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For mathematatics in Nyquist  Shannon type
Question

Hello, every one, i have two questions :
"In the type of Nyquist  Shannon there is a condition X(f)=0, f>B and the interval supp(X) ⊆ [−B, B]."
. If divide interval [B,B] to smallest intervals the aproach root of the equation would have less cutoff error with Series or with the Interval Bisection algorithm?
. The type of ShannonHartley which is C = Blog2(1+s/n), where :
C is the channel capacity in bits per second.
B is the bandwith of the channel in hertz (passband bandwidth in case of a modulated signal).
S is the total received signal power over the bandwidth (in case of a modulated signal, often denoted C, i.e. modulated carrier), measured in watt or volt (<sup>2</sup>;) in wiki Shannon–Hartley theorem .
N is the total noise or interference power over the bandwidth, measured in watt or volt (<sup>2</sup>;) in wiki Shannon–Hartley theorem and
S/N is the signal to noise ratio (SNR) or the carrier to noise ratio (CNR) of the communication signal to the Gaussian noise interference expressed as a linear power ratio (not as logarithmic decibels.)
is exactly the same with the above?
Thank you.
Mr. Leonidas Fengos Object Oriented Programmer Edited by Mr. Leonidas Fengos Saturday, April 28, 2012 1:34 AM
Monday, January 23, 2012 8:06 PM
All replies

Hello, every one, for the first above question i try it and i prefer algorithm and i try these steps :...k=10step_point = pi*0.1;x = z = Interval 1 [(k),k)]y(x)+log2(y*z)*z;Display_Graph(x,y)z = Interval 2 [(k*0.001),(k*0.001)]y(x)+log2(y*z)*z;Display_Graph(x,y)z = Interval 3 [(k*0.00001),(k*0.00001)]y(x)+log2(y*z)*z;Display_Graph(x,y)z = Interval 4 [(k*0.0000001),(k*0.0000001)]y(x)+log2(y*z)*z;Display_Graph(x,y)...Thank you.
Leonidas Fengos OO Programmer Edited by Mr. Leonidas Fengos Sunday, March 11, 2012 4:29 PM
 Proposed as answer by Leonidas Fengos Saturday, July 14, 2012 4:24 PM
Monday, January 30, 2012 11:13 AM 
Hello, every one, i try to do it for 3 dimentions :
k=10step_point = pi*0.1;x = z = Interval 1 [(k),k)]y(x)+log2(y*z)*z;Display_Graph(x,y,(variable depends on y))z = Interval 2 [(k*0.001),(k*0.001)]y(x)+log2(y*z)*z;Display_Graph(x,y,(variable depends on y))z = Interval 3 [(k*0.00001),(k*0.00001)]y(x)+log2(y*z)*z;Display_Graph(x,y,(variable depends on y and x))z = Interval 4 [(k*0.0000001),(k*0.0000001)]y(x)+log2(y*z)*z;Display_Graph(x,y,(variable depends on y and x))...Thank you. Edited by Mr. Leonidas Fengos Saturday, April 28, 2012 1:41 AM
 Proposed as answer by Leonidas Fengos Saturday, July 14, 2012 4:24 PM
Monday, January 30, 2012 12:10 PM 
Hello, every one, i take the above algorithms and the log2 and i approache the root with eps :
k=10step_point = pi*0.1;x = z = Interval 1 [(k),k)]
step_point = pi;
eps=x/n;
y(x)=sin(x)
y(x)+log2(y*z)*z;
Display_Graph(x,y,(variable depends on y))z = Interval 2 [(k*0.001),(k*0.001)]y(x)+log2(y*z)*z+abs(eps);Display_Graph(x,y,(variable depends on y))z = Interval 3 [(k*0.00001),(k*0.00001)]y(x)+log2(y*z)*z+eps;Display_Graph(x,y,(variable depends on y and x))z = Interval 4 [(k*0.0000001),(k*0.0000001)]y(x)+log2(y*z)*zeps;Display_Graph(x,y,(variable depends on y and x))Thank you.
Mr. Leonidas Fengos Object Oriented Programmer
 Edited by Mr. Leonidas Fengos Saturday, April 28, 2012 1:41 AM
 Proposed as answer by Leonidas Fengos Saturday, July 14, 2012 4:24 PM
Friday, February 10, 2012 11:47 AM