Find the solution for the following equation

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In summary: This can be achieved by choosing fc=0, which gives us the following equation: g_n(t) = H_n(t) * cos (0) = H_n(t) This means that the value of fc can be chosen as 0 in order to maintain the orthogonality property of Hermite polynomial. In summary, for the orthogonality property of Hermite polynomial to hold, the value of fc must be chosen as 0.
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T.Engineer
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Homework Statement



if H_n(t)= (-1)^n * e^(t^2) * d^n/dt^n * e^(-t^2)

where n=1,2,...,N

from the orthogonality property of Hermite polynomials we will have:

[tex]\int^{\infty}_{-\infty} e^{-t^2} H_n(t) H_m(t)dt [/tex] = [tex]\delta_n,m[/tex] 2^n n! [tex]\sqrt{}pi[/tex]

this gives


N_n= [tex]\frac{1}{\sqrt{}2^n n! \sqrt{}pi}[/tex]


consequently

[tex]\int^{\infty}_{-\infty}\varphi_n(t)\varphi_m(t) dt [/tex] = [tex]\delta_n,m[/tex]

let g_n(t) = H_n(t) cos (2 pi fc t)

the value of fc will be chosen in such a way that still keep the orthogonality property of Hermite polynomial.

find out if cos (2 pi fc t) is positive over the relevant range as follow


[tex]\int^{\infty}_{-\infty}\varphi_n(t)\varphi_m(t) dt [/tex] = [tex]\delta_n,m[/tex]



Homework Equations






The Attempt at a Solution

 
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  • #2
Let g_n(t) = H_n(t) * cos (2 pi fc t) \int^{\infty}_{-\infty} e^{-t^2} H_n(t) H_m(t) * cos (2 pi fc t) dt = \delta_n,m 2^n n! \sqrt{}pi * cos (2 pi fc t) From the above equation we can see that in order for the orthogonality property of Hermite polynomial to hold, the value of fc must be chosen such that cos (2 pi fc t) is positive over the relevant range. In order to find out if cos (2 pi fc t) is positive over the relevant range, we need to solve the following equation: 2 pi fc t = 0 The solution to this equation is t=0, which means that cos (2 pi fc t) is positive when t=0. Therefore, in order for the orthogonality property of Hermite polynomial to hold, the value of fc must be chosen such that cos (2 pi fc t) is positive at t=0.
 

1. What is the process for finding a solution to an equation?

The process for finding a solution to an equation involves identifying the variables, determining the operations needed to isolate the variable, and then performing the same operations on both sides of the equation until the variable is isolated on one side and the solution is on the other.

2. How do I know if my solution is correct?

You can check your solution by substituting it back into the original equation and seeing if both sides of the equation are equal. If they are, then your solution is correct.

3. Can an equation have more than one solution?

Yes, an equation can have more than one solution. This is particularly true for equations with multiple variables.

4. What do I do if there is no solution to an equation?

If there is no solution to an equation, it means that the equation is contradictory and cannot be solved. This could happen if the equation contains conflicting information or if there are no values that satisfy the equation.

5. What is the difference between an equation and an expression?

An equation is a mathematical statement that shows the equality between two expressions. An expression, on the other hand, is a mathematical phrase that may contain variables, constants, and operators, but it does not have an equal sign. Essentially, an equation shows a relationship between two expressions, while an expression is a stand-alone mathematical phrase.

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