Sunday, September 21, 2008

GATE EC2003 question

Ans :
Number of Mesh = B - N + 1 where B = No of Branches
N = number of Nodes
Node equation = N - 1
where N = number of Nodes

here B = 7 , N = 4

so minium number of equation need to analyze the circuit is = N - 1 = 3

Qu(2):

The Laplace transform of of i(t) is given by
I(s) = 2 / s(1+s)
As t --> infinity, The value of i( t) tends to


Ans:
Inverse of I(s) ,
I(s) = 2/s -2/(1+s)
i(t) = 2 -2(e^-t) because L(1) = 1/s and L(e^-t .1) = 1/(S+1)
As t --> infinity ,
==> i(t) = 2 -2/(e^infinity)
==> i(t) = 2 - 2/infinity
== > i(t) = 2 because 2/infinity is zero

This is the answer I got. I am not sure about 2/infinity . ,,...

i don't know these answers are correct or not .. I didn't get key for GATE these question.

Saturday, September 20, 2008

DIGITAL SIGNAL PROCESSING



I become a fan of Prof. S.C. Dutta Roy .....if you hear these videos then you will reach a new world. See some beautiful lectures ..

Brief Technical Biography of S C Dutta Roy


S C Dutta Roy , a Ph D from the Calcutta University, is currently an INSA Senior Scientist at the Electrical Engineering Department, IIT Delhi, where he served as a Professor for more than three decades. He has also held the positions of Head of the Department and Dean of Undergraduate Studies at IIT Delhi, and Visiting faculty positions at the University of Minnesota, USA ; University of Leeds, UK ; and the Iowa State University, USA. His research interests are in signal processing – analog as well as digital, and he has published extensively in IEEE, IEE and other professional journals of repute.


Professor Dutta Roy is a Fellow of the IEEE, the Acoustical Society of India, the Systems Society of India, and of all the Science and Engineering Academies of India, and is a Distinguished Fellow of the Institution of Electronics and Telecommunication Engineers (IETE), India. He has served on the Editorial Boards of a few IEEE and other prestigious international journals and all the national journals in his field. He received several prestigious national awards, including the Shanti Swarup Bhatnagar Prize, the Vikram Sarabhai Award and the Pandit Jawaharlal Nehru Award.



Thursday, September 18, 2008

why derivatives ?

To find an approximation of \sqrt[3]{25} one can do as follows.

  1. Consider the function  f(x)= x^{1/3}.\, Hence, the problem is reduced to finding the value of f(25).
  2. We have
     f\ '(x)= 1/3x^{-2/3}.
  3. According to linear approximation
     f(25) \approx f(27) + f\ '(27)(25 - 27) = 3 - 2/27.
  4. The result, 2.926, lies fairly close to the actual value 2.924…
Read some interesting and basic mathematics.. http://en.wikipedia.org/wiki/Linear_approximation