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

1. Aug 6, 2014

EngWiPy

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:

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

where

$$g(t)=\left\{\begin{array}{cc}1&t\in[0,T_s)\\0&\mbox{Otherwise}\end{array}\right.$$

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

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

and when

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

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.

Thanks

2. Aug 12, 2014

Greg Bernhardt

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?

3. Aug 12, 2014

D H

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

\begin{aligned} 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)}}, \frac{(l+1)T_s+\tau_q^{(u,n)}}{1+a^{(u)}}\right) \end{aligned}

That's a bit messy yet. I'll define four items to represent those endpoints:
\begin{aligned} 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}
With this, the intervals simplify to
\begin{aligned} a &= (a_1, a_2) \\ b &= (b_1, b_2) \end{aligned}

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 $d=\max(0, \min(a_2,b_2) - \max(a_1,b_1))$. 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
4. Aug 12, 2014

EngWiPy

I am really thankful of your effort, but I have a typo in my post where the integral is actually:

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

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?

5. Aug 12, 2014

D H

Staff Emeritus
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

$$\int_{\max(a_1, b_1)}^{\min(a_2,b_2)} \exp(j\omega t)\,dt$$

Here I've collapsed that big messy constant of yours into a single constant, $\omega$. 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.