Sunday, September 21, 2008
ascentengineers...
ascentengineers -----> Very interesting to see their work( http://ascentengineers.org/ ) and the way they are generating their business. I believe that they can be succeed in their business with electronics and free software . Because the market for these two are growing very fast in kerala. They found out their first customers as the college students. Giving consultancy to them .... is really a business.
I find that these ascentengineers are very young. And the courage they taking is really amazing.
computer graphics is not as simple as I expected
I have seen video from National Programme on Technology Enhanced Learning about COMPUTER GRAPHICS. I confused that whether I was hearing a Mathematics class or computer graphics. Real mathematics is used to map bits from image to CRT( screen).
see link
computer graphics ... videos
see link
computer graphics ... videos
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.
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.