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I don't know about the book reference, but the claim made (that the midpoint definitions is necessary to be the F.T. of sinc) is an easily disproven false claim. Tag: references removed |
→Demonstration of validity: use tfrac |
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It follows that:
<math display=block>\lim_{n\rightarrow \infty, n\in \mathbb(Z)} \frac{1}{(2t)^{2n}+1} = \frac{1}{0+1} = 1, |t|<\
Second, we consider the case where <math display=inline>|t|>\frac{1}{2}.</math> Notice that the term <math display=inline>(2t)^{2n}</math> is always positive for integer <math>n.</math> However, <math>2t>1</math> and hence <math display=inline>(2t)^{2n}</math> grows very large for large <math>n.</math>
It follows that:
<math display=block>\lim_{n\rightarrow \infty, n\in \mathbb(Z)} \frac{1}{(2t)^{2n}+1} = \frac{1}{+\infty+1} = 0, |t|>\
Third, we consider the case where <math display=inline>|t| = \frac{1}{2}.</math> We may simply substitute in our equation:
<math display=block>\lim_{n\rightarrow \infty, n\in \mathbb(Z)} \frac{1}{(2t)^{2n}+1} = \lim_{n\rightarrow \infty, n\in \mathbb(Z)} \frac{1}{1^{2n}+1} = \frac{1}{1+1} = \
We see that it satisfies the definition of the pulse function. Therefore,
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