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Can I write this integral in "closed-form" expression?

  1. Aug 6, 2014 #1
    Hi all,

    I have this integral, and I wish to write it in a closed-form in order to be able to program it. The integral is:

    [tex]\int_{-\infty}^{\infty}e^{j2\pi f_c[a^{(u)}-a^{(m)}]}g(t[1+a^{(m)}]-kT_s-\tau_p^{(m,n)}) g(t[1+a^{(u)}]-lT_s-\tau_q^{(u,n)})\,dt[/tex]



    Basically, I can write two different cases, namely, when:

    [tex]\frac{kT_s+\tau_p^{(m,n)}}{1+a^{(m)}}\leq \frac{lT_s+\tau_q^{(u,n)}}{1+a^{(u)}}<\frac{(k+1)T_s+\tau_p^{(m,n)}}{1+a^{(m)}}[/tex]

    and when

    [tex]\frac{lT_s+\tau_q^{(u,n)}}{1+a^{(u)}}\leq \frac{kT_s+\tau_p^{(m,n)}}{1+a^{(m)}}<\frac{(l+1)T_s+\tau_q^{(u,n)}}{1+a^{(u)}}[/tex]

    and solve the integral for these two separate cases. However, this requires me to check for every possible case separately which is taking too long time. I was hoping if I can write it as one expression which eases the programming part.

  2. jcsd
  3. Aug 12, 2014 #2
    I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
  4. Aug 12, 2014 #3

    D H

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    Staff Emeritus
    Science Advisor

    Simplifying that a bit, you have [itex]\int_{-\infty}^{\infty} c g_1(t) g_2(t)\, dt[/itex] where
    c &= e^{j2\pi f_c[a^{(u)}-a^{(m)}]} \\
    g_1(t) &= g(t[1+a^{(m)}]-kT_s-\tau_p^{(m,n)}) \\
    g_2(t) &= g(t[1+a^{(u)}]-lT_s-\tau_q^{(u,n)})
    That constant [itex]c=e^{j2\pi f_c[a^{(u)}-a^{(m)}]}[/tex] can be pulled out of the integral, and you're left with the integral of the product of two indicator functions. Each of those indicator functions is fairly simple (but messy): They describe a finite interval on the real number line. So the integral is just the constant c times the length of the intersection of these two intervals. I'll use [itex]a[/itex] to denote the interval corresponding to [itex]g_1(t)[/itex] and [itex]b[/itex] to denote the nterval corresponding to [itex]g_2(t)[/itex]:

    a &= \left(\frac{kT_s+\tau_p^{(m,n)}}{1+a^{(m)}},
    \frac{(k+1)T_s+\tau_p^{(m,n)}}{1+a^{(m)}}\right) \\
    b &= \left(\frac{lT_s+\tau_q^{(u,n)}}{1+a^{(u)}},

    That's a bit messy yet. I'll define four items to represent those endpoints:
    a_1 &= \frac{kT_s+\tau_p^{(m,n)}}{1+a^{(m)}} \\
    a_2 &= \frac{(k+1)T_s+\tau_p^{(m,n)}}{1+a^{(m)}} \\
    b_1 &= \frac{lT_s+\tau_q^{(u,n)}}{1+a^{(u)}} \\
    b_2 &= \frac{(l+1)T_s+\tau_q^{(u,n)}}{1+a^{(u)}}\end{aligned}[/tex]
    With this, the intervals simplify to
    a &= (a_1, a_2) \\
    b &= (b_1, b_2)

    You missed four cases. Here are all six, depicted in ASCII art. In all cases, I'm portraying interval a above and interval b below the real number line, and the intersection between intervals a and b below the line displaying interval b.
    Code (Text):

               (    a    )  
    Case 1:  ------------------------------------>
                               (    b   )  
             Intersection is null

                    (     a    )  
    Case 2:  ------------------------------------>
                         (     b     )  
                         ( a∩b )

                         (  a  )
    Case 3:  ------------------------------------>
                      (    b     )  
                         ( a∩b )

                      (    a     )  
    Case 4:  ------------------------------------>
                         (  b  )
                         ( a∩b )

                         (     a     )  
    Case 5:  ------------------------------------>
                    (     b    )  
                         ( a∩b )

                               (    a   )  
    Case 6:  ------------------------------------>
               (    b    )  
             Intersection is null
    All you need is the length of that intersection. This is actually quite easy: it's [itex]d=\max(0, \min(a_2,b_2) - \max(a_1,b_1))[/itex]. That outer max protects against a negative length, which is what would otherwise in the case of a null intersection.
    Last edited: Aug 12, 2014
  5. Aug 12, 2014 #4
    I am really thankful of your effort, but I have a typo in my post where the integral is actually:

    [tex]\int_{-\infty}^{\infty}e^{j2\pi f_c[a^{(u)}-a^{(m)}]t}g(t[1+a^{(m)}]-kT_s-\tau_p^{(m,n)}) g(t[1+a^{(u)}]-lT_s-\tau_q^{(u,n)})\,dt[/tex]

    where the exponential is a function of t, otherwise I wouldn't have put it inside the integral. This is actually what complicates the integration. Now, in its presence can I find something similar to what you had without it?

  6. Aug 12, 2014 #5

    D H

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    Staff Emeritus
    Science Advisor

    It doesn't complicate it by much. Those g(αt+β) are either zero or one, so instead of a length times a constant you get

    [tex]\int_{\max(a_1, b_1)}^{\min(a_2,b_2)} \exp(j\omega t)\,dt[/tex]

    Here I've collapsed that big messy constant of yours into a single constant, [itex]\omega[/itex]. That's an easy integral. There is one tricky issue now: You have to watch out for the two null cases. The above is correct only if there is a non-null intersection.
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