# 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 Shannon-Hartley 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
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=10
step_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

Monday, January 30, 2012 11:13 AM
• Hello, every one, i try to do it for 3 dimentions :

k=10
step_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.

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=10
step_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)*z-eps;
Display_Graph(x,y,(variable depends on y and x))

Thank you.

Mr. Leonidas Fengos Object Oriented Programmer

Friday, February 10, 2012 11:47 AM