Sunday, 1 October 2017

First fix $j$ and $k$, then we get

$$S_1=\sum_{\substack{i=0 \\ i\neq j,k}}^{\infty}\frac{1}{3^{i+j+k}}$$ where $j \ne k$

So we get $$S_1=\left(\frac{3}{2} \times\frac{1}{3^{j+k}}\right)-\frac{1}{3^{2j+k}}-\frac{1}{3^{j+2k}}$$

So now

$$S_2=\sum_{\substack{j=0 \\ j\neq k}}^{\infty}\left(\left(\frac{3}{2} \times\frac{1}{3^{j+k}}\right)-\frac{1}{3^{2j+k}}-\frac{1}{3^{j+2k}}\right)$$

So $$S_2=\left(\frac{9}{4} \times \frac{1}{3^k}\right)-\left(\frac{9}{8} \times \frac{1}{3^k}\right)-\left(\frac{3}{2}\times \frac{1}{9^k}\right)-\left(\frac{3}{2}\times \frac{1}{3^{2k}}\right)+\left(\frac{2}{3^{3k}}\right)$$


$$S=\sum_{k=0}^{\infty}\left(\frac{9}{4} \times \frac{1}{3^k}\right)-\left(\frac{9}{8} \times \frac{1}{3^k}\right)-\left(\frac{3}{2}\times \frac{1}{9^k}\right)+\sum_{k=0}^{\infty}\frac{2}{3^{3k}}-\sum_{k=0}^{\infty}\frac{\frac{3}{2}}{3^{2k}}$$

we get

$$S=\frac{27}{16}-\frac{27}{16}+\frac{54}{26}-\frac{27}{16}$$  hence


Thursday, 18 April 2013


Can we Find any two Distinct Functions
$$ f:\Re \rightarrow \Re $$ such that

$$ f\left ( x \right )=f\left (\frac{x}{2}\right ) $$
Necessary and Sufficient Conditions for the General Second Degree Equation
$$ Ax^2+2Hxy+By^2+2Gx+2Fy+C=0 $$
 to Represent Pair of Straight Lines:

\Delta =\begin{vmatrix}
A &  H &G\\
 H& B & F\\
 G& F& C
\end{vmatrix} =0

Case 1 :if $$ H^2 > AB $$ Then they are Intersecting pair of Straight Lines

Case 2: if $$ H^2 = AB $$ Then they are pair of Parallel Straight Lines

Case 3: if $$ H^2 < AB $$ 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

$$ N=2222^{5555}+5555^{2222} $$

Thursday, 12 July 2012

Multi User CDMA System Model

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

$$ S_j(t)= \sqrt{2 P_j}\,c_j(t)\,b_j(t)\,Cos(\omega_c t+\theta_j) $$ Where,

$$c_j(t)=\sum_{n=-\infty}^{\infty}\,c_j^{(n)}\,p_{T_c}(t-n\,T_c) $$ is the $j$th User Signature Waveform, with the Signature bits

$$c_j^{(n)} \in {-1,+1} $$

$$ b_j(t)=\sum_{n=-\infty}^{\infty}\,b_j^{(n)}\,p_T(t-n \, T) $$ is the BPSK Modulated Signal for User $j$

$$b_j^{(n)} \in {-1,+1} $$ and

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

$$ r(t)= \sum_{j=1}^{K} \sqrt{2 P_j}\,c_j(t-\tau_j)\,b_j(t-\tau_j)\,Cos(\omega_c t+\phi_j) + \eta(t) $$ Where $\tau_j$ is Relative Time offset and $$\tau_j \in \left [ 0 \: T \right ] $$ $\phi_j$ is Phase Offset such that $$\phi_j  \in \left [ 0 \: 2 \pi \right ]$$ and $$\phi_j=\theta_j-\omega_c \tau_j $$ $\eta(t)$ is Zero Mean AWGN Process with PSD $N_0$

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

$$ \Phi_X(\omega)= \int_{-\infty}^{\infty} f_X(x) e^{j \omega x} dx $$, Then

$$ F_X(x)=0.5-\frac{1}{j\,\pi}\int_0^{\infty} \frac{\Phi_X(\omega)}{\omega} d \omega $$