Figure 1: Ocean Tide Example of a Random Process

A random process is a random variable which is a function of time. Thus, a random process is a waveform which is made up of an infinite number of random variables sampled over a period of time. An example of a random process is the tide of the ocean. Say the tide of the ocean is measured in feet at a certain beach. The tide would be measured continuously over time and graphed to show a waveform much like the one shown in Figure 1[1].
A bandpass random process is a random process whose power spectrum is contained in a band of frequencies centered around a center frequency $ f_{c}< $ . Bandpass random processes take on the form of bandpass signals, and can be writtern in terms of in phase and quadrature components as follows:

$ X(t)=X_{I}cos(\omega_{c}t+\theta)+X_{Q}sin(\omega_{c}t+\theta) $

Figure 2 displays a system which proves this equation.

Figure 2: Ideal Bandpass Filter System

This system is an ideal bandpass filter with the following transfer function:

$ H(\omega)=H_{LP}(\omega+\omega_{c})+H_{LP}(\omega-\omega_{c}) $

Thus any bandpass process applied to the input of the system, results in the same bandpass process at the output. If a bandpass process $ x(t) $ is applied, it first modulates carrier sine and cosine waves. It is then applied to a lowpass filter which results in only the baseband signal. The signal which modulated the cosine wave is called the quadrature component while the signal which modulated the sine wave is called the in phase component. $ \theta $ is a random variable which is uniformly distributed between $ 0 $ and $ 2\pi $.

These components are lowpass random processes which are bandlimited to $ B Hz $ due to the lowpass filter. Thus as shown, the bandpass filter is the summation of the quadrature and in phase components multiplied by a cosine and sine wave respectively.

PSD of Bandpass Random ProcessesEdit


Figure 3: PSD of a bandpass random process


Figure 4: PSD of quadrature components

Consider a Bandpass Random process with a PSD shown in Figure 3.

The PSD of the bandpass random process modulates the carrier waves resulting in the PSD being shifted up and down by $ \omega_{c} $. The lowpass filter filters out frequencies above $ 2\pi B $.

The result is the overlapped PSD shown in Figure 4. Since multiplying by cosine and sine obtains the same result, the PSDs of the quadrature components are equal. In fact, the areas under the PSDs $ S_{x}(\omega) $, $ S_{xc}(\omega) $ and $ S_{xs}(\omega) $ are all equaivalent, as well as the mean square values of $ x(t) $, $ x_{c}(t) $ and $ x_{s}(t) $

There are an infinite number of frequencies that could be chosen to be the center frequency of a bandpass signal. For each center frequency choice, there is a different quadrature representation of the bandpass process.

White Gaussian Bandpass Random ProcessesEdit

A random variable sampled at time t of a gaussian process is gaussian and has a gaussian PDF of the following form:

$ p_{x}(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-(x-m)^{2}/2\sigma^{2}} $

When a gaussian process has a uniform PSD it is called a white gaussian random process. A bandpass random process with PSD $ N/2 $ of bandwidth $ 4\pi B $ centered at $ \omega_{c} $ can also be expressed in terms of quadrature components as was shown earlier. The PSD and mean square value of the quadrature components are as follows:

$ S_{nc}(\omega)=S_{ns}(\omega)= N $ for $ |\omega|\leq 2\pi B $
and $ 0 $ otherwise
$ \bar{n_{c}^{2}}=\bar{n_{s}^{2}}=2 N B $

Bandpass signals can also be expressed in polar form using the following relationships:

$ n(t)=E(t)cos(\omega_{c}t+\theta) $
$ E(t)=\sqrt{n_{c}^{2}(t)+n_{s}^{2}(t)} $
$ \Theta(t)=-\tan^{-1}\frac{n_{s}(t)}{n_{c}(t)} $

where E(t) is the envelope function.

$ n_{c}(t) $ and $ n_{s}(t) $ are uncorrelated gaussian random vairbles with zero mean, variance of $ 2 N B $ and identical gaussian PDFs. If two gaussian random variables are uncorrelated and consequently independent, E(t) will have a Rayleigh density:

$ p_{E}(E)=\frac{E}{\sigma^{2}}e^{-E^{2}/2\sigma^{2}} $

Sinusoids in Noise Edit

If a certain signal is a sinusoid mixed in with narrow band gaussian noise, it will be of the form

$ y(t)=Acos(\omega_{c}t+\phi)+n(t) $
writing this in quadrature component form we have:
$ y(t)=[A+n_{c}(t)]cos(\omega_{c}t+\phi)+n_{s}(t)sin(\omega_{c}t+\phi) =E(t)cos[\omega_{c}t+\Theta(t)+\phi] $

$ E(t) $ and $ \Theta(t) $ are calculated as follows:

$ E(t)=\sqrt{[A+n_{c}(t)]^{2}+n_{s}^{2}(t)} $
$ \theta(t)=-tan^{-1}\frac{n_{s}(t)}{A+n_{c}(t)} $

Further derivation shows that the joint PDF of $ E $ and $ \Theta $ becomes:

$ p_{E\Theta}(E,\Theta)=\frac{E}{2\pi\sigma^{2}}e^{-(E^{2}-2A E cos\theta+A^{2})/2\sigma^{2}} $
Rice distributiona PDF

Figure 5: Rician PDF

where $ \sigma^{2} $ is equal to $ 2 N B $ for white noise. We can obtain the PDF of $ E $ by integrating the joint PDF with respect to $ \theta $ from $ -\pi $ to $ \pi $. The result is as follows:

$ p_{E}(E)=\frac{E}{\sigma^{2}}e^{-(E^{2}+A^{2})/2\sigma^{2}}I_{0}(\frac{A E}{\sigma^{2}}) $

This is called the Rician Density. Futher simplification of the PDF shows that it is very close to a gaussian density with a mean $ A $ and variance $ \sigma $, when $ A>>\sigma $ and $ E\simeq $A. Figure 5[2] shows a Rician Density plotted for equal variances but different means. The figure uses $ v $ as the mean instead of A. When the mean is 0, we have a Rayleigh density. The PDF of $ \theta $ can also be obtained from the joint PDF by integrating with respect to E. The process is involved and will not be discussed here.[3]

Applications in Mobile Communications: "fading" Edit

Mobile communication deals with radio communication between two stations, one or more of which are in motion. The most obvious example of this are cell phones, in which instant communication can be made between two people at any time, anywhere... as long as there is a signal. The strength of a signal has alot to do with how close the phone is to a base station, most of which are placed in densely populated areas. Everyone knows if you go out into the desert or somewhere barren, you are most likely going to lose your signal.

Proximity to a mobile base station is definitely not the only factor in mobile communication signal strength. Many problems arise as electromagnetic waves move through space, trying to reach their destination. Losses can occur do to atmospheric propagation, and topography of terrain also plays a big part. Certain textures such as roughness tends to dissipate propagating energy, reducing signal strength. Trees and other natural and man made obstacles cause what is called "shadowing" which also reduces signal strength[4]. More relevant to our discussion is the problem of fading, which is "the deviation of the attenuation that a carrier-modulated telecommunication signal experiences over certain propagation media."[5] Fading could vary due to multiple things such as time or position. Thus, fading is often modeled as a random process. One specific type of fading is multipath fading. This is caused by multiple paths being created in an environment due to the presence of reflectors. This gives the propagating wave different options in terms of which path it takes to get to the receiver. Because of this, the receiver will receive a superposition of various copies of the same signal traveling different paths. Each copy is affected differently by attentuation, delay and phase shifts which may contribute to either constructive or destructive interference. Destructive interference leads to fading.[6]

Both Rayleigh and Rician Densities are used to model fading. Here I will focus on the Rayleigh model.
Rayleigh distributionPDF

Figure 6: Rayleigh PDF

The Rayleigh fading model is used only for fading of signals that do not have direct line-of-sight between transmitter and receiver. For dominant line-of-sight, the Rician model is used. A Rayleigh model assumes that the magnitude of the signal passing through a communication channel will vary randomly according to a Rayleigh distribution[7]. Figure 6[8] illustrates the Rayleigh distribution for different variances. As discussed previously, the rayleigh distribution is the radial component of the sum of two uncorrelated gaussian random variables. An example of where the Rayleigh fading model would apply is a condensed city such as New York, where there are are many skyscrapers providing obstacles for the signals. Obstacles such as buildings provide attenuation, reflection and refraction, leading to the scattering of signals before they reach the receiver. According to the central limit theorem, sufficient scattering will lead to the channel impulse response to take on the likeliness of a gaussian process. Without a dominate component to the scatter (contributed by line-of-sight between transmitter and receiver), it will have a zero mean. As discussed previously, these characteristics show that the envelope function of the channel response will have a Rayleigh density PDF.

With the ability to model fading using a Rayleigh distribution, analysis can be performed on the key contributors that affect the performance of a wireless network.[9]

Numerical Example Edit


Figure 7: PSD of a bandpass random process


Figure 8: PSD of quadrature components

The PSD of a bandpass random process is shown in Figure 7. The center frequency of this PSD is 100 kHz. To represent this process in terms of quadrature components, each band of frequencies is shifted up and down by a 100 kHz. The baseband signal centered at zero is then low pass filtered to get rid of the outerlying spectrum. The PSD of the quadrature componenets $ S_{xc}(\omega) $ and $ S_{xs}(\omega) $ are displayed in Figure 8.

The mean square value of the quadrature components is calculated by taking the areas of the PSD as follows:

$ \bar{x^{2}}=\bar{x_{c}^{2}}=\bar{x_{s}^{2}}=2*20 kHz+4*10 kHz = 80 kHz $

References Edit

  3. Lathi, B.P. "Modern Digital and Analog Communication Systems" 3rd Ed.