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What is mathematically wrong with this integration?

  1. Apr 7, 2014 #1
    Hello all,

    I have the following integration:

    [tex]\int_{-\infty}^{\infty}e^{-j2\pi f_ct[a_p-a_q]}g(t[1+a_p]-kT_s-\tau_p)g(t[1+a_q]-mT_s-\tau_q)\,dt[/tex]

    where g(t) is 1 in the interval [0,Ts]. This means that the integration has value when both function g(t[1+a_p]-kT_s-\tau_p) and g(t[1+a_q]-mT_s-\tau_q) are 1. Both are 1 when:

    [tex]0\leq t[1+a_p]-kT_s-\tau_p\leq T_s[/tex]

    and

    [tex]0\leq t[1+a_q]-mT_s-\tau_q\leq T_s[/tex]

    Which implies that both are 1 when:

    [tex]t=\frac{(\tau_p-\tau_q)+(k-m)T_s}{a_p-a_q}[/tex]

    But the integration over a point is zero, which can be the answer of the physical problem I have in hand. Where did I go wrong in the process?

    Thanks.
     
  2. jcsd
  3. Apr 7, 2014 #2

    This is where you went wrong. You can't use the two inequalities to evaluate t directly. You need to calculate the range of t where the two functions are both nonzero.

    For example le[itex] \tau_p=\tau_q=a_p=a_q=0[/itex],
    [itex] k=-1[/itex],
    and
    [itex] m=-2[/itex]

    Now
    [itex]g(t[1+ap]−kTs−τp) =1 [/itex] for [itex]t\in(0,1)[/itex]
    and
    [itex]g(t[1+aq]−mTs−τq) =1 [/itex] for [itex]t\in(0,.5)[/itex]

    As you can see there is still a range in [itex]t[/itex] where both g are nonzero.
     
  4. Apr 7, 2014 #3
    When I subtract the ranges of both functions I got something like:

    [tex]0\leq x\leq 0[/tex]

    which implies that x=0. Right?
     
  5. Apr 7, 2014 #4
    But if I add the ranges I'll get a range of t!! Which one is more correct? and why?
     
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