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## Exponential Distribution |

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

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 :: 16.08.2013 03:12 :: albert_new :: Replies: **2** :: Views: **244**

Digital communication :: 16.04.2007 09:24 :: mathuranathan :: Replies: **1** :: Views: **986**

thanks very muchxulfee, but i think i'm looking for a random sequence with **exponential** pdf...And the problem is, such a sequence with zero mean returns all zeros as i already mentioned.

Digital Signal Processing :: 12.09.2008 02:25 :: vahidkh6222 :: Replies: **10** :: Views: **1149**

This has been a while, so check my work ...
I presume you have a Gaussian RV with mean ?, and variance σ?.
The integral with |x| is awkward to think of - break it into two pieces.
With a little thought, the integral from 0 to ∞ is always positive - no need for |x|.
Similarly, the integral from -∞ to 0 is always negative,

Mathematics and Physics :: 05.03.2007 01:56 :: LouisSheffield :: Replies: **1** :: Views: **925**

In order to find the inverse cdf of the **exponential** **distribution** I have solved Y=1-exp(-X) for X. This is the point. You have to find algebric inverse.
Also if you jenerate the data using the line you have here, I think you can see that the drived data are not **exponential**ly distributed.
Yours,
dspman.

Mathematics and Physics :: 12.04.2007 13:19 :: dspman :: Replies: **5** :: Views: **700**

The built-in potential is generated because of the diffusion of the mobile charges across the junction which leaves back the uncompensated immobile charges near the junction. The number of charges diffusing across the junction is equal to the number of immobile charges left back. The mobile charges diffuse in the p or n type material and follow a n

Electronic Elementary Questions :: 03.03.2008 01:58 :: subharpe :: Replies: **2** :: Views: **1566**

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 :: 13.11.2008 04:04 :: urwelcome :: Replies: **2** :: Views: **1488**

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 :: 16.06.2009 04:16 :: aliazmat :: Replies: **2** :: Views: **787**

Hi,
I want to solve this problem for my friend.But I have some questions to ask you.
"Paris Wholesale fruit distributors employ one worker whose job is to load fruit on outgoing company trucks. Trucks arrive at the loading gate at an average of 24 per day , or 3 per hour, according to a Poisson **distribution**. The worker loads them at a rate of 4

Mathematics and Physics :: 05.01.2010 01:27 :: sivamit :: Replies: **0** :: Views: **720**

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 :: 19.10.2011 18:38 :: David83 :: Replies: **5** :: Views: **3969**

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 :: 20.09.2011 16:00 :: samir67 :: Replies: **1** :: Views: **1883**

There are many trends exhibited by some research groups:
1. multiple or variable voltage supplies (transistor-level)
2. application transformations that infer differentiations at the system level
e.g. the memory subsystem organization and the update rules for these memories. (RTL and system level)
3. adiabatic switching (transistor level)

ASIC Design Methodologies and Tools (Digital) :: 19.01.2003 19:11 :: the_penetrator :: Replies: **4** :: Views: **1208**

Random **distribution** is normally characterized by its probability **distribution** function (pdf). The area covered by the pdf is always one.
Uniform is one where the probability of any value occuring within a specified range (say from a to b) is the same, ie p(x)=1/(b-a).
Normal, or gaussian **distribution**s follow a symmetrical bell-shape curve. (...)

Electronic Elementary Questions :: 31.08.2005 06:26 :: checkmate :: Replies: **2** :: Views: **536**

Hi,
you have many elements, but exactly ?
it is instable if you use a direct approach (till 40 elements in double precison it works). This is beacuse the close formula has products of **exponential** and factorial number that becames quckly very intractable.
You need a factorization scheme to make this stable (tenth of thousands of elements !!

RF, Microwave, Antennas and Optics :: 28.03.2008 14:42 :: juppydu :: Replies: **3** :: Views: **512**

Hi all!
I have simulated a rayleigh flat fading channel (ofdm) and a frequency selective **exponential** channel in which every tap follows rayleigh **distribution**,thus every tap 'sees' a flat rayleigh channel (due to this right? However i wonder what the plot would be? The plot (BER vs SNR) is almost the same for both cases?
This is right,i

Digital communication :: 19.08.2010 10:26 :: anta :: Replies: **5** :: Views: **2733**

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 :: 25.08.2010 11:45 :: Communications_Engineer :: Replies: **6** :: Views: **1355**

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 :: 28.02.2011 00:10 :: klystron :: Replies: **1** :: Views: **373**

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