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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 19:15 :: David83 :: Replies: **3** :: Views: **328**

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 10:15 :: panda1234 :: Replies: **7** :: Views: **974**

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 03:12 :: albert_new :: Replies: **2** :: Views: **584**

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 18:38 :: David83 :: Replies: **5** :: Views: **7207**

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 16:00 :: samir67 :: Replies: **1** :: Views: **3082**

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 00:10 :: klystron :: Replies: **1** :: Views: **705**

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 answer will be he

Digital communication :: 08-24-2010 19:32 :: anta :: Replies: **6** :: Views: **1714**

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 04:16 :: aliazmat :: Replies: **2** :: Views: **1059**

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 04:04 :: urwelcome :: Replies: **2** :: Views: **2071**

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

Digital communication :: 04-02-2007 09:30 :: vamsi charan :: Replies: **0** :: Views: **1198**

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