$x_m^*$$ is its complex conjugate, which shows up as the negative frequency component in a symmetrical spectrum. We are now able to define the discrete Fourier transform (DFT) as \begin{equation*} F(n) = X_n = \sum \limits_{n=0}^{N-1} x_k e^{-j (2 \pi \frac{n}{N} k)} \end{equation*} the DFT is a transformation that maps a set of sample values in the time domain onto complex Fourier coefficients in the frequency domain. Here $$N$$ is the sample rate, denoted as well as symbol rate, the fundamental has a frequency of $$1 Hz$$, meaning that all harmonics have an equally spaced distance of $$1 Hz$$. As before $$n = 0$$ marks the direct current component. The frequencies are normalized by the sample frequency, which makes the DFT periodic. The period of $$2 \pi$$ is divided into equal phase intervalls of size $$\frac{2 \pi}{N}$$. It is \begin{equation*} X_{n + N} = \sum \limits_{k=0}^{N-1} x_k e^{j 2 \pi \frac{k}{N} (n+N)} \end{equation*} \begin{equation*} = \sum \limits_{k=0}^{N-1} x_k e^{j 2 \pi \frac{k}{N} n} e^{j 2 \pi \frac{k}{N} N)} \end{equation*} \begin{equation*} = \sum \limits_{k=0}^{N-1} x_k e^{j 2 \pi \frac{k}{N} n} e^{j 2 \pi k} \end{equation*} \begin{equation*} = \sum \limits_{k=0}^{N-1} x_k e^{j 2 \pi \frac{k}{N} n} \end{equation*} since the function was said to be $$2 \pi$$ periodic and $$k$$ is integer valued. The inverse DFT (iDFT) is given as \begin{equation*} f(k) = x_k = \frac{1}{N} \cdot \sum \limits_{n=0}^{N-1} X_n e^{j (2 \pi \frac{n}{N} k)} \end{equation*} It maps a set of complex fourier coefficients in the frequency domain onto a set of sample values in the time domain. Since, as we can see from the definition of the DFT, $$f(m) = F(-m)$$ for some integer value , normally a proof for one of the transforms suffices.

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