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## Exponential Distribution |

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exponential , exponential filter , exponential algorithm , exponential and voltage

10 Threads found on edaboard.com: **Exponential Distribution**

Rayleigh channels are almost worst case scenario. In Nakagami-m channel modeling the worst channel has a parameter m = 0.5, while Rayleigh has a parameter m = 1. Also, Rayleigh channel is mathematically more convenient; the SNR **distribution** is **exponential**.

Digital communication :: 10-19-2016 23:15 :: David83 :: Replies: **3** :: Views: **458**

Hi,
if we create a sequence and call these w1...wN with **exponential** **distribution** with parameter lambda then we create:
118444
what's name of this **distribution**?
Thanks

Mathematics and Physics :: 06-14-2015 14:15 :: panda1234 :: Replies: **7** :: Views: **1084**

Hi all,
I want to generate random signal with **exponential** **distribution** (exp(T/tau) tau is my constant) in verilog. Could you please advice me how I should do it?
Sincerely, Albert

Microcontrollers :: 08-16-2013 07:12 :: albert_new :: Replies: **2** :: Views: **663**

Basically, if the underlying fading channels are Rayleigh, then SNR follows **exponential** **distribution**. That is y0, y1, and y2 are all **exponential**s, where y0, y1, and y2 are the instantaneous SNR of the direct path, the first hop of the relay path, and the second hop of the relay path, respectively. After generating them, calculate the (...)

Digital communication :: 10-19-2011 22:38 :: David83 :: Replies: **5** :: Views: **7695**

Here is the source file of exprnd function
function r = exprnd(mu,varargin)
%EXPRND Random arrays from **exponential** **distribution**.
% R = EXPRND(MU) returns an array of random numbers chosen from the
% **exponential** **distribution** with mean parameter MU. The size of R is
% the size of MU.
%
% R = (...)

Digital communication :: 09-20-2011 20:00 :: samir67 :: Replies: **1** :: Views: **3351**

IF X and Y are statistically independent (or orthogonal) then the RMS of the combined signal (sum) will be sqrt(X^2 + Y^2) and the **distribution** will be **exponential**

Digital communication :: 02-28-2011 05:10 :: klystron :: Replies: **1** :: Views: **724**

Hi all! I would like to ask if there's any advantage in doing OFDM in a flat fading Rayleigh environment?
Also can anyone tell me if there would be any difference in the BERvsSNR results , when we simulate a flat Rayleigh channel and a frequency selctive **exponential** channel, where each tap follows Rayleigh **distribution**...
Any answe

Digital communication :: 08-25-2010 15:45 :: Communications_Engineer :: Replies: **6** :: Views: **1773**

I want to find out the cdf
Fx(x)=Pr(X ≤ k)
where "X" is a Rayleigh **distribution** and "k" is some other Rayleigh **distribution** with may be different mean and variance.
How can I find it out, somebody help me please.

Digital communication :: 06-16-2009 08:16 :: aliazmat :: Replies: **2** :: Views: **1090**

If you look in the book of David Tse, Fundamentals of wireless communications.
instantaneous SNR is defined as, |h|^2 SNR, therefore if we take square of Rayleigh it becomes **exponential** **distribution** therefore the overal instantaneous recieved SNR becomes exp. with mean SNR.
hope this helps.

Digital communication :: 11-13-2008 09:04 :: urwelcome :: Replies: **2** :: Views: **2166**

please give me the charecteristic functions of
1)weibull **distribution**
2)rayleigh **distribution**
3)pareto **distribution**
4)**exponential** **distribution**

Digital communication :: 04-02-2007 13:30 :: vamsi charan :: Replies: **0** :: Views: **1233**

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transistor vds | pic keypad display | frequnecy | matching vds | optical laser | counters using vhdl | pic16f877 mikroc | mcu vhdl | drill altium | altium mechanical layers