Evaluation of a Sinc Integral: Solving a Common Problem in Communications Theory

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In summary, the Integrate command runs forever on this integral, but I believe it can be done by hand. The identity given solves the problem.
  • #1
chroot
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Can anyone show me how to evaluate an integral like this by hand? I believe such integrals have an analytic solution, but I can't figure out how to find them. Mathematica seems unable to help (the Integrate command runs forever) but I believe this can be done by hand. It's a sort of integral commonly found in communications theory. I actually don't think it's supposed to be very hard...

[itex]
\int_{ - \infty }^\infty {\left[ {\frac{1}
{{\sqrt T }}\operatorname{sinc} \left( {\frac{t}
{T}} \right) - \frac{1}
{{2\sqrt T }}\operatorname{sinc} \left( {\frac{{t - T}}
{T}} \right)} \right]^2 dt}
[/itex]

where

[itex]
\operatorname{sinc} \left( t \right) \triangleq \frac{{\sin \left( {\pi t} \right)}}
{{\pi t}}
[/itex]

Obviously I can expand the binomial out, but I'm left with products of sinc's with different arguments, and I don't know how to continue.

- Warren
 
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  • #2
Actually, the identity

[itex]
\operatorname{sinc} \left( {\frac{{t + mT}}
{T}} \right) \cdot \operatorname{sinc} \left( {\frac{{t + nT}}
{T}} \right) = T\operatorname{sinc} \left( {\frac{{t + (m + n)T}}
{T}} \right)
[/itex]

makes life much sweeter. I got 1/4. Anyone else care to check my work?

- Warren
 
  • #3
Using the identities you posted I got [itex]\pi T / 4[/itex]. I squared out the integrand and split up the integral into 3 integrals. The first two exactly canceled each other out, leaving only the third one:

[tex]\frac{1}{4}\int_{ - \infty }^\infty \operatorname{sinc}\left(\frac{t-2T}{T}\right)dt [/tex]

U-substitution yields:

[tex]\frac{T}{4}\int_{ - \infty }^\infty \operatorname{sinc}(u)du [/tex]

The integral (sans outside coefficient) evaluates to [itex]\pi[/itex].

Regarding Mathematica, MathWorld has this to say in its article on the sinc function:

MathWorld said:
This function will be implemented in a future version of Mathematica as Sinc[x].

So I guess you'll have to wait a bit longer to get the computer to do it for you.
 
Last edited:
  • #4
You can define your own piecewise Sinc function in Mathematica quite easily.

As it turns out, though, Tom, we're both wrong, but I can't exactly say why. I followed your method similarly, finding two terms which cancelled, leaving the third.

However, there's a much simpler way, subject to fewer mistakes. Note that all terms of the form

[itex]
\frac{a}
{{\sqrt T }}\operatorname{sinc} \left( {\frac{{t - nT}}
{T}} \right)
[/itex]

are normalized, and integrate to a.

You can literally read the value of this integral directly off the coefficients a, if you consider the integrand to be a vector multiplication like this:

[itex]
\frac{{a_1 }}
{{\sqrt T }}\operatorname{sinc} \left( {\frac{t}
{T}} \right) + \frac{{a_2 }}
{{\sqrt T }}\operatorname{sinc} \left( {\frac{{t - T}}
{T}} \right) \Leftrightarrow \left[ {\begin{array}{*{20}c}
{a_1 } & {a_2 } \\

\end{array} } \right] \cdot \left[ {\begin{array}{*{20}c}
1 \\
1 \\

\end{array} } \right]
[/itex]

The value of the integral, then, is the squared norm of this vector. In this case, the result is 5/4. I'm still not exactly sure why integrating this long-hand produces a different value, but I'm pretty sure I just flubbed some algebra.

I've learned something in this communications class: if anything looks hard, you're doing it wrong. Thanks for the help though!

- Warren
 
  • #5
chroot said:
As it turns out, though, Tom, we're both wrong, but I can't exactly say why. I followed your method similarly, finding two terms which cancelled, leaving the third.

I am certain that I integrated correctly. That being the case I decided to take a closer look at the identity you posted, since I started from that.

[itex]
\operatorname{sinc} \left( {\frac{{t + mT}}
{T}} \right) \cdot \operatorname{sinc} \left( {\frac{{t + nT}}
{T}} \right) = T\operatorname{sinc} \left( {\frac{{t + (m + n)T}}
{T}} \right)
[/itex]

Let m=0 and n=-1. Then you get:

sinc(t/T)sinc((t-T)/T)=Tsinc((t-T)/T)

Now let t=0:

sinc(0)sinc(-1)=Tsinc(-1)

Since T is arbitrary and sinc(-1) does not equal zero, the equation above is not an identity.
 
  • #6
Whoops! I posted my identity incorrectly.

It should be

[itex]
\operatorname{sinc} \left( {\frac{{t + mT}}
{T}} \right) \,\star\, \operatorname{sinc} \left( {\frac{{t + nT}}
{T}} \right) = T\operatorname{sinc} \left( {\frac{{t + (m + n)T}}
{T}} \right)
[/itex]

Convolution, not multiplication. :blushing:

- Warren
 

What is a sinc integral?

A sinc integral is a mathematical function that is used to evaluate the integral of the sinc function, which is defined as sin(x)/x. It is commonly used in signal processing and Fourier analysis.

How is a sinc integral calculated?

The sinc integral is calculated by using a mathematical formula that involves taking the definite integral of the sinc function over a given interval. This can be done analytically or numerically using numerical integration methods.

What are the applications of a sinc integral?

A sinc integral has many applications in signal processing, such as in the reconstruction of signals from their Fourier transforms. It is also used in the analysis of digital filters and in the interpolation of sampled data.

What is the relationship between the sinc function and the sinc integral?

The sinc integral is the antiderivative of the sinc function. This means that the sinc integral is the function that, when differentiated, gives the sinc function. The two are closely related and often used together in mathematical calculations.

Are there any limitations or special cases when evaluating a sinc integral?

Yes, there are some limitations and special cases when evaluating a sinc integral. One limitation is that the sinc function is undefined at x=0, so the sinc integral cannot be evaluated at that point. Another special case is when the interval of integration includes the point x=0, in which case the sinc integral can be evaluated using a different formula.

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