Understanding the sinc function

[grandpa2390]

Homework Statement

My problem is that I need to be able to convert my given sine functions to sinc functions in order to do my Fourier transforms. But I can't find anything that really explains the sinc function besides the given formulas.

but this doesn't really help me in solving my problems because it doesn't seem to work out right. What are the rules for changing sin(ax)/ax into sinc(x)?

for example I have a simple problems

sin(x)/x ⊃ πΠ(πs)
sin(Ax)/Ax ⊃ π/A * Π(πs/A)
-sin(π(x-1/2)) / (π(x-1/2)) ⊃ -e^-iπs * Π(s)

Homework Equations

sin(x)/x = sinc(x)
sin(πx)/πx = sinc(x)
sin(ax)/ax = sinc(x)

The Attempt at a Solution

All of these based on my googling would simplify to sinc(x), but that doesn't work.

i want to transform the sin(x)/x to sinc(x), but that would just give me the Π(s). if I say wikipedia is wrong and that it has to be πx, I can say sinc(x/π) and the similarity theorem will give me the desired result.

for the second one I would have to say that simplify the left side gives me sinc (A/π * x) but it doesn't fit with the formulas that I found on the internet for sinc.

the third one seems like it should also be sinc(x) but in order to get the transform on the right, it needs to simplify to sinc(x-1/2)

 

[blue_leaf77]

$\textrm{sinc}(x) \equiv \frac{\sin x}{x}$. Sinc function is just a shorter way to write (sin x)/x.

 

[grandpa2390]

$\textrm{sinc}(x) \equiv \frac{\sin x}{x}$. Sinc function is just a shorter way to write (sin x)/x.

so then what is sin(x) / x ? why is it equal to sinc (x/pi) the similarity theorem states that f(ax) ⊃ 1/a * F(s/a)

so when I change sin into sinc, a must = 1/pi

because the transform of sinc(x) = rect(s)
and the transform of sin(x)/x = pi*rect(pi*s)

 

[blue_leaf77]

Sometimes, a different definition of sinc function is also used, for example see equation (5) in http://mathworld.wolfram.com/SincFunction.html.

the similarity theorem states that f(ax) ⊃ 1/a * F(s/a)

This has nothing to do with Fourier transform, it's just a mere definition.

 

[grandpa2390]

Sometimes, a different definition of sinc function is also used, for example see equation (5) in http://mathworld.wolfram.com/SincFunction.html.

This has nothing to do with Fourier transform, it's just a mere definition.

I don't understand what you mean by it having nothing to do with the Fourier Transform. my textbook says, "Theorems for the Fourier Transform"
and that is the similarity theorem.

but regardless of that:

the similarity theorem states that f(ax) ⊃ 1/a * F(s/a)

so when I change sin(x)/(x) into sinc(ax), a must = 1/π
and when I change sin(Ax)/Ax into sinc(ax), a must = A/π

because the transform of sinc(x) = rect(s)
and the transform of sin(x)/x = pi*rect(pi*s)
and the transform of sin(Ax)/Ax = π/A * Π(π/A * s)

 

[grandpa2390]

is the definition used by Bracewell simply that sin(πx)/πx = sinc(x) and only when it is πx?
and that any other coefficient does not work?

sin(2x)/2x ≠ sinc(x) it is equal to sinc(2x)?

 

[blue_leaf77]

but regardless of that:

the similarity theorem states that f(ax) ⊃ 1/a * F(s/a)

so when I change sin(x)/(x) into sinc(ax), a must = 1/π
and when I change sin(Ax)/Ax into sinc(ax), a must = A/π

This is why I said it has nothing to do with Fourier transform. $\textrm{sinc}(x)$ being equivalent to $\frac{\sin x}{x}$ is just a definition, there was no theorem from Fourier transform involved when such notation was invented. You must be confusing the two different definitions of sinc function, which are covered in the link I gave above.

is the definition used by Bracewell simply that sin(πx)/πx = sinc(x) and only when it is πx?

Only when you follow those guys' definition and its $\pi x$.

sin(2x)/2x ≠ sinc(x)

No, unless you want to invent your own definition.