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Sunday, 1 October 2017

First fix j and k, then we get

S1=i=0ij,k13i+j+k where jk

So we get S1=(32×13j+k)132j+k13j+2k

So now

S2=j=0jk((32×13j+k)132j+k13j+2k)

So S2=(94×13k)(98×13k)(32×19k)(32×132k)+(233k)

Finally

S=k=0(94×13k)(98×13k)(32×19k)+k=0233kk=03232k

we get

S=27162716+54262716  hence

S=81208

Thursday, 18 April 2013

Functions

Can we Find any two Distinct Functions
f: such that

f(x)=f(x2)
Necessary and Sufficient Conditions for the General Second Degree Equation
Ax2+2Hxy+By2+2Gx+2Fy+C=0
 to Represent Pair of Straight Lines:

Δ=|AHGHBFGFC|=0


Case 1 :if H2>AB Then they are Intersecting pair of Straight Lines

Case 2: if H2=AB Then they are pair of Parallel Straight Lines

Case 3: if H2<AB Then they Represent a Point in a Plane

Friday, 8 February 2013

Remainder Problem

Find the Remainder when the number N is divided by 7, where

N=22225555+55552222

Thursday, 12 July 2012

Multi User CDMA System Model

Let's have CDMA System with K Users indexed by j=1K. Let Each User Transmits BPSK Modulated Signal Simultaneously. Then the Transmitted Signal for one such User j can be Represented as:

Sj(t)=2Pjcj(t)bj(t)Cos(ωct+θj) Where,

cj(t)=n=c(n)jpTc(tnTc) is the jth User Signature Waveform, with the Signature bits

c(n)j1,+1

bj(t)=n=b(n)jpT(tnT) is the BPSK Modulated Signal for User j


b(n)j1,+1 and

T=NTc Where N is the Spreading Factor or Processing Gain. The Received Signal at the Receiver can be Represented as

r(t)=Kj=12Pjcj(tτj)bj(tτj)Cos(ωct+ϕj)+η(t) Where τj is Relative Time offset and τj[0T] ϕj is Phase Offset such that ϕj[02π] and ϕj=θjωcτj η(t) is Zero Mean AWGN Process with PSD N0



Monday, 2 July 2012

Gil-Palaez Theorem

It Helps us to Find the CDF of a Random Variable Directly from MGF or Characteristic Function .

if X is any Random Variable with Characteristic Function (CF) given by


ΦX(ω)=fX(x)ejωxdx, Then


FX(x)=0.51jπ0ΦX(ω)ωdω